Modélisation moléculaire PPH-1003 Exercices, HIV-2012€¦ · Tutorial and User's Guide Wavefunction, Inc. 18401 Von Karman Avenue, Suite 370 Irvine, CA 92612 U.S.A. Wavefunction,

Sylvain Robert

Modélisation moléculaire

PPH-1003

Exercices, HIV-2012

Exercices: Simulation, modélisation et traitement des données, PPH-1003

Copyright © 2003-2012 par Sylvain Robert

Tous droits réservés. Ces notes ne peuvent être reproduites, mécaniquement ou

électroniquement, autrement que pour des fins pédagogiques individuelles. Toute référence à

cet ouvrage, ou aux livres de références qui y sont cités doit clairement faire mention de leur

source. Pour informations, contactez:

Dr. Sylvain Robert

Département de Chimie-Biologie

Université du Québec à Trois-Rivières

C.P. 500, Trois-Rivières, Qc, G9A 5H7

Tél.: 819.376.5011, poste 4509

Fax: 819.376.5148

@: [email protected]

URL: http://www.uqtr.ca/Cours.Sylvain.Robert

La compagnie Wavefunction permet l’usage gratuit des manuels suivants à des fins

d’enseignement.

Spartan 10 for Windows and Linux – Tutorial and user’s guide,

http://http://downloads.wavefun.com/Spartan10Manual.pdf

Spartan for students for Windows and Linux – Tutorial and user’s guide,

http://http://downloads.wavefun.com/Spartan10Manual.pdf

Qu’pla E

mailto:[email protected]

http://www.uqtr.ca/Cours.Sylvain.Robert

http://http/downloads.wavefun.com/Spartan10Manual.pdf




Exercices: Simulation, modélisation et traitement des données, PPH-1003

Copyright © 2003-2012 par Sylvain Robert

Les pages suivantes sont tirées du manuel de Spartan.

Bien que la version que vous possédiez ne soit que la version étudiante qui est plus limitée, il y

beaucoup d’exercices qui peuvent être faits.

Vous pouvez comparer les différentes versions dans les pages qui suivent.

Je vous invite à faire ces exercices à votre gré. Il n’y en a aucun d’obligatoire, mais sachez

qu’il arrive que je m’inspire de ceux-ci pour vous poser des questions d’examen…

Les temps indiqués représentent un estimé du temps requis pour un Pentium portable de 2

GHz. Lorsque 2 temps sont indiqués, le second tient compte des sections optionnelles du

tutoriel.

Qu’pla E

Tutorialand

User's Guide

Wavefunction, Inc.18401 Von Karman Avenue, Suite 370

Irvine, CA 92612 U.S.A.www.wavefun.com

Wavefunction, Inc., Japan Branch Office3-5-2, Kouji-machi, Suite 608

Chiyoda-ku, Tokyo, Japan [email protected] • www.wavefun.com/japan

WAVEFUNCTION

First Page new 7/20/06, 1:13 PM1

Table of Contents iii

Table of ContentsSection I INTRODUCTION ............................................. 1

Chapter 1 Spartan’06 ........................................................... 9

Section II GETTING STARTED ....................................... 15

Chapter 2 Operating Spartan ............................................. 17 Starting and Quitting Spartan ............................. 17 Menus .................................................................. 17 Keystroke Equivalents ......................................... 18 Icons..................................................................... 19 Mouse and Keyboard Functions .......................... 19 Selecting Molecules, etc ...................................... 21 Dialogs ................................................................. 22 3-Dimensional Displays....................................... 23 Changing Colors and Setting Preferences ........... 23 Monitoring and Killing Jobs ................................ 23 Iconifying Spartan............................................... 23

Chapter 3 Basic Operations ................................................ 25

Chapter 4 Organic Molecules ............................................. 37 Acrylonitrile ......................................................... 39 Cyclohexanone .................................................... 47 Camphor .............................................................. 50 Ethinamate ........................................................... 53 3-Cyano-4-methylcyclohexenyl Radical ............. 55 Infrared Spectrum of Benzyne............................. 58 13C Spectrum of cis-1,2-Dimethylcyclohexane ... 61 Androsterone........................................................ 64

Chapter 5 Groups of Organic Molecules ........................... 67 Dienophiles in Diels-Alder Cycloadditions......... 69 Allyl Vinyl Ether.................................................. 75 Internal Rotation in Dimethylperoxide ................ 78 Hydration of Carbonyl Compounds..................... 82 Acidities of Carboxylic Acids .............................. 85

TOC 7/20/06, 1:42 PM3

iv Table of Contents Table of Contents v

Stereochemical Assignments from 13C Spectra ... 88 Tautomers of Nucleotide Bases ........................... 90

Chapter 6 Organic Reactions.............................................. 93 Ene Reaction of 1-Pentene................................... 95 SN2 Reaction of Bromide and Methyl Chloride... 99 Carbene Additions to Alkenes ............................. 103 Stereospecific Diels-Alder Reactions .................. 107 Thermodynamic vs. Kinetic Control ................... 110 Activation Energies of Diels-Alder Reactions..... 113

Chapter 7 Medicinal Chemistry ......................................... 119 Anticipating Blood-Brain Transport .................... 121 Terfenadine. A Potassium Channel Blocker?....... 124 Morphine. Structure vs. Pharmacophore ............. 129

Chapter 8 Polypeptides to Proteins .................................... 137 Polyglycine .......................................................... 139 Gleevec. Protein Bound vs. Free Conformer....... 143 Gleevec. Making a Pharmacophore from PDB ... 147

Chapter 9 Inorganic and Organometallic Molecules ....... 151 Sulfur Tetrafluoride.............................................. 153 Reactivity of Silicon-Carbon Double Bonds ....... 155 Benzene Chromium Tricarbonyl.......................... 158 Ziegler-Natta Polymerization of Ethylene........... 160

Section III. FEATURES AND FUNCTIONS ...................... 163

Chapter 10 The File Menu .................................................... 165 New...................................................................... 165 Open..................................................................... 165 Close .................................................................... 166 Save...................................................................... 166 Save As ................................................................ 166 New Molecule...................................................... 167 Delete Molecule................................................... 167 Append Molecule(s)............................................. 167 Print...................................................................... 167

TOC 7/20/06, 1:42 PM4

Chapter 5 67

Chapter 5Groups of Organic Molecules

This chapter introduces and illustrates a number of basic operations involved in processing groups of molecules, as well as the associated spreadsheet for organizing and fitting data and facilities for making plots.

Computational investigations like experimental investigations are rarely restricted to single molecules, but rather may involve a series of related molecules. Here, it may be of interest to compare geometries, energies or other calculated properties, or to compare trends in calculated and measured properties. Spartan provides facilities for these purposes. In particular, it allows molecules to be grouped, either manually, or automatically as a result of a conformational search, from following a particular vibrational motion, or from a scan of one or more geometrical variables. Once put into a group, molecules may be aligned based either on their structures or chemical functionalities. Calculations may be performed either on individual molecules or, just as simply, on the complete group of molecules.

Associated with a group (including a group of one molecule) is a spreadsheet. This allows convenient access to virtually any calculated quantity. Additionally, data may be entered manually into the spreadsheet or transferred from another application such as Excel. Data in the spreadsheet may be manipulated, linear regression analyses performed and both 2D and 3D plots constructed. Alternatively, the data in a Spartan spreadsheet may be transferred to Excel for further analysis.

The tutorials in this chapter introduce a number of the basic group operations available in Spartan. These include building a group from scratch, and processing groups resulting from a conformational search and from varying a torsion angle. Also provided is an example of fitting an experimental observable to one or more calculated properties by way of linear regression.

Chapter 5.1 7/20/06, 2:33 PM67

Chapter 5 69

Dienophiles in Diels-Alder Cycloadditions

The most common Diels-Alder reactions involve electron-rich dienes and electron-deficient dienophiles.

X

Y

X = R, OR

Y = CN, CHO, CO2HX

Y

+

The rate of these reactions generally increases with increasing π-donor ability of the diene substituent, and with increasing π−acceptor ability of the dienophile substituent. This can be rationalized by noting that donor groups raise the energy of the highest-occupied molecular orbital (HOMO) on the diene, while acceptor groups lower the energy of the lowest-unoccupied molecular orbital (LUMO) on the dienophile. Thus, the HOMO-LUMO gap is reduced, leading to enhanced stabilization.

HOMO

LUMO

Orbital Energy

diene dienophile

better acceptor groups

better donor groups

To test such an hypothesis, you will examine the extent to which experimental relative rates of Diels-Alder cycloadditions involving cyclopentadiene and a variety of cyanoethylenes correlate with dienophile LUMO energies.

1. Bring acrylonitrile that you previously constructed onto the screen. Select Open... from the File menu (or click on ), and double click on acrylonitrile in the file browser which results.* Place acrylonitrile onto the clipboard, by first drawing a selection box. Position the cursor slightly above and slightly to the left of the molecule then, while holding down on both left and

2 mins

* If the electrostatic potential map is displayed, enter the Surfaces dialog (Display menu) and click on the line density potential... to turn it “off”.

Chapter 5.1 7/20/06, 2:33 PM69

70 Chapter 5 Chapter 5 71

right buttons, drag the mouse to a position slightly below and slightly to the right of the molecule. Finally, release both buttons. Acrylonitrile will be surrounded by dotted lines (a selection box).

Select Copy from the Edit menu. Close acrylonitrile ( ).

2 Bring up the organic model kit ( ). Click on Clipboard at the bottom of the model kit. The structure of acrylonitrile that you copied to the clipboard will appear in a box at the top of the model kit. (You can manipulate this structure by positioning the cursor inside the box and using the usual mouse/keyboard commands.) Click anywhere on screen and acrylonitrile (the contents of the clipboard) will be drawn. Click on to produce a refined structure.

3. Select New Molecule from the File menu. This requests that a molecule that has yet to be built is to be appended onto the end of a group of molecules in which the currently selected molecule (acrylonitrile) is a member. The screen will be cleared. Click on Clipboard at the bottom of the model kit and click on screen. Acrylonitrile will appear. Click on Groups in the model kit, select Cyano and add to the appropriate free valence on acrylonitrile to make 1,1-dicyanoethylene. Click on .

4. Repeat this procedure (New Molecule, followed by Clipboard, followed by Groups, followed by ) four more times to build cis and trans-1,2-dicyanoethylene, tricyanoethylene and tetracyanoethylene. When you are all done (six molecules in total), click on to remove the model kit.

Chapter 5.1 7/20/06, 2:33 PM70


5. The molecules have been put into a single group, allowing calculated properties to be accessed via a spreadsheet. Select Spreadsheet from the Display menu.

To select individual molecules, click on their labels (M001, ...) in the left hand column, or use the and keys at the bottom left of the screen. You may change the default labels to proper names (acrylonitrile, ...) by directly typing in new names into cells under the Label column. Press the Enter key following each entry.

6. Enter the Calculations dialog (Setup menu), and specify an energy calculation using the Hartree-Fock 3-21G model. Select AM1 from the Start from menu. This indicates that the 3-21G energy calculation is to be preceded by calculation of geometry using the AM1 model. Note, that Global Calculations is checked. This ensures that the calculations are to be applied to all members of the group (and not just the selected member).

7. Submit the job. Name it Diels-Alder dienophiles. While you are waiting for it to complete, enter the experimental relative rates from the table below into the spreadsheet.

Experimental relative rates for

Diels-Alder cycloadditions of cyclopentadiene*

log10 log10 dienophile (relative rate) dienophile (relative rate)

acrylonitrile 0 1,1-dicyanoethylene 4.64trans-1,2-dicyanoethylene 1.89 tricyanoethylene 5.66cis-1,2-dicyanoethylene 1.94 tetracyanoethylene 7.61

* J. Sauer, H. Weist and A. Mielert, Chem. Ber., 97, 3183 (1964).

Chapter 5.1 7/20/06, 2:33 PM71


First, type a title Log(rate)= into the header cell at the top of a blank column in the spreadsheet and press the Enter key. Then enter the rate data in the appropriate cells. Press the Enter key after each entry. Sort the group according to relative rate. Click on the column header Log(rate), and then click on Sort at the bottom of the spreadsheet.

8. After all calculations are complete, click on a header cell at the top of a blank column. Then, click on Add... at the bottom of the spreadsheet. Alternatively, hold down the right mouse button while the cursor is positioned inside the spreadsheet, and select Add... from the menu which results.

Select E LUMO from the list in the dialog that results and eV from the Energy menu. Click on OK.

Chapter 5.1 7/20/06, 2:33 PM72


The spreadsheet has now served its purpose by collecting both calculated (LUMO energy) and experimental (relative rate) data for all six dienophiles. You can, if you wish, now remove it from the screen (click on at the top right).

9. Select Plots... from the Display menu. This leads to the Plots dialog.

Make certain that the XY Plot tab at the top of the dialog is selected. Select Log(rate) from the list of items in the X Axis menu and E LUMO(eV) from the Y Axes list. Click on OK to remove the dialog and show the plot.

The default curve, a so-called cubic spline, smoothly connects the data points, but is of little value in establishing a correlation. A better representation is provided by a simple linear least-squares fit. Select Properties from the Display menu and click anywhere

Chapter 5.1 7/20/06, 2:33 PM73


on the curve that you have just drawn. The Curve Properties dialog appears.

Select Linear LSQ from the Fit menu. A new plot appears.

Note that there is a correlation between relative rates and LUMO energy.

10. Remove all molecules and any remaining dialogs from the screen.

Chapter 5.1 7/20/06, 2:33 PM74


Allyl Vinyl Ether

O

Allyl vinyl ether undergoes Claisen rearrangement, the mechanism for which demands a chair arrangement of the reactant.

O O

O

‡

Is this the lowest-energy conformer (global minimum) or is additional energy required to properly orient the molecule for reaction? To find out, you need to locate all the conformers of allyl vinyl ether, identify the best chair structure and evaluate its energy relative to that of the actual global minimum. You will first carry out a conformational search using the MMFF molecular mechanics model, and then obtain relative conformer energies using single-point 6-31G* Hartree-Fock calculations based on AM1 equilibrium geometries.

1. Bring up the organic model kit and construct allyl vinyl ether (in any conformation) from a sequence of sp2 and sp3 carbons and an sp3 oxygen. Click on to produce a refined geometry. Click on .

2. Bring up the Calculations dialog and select Conformer Distribution from the top menu to the right of Calculate and Molecular Mechanics and MMFF from the two bottom menus. Click on OK.

3. Submit the job. Name it allyl vinyl ether. When completed, it will give rise to a series of low-energy conformers* placed in a new file allyl vinyl ether.M001.** Open this file.*** Select Spreadsheet from the Display menu.

* By default, only conformers within 40 kJ/mol of the global minimum will be kept. This can be changed (see Conformational Search in Appendix C).

** M001 is the default label of the molecule you built. You can change it by altering the Labels field in the Molecule Properties dialog (Properties under the Display menu; Chapter 15).

*** To avoid confusion, it is a good idea to close the original file allyl vinyl ether.

5 mins

Chapter 5.1 7/20/06, 2:33 PM75


Size the spreadsheet such that all rows (corresponding to different conformers) are visible at one time. Click on Add... at the bottom of the spreadsheet. Select E from the list at the top of the dialog which appears, kJ/mol from the Energy menu and click on OK. Energies for each of the different conformers will be added to the spreadsheet. Examine the lowest-energy conformer (the top entry). Is it in a chair conformation suitable for Claisen rearrangement? If not, identify the lowest-energy conformer which is suitable. You can keep two or more conformers on screen at the same time by checking the boxes immediately to the right of the molecule labels (the leftmost column) in the spreadsheet. To get a clearer idea of structural similarities and differences, align the conformers. Select Align from the Geometry menu (or click on the icon at the top of the screen) and select Structure from the Align by menu at the bottom right of the screen. A message will appear at the bottom left of the screen.

Click on the two carbons of the vinyl group and on the oxygen to designate them as alignment centers. Each will be marked by a red circle. If you make a mistake, click on the circle and it will disappear. When you are done, click on the Align by button at the bottom right of the screen. Click on .

4. To obtain a better estimate of conformational energy differences, perform 6-31G* energy calculations using AM1 geometries. Make a copy of allyl vinyl ether.M001 ( ), and name it allyl vinyl ether Hartree-Fock. Using the copy, delete all conformers except the global minimum and the lowest-energy conformer poised for Claisen rearrangement. Select each conformer to be discarded, and then select Delete Molecule from the File menu. Alternatively, click on the cell in the spreadsheet containing the label for the molecule to be deleted, and then click on Delete at the bottom of the spreadsheet. When you are done, the spreadsheet should contain only two entries.

Chapter 5.1 7/20/06, 2:33 PM76


5. Enter the Calculations dialog, and specify calculation of an energy using the Hartree-Fock 6-31G* model. Also, specify AM1 from the Start from menu. Make certain that Global Calculations at the bottom of the dialog is checked to specify that the dialog settings apply to both conformers.

6. Submit the job. When completed, examine the conformer energies. How much energy is needed to go from the global minimum to a conformer poised to undergo Claisen rearrangement?


Chapter 5.1 7/20/06, 2:33 PM77


Internal Rotation in Dimethylperoxide

OOCH3

CH3

Quantum chemical calculations, in particular, Hartree-Fock molecular orbital calculations, density functional calculations and MP2 calculations, may be called on to furnish data to parameterize empirical energy functions for use in molecular mechanics and/or molecular dynamics calculations. Most important are data relating to torsional motions, for it is here that experimental data are most scarce. Note that the empirical energy function needs to reflect the inherent periodicity. One suitable choice is a truncated Fourier series.

Etorsion (ω) = ktorsion1 (1 - cos(ω - ωeq)) + ktorsion2 (1 - cos2(ω - ωeq))+ ktorsion3 (1 - cos3(ω - ωeq))

Here, ωeq is the ideal dihedral angle and ktorsion1, ktorsion2 and ktorsion3 are parameters. The first (one-fold) term accounts for the difference in energy between syn and anti conformers, the second (two-fold) term for the difference in energy between planar and perpendicular conformers, and the third (three-fold) term for the difference in energy between eclipsed and staggered conformers.

In this tutorial, you will generate a potential energy function for rotation about the oxygen-oxygen bond in dimethylperoxide using 6-31G* Hartree-Fock calculations (and optionally from B3LYP/6-31G* density functional calculations) and then fit this potential to a truncated Fourier series.

1. Build dimethylperoxide. If the molecule is not already in an anti conformation, click on and set the COOC dihedral angle to 180 (180°) by typing 180 in the box at the lower right of the screen and pressing the Enter key. Do not minimize.

2. Select Constrain Dihedral from the Geometry menu (or click on the icon at the top of the screen). Select the COOC torsion, and then click on at the bottom right of the screen. The icon will change to indicating that a dihedral constraint

30 mins/40 mins

Chapter 5.2 7/20/06, 2:34 PM78


is to be applied. Select Properties (Display menu) and click on the constraint marker on the model on screen. This leads to the Constraint Properties dialog.

3. Check Dynamic inside the dialog. This leads to an extended form of the Constraint Properties dialog that allows the single (dihedral angle) constraint value to be replaced by a sequence of constraint values.

Leave 180 (180°) in the box to the right of Value alone, but change the contents of the box to the right of to to 0 (0°). You need to press the Enter key after you type in the value. Steps should be 10. (If it is not, type 10 and press the Enter key.) What you have specified is that the dihedral angle will be constrained first to 180°, then to 160°*, etc. and finally to 0°. Click on to dismiss the dialog.

4. Bring up the Calculations dialog and select Energy Profile from the top menu to the right of Calculate, and Hartree-Fock and 6-31G* from the two bottom menus. Click on OK and submit the job. Name it dimethylperoxide.

* The difference between constraint values is given by: (final-initial)/(steps-1).

Chapter 5.2 7/20/06, 2:34 PM79


5. When the calculations have completed, they will go into a file named dimethylperoxide.M001. Open this file. Align the conformers. Click on , select Structure from the Align by menu at the bottom right of the screen and, one after the other, click on both oxygens and on one of the carbons. Then click on the Align by button at the bottom right of the screen. Bring up the spreadsheet (Display menu), and enter both the energies relative to the 180° conformer, and the COOC dihedral angles. First click on the label (M001) for the top entry in the spreadsheet (the 180° conformer), then click on the header cell for the leftmost blank column, and finally, click on Add... at the bottom of the spreadsheet. Select rel. E from the quantities in the dialog that results, kJ/mol from the Energy menu and click on OK. To enter the dihedral angle constraints, click on , click on the constraint marker attached to dimethylperoxide and click on at the bottom of the screen (to the right of the value of the dihedral angle). Click on .

6. Select Plots... (Display menu). Select Constraint(Con1) from the X Axis menu and rel. E(kJ/mol) from the Y Axes list. Click on OK.

The curve (a so-called cubic spline) smoothly connects the data points. To fit the points to a Fourier series, bring up the Properties dialog (Display menu) and click anywhere on the

Chapter 5.2 7/20/06, 2:34 PM80


plot. Select Fourier LSQ from the Fit menu and click on to remove the dialog. A new plot appears.

The actual fit expression appears at the bottom of the plot.

7 and 8 optional (cannot be completed with the Essential Edition)

7. Use energies from B3LYP/6-31G* density functional calculations, together with the Hartree-Fock geometries to provide a better fitting function. First, make a copy of dimethylperoxide.M001. Name it dimethylperoxide density functional. Next, enter the Calculations dialog and specify calculation of energies using the B3LYP/6-31G* density functional model. Make certain that Global Calculations (at the bottom of the dialog) is checked to signify that energy calculations are to be performed on all conformers.

8. Submit the job. It will require several minutes to complete. When it is done, draw a new energy plot and compare it to the energy plot produced earlier.


Chapter 5.2 7/20/06, 2:34 PM81


Hydration of Carbonyl Compounds

The hydration of carbonyl compounds has been extensively studied primarily because it serves as a model for a number of important reactions, nucleophilic addition to carbonyl compounds among them.

OR

R‘ H2O-H2O

R

R‘ OH

OHKeq =

[hydrate][H2O] [carbonyl]

=[hydrate]

55.5 [carbonyl]

Experimental Keq for hydration of carbonyl compounds*

log(keq/55.5) log(keq/55.5)

PhCOCH3 -6.8 CF3COCH3 -0.2 CH3COCH3 -4.6 PhCOCF3 0.1 PhCHO -3.8 H2CO 1.6 t-BuCHO -2.4 CF3CHO 2.7 CH3CHO -1.7 CF3COCF3 4.3 * J.P. Guthrie, Can. J. Chem., 53, 898 (1975); 56, 962 (1978).

In this tutorial, you will use Spartan’s linear regression analysis tool to correlate calculated properties of carbonyl compounds with measured equilibrium constants for their hydration.

1. Build the compounds* listed above. Use Carbonyl (Groups menu) as a starting point for each to ensure consistent atom numbering. Put in one list; select New Molecule following construction of each molecule, that is, build and minimize acetophenone, select, New Molecule, build and minimize acetone, select New Molecule, etc. Click on .

2. Enter the Calculations dialog and specify calculation of equilibrium geometry using the Hartree-Fock 3-21G model.**

* Keep the phenyl ring and the carbonyl group coplanar for benzaldehyde, acetophenone and trifluoroacetophenone.

* * You can save computer time by specifying calculation of an energy using the Hartree-Fock 3-21G model, and then selecting AM1 from the Start from menu. This requests a semi-empirical AM1 geometry in place of the 3-21G geometry. As all of these molecules are in SMD, you can save even more time by bringing up the SMD Preview dialog (click on at the bottom right of the screen), clicking on the yellow square to the left of 3-21G and then clicking on Replace All at the bottom of the dialog. In this case, skip the calculation step as all the results you require are already available.

30 mins

Chapter 5.2 7/20/06, 2:34 PM82


Make certain that Global Calculations is checked. Click on Submit. Name the job carbonyl compounds.

3. While you are waiting for the calculations to complete, bring up the spreadsheet. Click inside the header cell of an empty column, type Log(Keq) and press the Enter key. Then enter the experimental equilibrium constants into the appropriate cells. You need to press the Enter key following each data entry. Sort the list according to the value of Log(Keq). Click inside the header cell LogKeq and then click on Sort at the bottom of the spreadsheet. If you like, replace the labels in the spreadsheet (M001, ...) by correct names (acetophenone, ... ).

4. After the calculations have completed, click inside the header cell for an empty column, and then click on Add... at the bottom of the spreadsheet. Select (click on) the following quantities in the dialog: E LUMO, E HOMO, Dipole, CPK Volume and Molecular Weight. Click on OK.

5. Select Properties (Display menu) and click on the oxygen atom for whatever compound is displayed. Click on to the left of Electrostatic (under Charges) in the Atom Properties dialog (that has replaced the Molecular Properties dialog), to place oxygen charges into the spreadsheet.

6. Click on Add... at the bottom of the spreadsheet, and then click on the Linear Regression tab at the top of the dialog that results. This brings up the Add Linear Regression dialog.

Chapter 5.2 7/20/06, 2:34 PM83


Select Log(Keq) from the Fit menu and Electrostatic (OX)* from the Using list. Click on OK. Two new rows will be added at the bottom of the spreadsheet. With a Properties dialog on screen, click anywhere on the row in the spreadsheet labelled Fit1 (the first fit performed on this data set). The Regression Properties dialog appears.

This provides information about the fit of Log(Keq) to the charge on oxygen, in particular, the value of R2. This will tend to unity as the quality of the fit improves.

7. Inside the Regression Properties dialog, try fitting to different (single) properties in the Using list. (The entries toggle “on” and “off” with repeated clicking.) Maximize the value of R2. Next, try all combinations of two properties. The combination of E LUMO and Electrostatic (OX) should give the highest value of R2. With this combination selected, bring up the Plots dialog (Display menu) and select Log(Keq) from the X Axis menu, and FitVals (Fit1) from the Y Axes list and click on OK. Next, click on the curve (not on the axes) to bring up the Curve Properties dialog (which replaces the Regression Properties dialog). Select Linear LSQ (to replace Cubic Spline) from the Fit menu. The resulting plot should show good correlation.

8. Close carbonyl compounds and remove any remaining dialogs from the screen.

* Numbering depends on the order that atoms were introduced during building, but the reference is to the carbonyl oxygen.

Chapter 5.2 7/20/06, 2:34 PM84


Acidities of Carboxylic Acids

Acid strength is among the most important molecular properties. It is readily available from calculation, either in terms of absolute deprotonation energy,

AH A– + H+

or, more commonly, as the deprotonation energy relative to that of some standard acid (A°H).

AH + A°– A– + A°H

An alternative measure of (relative) acid strength is the value of the electrostatic potential in the vicinity of the acidic hydrogen in the neutral acid. Electrostatic potential maps are known to uncover gross trends in acidity, for example, the acidic hydrogen in a strong acid, such as nitric acid, is more positive than that in a weak acid, such as acetic acid, which in turn is more positive than that in a very weak acid, such as ethanol.

ethanol acetic acid nitric acid

In this tutorial, you will explore the use of electrostatic potential maps to quantify changes in acid strength due to subtle variations in structure. You will use the Spartan Molecular Database of calculated structures, energies and properties may be used to shorten the time required for an investigation.

1. One after the other, build trichloroacetic, dichloroacetic, chloroacetic, formic, benzoic, acetic and pivalic acids. Put all molecules into the same document (use New Molecule from the File menu following the first molecule). Click on when you are done.

10 mins

Chapter 5.2 7/20/06, 2:34 PM85


2. All of the molecules that you have built are available in the Spartan Molecular Database. Click on to the left of the name of the presently selected molecule at the bottom of the screen. The SMD Preview dialog will appear. This displays the structure of the selected molecule on the left and a list of theoretical models for which calculated results are available on the right. Check the yellow square to the left of 3-21G and click on Replace All at the bottom of the dialog. Structures obtained from Hartree-Fock 3-21G calculations will replace those you have built. However, you need to calculate a wavefunction for each of the molecules in order to produce electrostatic potential maps.

3. Enter the Calculations dialog (Setup menu). Request a Hartree-Fock 3-21G energy calculation, and make certain that Global Calculations is checked. Click on OK. Enter the Surfaces dialog (Setup menu). Make certain that Global Surfaces at the bottom of the dialog is checked. Click on Add... at the bottom of the dialog, select density from the Surface menu and potential from the Property menu (medium from the Resolution menu) in the Add Surface dialog that appears and then click on OK. Leave the Surfaces dialog on screen. Submit the job (Submit from the Setup menu). Name it carboxylic acids.

4. While you are waiting for the calculations to complete, bring up the spreadsheet (Spreadsheet from the Display menu). Expand it so that you can see all seven molecules, and that three data columns are available. Click inside the header cell of a blank column. Click on Add... at the bottom of the spreadsheet, select Name from the list of entries and click on OK. The name of each molecule will appear. Next, click inside the header cell of the next available data column, type pKa and press the Enter key. Enter the experimental pKa’s (given on the next page) into the appropriate cells under this column. You need to press the Enter key following each entry.

Chapter 5.2 7/20/06, 2:34 PM86


acid pKa acid pKa

trichloroacetic (Cl3CCO2H) 0.7 benzoic (C6H5CO2H) 4.19 dichloroacetic (Cl2CHCO2H) 1.48 acetic (CH3CO2H) 4.75 chloroacetic (ClCH2CO2H) 2.85 pivalic ((CH3)3CCO2H) 5.03 formic (HCO2H) 3.75

Experimental data from: E.P. Sargeant and B. Dempsey, Ionization Constants of Organic Acids in Aqueous Solution, IUPAC no. 23, Permagon Press, 1979.

5. After all calculations have completed, arrange the seven molecules on screen such that you can clearly see the acidic hydrogen on each. To display all molecules at once, check the box on the right hand side of the Label column (in the spreadsheet) for each entry. To manipulate them independently of one another, select Coupled from the Model menu.

6. Bring up the Surfaces dialog (Setup or Display menu) and select density... to turn on the electrostatic potential map for each molecule. Bring up the Properties dialog (Display menu) and click on to the right of Maximum. The maximum value of the electrostatic potential (corresponding to the acidic proton) will be pasted to the spreadsheet.

7. Plot experimental pKa vs. maximum in the electrostatic potential. Bring up the Plots dialog (Display menu), select pKa under the X Axis menu and Property Max (Surface) from the Y Axes list, and click on OK. The data points are connected by a smooth curve (a so-called cubic spline). To get a least squares fit, select Properties from the Display menu, click on the curve, and select Linear LSQ from the Fit menu in the Curve Properties dialog.

8. Close carboxylic acids and any remaining dialogs from the screen.

Chapter 5.2 7/20/06, 2:34 PM87


30 mins

Stereochemical Assignments from 13C Spectra(cannot be completed with the Essential Edition)

NMR spectroscopy, in particular 13C spectroscopy, is without doubt the method of choice to establish the three-dimensional structure of organic molecules. Only X-ray diffraction provides more definitive results, although the requirement of a crystalline sample severely limits its application. It is now practical to routinely calculate the NMR spectra of organic molecules. In time, the availability of a “virtual NMR spectrometer” will offer organic chemists an entirely new paradigm for structure determination, that is direct comparison of a measured spectrum with calculated spectra for one or more chemically reasonable candidates.

Here, you will obtain 13C chemical shifts for cis and trans stereoisomers of 2-hexene and endo and exo stereoisomers of 2-methylnorbornane, and compare these to experimental shifts. The Hartree-Fock 6-31G* model will be employed, but you will get all structures from the Spartan Molecular Database. You will establish the extent to which the calculations are able to reproduce differences in chemical shifts as a function of stereochemistry.

1. Build cis-2-hexene. Click on at the bottom of the screen (to the left of the molecule name), check the box to the left of the 6-31G* in the SMD Preview dialog that appears and click on Replace. Do not minimize. Select New Molecule from the File menu, build trans-2-hexene, check the box to the left of 6-31G* in the preview dialog and click on Replace. Do not minimize. Click on .

2. Enter the Calculations dialog and specify a Hartree-Fock energy calculation using the 6-31G* basis set. Check NMR to the right of Compute. Submit the job with the name 2-hexene.

3. When completed, examine the 13C chemical shifts, either from the Atom Properties dialog or as atom labels (Configure... under the Model menu and check Chem Shift under Atom in the dialog which results). Compare them to measured values with particular attention to the difference in shifts between cis and trans stereoisomers.

Chapter 5.2 7/20/06, 2:34 PM88


cis trans ∆

C1 11.4 16.5 5.1 C2 122.8 123.9 1.1 C3 129.7 130.6 0.9 C4 28.2 34.1 5.9 C5 21.4 22.1 0.7 C6 12.5 12.5 0.0

Do the calculations distinguish between the two isomers?

4. Close 2-hexene.

5. Build endo-2-methylnorbornane. As you did for 2-hexene, replace your structure with the (6-31G*) structure in the Spartan Molecular Database. Select New Molecule (File menu) and repeat the process for the exo isomer. Click on .

6. Inside the Calculations dialog, specify a Hartree-Fock 6-31G* energy calculation, and check NMR (to the right of Compute). Submit the job with the name 2-methylnorbornane.

7. When completed, compare the 13C chemical shifts with experimental values with particular attention to differences between endo and exo stereoisomers.

endo exo ∆

C1 11.4 16.5 5.1

C1 11.4 16.5 5.1

C1 43.5 42.2 -1.3 C2 36.8 34.6 -2.2 C3 40.2 40.7 0.5 C4 37.3 38.2 0.9 C5 30.3 30.6 0.3 C6 29.0 22.4 -6.6 C7 35.0 38.9 3.9 CH3 22.3 17.4 -4.9

Do the calculations distinguish between the two isomers?

8. Close 2-methylnorbornane.

CH3 CH CHCH2CH2CH3

CH3

4

1 2

35

6

7

Chapter 5.2 7/20/06, 2:34 PM89


Tautomers of Nucleotide Bases

Protons bound to heteroatoms in heterocyclic compounds are likely to be very mobile in solution and, where two or more heteroatoms are present in a structure, different isomers (tautomers) may be in equilibrium. As a case in point, consider the nucleotide base cytosine (where a methyl group has replaced the sugar-phosphate backbone).

N

N

O

CH3

N

NH

O

CH3

N

N

OH

CH3

NH2 NH NH

The existence of a low-energy tautomer could have far-reaching consequences, given that the valence structure of cytosine is key to hydrogen bonding in DNA. In this tutorial, you will examine the possible tautomers of 1-methylcytosine for evidence of low-energy structures.*

1. Build 1-methylcytosine. Start with Amide from the Groups menu. Select planar trigonal nitrogen for the external amino group. Minimize (to Cs symmetry) and click on .

2. Note that the word Tautomer appears at the bottom right of the screen, indicating that tautomers exist. Select Tautomers from the Search menu (or click on the icon at the top of the screen). Step through the tautomers using the and keys that appear at the bottom right of the screen. To put the tautomers in a list, click on at the bottom right of the screen and click on OK in the dialog that results.

3. Enter the Calculations dialog with the list of tautomers (close the molecule you built to get the tautomers). Specify calculation

5 mins/30 mins

* Note, however, that were the energy of an alternative tautomer only 10 kJ/mol higher than that for the normal structure, this would translate into a relative abundance of only about 1% at room temperature. Thus, any alternative tautomers would need to be very close in energy to the lowest-energy tautomer to have noticeable effect.

Chapter 5.2 7/20/06, 2:34 PM90


of equilibrium geometry using the 3-21G Hartree-Fock model. Submit the job with the name cytosine tautomers.

4. After the calculations have completed, bring up the spreadsheet and click on the line corresponding to the structure for cytosine that you originally built. Click on Add at the bottom of the spreadsheet, select rel.E and Boltzmann from the available quantities and kJ/mol from the Energy menu and click on OK. Are either of the alternatives close in energy to the normal form of cytosine?

5 and 6 optional (cannot be completed with the Essential Edition)

To get a better estimate of relative tautomer energies, perform energy calculations using the RI-MP2/6-31G* model.

5. Make a copy of cytosine tautomers. Name it cytosine tautomers RI-MP2. Enter the Calculations dialog (Setup menu) with this copy. Specify Energy from the top menu to the right of Calculate and RI-MP2 and 6-31G* from the two bottom menus. Submit the job.

6. When the calculations have completed, bring up the spreadsheet and again examine relative tautomer energies and Boltzmann populations. Are either of the alternative structures competitive?

7. Remove all molecules and remaining dialogs from the screen.

Chapter 5.2 7/20/06, 2:34 PM91

Chapter 6 93

Chapter 6Organic Reactions

This chapter outlines and illustrates strategies for locating and verifying transition states for organic reactions as well as for exploring changes in product distributions as a function of substituents and reactant stereochemistry.

The treatment of chemical reactions adds an entirely new dimension to the application of quantum chemical models. Whereas unique valence structures may generally be written for most molecules and, based on these structures, reasonable guesses at bond lengths and angles may be made, it is often difficult to designate appropriate valence structures for transition states, let alone specify detailed geometries. This is, for the most part, due to a complete absence of experimental data on the structures of transition states. However, calculated transition-state geometries are now commonplace. Spartan provides both an extensive library of calculated transition states and a facility for automatically matching an entry in this library with the reaction at hand.* This library is also available as the Spartan Reaction Database (SRD), and may be searched by substructure to yield all available transition states of reactions related to the one of interest.

Spartan also provides a procedure for driving user-defined coordinates. Aside from conformational analysis (see discussion in previous chapter), the major application of this is to force reactions, thereby permitting identification of transition states.

The first four tutorials in this chapter illustrate Spartan’s automatic procedure for guessing transition state geometries based on its library of organic reactions. The last two of these show how a transition state for one reaction may be used to provide a good guess at the transition state of a closely-related reaction. These tutorials also illustrate the

* Where a reaction is unknown to Spartan’s library, a fallback technique which averages reactant and product geometries (similar to the so-called linear synchronous transit method) is invoked.

Chapter 6 df 7/20/06, 2:57 PM93


use of vibrational analysis (infrared spectroscopy) to verify that a particular structure corresponds to a transition state and to show the motion connecting it to reactants and products. The fifth tutorial illustrates how a reaction may be “driven” through a transition state. This tutorial, along with the fourth tutorial, draw the connection between relative activation energies and kinetic product distributions. The last tutorial in this chapter illustrates how transition states may be extracted from the Spartan Reaction Database.

An example of a transition state calculation for an organometallic reaction is provided in Chapter 9.

Chapter 6 df 7/20/06, 2:57 PM94


Ene Reaction of 1-Pentene

C

CC

C

C C

CH

HC

CHHH

H

1-pentene ethylene propene

H

C

H

H

∆ +

H H

H HH

H

H

HHH H

The proposed mechanism of the ene reaction involves simultaneous transfer of a hydrogen atom and CC bond cleavage. Here, you will obtain the transition state for the ene reaction of 1-pentene from 3-21G Hartree Fock calculations, and examine the reaction coordinate for evidence of concerted motion. Optionally, you will perform B3LYP/6-31G* density functional energy calculations to estimate the activation energy for the reaction.

1. Bring up the organic model kit and build 1-pentene in a conformation in which one of the terminal hydrogens on the ethyl group is poised to transfer to the terminal methylene group. Minimize (click on ) and click on .

2. First, save 1-pentene as 1-pentene density functional for optional use later ( )*. Also make a copy for immediate use; name it ene reaction 1-pentene.

3. Select Transition States from the Search menu (or click on the icon at the top of the screen). Click on bond a in the figure

on the following page and then click on bond b. A curved arrow from bond a to bond b will be drawn.

C

CC

C

C

HHH

H H

HHHH H

a

bc

d

e

Next, click on bond c and then on bond d. A second curved arrow from bonds c to d will be drawn. Finally, click on

10 mins/20 mins

* This is not necessary for users of the essential edition of Spartan, where density functional models are not available.

Chapter 6 df 7/20/06, 2:57 PM95


bond e and, while holding down the Shift key, click on the (methyl) hydrogen to be transferred and on the terminal (methylene) carbon to receive this hydrogen. A third curved arrow from bond e to the center of a dotted line that has been drawn between the hydrogen and oxygen will appear.

If you make a mistake, you can remove an arrow by selecting Delete from the Build menu (click on ) and then clicking on the arrow. (You will need to select to continue.) Alternatively, hold down the Delete key as you click on an arrow. With all three arrows in place, click on at the bottom right of the screen. Your structure will be replaced by a guess at the ene transition state. If the resulting structure is unreasonable, then you have probably made an error in the placement of the arrows. In this case, select Undo from the Edit menu to return to the model with the arrows and modify accordingly.

4. Enter the Calculations dialog (Setup menu), and specify calculation of transition-state geometry using the 3-21G Hartree-Fock model. Select Transition State Geometry from the top menu to the right of Calculate and Hartree-Fock and 3-21G from the two bottom menus. Also, check IR to the right of Compute. This will allow you to confirm that you have found a transition state, and that it smoothly connects reactant and product. Click on Submit.

5. When the job completes, animate the motion of atoms along the reaction coordinate. Select Spectra from the Display menu and click on the IR tab. Click on the top entry in the list in the IR dialog that results. It corresponds to an imaginary frequency, and will be designated with an i in front of the number.

A vibrational frequency is proportional to the square root of a quantity that reflects the curvature of the potential surface divided by a mass. At a transition state (the “top of a hill”), the curvature is negative (it “points down”). Since mass is positive, the quantity inside the square root is negative and the frequency is an imaginary number.

Chapter 6 df 7/20/06, 2:57 PM96


Is the vibrational motion consistent with an ene reaction of interest and not with some other process?

6. Controls at the bottom of the IR dialog allow for changing both the amplitude of vibration (Amp) and the number of steps that make up the motion (Steps). The latter serves as a speed control. Change the amplitude to 0.3 (type 0.3 in the box to the right of Amp and press the Enter key). Next, click on Make List at the bottom of the dialog. This will give rise to a group of structures that follow the reaction coordinate down from the transition state both toward reactant and product. Remove the original transition state from the screen. Click on ene reaction 1-pentene (the vibrating molecule) and close it, along with the IR dialog.

7. Enter the Calculations dialog and specify calculation of energy using the 3-21G Hartree-Fock model (the same level of calculation used to obtain the transition state and calculate the frequencies). Make certain that Global Calculations is checked. Next, enter the Surfaces dialog and specify evaluation of two surfaces: a bond density surface and a bond density surface onto which the electrostatic potential has been mapped. Click on Add . . ., select density (bond) for Surface and none for Property and click on Apply. Select density (bond) for surface and potential for Property and click on OK. Make certain that Global Surfaces is checked.

8. Submit for calculation. Name it ene reaction 1-pentene sequence. Once the job has completed, enter the Surfaces dialog and, one after the other, select the surfaces that you have calculated. For each, step through the sequence of structures ( and ) keys at the bottom of the screen) or animate the reaction ( ). Note, in particular, the changes in bonding revealed by the bond density surface. Also pay attention to the value of the potential on the migrating atom. This reflects its charge. Is it best described as a proton (blue), hydrogen atom (green) or hydride anion (red)?

9. Close ene reaction 1-pentene sequence as well as any remaining dialogs.

Chapter 6 df 7/20/06, 2:57 PM97


10 to 14 optional (cannot be completed with the Essential Edition)

Methods that account for electron correlation are generally needed to furnish accurate estimates of absolute activation energies. Perform B3LYP/6-31G* density functional energy calculations on both 1-pentene and on the ene reaction transition state (using 3-21G geometries).

10. Open ene reaction 1-pentene ( ) and make a copy ( ). Name it ene reaction 1-pentene density functional.

11. Enter the Calculations dialog, and specify calculation of energy using the B3LYP/6-31G* density functional model. Remove the checkmark from IR. Submit the job.

12. Open 1-pentene density functional ( ). Specify calculation of energy using the B3LYP/6-31G* density functional model. Also select 3-21G from the Start from menu. This designates that a 3-21G equilibrium structure is to be used (that is, the same as employed for the transition state geometry). Submit the job.

13. Obtain the activation energy (difference in total energies between 1-pentene and the ene reaction transition state).

14. Remove any molecules and dialogs from the screen.

Chapter 6 df 7/20/06, 2:57 PM98


SN2 Reaction of Bromide and Methyl Chloride

Br– + C Cl

H

H

H+ Cl–CBr

H

H

HC

H

H H

ClBr

1. The SN2 reaction passes through a transition state in which carbon is in a trigonal bipyramid geometry and the entering and leaving groups are colinear. To build it, first construct methyl chloride. Then click on the Insert button at the bottom right of the screen (or hold down the Insert key), select bromine from the palette of icons in the model kit and click anywhere on screen. Two detached fragments, methyl chloride and hydrogen bromide, appear on screen. Click on and then click on the free valence on bromine. Click on . Alternatively, hold down the Delete key and click on the free valence on bromine. You are left with methyl chloride and bromine atom (bromide). Manipulate the two such that bromide is poised to attack methyl chloride from the backside (as in the transition state above). (Recall that translations and rotations normally refer to both fragments, but can be made to refer to a single fragment by first clicking on the fragment and then holding down on the Ctrl key while carrying out the manipulations.) Do not minimize. Click on .

2. Click on . Click on bromide and, while holding down the Shift key, click again on bromide and then on carbon. A dotted line will be drawn from bromine to carbon, together with an arrow from bromine to the center of this line. Next, click on the CCl bond and then click on the chlorine. A second arrow from the carbon-chlorine bond to the chlorine will be drawn.

Br C Cl

Click on at the bottom right of the screen. Your structure will be replaced by a guess at the transition state.

2 mins/5mins

Chapter 6 df 7/20/06, 2:57 PM99


3. Click on and then on the CBr bond. Replace the current CBr distance in the box at the bottom right of the screen by 3.8 (3.8Å) and press the Enter key. You have now made a complex representing the reactant.

4. Select Constrain Distance from the Geometry menu (or click on the icon at the top of the screen). Click on the CBr bond, and then click on at the bottom right of the screen. The icon will change to indicating a constraint is to be applied to this distance. Next, bring up the Properties dialog and click on the constraint marker. The Constraint Properties dialog appears. Click on Dynamic. Leave the value 3.8 (3.8Å) in the box to the right of Value alone, but change the number in the box to the right of to to 1.9 (1.9Å) and press the Enter key. You have requested 10 (the default number) structures with CBr bond lengths from 3.8Å (the starting point) to 1.9Å (the ending point). The transition state should have a CBr distance in between these values. Dismiss the Constraint Properties dialog.

5. Enter the Calculations dialog, and select Energy Profile, Semi-Empirical and AM1 from the appropriate menus to the right of Calculate. You need to change Total Charge to Anion.

6. Submit the job. Name it bromide+methyl chloride. When completed, it will give rise to a sequence of calculations placed in bromide+methyl chloride.M001. Open this file, and align the molecules. Click on , select Structure from the Align by menu at the bottom right of the screen and, one after the other, click on the three hydrogens. Finally press the Align by button at the bottom right of the screen. Click on .

7. Bring up the Spreadsheet and click on Add.... Select E from among the quantities listed at the top of the dialog, kJ/mol from the Energy menu, and click on OK. Next, enter the CBr distances and bromine charges in the spreadsheet. Click on , select the CBr distance and click on at the bottom right of the screen. Click on . Bring up the Properties dialog. Click on bromine and click on to the left of Electrostatic under Charges in the

Chapter 6 df 7/20/06, 2:57 PM100


Properties dialog. Finally, bring up the Plots dialog, and select Distance (C1,Br1) from the X Axis menu, and both E (kJ/mol) and Electrostatic (Br1) from the Y Axes list. Click on OK.

One plot gives the energy as the reaction proceeds and the other gives charge on bromine. Are the two related? Explain.

SN2 reactions involving charged species normally need to be carried out in highly-polar media, for example, water. Spartan accounts for the solvent by way of the SM5.4 model of Cramer and Truhlar (see Appendix A) in which the solvent is represented by a reaction field.

8. Add aqueous phase data to the spreadsheet. Click on an empty column header, click on Add..., select Eaq from the list of available quantities (kJ/mol from the Energy menu), and click on OK. Bring up the Plots dialog and select Distance (C1,Br1) from the X Axis and Eaq(kJ/mol) from the Y Axes list. Click on OK.

9 to 12 optional

9. With bromide+methyl chloride.M001 selected, enter the Calculations dialog. Specify Energy (Semi-Empirical and AM1 should be selected) and click on OK.

10. Enter the Surfaces dialog. Click on Add.... Select density (bond) from the Surface menu and none from the Property menu and

Chapter 6 df 7/20/06, 2:57 PM101


click on Apply. Select density (bond) from the Surface menu, but this time potential from the Property menu. Click on OK.

11. Submit the job. When completed, select density... inside the Surfaces dialog. Click on at the bottom left of the screen to animate the display. Note, that bonds are smoothly broken and formed during the course of reaction. Click on at the bottom of the screen when you are done.

12. Reenter the Surfaces dialog. Turn “off” display of the bond density (select density...), and turn “on” display of the electrostatic potential mapped onto the bond density (select density potential...). Click on . Relate the migration of negative charge during reaction as indicated by colors in the electrostatic potential map to the plot you constructed earlier in step 7. Recall, that colors near red indicate maximum negative potential.


Chapter 6 df 7/20/06, 2:57 PM102


Carbene Additions to Alkenes

C C HH

H H

C

C C

X

H H

X

H H

+ :CX2

Singlet carbenes add to alkenes to yield cyclopropanes. Since a singlet carbene possesses both a high-energy filled molecular orbital in the plane of the molecule, and a low-energy, out-of-plane unfilled molecular orbital, this reaction presents an interesting dilemma. Clearly it would be more advantageous for the low-lying vacant orbital on the carbene, and not the high-lying filled orbital, to interact with the olefin π system during its approach.

C

C CHH

H H

However, this leads to a product with an incorrect geometry. The carbene must “twist” by 90° during the course of reaction.

C

C C

XX

H HH H

C

C C

XX

H HH H

C

C C

XX

H HH H

In this tutorial, you will use the Hartree-Fock 3-21G model to find the transition state and to analyze the motion of the fragments.

1. Bring up the organic model kit and build ethylene.

2. Select from the model kit. Click on the Insert button at the bottom right of the screen (or hold down the Insert key) and then click anywhere on screen. Next, select from the model kit and click on two of the free valences on the sp3 carbon. Next,

5 mins/15 mins

Chapter 6 df 7/20/06, 2:57 PM103


click on and, one after the other, click on the remaining two free valences on the sp3 carbon. Finally, click on . You are left with two fragments, ethylene and difluorocarbene. Orient the two as to be poised for reaction.*

C

C C

FF

Translations and rotations normally refer to the complete set of fragments, but if you click on a fragment (not on a free valence) to select it, and then hold down the Ctrl key they will refer to an individual fragment.

3. Click on . Click on the carbon on the CF2 fragment and then, while holding down the Shift key, click first on the CF2 carbon and then on one of the carbons on the ethylene fragment. A dotted line between the two carbons that are to be bonded will be drawn and an arrow will be drawn from the methylene carbon to the center of this line.

C

C C

FF

Next, click on the CC double bond and then, while holding down the Shift key, click on the other ethylene carbon and then on the CF2 carbon. A second dotted line and second arrow will be added to the structure.

C

C C

FF

* Proper orientation of the two fragments is not crucial in this case, but is primarily to allow you to associate the arrows with the intended reaction. Proper orientation is, however, essential where different stereochemical outcomes are possible.

Chapter 6 df 7/20/06, 2:57 PM104


Click on at the bottom right of the screen. Your structure will be replaced by a guess at the transition state.

4. Enter the Calculations dialog, and specify calculation of a transition-state geometry using the 3-21G Hartree-Fock model. Check IR to the right of Compute, and click on Submit. Name the job difluorocarbene+ethylene.

5. When the job is complete, examine the geometry of the transition state. In light of the previous discussion, would you describe your structure as corresponding to an early or late transition state? Animate the vibration corresponding to the reaction coordinate. Bring up the IR dialog (Spectra under the Display menu and click on the IR tab). Click on the imaginary frequency at the top of the list of frequencies. Does the animation show that the carbene reorients as it approaches the double bond? Turn “off” the animation by again clicking on the imaginary frequency.

6. Select Properties (Display menu) and, in turn, click on each of the four hydrogens in the transition state. Change the value in the Mass Number menu in the Atom Properties dialog from 1 to 2 Deuterium. Resubmit the job. (No additional quantum chemical calculations are involved, but the vibrational analysis needs to be repeated to account for change in atomic masses.) When complete, examine the new set of vibrational frequencies. Note that they are uniformly smaller than those for the undeuterated system, and that the largest changes are for vibrational motions where hydrogens are involved.

7 to 10 optional

7. Make a copy of difluorocarbene+ethylene name it difluoro-carbene+cyclohexene.

8. Bring up the organic model kit with the copy (click on ). Select Freeze Center from the Geometry menu (or click on the icon at the top of the screen). One after the other, click on all atoms and free valences in the transition state except two of the free valences

Chapter 6 df 7/20/06, 2:57 PM105


on one side of the ethylene fragment.* Each of the selected atoms will be given a magenta colored marker, indicating that it is to be frozen (not moved during optimization).** Using sp3 carbon from the model kit and then , connect the two free valences that you have not frozen by a chain of four methylene groups to make a cyclohexene ring. Click on . Note that the only four methylene groups move and not the underlying transition state. You have produced an excellent guess at the transition state for difluorocarbene addition to cyclohexene based on your previous transition state for the simpler process.***

9. Enter the Calculations dialog. It should already specify calculation of transition-state geometry using the 3-21G Hartree-Fock model. Remove the checkmark on IR (to the right of Compute). Click on Submit.

10. When completed, compare the geometries of the two transition states (corresponding to addition of difluorocarbene to ethylene and cyclohexene, respectively). To do this, you need to put the two transition states into the same group. Select Append Molecule(s)... from the File menu and double click on difluorocarbene+ethylene in the browser that results. Then bring up the spreadsheet (Spreadsheet from the Display menu) and check each of the boxes to the right of the molecule names. Click on , select structure from the Align by menu at the bottom right of the screen and, one after another, click on the three carbons which make up the cyclopropane ring. Finally, click on Align by at the bottom right of the screen, and then click on

. Are the two transition states very similar as expected?

11. Remove all molecules and dialogs from the screen.

* You could do this more quickly by first clicking on , then drawing a selection box around all atoms to be frozen, and then clicking inside this box.

** You can remove the frozen atom markers from the model by bringing up the Molecule Properties dialog (Display menu), clicking on at the bottom right of the dialog and then removing the checkmark from Frozens in the Molecule Utilities dialog that results.

*** It is not necessary to replace the two deuteriums by hydrogens. You will not request an IR spectrum and calculated geometry does not depend on atomic mass.

Chapter 6 df 7/20/06, 2:57 PM106


Stereospecific Diels-Alder Reactions

X

HCN

+

H X

CN

H X X H

CN

CN

X H

CN

syn endo syn exo anti endo anti exo

Diels-Alder cycloaddition of 5-substituted cyclopentadienes with acrylonitrile can lead to four stereoproducts, in which the substituent, X, at the 5 position is syn or anti to the dienophile, and the nitrile is endo or exo. Anti products are preferred when X is alkyl (consistent with steric dictates), while syn products are favored when X is halogen or alkoxy. In general, endo adducts are kinetically favored over exo adducts (see following tutorial). In this tutorial, you will use AM1 calculations to obtain both syn and anti transition states for endo addition of both 5-methylcyclopentadiene and 5-fluorocyclopentadiene and acrylonitrile, and then use Hartree-Fock 3-21G calculations to estimate relative activation energies. As the four transition states are likely to be very similar that for the parent cycloaddition (cyclopentadiene and acrylonitrile), you will first obtain a transition state for this reaction, and then to use it as a starting point for locating transition states for the substituted systems.

1. Construct the following substituted norbornene (the product of endo addition of cyclopentadiene and acrylonitrile).

a

CN

b

cde

f

2. Click on . Click on bond a (see figure above) and then on bond b. A curved arrow will be drawn from a to b. Next, click on bond c and then on bond d; a second curved arrow from c to d be drawn. Finally, click on bond e and then on bond f, leading to a third curved arrow. The model on screen should now appear as follows.

5 mins/10 mins

Chapter 6 df 7/20/06, 2:58 PM107


CN

Click on at the bottom right to produce a guess at the transition state.

3. Inside the Calculations dialog, specify calculation of transition state geometry using the semi-empirical AM1 model. Check IR under Compute to specify calculation of vibrational frequencies. This will allow you to verify that you have indeed located a reasonable transition state.

4. Submit the job. Name it cyclopentadiene+acrylonitrile. When completed, examine the resulting structure and calculated frequencies. Make certain that there is one and only one imaginary frequency. Animate the motion associated with the imaginary frequency (Spectra under the Display menu, click on the IR tab and check the box to the left of the imaginary frequency in the dialog that results) to verify that it corresponds to the expected motion along the reaction coordinate.

5. Place the transition-state structure onto the clipboard. Position the cursor slightly above and slightly to the left of the model, then, with both buttons depressed, drag the mouse to a position slightly below and slightly to the right of the model, and finally release both buttons. Dotted lines will enclose the model. Select Copy from the Edit menu. Close cyclopentadiene+ acrylonitrile ( ).

6. Bring up the organic model kit ( ). Click on Clipboard (at the bottom) and then click anywhere on screen. The transition-state structure will appear. Add a fluorine to the methylene group of the cyclopentadiene fragment of the transition state either syn or anti. Do not minimize. Your starting structure should provide an excellent guess at the transition state for the substituted system.

Chapter 6 df 7/20/06, 2:58 PM108


7. Select New Molecule from the File menu. The screen will clear. Click on Clipboard and then click anywhere on screen. Add fluorine to the other methylene group position.

Repeat the process two more times (starting with New Molecule), to add methyl both syn and anti on the methylene group. When you are done (four substituted transition states in total), click on .

8. Enter the Calculations dialog and specify calculation of transition state geometry using the semi-empirical AM1 model. Check IR to the right of Compute. Make certain that Global Calculations is checked. Submit the job (name it Diels-Alder stereochemistry. When it has completed, verify that all four structures correspond to transition states (using Spectra from the Display menu).

9. Make a copy of Diels-Alder stereochemistry ( ). Name it Diels-Alder stereochemistry Hartree-Fock. Enter the Calculations dialog with this copy and specify a 3-21G Hartree-Fock energy calculation. Make certain that you remove the checkmark on IR and that Global Calculations is checked before you exit the dialog.

10. Submit the job. When it completes, bring up the spreadsheet (Display menu) and enter the total energies. Click on the header cell of a blank column and then on Add... at the bottom of the spreadsheet. Select E from the menu of available quantities and au (atomic units) from the Energy menu. Click on OK. Are your results consistent with the observed syn/anti selectivity in these reactions?*


* You can use the spreadsheet to calculate relative energies. Select (click on) the label for one of the fluorine substituted transition states in the spreadsheet, then click on Add... and select rel. E from the menu of available quantities and select appropriate energy units. Click on OK. Repeat for the methyl substituted transition states.

Chapter 6 df 7/20/06, 2:58 PM109


Thermodynamic vs. Kinetic Control

Chemical reactions may yield different products depending on the conditions under which they are carried out. High temperatures and long reaction times favor the most stable (thermodynamic) products, whereas low temperatures and short reaction times favor the most easily formed (kinetic) products. For example, the kinetic product in Diels-Alder cycloaddition of cyclopentadiene and maleic anhydride is the endo adduct, whereas it seems likely that the less-crowded exo adduct is the thermodynamic product.

O

O

O

H

H

O

O

O

O

O

O

H+

D+

endo exo

H

b b

aa

Here you will first obtain pathways for reactions leading to both endo and exo adducts using the PM3 semi-empirical model and then follow these by 3-21G Hartree-Fock energy calculations to get a better estimate of the difference in activation energies.

1. Build the endo Diels-Alder adduct of cyclopentadiene and maleic anhydride and minimize ( ).

2. Select Constrain Distance from the Geometry menu (or click on ), and click on bond a in the figure above. Click on the icon

at the bottom right of the screen (it will then turn to ). Click on .

3. Select Properties from the Display menu, and click on the constraint marker for bond a. The Constraint Properties dialog appears, with the value of the constraint for bond distance a (1.54Å) given in a box to the right of Value. Change this to 1.5Å by typing 1.5 inside the box and then pressing the Enter key. Check Dynamic inside the dialog. This leads to an extended form of the Constraint Properties dialog, that allows the single constraint value to be replaced by a sequence of constraint values.

5 mins/15 mins

Chapter 6 fd 7/20/06, 3:02 PM110


Type 2.7 (2.7Å) into the box to the right of to and press the Enter key. Set Steps to 13. Type 13 and press the Enter key. You have specified that bond a will be constrained first to 1.5Å, then to 1.633Å, then to 1.767Å, etc. and finally to 2.7Å.

You need to repeat the process starting with step 2 for bond b. When you are done, both bonds a and b will be constrained from 1.5Å to 2.7Å in 13 equal steps. Close the Constraint Properties dialog.

4. Bring up the Calculations dialog and select Energy Profile from the top menu to the right of Calculate, and Semi-Empirical and PM3 from the two bottom menus. Click on Submit at the bottom of the dialog. Name it cyclopentadiene+ maleic anhydride endo.

5. When completed the job will give rise to a new file cyclopentadiene+maleic anhydride endo.M001 containing the 10 steps that make up the energy profile. Open this file.*

6. Bring up the Spreadsheet, and click on Add... at the bottom. Select E from among the entries, kJ/mol from the Energy menu, and click on OK. Next, click on and click on one of the two CC bonds varied in the energy profile. Click on at the bottom right of the screen. Click on . Finally, bring up the Plots dialog, and select Constraint (BondX)** from among the items in the X Axis menu and E (kcal/mol) from the Y Axes list. Click on OK.

* To avoid confusion, it is a good idea to close the original file cyclopentadiene+maleic anhydride endo.

** Bonds are numbered in the order they were formed upon initial construction of the molecule.

Chapter 6 df 7/20/06, 2:58 PM111


Identify both the reactant and transition state from the plot and estimate the activation energy for the cycloaddition reaction.

7. Repeat steps 1 to 6 for the exo adduct. Compare the activation energy for exo addition to that for endo addition (above). What is the kinetic product?


9 to 11 optional

9. Open cyclopentadiene+maleic anhydride endo.M001 ( ) and make a copy ( ). Name it cyclopentadiene +maleic anhydride endo Hartree-Fock. Enter the Calculations dialog and specify a Hartree-Fock 3-21G energy calculation. Submit. When completed, perform the same spreadsheet and plot operations you did for the PM3 calculations.

10. Repeat the above procedure for the exo adduct and compare the two activation energies. What is the kinetic product?


Chapter 6 df 7/20/06, 2:58 PM112


Activation Energies of Diels Alder Reactions

In an earlier tutorial Dienophiles in Diels-Alder Reactions (Chapter 5), you examined the extent to which LUMO energies for a series of related dienophiles correlated with relative rates of Diels-Alder cycloadditions involving cyclopentadiene. In this tutorial, you will compare calculated activation energies for this same set of reactions with the experimental rates. You will use the previous set of calculations on the dienophiles and obtain transition states from the Spartan Reaction Database (SRD). The only quantum chemical calculation that is required is for cyclopentadiene.

1. Build cyclopentadiene and minimize ( ). Before you proceed with the next step, copy cyclopentadiene onto the clipboard. Position the cursor above and to the left of cyclopentadiene, hold down both mouse buttons, drag the mouse below and to the right of cyclopentadiene and finally release the buttons. A selection box will be drawn. Select Copy from the Edit menu to put cyclopentadiene onto the clipboard.

2. Enter the Calculations dialog (Setup menu) and specify an equilibrium geometry calculation using the Hartree-Fock 3-21G model. Submit the job with the name cyclopentadiene. Move it off to the side of the screen for use later on.

3. Enter the builder ( ). Click on Clipboard. The structure of cyclopentadiene will appear at the top of the model kit. Click anywhere on screen and cyclopentadiene will appear.

4. Select Ethylene from the Groups menu, press the Insert key (at the bottom right of the model kit) and click on screen. Both cyclopentadiene and ethylene will appear on screen, but they will not be connected. Select Cyano from the Groups menu and click on one of the free valences on ethylene. Both cyclopentadiene and acrylonitrile will appear on screen.

5. Orient the two molecules such that they are poised for a Diels-Alder reaction leading to an endo product.

5 mins

Chapter 6 df 7/20/06, 2:58 PM113


NC

a

fe

d

cb

Select Transition States from the Search menu (click on ). Click on the bond marked a in the above figure and, while

holding down the Shift key, click on the two carbons that when connected will lead to the bond marked b. A dotted line will be drawn between these carbons and an arrow drawn from the center of bond a to the center of bond b. Next, click on bond c and then on bond d. A second arrow will be drawn. Finally, click on bond e and, while holding down the Shift key, click on the two carbons that when connected will lead to the bond marked f. A second dotted line and a third arrow will be drawn.

6. Select Databases from the Search menu (click on ) and click on the SRD tab at the top of the dialog that results. This leads to the Spartan Reaction Database (SRD) dialog.

Click on all three free valences on acrylonitrile. They will be replaced by orange cones, meaning that anything (including hydrogen) may be attached to these positions. Before you begin the search, click on to the right the Search button at the bottom of the dialog. This brings up the SRD Search Options dialog.

Chapter 6 df 7/20/06, 2:58 PM114


Check 3-21G under Method Filters, and make certain that Structure is selected under Search by. Click on OK. Click on the Search button to search the Spartan Reaction Database for Diels-Alder transition states between cyclopentadiene and substituted acrylonitriles.

7. When the search completes, a listing of hits appears at the right of the dialog.

This should contain all six Diels-Alder transition states for endo additions of cyclopentadiene and acrylonitrile, trans-1,2-dicyanoethylene, cis-1,2-dicyanoethylene, 1,1-dicyanoethylene, tricyanoethylene and tetracyanoethylene. All need to be retrieved into a single document. Click on to the right of the Retrieve button at the bottom of the dialog to bring up the Retrieve Options dialog. Click on New Document inside this dialog

Chapter 6 df 7/20/06, 2:58 PM115


and click on OK. Click on the transition state corresponding to endo addition of cyclopentadiene and acrylonitrile in the list of available transition states. Its structure will be shown in the window at the left of the Spartan Reaction Database dialog and may be manipulated in the usual way. Click on the Retrieve button to transfer this structure into Spartan’s file system.

8. Again click on to the right of the Retrieve button at the bottom of the Spartan Reaction Database dialog. Change New Document to Current Document in the Retrieve Options dialog that results. This ensure that the remaining transition states will go into the same document as the transition state for reaction of cyclopentadiene and acrylonitrile.

9. One after the other and in proper order, select (click on) end transition states for reactions of cyclopentadiene and trans-1,2-dicyanoethylene, cis-1,2-dicyanoethylene, 1,1-dicyanoethylene, tricyanoethylene and tetracyanoethylene (five transition states), and after each selection click on the Retrieve button. You are left with a single document containing all six Diels-Alder transition states.

10. Bring up a spreadsheet (Spreadsheet under the Display menu) and size it to contain all six transition states. Click on Add at the bottom of the dialog, select Name and E from the list of the available quantities and au from the Energy menu and click on OK. Transition state names and energies will populate the spreadsheet. Change the title of the column of energies to Ets (from E (au)). Also provide titles for three empty columns: Ecyclopentadiene, Edienophile and log(rate). Save the document ( ) with the name Diels-Alder transition states.

11. Open Diels-Alder dienophiles that you created in an earlier tutorial.* Bring up a spreadsheet for this document. This should contain the experimental relative rate data. If it does not already contain calculated energies, click on Add, select E from the list of

* If you have not completed this tutorial, a copy of the document Diels-Alder dienophiles may be found in the organic reactions sub-directory (tutorials directory).

Chapter 6 df 7/20/06, 2:58 PM116


quantities and au from the Energy menu and click on OK. Verify that the ordering of dienophiles in this document is the same as the ordering of transition states in Diels-Alder transition states.

Click on the top entry in the column under the log(rate) in the spreadsheet for Diels-Alder dienophiles and then, while holding down the Shift key, click on the bottom entry. Select Copy from the Edit menu. Move to the spreadsheet for Diels-Alder transition states, click inside the top cell under log(rate) and select Paste from the Edit menu. The experimental rate data will be added to the spreadsheet for Diels-Alder transition states, Repeat for energy values in Diels-Alder dienophiles (E (au)), copying them to Edienophile in Diels-Alder transition states spreadsheet. Close Diels-Alder dienophiles ( ).

12. Locate cyclopentadiene on screen and bring up a spreadsheet. Click on Add, select E from the list of quantities and au from the Energy menu and click on OK. Click inside the cell containing the energy and select Copy from the Edit menu. Move to the spreadsheet for Diels-Alder transition states and paste this value into all six cells below Ecyclopentadiene (Paste from the Edit menu). Close cyclopentadiene ( ).

13. Calculate activation energies (in kJ/mol). Inside the header cell of a an empty column (in the spreadsheet for Diels-Alder transition states) type Eact=(Ecyclopentadiene+Edienophile-Ets)*@hart2kJ (The last quantity provides conversion from atomic units to kJ/mol; see Table 16-4 in Chapter 16).

14. Plot calculated activation energies vs. experimental relative reaction rates. Select Plots from the Display menu. Select log(rate) from the list of items in the X Axis menu and Eact from the Y Axis list. Click on OK. The curve displayed smoothly connects the data points, but is of little value for establishing the quality of the correlation. A better representation is provided by a simple least-squares fit. Select Properties from the Display menu and click anywhere on the curve. Select Linear LSQ from the Curve Properties dialog that appears. This leads to

Chapter 6 df 7/20/06, 2:58 PM117

118 Chapter 6

a conventional least squares plot relating calculated activation energies and experimental relative rates.

15. Close all molecules and any open dialogs.

Chapter 6 df 7/20/06, 2:58 PM118

Chapter 9 151

Chapter 9Inorganic and

Organometallic MoleculesThis chapter shows how to construct inorganic and organometallic molecules using Spartan’s inorganic model kit. It also identifies quantum chemical models that appear to be suitable for calculations on molecules incorporating transition metals.

The majority of organic molecules are made up of a relatively few elements and obey conventional valence rules. They may be easily built using the organic model kit. However, many molecules incorporate other elements, or do not conform to normal valence rules, or involve ligands. Most important among them are inorganic and organometallic compounds involving transition metals. These need to be constructed using the inorganic model kit.

Transition-metal inorganic and organometallic compounds may also require different quantum chemical methods from those that have proven to be satisfactory for organic molecules. In particular, Hartree-Fock and MP2 models are known to produce poor results where transition metals are involved. The PM3 semi-empirical model, that has been parameterized for most transition metals, generally provides a good account of equilibrium geometries, as do density functional models. The latter are also believed to provide a satisfactory account of the thermochemistry of reactions involving transition-metal systems, although there are very little experimental data with which to compare.

The tutorials in this chapter illustrate construction of inorganic and organometallic molecules, as well as the selection of suitable quantum chemical models.

Chapter 9 7/20/06, 3:53 PM151

Chapter 9 153

Sulfur Tetrafluoride

SF

F

F

F

Sulfur tetrafluoride cannot be constructed using Spartan’s organic model kit. This is because sulfur is not in its normal bent dicoordinate geometry, but rather in a trigonal bipyramid geometry with one of the equatorial positions vacant. However, the molecule can easily be made using the inorganic model kit.

1. Bring up the inorganic model kit by clicking on and then clicking on the Inorganic tab at the top of the (organic) model kit.

The inorganic model kit comprises a Periodic Table* followed by a selection of atomic hybrids, then bond types, and finally Rings, Groups and Ligands menus (the first two of which are the same as found in the organic model kit).

* Not all methods are available for all elements listed. Elements for which a specific method (selected in the Calculations dialog) are available will be highlighted in the Periodic Table if Element Overlay in the Settings Preferences dialog (Preferences... under the Options menu; see Chapter 18) is turned “on”. Note also that elements beyond those listed in the Periodic Table may be specified. For a discussion, see Atom Properties under the Display menu (Chapter 16).

1 min

Chapter 9 7/20/06, 3:53 PM153


2. Select (click on) S in the Periodic Table and the five coordinate trigonal bipyramid structure from the list of atomic hybrids. Click on screen. A trigonal bipyramid sulfur will appear.

3. Select F in the Periodic Table and the one-coordinate entry from the list of atomic hybrids. One after the other, click on both axial free valences of sulfur, and two of the three equatorial free valences.

4. It is necessary to delete the remaining free valence (on an equatorial position); otherwise it will become a hydrogen. Click on and then click on the remaining equatorial free valence.

5. Click on . Click on to remove the model kit.

6. Select Calculations... from the Setup menu. Specify calculation of equilibrium geometry* using the Hartree-Fock 3-21G model.**

7. Submit the job. Name it sulfur tetrafluoride.When completed, select Properties (Display menu) and click on an atom, for example, sulfur. Three different atomic charges will appear in the (Atom Properties) dialog (corresponding to different methods for establishing atomic charge). Of these, the procedure based on fitting the electrostatic potential is generally considered to be the best. Are the electrostatic charges consistent with covalent or ionic bonding?

8. Remove sulfur tetrafluoride and any remaining dialogs from the screen.

* It should be noted that were an incorrect geometry specified at the outset, optimization would lead to the correct structure, as long as the starting geometry possessed no symmetry (C1 point group). Thus, square planar SF4 in D4h symmetry would remain square planar, while an almost square planar structure (distorted only slightly from D4h symmetry to C1 symmetry) would collapse to the proper structure.

** The 3-21G(*) basis set, that incorporates a set of unoccupied d-type functions, is used in place of 3-21G for the sulfur atom.

Chapter 9 7/20/06, 3:53 PM154


Reactivity of Silicon-Carbon Double Bonds

C C

CH3

CH3H3C

H3C

Si C

CH3

CH3H3C

H3C

2,3-dimethyl-2-butene tetramethylsilaethylene

With the exception of so-called phosphorous ylides, compounds incorporating a double bond between carbon and a second-row element are quite rare. Most curious perhaps is the absence of stable compounds incorporating a carbon-silicon double bond. This can be rationalized by using local ionization potential and LUMO maps to compare the reactivities of olefins and silaolefins.

1. Bring 2,3-dimethyl-2-butene, minimize ( ) and copy onto the clipboard. Select New Molecule (not New) from the File menu, click on Clipboard inside the organic model kit and click anywhere on screen. Click on the Inorganic tab (at the top of the organic model kit) to bring up the inorganic model kit. Click on Si in the Periodic Table and double click on one of the sp2 carbons to substitute a silicon atom for a carbon atom and to make tetramethylsilaethylene. Click on and then on . You are left with 2,3-dimethyl-2-butene and tetramethylsilaethylene in the same document.

2. Use the structures in SMD to save computer time. Click on to the left of the molecule name (either 2,3-dimethyl-2-butene or tetramethylsilaethylene) at the bottom of the screen, select 3-21G from the entries in the SMD Preview dialog and click on Replace All.

3. Select Calculations from the Setup menu. Specify calculation of energy using the Hartree-Fock 3-21G model. Click on OK.

4. Select Surfaces from the Setup menu. First request a local ionization potential map. Click on Add, select density from the Surface menu and ionization from the Property menu and click on OK. Then request a LUMO map. Click on Add a second

5 mins

Chapter 9 7/20/06, 3:53 PM155


time and again select density from the Surface menu. Select LUMO from the Property menu and click on OK.

5. Submit the job. Name it tetramethylsilaethylene. When completed, bring up the spreadsheet (Spreadsheet under the Display menu) and check the box to the right of the label for both entries. This allows simultaneous display of both molecules. Select Coupled (Model menu) to uncouple the motions of the molecules. Position the two molecules side by side on screen.

6. Inside the Surfaces dialog, select density ionization.... Compare local ionization potential maps for the olefin and silaolefin, recognizing that the more red the color, the lower the ionization potential and the more susceptible toward electrophilic attack. Which molecule do you conclude is likely to be more reactive? Turn “off” local ionization potential maps by again selecting density ionization.... Select density |LUMO|... to turn on LUMO maps. Here, the more blue the color, the greater the concentration of the LUMO and the more susceptible toward nucleophilic attack. Which molecule do you conclude is likely to be more reactive? Close tetramethylsilaethylene when you are finished.

7 - 10 Optional (Requires Access to the Cambridge Structural Database)

Identify molecules in the Cambridge Structural Database (CSD) that incorporate a carbon-silicon double bond.

7. Build parent silaethylene, H2Si=CH2, by starting from ethylene and substituting a silicon for one of the carbons.

8. Bring up the Cambridge Structural Database (CSD) dialog by first selecting Databases from the Search menu and then clicking on the CSD tab. Click on all four free valences in silaethylene. They will be marked by orange cones. Click on the Search button at the bottom of the dialog. In a few seconds, hits will begin to appear in the box at the right of the dialog. Examine each in turn (in the window at the left of the dialog) by clicking on its CSD

Chapter 9 7/20/06, 3:53 PM156


reference code (REFCODE). Identify a structure that appears to be representative for closer scrutiny. Before you retrieve it, click on to the right of the Retrieve button at the bottom of the dialog. Check all four boxes under Post Retrieval Filters in the Retrieve Options dialog that results and click on OK. Click on the selected REFCODE and then click on the Retrieve button. Dismiss the CSD dialog.

9. Click on . Except for hydrogens, all atoms in the structure you have retrieved will be fixed and energy minimization will only change hydrogen positions. This is to compensate for the fact that hydrogens are generally not well located in X-ray crystal structure determinations.

Change the display to a space-filling model. Pay particular attention to the incorporated silicon-carbon double bond. Is it accessible in all of the compounds or is it encumbered by the surrounding groups? What, if anything, do your observations suggest about why this particular compound can be (has been) made?

10. Close silaethylene (without saving) and any open dialogs.

Chapter 9 7/20/06, 3:53 PM157


2 mins/30 mins

Benzene Chromium Tricarbonyl

CrOC CO

CO

benzene chromium tricarbonyl

Comparison of electrostatic potential maps for this system and that of free benzene will allow you to classify Cr(CO)3 as an electron donor or an electron acceptor substituent.

1. Click on and bring up the inorganic model kit. Select Cr from the Periodic Table and the four-coordinate tetrahedral structure

from the list of atomic hybrids. Click anywhere on screen.

2. Click on Ligands in the model kit, select Benzene from the menu of available ligands.

Click on one of the free valences on the four-coordinate chromium center.

3. Select Carbon Monoxide from the Ligands menu, and click on the remaining (three) free valences on chromium. Click on

to produce a refined structure.

4. Select New Molecule from the File menu. The screen will blank. Build benzene and click on . Click on .

5. Select Calculations... (Setup menu). Specify calculation of equilibrium geometry with the semi-empirical PM3 model. Make certain that Global Calculations (at the bottom of the dialog) is checked. You want the calculations to apply to both benzene chromium tricarbonyl and benzene. Click on OK.

Chapter 9 7/20/06, 3:53 PM158


6. Select Surfaces (Setup or Display menu). Click on Add.... Specify density from the Surface menu, and potential from the Property menu, and click on OK. Make certain that Global Surfaces is checked.

7. Submit the job (two separate calculations). Name it benzene chromium tricarbonyl. When completed bring up the spreadsheet (Spreadsheet from the Display menu), and check the box to the right of the label for both entries. This allows the two molecules to be displayed simultaneously on screen. If Coupled (Model menu) is checked, remove the checkmark by selecting it. The two molecules may now be moved independently. Orient each molecule so that you can clearly see the benzene face (exposed face in the case of the organometallic).

8. Inside the Surfaces dialog, select density potential.... Compare electrostatic potential maps for both free and complexed benzene, with attention to the exposed face on benzene.* Is the effect of the Cr(CO)3 group to donate or to withdraw electrons from the ring? Would you expect the aromatic ring in benzene chromium tricarbonyl to be more or less susceptible to electrophilic attack than free benzene? More or less susceptible to nucleophilic attack?

9 optional (cannot be completed with the Essential Edition)

9. Repeat the two calculations using density functional theory. Make a copy of benzene chromium tricarbonyl ( ); name it benzene chromium tricarbonyl density functional. Inside the Calculations dialog, specify an energy calculation using the BP/6-31G* model. Submit. When completed, examine the electrostatic potential maps. Are they qualitatively similar to those from the PM3 calculations?


* Electrostatic potential maps (as well as other maps) for molecules in a group will be put onto the same (color) scale. This allows comparisons to be made among different members.

Chapter 9 7/20/06, 3:53 PM159


5 mins/90 mins

Ziegler-Natta Polymerization of Ethylene

Ziegler-Natta polymerization involves a metallocene. This first complexes an olefin, which then inserts into the metal-alkyl bond.

Cp2Zr+R Cp2Zr+RH2C CH2 Cp2Zr+CH2CH2R

H2C CH2 H2C CH2

RCp2Zr+

In this tutorial, you will use PM3 calculations to obtain a transition state for insertion of ethylene into Cp2ZrCH3

+ and, (optionally), estimate the activation energy using density functional calculations.

1. Bring up the inorganic model kit. Select Zr and , and click on screen. Select Cyclopentadienyl from the Ligands menu and click on two of the free valences on zirconium.

2. Move to the organic model kit. Select sp3 carbon and click on the remaining free valence on zirconium. Select Alkene from the Groups menu, click on the Insert button at the bottom of the model kit and click anywhere on screen.

3. Orient the two fragments (Cp2ZrCH3 and ethylene) as shown below:

Zr CH3

Cp

Cp

H2C CH2

(To move the two fragments independently, hold down the Ctrl key.)

4. Select Transition States from the Search menu ( ). Click on the ZrC (methyl) bond, hold down the Shift key and, one after another, click on the methyl carbon and on one of the ethylene carbons. Next, click on the ethylene double bond, hold down the Shift key and, one after another, click on the other ethylene carbon and on zirconium. Click on at the bottom right of the screen. In a few seconds, a guess at the transition state appears.

5. Bring up the Calculations dialog. Specify a transition state geometry calculation using the PM3 semi-empirical model.

Chapter 9 7/20/06, 3:53 PM160


Change Total Charge to Cation and check IR to the right of Compute. Click on Submit and name the job Cp2ZrMe cation + ethylene.

6. When the job has completed, bring up the IR dialog (Spectra from the Properties menu and click on the IR tab) and click on the imaginary frequency. Would you describe the process as concerted or occurring in discrete steps?

7 to 9 optional (cannot be completed with the Essential Edition)

7. Perform single-point BP/6-31G* density functional calculations to obtain an estimate for the energy barrier for ethylene insertion. Make a copy of Cp2ZrMe cation+ ethylene ( ); name it Cp2ZrMe cation+ethylene density functional. Enter the Calculations dialog with this copy, and specify calculation of an energy using the BP/6-31G* density functional model. Check Pseudopotential at the far lower right of Calculate to specify use of a pseudopotential for Zr associated with the 6-31G* basis set. Remove the checkmark on IR (to the right of Compute). Total Charge should still be set to Cation. Click on Submit.

8. Build both ethylene and Cp2ZrCH3+ (name them ethylene

density functional and Cp2ZrMe cation, respectively). For Cp2ZrCH3

+, start with three-coordinate trigonal Zr, and then add two cyclopentadienyl ligands and a four-coordinate tetrahedral carbon. For each, enter the Calculations dialog, and specify calculation of energy using the BP/6-31G* density functional model. Select PM3 from the Start from menu to designate use of a PM3 geometry. For Cp2ZrMe cation density functional (only) check Pseudopotential, and set Total Charge to Cation.

9. Submit both jobs. When they (and the transition-state calculation) have completed, calculate an activation energy for the insertion reaction.


Chapter 9 7/20/06, 3:53 PM161

Appendix D 349

Appendix DUnits

Geometries

Cartesian coordinates are given in Ångstroms (Å), but are also available in atomic units (au).

Bond distances are given in Å but are also available in au. Bond angles and dihedral angles are given in degrees (°).

Surface areas, accessible surface areas and polar surface areas are given in Å2 and volumes in Å3, but are also available in au2 (au3).

1 Å = 0.1 nm=1.889762 au

Energies, Heats of Formation and Strain Energies, Enthalpies and Gibbs Energies and Entropies

Total energies from Hartree-Fock calculations are given in au, but are also available in kcal/mol, kJ/mol and electron volts (eV).

Experimental heats of formation as well as those from semi-empirical calculations and from thermochemical recipes are given in kJ/mol, but are also available in au, kcal/mol and eV.

Strain energies from molecular mechanics calculations are given in kJ/mol, but are also available in au, kcal/mol and eV.

Energies, heats of formation and strain energies corrected empirically for the effects of aqueous media are given in the same units as the corresponding gas-phase quantities.

Enthalpies and Gibbs Energies are given in kJ/mol. Entropies are given in kJ/mol•degree.

Appendix D 7/21/06, 8:44 AM349

350 Appendix D

Orbital Energies

Orbital energies are given in eV, but are also available in kcal/mol, kJ/mol and au.

Energy Conversions

au kcal/mol kJ/mol eV

1 au - 627.5 2625 27.211 kcal/mol 1.593 (-3) - 4.184 4.337 (-2)1 kJ/mol 3.809 (-4) 2.390 (-1) - 1.036 (-2)1 eV 3.675 (-2) 23.06 96.49 -

a) exponent follows in parenthesis, e.g., 1.593 (-3) = 1.593 x 10-3

Electron Densities, Spin Densities, Dipole Moments, Charges, Electrostatic Potentials and Local Ionization Potentials

Electron densities and spin densities are given in electrons/au3.

Dipole moments are given in debyes.

Atomic charges are given in electrons.

Electrostatic potentials are given in kJ/mol.

Local ionization potentials are given in eV.

Vibrational Frequencies

Vibrational frequencies are given in wavenumbers (cm-1).

Chemical Shifts

Chemical shifts are given in parts-per-million (ppm) relative to appropriate standards: hydrogen, tetramethylsilane; carbon, tetramethylsilane; nitrogen, nitromethane; fluorine, fluorotrichloromethane; silicon, tetramethylsilane, phosphorous, phosphoric acid.

UV/visible Spectra

λmax is given in wavenumbers (cm-1).

Appendix D 7/21/06, 8:45 AM350

Getting Started

with Spartan 3rd Edition

WAVEFUNCTION

Wavefunction, Inc.

18401 Von Karman Avenue, Suite 370

Irvine, CA 92612 U.S.A.www.wavefun.com

Wavefunction, Inc., Japan Branch Office

Level 14 Hibiya Central Building, 1-2-9 Nishi-ShinbashiMinato-ku, Tokyo, Japan 105-0003

+81-3-5532-7335 • +81-3-5532-7373 fax

[email protected] • www.wavefun.com/japan

First Page 7/1/04, 1:42 PM1

7

1WBasic Operations

This tutorial introduces a number of basic operations in the Windowsversion of Spartan required for molecule manipulation and propertyquery. Specifically it shows how to: i) open molecules, ii) view differentmodels and manipulate molecules on screen, iii) measure bonddistances, angles and dihedral angles, iv) display energies, dipolemoments, atomic charges and infrared spectra, and v) displaygraphical surfaces and property maps. Molecule building is notillustrated and no calculations are performed.

1. Start Spartan. Click on the Start button, then click on Programsand finally click on Spartan. Spartan’s window will appear witha menu bar at the top of the screen.

File Allows you to create a new molecule or read in a moleculewhich you have previously created.

Model Allows you to control the style of your model.Geometry Allows you to measure bond lengths and angles.Build Allows you to build and edit molecules.Setup Allows you to specify the task to be performed and the

theoretical model to be employed, to specify graphicalsurfaces and property maps and to submit jobs forcalculation.

Display Allows you to display text output, molecular and atomicproperties, surfaces and property maps and infraredspectra. Also allows data presentation in a spreadsheet andplots to be made from these data.

Search Allows you to “guess” a transition-state geometry basedon a library of reactions. This guess may then be used asthe basis for a quantum chemical calculation of the actualreaction transition state.

Tutorial 1W 7/1/04, 2:38 PM7

8

2. Click with the left mouse button on File from the menu bar.

Click on Open.... Alternatively, click on the icon in the toolbar.

Several important functions provided in Spartan’s menus may also beaccessed from icons in the toolbar.

New Add Fragment Measure Distance

Open Delete Measure Angle

Close Make Bond Measure Dihedral

Save As Break Bond Transition States

View Minimize

Locate the “tutorials” directory in the dialog which appears, clickon “tutorial 1” and click on Open (or double click on “tutorial1”). Ball-and-spoke models for ethane, acetic acid dimer, propene,ammonia, hydrogen peroxide, acetic acid, water, cyclohexanone,ethylene, benzene and aniline appear on screen. You can select amolecule by clicking on it with the left mouse button. Once selected,a molecule may be manipulated (rotated, translated and scaled).

select molecule click (left mouse button)rotate molecule press the left button and move the mousetranslate molecule press the right button and move the mousescale molecule press both the Shift key and the right button and

move the mouse “up and down”

Tutorial 1W 7/1/04, 2:38 PM8

9

3. Identify ethane on the screen, and click on it (left button) to makeit the selected molecule. Practice rotating (move the mouse whilepressing the left button) and translating (move the mouse whilepressing the right button) ethane. Click on a different molecule,and then rotate and translate it.

4. Return to ethane. Click on Model from the menu bar.

Wire Ball-and-Wire Tube Ball-and-Spoke

One after the other, select Wire, Ball and Wire, Tube and finallyBall and Spoke from the Model menu. All four models for ethaneshow roughly the same information. The wire model looks like aconventional line formula, except that all atoms, not just thecarbons, are found at the end of a line or at the intersection oflines. The wire model uses color to distinguish different atoms,and one, two and three lines between atoms to indicate single,double and triple bonds, respectively.

Tutorial 1W 7/1/04, 2:38 PM9

10

Atoms are colored according to type:

Hydrogen whiteLithium tan Sodium yellowBeryllium green Magnesium dark blueBoron tan Aluminum violetCarbon black Silicon greyNitrogen blue-gray Phosphorous tanOxygen red Sulfur sky blueFluorine green Chlorine tan

Atom colors (as well as bond colors, the color of the background, etc.)may be changed from their defaults using Colors under the Optionsmenu. An atom may be labelled with a variety of different quantitiesusing Configure... under the Model menu. Labels are thenautomatically turned “on” and may be turned “off” by selecting Labelsunder the Model menu.

The ball-and-wire model is identical to the wire model, exceptthat atom positions are represented by small spheres. This makesit easy to identify atom locations. The tube model is identical tothe wire model, except that bonds, whether single, double or triple,are represented by solid cylinders. The tube model is better thanthe wire model in conveying the three-dimensional shape of amolecule. The ball-and-spoke model is a variation on the tubemodel; atom positions are represented by colored spheres, makingit easy to see atom locations.

Select Space Filling from the Model menu.

Space-Filling

This model is different from the others in that bonds are not shown.Rather, each atom is displayed as a colored sphere that representsits approximate “size”. Thus, the space-filling model for a moleculeprovides a measure of its size. While lines between atoms are notdrawn, the existence (or absence) of bonds can be inferred from

Tutorial 1W 7/1/04, 2:38 PM10

11

the amount of overlap between neighboring atomic spheres. If twospheres substantially overlap, then the atoms are almost certainlybonded, and conversely, if two spheres hardly overlap, then theatoms are not bonded. Intermediate overlaps suggest “weakbonding”, for example, hydrogen bonding (see the activity “Water”).

Select acetic acid dimer. Switch to a space-filling model and lookfor overlap between the (OH) hydrogen on one acetic acid moleculeand the (carbonyl) oxygen on the other. Return to a ball-and-spokemodel and select Hydrogen Bonds from the Model menu.

Ball-and-Spoke model for acetic acid dimerwith hydrogen bonds displayed

The two hydrogen bonds, which are responsible for holding theacetic acid molecules together, will be drawn.

Use the 3 key to toggle between stereo 3D and regular display. Toview in 3D you will need to wear the red/blue glasses providedwith Spartan.

5. Distances, angles, and dihedral angles can easily be measuredwith Spartan using Measure Distance, Measure Angle, andMeasure Dihedral, respectively, from the Geometry menu.

Tutorial 1W 7/1/04, 2:38 PM11

12

Alternatively, the measurement functions may be accessed fromthe , and icons in the toolbar.

a) Measure Distance: This measures the distance between twoatoms. First select propene from the molecules on screen, andthen select Measure Distance from the Geometry menu (orclick on the icon in the toolbar). Click on a bond or on twoatoms (the atoms do not need to be bonded). The distance (inÅngstroms) will be displayed at the bottom of the screen.Repeat the process as necessary. When you are finished, selectView from the Build menu.

Alternatively, click on the icon in the toolbar.

b) Measure Angle: This measures the angle around a central atom.Select ammonia from the molecules on screen, and then selectMeasure Angle from the Geometry menu (or click on the icon in the toolbar). Click first on H, then on N, then on anotherH. Alternatively, click on two NH bonds. The HNH angle (indegrees) will be displayed at the bottom of the screen. Clickon when you are finished.

c) Measure Dihedral: This measures the angle formed by twointersecting planes, the first containing the first three atomsselected and the second containing the last three atoms selected.Select hydrogen peroxide from the molecules on screen, thenselect Measure Dihedral from the Geometry menu (or clickon the icon in the toolbar) and then click in turn on the fouratoms (HOOH) which make up hydrogen peroxide. The HOOHdihedral angle will be displayed at the bottom of the screen.Click on when you are finished.

Tutorial 1W 7/1/04, 2:38 PM12

13

6. Energies, dipole moments and atomic charges among othercalculated properties, are available from Properties under theDisplay menu.

a) Energy: Select acetic acid from the molecules on screen andthen select Properties from the Display menu. The MoleculeProperties dialog appears.

This provides the total energy for acetic acid in atomic units(au). See the essay “Total Energies and Thermodynamic andKinetic Data” for a discussion of energy units.

b) Dipole Moment: The magnitude of the dipole moment (indebyes) is also provided in the Molecule Properties dialog. Alarge dipole moment indicates large separation of charge. Youcan attach the dipole moment vector, “ ” where the lefthandside “+” refers to the positive end of the dipole, to the modelon the screen, by checking Dipole near the bottom of the dialog.The vector will not be displayed if the magnitude of the dipolemoment is zero, or if the molecule is charged.

c) Atomic Charges: To display the charge on an atom, click onit with the Molecule Properties dialog on the screen. The AtomProperties dialog replaces the Molecule Properties dialog.

Tutorial 1W 7/1/04, 2:38 PM13

14

Atomic charges are given in units of electrons. A positive chargeindicates a deficiency of electrons on an atom and a negativecharge, an excess of electrons. Repeat the process as necessaryby clicking on other atoms. Confirm that the positively-chargedatom(s) lie at the positive end of the dipole moment vector.When you are finished, remove the dialog from the screen byclicking on the in the top right-hand corner.

d) Infrared Spectra: Molecules vibrate (stretch, bend, twist) evenif they are cooled to absolute zero. This is the basis of infraredspectroscopy, where absorption of energy occurs when thefrequency of molecular motions matches the frequency of the“light”. Infrared spectroscopy is important in organic chemistryas different functional groups vibrate at noticeably differentand characteristic frequencies.

Select water from the molecules on screen. To animate avibration, select Spectra from the Display menu. This leadsto the Spectra dialog.

Tutorial 1W 7/1/04, 2:38 PM14

15

This displays the three vibrational frequencies for the watermolecule, corresponding to bending and symmetric andantisymmetric stretching motions. One after the other, doubleclick on each frequency and examine the motion. Turn “off”the animation when you are finished.

No doubt you have seen someone “act out” the three vibrationsof water using his/her arms to depict the motion of hydrogens.

While this is “good exercise”, it provides a poor account of theactual motions. Equally important, it is clearly not applicable tolarger molecules (see below).

Select cyclohexanone. The Spectra dialog now lists its 45vibrational frequencies. Examine each in turn (double click onthe entry in the dialog) until you locate the frequencycorresponding to the CO (carbonyl) stretch. Next, click onDraw IR Spectrum at the bottom of the dialog. The infraredspectrum of cyclohexanone will appear.

You can move the spectrum around the screen by first clickingon it to select it and then moving the mouse while pressing theright button. You can size it by moving the mouse “up anddown” while pressing both the Shift key and the right button.

Tutorial 1W 7/1/04, 2:38 PM15

16

Identify the line in the spectrum associated with the CO stretch(a small gold ball moves from line to line as you step throughthe frequencies in the Spectra dialog). Note that this line isseparated from the other lines in the spectrum and that it isintense. This makes it easy to find and is the primary reasonwhy infrared spectroscopy is an important diagnostic forcarbonyl functionality. When you are finished, click on atthe top of the Spectra dialog to remove it from the screen.

7. Spartan permits display, manipulation and query of a number ofimportant quantities resulting from a quantum chemical calculationin “visual” format. Most important are the electron density (whichreveals “how much space” a molecule actually takes up; see theessay “Electron Densities: Sizes and Shapes of Molecules” for adiscussion), the bond density (which reveals chemical “bonds”;see the essay on electron densities), and key molecular orbitals(which provide insight both into bonding and chemical reactivity;see the essay “Atomic and Molecular Orbitals”). In addition, theelectrostatic potential map, an overlaying of a quantity called theelectrostatic potential (the attraction or repulsion of a positivecharge for a molecule) onto the electron density, is valuable fordescribing overall molecular charge distribution as well asanticipating sites of electrophilic addition. Further discussion isprovided in the essay “Electrostatic Potential Maps: ChargeDistributions”. Another indicator of electrophilic addition isprovided by the local ionization potential map, an overlaying ofthe energy of electron removal (“ionization”) onto the electrondensity. Finally, the likelihood of nucleophilic addition can beascertained using a LUMO map, an overlaying of the lowest-unoccupied molecular orbital (the LUMO) onto the electrondensity. Both of these latter graphical models are described in theessay “Local Ionization Potential Maps and LUMO Maps”.

Select ethylene from among the molecules on screen, and then selectSurfaces from the Display menu. The Surfaces dialog appears.

Tutorial 1W 7/1/04, 2:38 PM16

17

Double click on the line “homo...” inside the dialog. This willresult in the display of ethylene’s highest-occupied molecularorbital as a solid. This is a π orbital, equally concentrated aboveand below the plane of the molecule. The colors (“red” and “blue”)give the sign of the orbital. Changes in sign often correlate withbonding or antibonding character. You can if you wish, turn “off”the graphic by again double clicking on the line “homo . . .”.

Next, select benzene from among the molecules on screen anddouble click on the line “density potential...” inside the Surfacesdialog. An electrostatic potential map for benzene will appear.Click on the map. The Style menu will appear at the bottom rightof the screen. Select Transparent from this menu to present themap as a translucent solid. This will allow you to see the molecularskeleton underneath. The surface is colored “red” in the π system(indicating negative potential and the fact that this region isattracted to a positive charge), and “blue” in the σ system(indicating positive potential and the fact that this region is repelledby a positive charge).

Select aniline from the molecules on screen, and double click onthe line “density ionization...” inside the Surfaces dialog. Thegraphic which appears, a so-called local ionization potential map,colors in red regions on the density surface from which electronremoval (ionization) is relatively easy, meaning that they are subjectto electrophilic attack. These are easily distinguished from regionswhere ionization is relatively difficult (colored in blue). Note thatthe ortho and para ring carbons are more red than the meta carbons,consistent with the known directing ability of the amino substituent.

Tutorial 1W 7/1/04, 2:38 PM17

18

Finally, select cyclohexanone from the molecules on screen , anddouble click on the line “lumo...” in the Surfaces dialog. Theresulting graphic portrays the lowest-energy empty molecularorbital (the LUMO) of cyclohexanone. This is a so-called π* orbitalwhich is antibonding between carbon and oxygen. Note that theLUMO is primarily localized on carbon, meaning that this is wherea pair of electrons (a nucleophile) will “attack” cyclohexanone.

A better portrayal is provided by a LUMO map, which displaysthe (absolute) value of the LUMO on the electron density surface.Here, the color blue is used to represent maximum value of theLUMO and the color red, minimum value. First, remove theLUMO from your structure (double click on the line “lumo...” inthe Surfaces dialog) and then turn on the LUMO map (doubleclick on the line “density lumo...” in the dialog). Note that theblue region is concentrated directly over the carbonyl carbon. Also,note that the so-called axial face shows a greater concentration ofthe LUMO than the equatorial face. This is consistent with theknown stereochemistry of nucleophilic addition (see the activity“Molecular Shapes V. Which Conformer Leads to Product?”).

8. When you are finished, close all the molecules on screen by selectingClose from the File menu or alternatively by clicking on .

Tutorial 1W 7/1/04, 2:38 PM18

19

2WAcrylonitrile:

Building an Organic MoleculeThis tutorial illustrates use of the organic model kit, as well as thesteps involved in examining and querying different molecular modelsand in carrying out a quantum chemical calculation.

The simplest building blocks incorporated into Spartan’s organic modelkit are “atomic fragments”. These constitute specification of atomtype, e.g., carbon, and hybridization, e.g., sp3. The organic model kitalso contains libraries of common functional groups and hydrocarbonrings, the members of which can easily be extended or modified. Forexample, the carboxylic acid group in the library may be modified tobuild a carboxylate anion (by deleting a free valence from oxygen),or an ester (by adding tetrahedral carbon to the free valence at oxygen).

RC

OH

C

carboxylic acid

RC

O–

O

carboxylate anion

RC

OCH3

O

ester

Acrylonitrile provides a good first opportunity to illustrate the basicsof molecule building, as well as the steps involved in carrying outand analyzing a simple quantum chemical calculation.

C CH

H H

CN

acrylonitrile

1. Click on File from the menu bar and then click on New from themenu which appears (or click on the icon in the File toolbar).The “organic” model kit appears.

Tutorial 2W 7/1/04, 2:47 PM19

20

Click on trigonal planar sp2 hybridized carbon from the libraryof atomic fragments. The fragment icon is shown in reverse video,and a model of the fragment appears at the top of the model kit.Bring the cursor anywhere on screen and click. Rotate the carbonfragment (move the mouse while holding down the left button) sothat you can clearly see both the double free valence (“=”) andthe two single free valences (“-”).

Spartan’s model kits connect atomic fragments (as well as groups, ringsand ligands) through free valences. Unless you “use” them or deletethem, free valences will automatically be converted to hydrogen atoms.

2. sp2 carbon is still selected. Click on the double free valence. Thetwo fragments are connected by a double bond, leaving you withethylene.

Spartan’s model kits allows only the same type of free valences to beconnected, e.g., single to single, double to double, etc.

3. Click on Groups in the model kit, and then select Cyano fromamong the functional groups available from the menu.

Tutorial 2W 7/1/04, 2:47 PM20

21

Click on one of the free valences on ethylene, to make acrylonitrile.*

If you make a mistake, you can select Undo from the Edit menuto “undo” the last operation or Clear (Edit menu) to start over.

4. Select Minimize from the Build menu (or click on the icon inthe toolbar). The “strain energy” and symmetry point group (Cs)for acrylonitrile are provided at the bottom right of the screen.

5. Select View from the Build menu (or click on the icon in thetoolbar). The model kit disappears, leaving only a ball-and-spokemodel of acrylonitrile on screen.

6. Select Calculations... from the Setup menu.

The Calculations dialog appears. This will allow you to specifywhat task is to be done with your molecule and what theoreticalmodel Spartan will use to accomplish this task.

* You could also have built acrylonitrile without using the Groups menu. First, clear the screenby selecting Clear from the Edit menu. Then build ethylene from two sp2 carbons (as above),select sp hybridized carbon from the model kit and then click on the tip of one of the freevalences on ethylene. Next, select sp hybridized nitrogen from the model kit and click onthe triple free valence on the sp carbon. Alternatively, you could have built the moleculeentirely from groups. First, clear the screen. Then click on Groups, select Alkene from themenu and click anywhere on screen. Then select Cyano from the same menu and click onone of the free valences on ethylene. In general, molecules can be constructed in many ways.

Tutorial 2W 7/1/04, 2:47 PM21

22

Select Equilibrium Geometry from the menu to the right of“Calculate”. This specifies optimization of equilibrium geometry.Next, select Hartree-Fock/3-21G from the menu to the right of“with”. This specifies a Hartree-Fock calculation using the 3-21Gbasis set (referred to as an HF/3-21G calculation). This methodgenerally provides a reliable account of geometries (see the essay“Choosing a Theoretical Model”).

7. Click on Submit at the bottom of the dialog. The Calculationsdialog is replaced by a file browser.

Type “acrylonitrile” in the box to the right of “File name”, andclick on Save*. You will be notified that the calculation has beensubmitted. Click on OK to remove the message.

* You can use default names (spartan1, spartan2, . . .) simply by clicking on Save.

Tutorial 2W 7/1/04, 2:47 PM22

23

After a molecule has been submitted, and until the calculation hascompleted, you are not permitted to modify information associatedwith it. You can monitor your calculation as well as abort it if necessaryusing Monitor under the Options menu.

8. You will be notified when the calculation has completed. ClickOK to remove the message. Select Output from the Display menu.A window containing “text output” for the job appears.

You can scan the output from the calculation by using the scrollbar at the right of the window. Information provided includes thetask, basis set, number of electrons, charge and multiplicity, as wellas the point group of the molecule. A series of lines, each beginningwith “Cycle no:”, tell the history of the optimization process. Eachline provides results for a particular geometry; “Energy” gives theenergy in atomic units (1 atomic unit = 2625 kJ/mol) for thisgeometry, “Max Grad.” gives the maximum gradient (“slope”), and“Max Dist.” gives the maximum displacement of atoms betweencycles. The energy will monotonically approach a minimum valuefor an optimized geometry, and Max Grad. and Max Dist. will eachapproach zero. Near the end of the output is the final total energy(-168.82040 atomic units). Click on in the top right-hand cornerof the dialog to remove the dialog from the screen.

9. You can obtain the final total energy and the dipole moment from

Tutorial 2W 7/1/04, 2:47 PM23

24

the Molecule Properties dialog, without having to go throughthe text output. Select Properties from the Display menu. Youcan “see” the dipole moment vector (indicating the sign and overalldirection of the dipole moment), by checking Dipole near thebottom of this dialog. (A tube model provides the clearest picture.)

When you are finished, turn “off” display of the dipole momentvector by unchecking the box.

10.Click on an atom. The (Molecule Properties) dialog will bereplaced by the Atom Properties dialog. This gives the chargeon the selected atom. To obtain the charge on another atom, simplyclick on it. Click on at the top right of the Atom Propertiesdialog to remove it from the screen.

11.Atomic charges can also be attached as “labels” to your model.Select Configure... from the Model menu, and check Chargeunder “Atom” in the Configure dialog which appears.

Click OK to remove the dialog.

12.Click on to remove “acrylonitrile” from the screen. Also, closeany dialogs which may still be open.

Tutorial 2W 7/1/04, 2:47 PM24

25

3WSulfur Tetrafluoride:

Building an Inorganic MoleculeThis tutorial illustrates the use of the inorganic model kit for moleculebuilding. It also shows how molecular models may be used to quantifyconcepts from more qualitative treatments.

Organic molecules are made up of a relatively few elements andgenerally obey conventional valence rules. They may be easily builtusing the organic model kit. However, many molecules incorporateother elements, or do not conform to normal valence rules, or involveligands. They cannot be constructed using the organic model kit.Sulfur tetrafluoride is a good example.

SF

F

F

F

sulfur tetrafluoride

The unusual “see-saw” geometry observed for the molecule is aconsequence of the fact that the “best” (least crowded) way to positionfive electron pairs around sulfur is in a trigonal bipyramidalarrangement. The lone pair assumes an equatorial position so as toleast interact with the remaining electron pairs. The rationale behindthis is that a lone pair is “bigger” than a bonding electron pair.

Sulfur tetrafluoride provides the opportunity to look at the bondingand charges in a molecule which “appears” to have an excess ofelectrons around its central atom (ten instead of eight), as well as tolook for evidence of a lone pair. Further attention is given to sulfurtetrafluoride in the activity “Beyond VSEPR Theory” later in this guide.

Tutorial 3W 7/1/04, 3:12 PM25

26

1. Bring up the inorganic model kit by clicking on and thenclicking on the Inorganic tab at the top of the organic model kit.

The inorganic model kit comprises a Periodic Table followed bya selection of “atomic hybrids” and then bond types. Further downthe model kit are the Rings, Groups and Ligands menus, thefirst two of which are the same as found in the organic model kit.

2. Select (click on) S in the Periodic Table and the five-coordinatetrigonal bipyramid structure from the list of atomic hybrids.Click anywhere in the main window. A trigonal bipyramid sulfurwill appear.

3. Select F in the Periodic Table and the one-coordinate entry fromthe list of atomic hybrids. One after the other, click on both axialfree valences of sulfur, and two of the three equatorial free valences.*

4. It is necessary to delete the remaining free valence (on anequatorial position); otherwise it will become a hydrogen uponleaving the builder. Select Delete from the Build menu (or click

* This step and the following two steps could also be accomplished from the organic modelkit. To bring it up, click on the Organic tab at the top of the inorganic model kit.

Tutorial 3W 7/1/04, 3:12 PM26

27

on the icon in the toolbar) and then click on the remainingequatorial free valence.

5. Click on . Molecular mechanics minimization will result in astructure with C2v symmetry. Click on .

6. Select Calculations... from the Setup menu. Specify calculationof equilibrium geometry using the HF/3-21G model. Click on OK.

7. Select Surfaces from the Setup menu. Click on Add... at thebottom of the Surfaces dialog and select HOMO from the Surfacemenu in the (Add Surface) dialog which appears.

Click on OK. Leave the Surfaces dialog on screen.

8. Select Submit from the Setup menu, and supply the name “sulfurtetrafluoride see-saw”.

9. After the calculations have completed, select Properties from theDisplay menu to bring up the Molecule Properties dialog. Next,click on sulfur to bring up the Atom Properties dialog. Is sulfurneutral or negatively charged, indicating that more than the normalcomplement of (eight) valence electron surrounds this atom, or isit positively charged, indicating “ionic bonding”?

SF–

F

F

F

+

10.Click on the line “homo...” inside the Surfaces dialog to examinethe highest-occupied molecular orbital. Does it “point” in the

Tutorial 3W 7/1/04, 3:12 PM27

28

expected direction? It is largely localized on sulfur or is theresignificant concentration on the fluorines? If the latter, is the orbital“bonding” or “antibonding”? (For a discussion of non-bonding,bonding and antibonding molecular orbitals, see the essay “Atomicand Molecular Orbitals”.)

11.Build square planar SF4 as an alternative to the “see-saw” structure.Bring up the inorganic model kit ( ), select S from the PeriodicTable and the four-coordinate square-planar structure from thelist of atomic hybrids. Click anywhere on screen. Select F in thePeriodic Table and the one-coordinate entry from the list ofatomic hybrids. Click on all four free valences on sulfur. Click on

and then on .

12.Enter the Calculations dialog (Setup menu) and specifycalculation of equilibrium geometry using the HF/3-21G model(the same level of calculation as you used for the “see-saw”structure*). Click on Submit at the bottom of the dialog, with thename “sulfur tetrafluoride square planar”.

13.After the calculation has completed, bring up the MoleculeProperties dialog (Properties from the Display menu) and notethe energy. Is it actually higher (more positive) than that for the“see-saw” structure?

14.Close both molecules as well as any remaining dialogs.

* You need to use exactly the same theoretical model in order to compare energies or otherproperties for different molecules.

Tutorial 3W 7/1/04, 3:12 PM28

29

4WInfrared Spectrum of Acetone

This tutorial illustrates the steps required to calculate and displaythe infrared spectrum of a molecule.

Molecules vibrate in response to their absorbing infrared light.Absorption occurs only at specific wavelengths, which gives rise tothe use of infrared spectroscopy as a tool for identifying chemicalstructures. The vibrational frequency is proportional to the square rootof a quantity called a “force constant” divided by a quantity called the“reduced mass”.

frequency α force constant

reduced mass

The force constant reflects the “flatness” or “steepness” of the energysurface in the vicinity of the energy minimum. The steeper the energysurface, the larger the force constant and the larger the frequency.The reduced mass reflects the masses of the atoms involved in thevibration. The smaller the reduced mass, the larger the frequency.

This tutorial shows you how to calculate and display the infraredspectrum of acetone, and explore relationships between frequencyand both force constant and reduced mass. It shows why the carbonylstretching frequency is of particular value in infrared spectroscopy.

1. Click on to bring up the organic model kit. Select sp2 carbon( ) and click anywhere on screen. Select sp2 oxygen ( ) andclick on the double free valence on carbon to make the carbonylgroup. Select sp3 carbon ( ) and, one after the other, click on thetwo single free valences on carbon. Click on and then on .

2. Enter the Calculations dialog (Setup menu) and requestcalculation of an equilibrium geometry using the HF/3-21G model.

Tutorial 4W 7/1/04, 3:15 PM29

30

Check IR to the right of “Compute” to specify calculation ofvibrational frequencies. Finally, click on the Submit button at thebottom of the dialog, and provide the name “acetone”.

3. After the calculation has completed, bring up the Spectra dialog(Display menu). This contains a list of vibrational frequenciesfor acetone. First click on the top entry (the smallest frequency)and, when you are done examining the vibrational motion, clickon the bottom entry (the largest frequency).

The smallest frequency is associated with torsional motion of themethyl rotors. The largest frequency is associated with stretchingmotion of CH bonds. Methyl torsion is characterized by a flat potentialenergy surface (small force constant), while CH stretching ischaracterized by a steep potential energy surface (large force constant).

Display the IR spectrum (click on Draw IR Spectrum at thebottom of the dialog). Locate the frequency corresponding to theCO stretch. The experimental frequency is around 1740 cm-1, butthe calculations will yield a higher value (around 1940 cm-1).

The CO stretching frequency is a good “chemical identifier” becauseit “stands alone” in the infrared spectrum and because it is “intense”.

4. Change all the hydrogens in acetone to deuteriums to see the effectwhich increased mass has on vibrational frequencies. First makea copy of “acetone” (Save As... from the File menu or click onthe icon in the toolbar). Name the copy “acetone d6” SelectProperties from the Display menu and click on one of thehydrogens. Select 2 deuterium from the Mass Number menu.Repeat for the remaining five hydrogens.

5. Submit for calculation. When completed, examine the vibrationalfrequencies. Note that the frequencies of those motions whichinvolve the hydrogens (in particular, the six vibrational motionscorresponding to “CH stretching”) are significantly reduced overthose in the non-deuterated system.

6. Close all molecules on screen in addition to any remaining dialogs.

Tutorial 4W 7/1/04, 3:15 PM30

31

5WElectrophilic Reactivity of

Benzene and PyridineThis tutorial illustrates the calculation, display and interpretation ofelectrostatic potential maps. It also illustrates the use of “documents”comprising two or more molecules.

While benzene and pyridine have similar geometries and while bothare aromatic, their “chemistry” is different. Benzene’s chemistry isdictated by the molecule’s π system, while the chemistry of pyridineis a consequence of the lone pair on nitrogen. This tutorial showshow to use electrostatic potential maps to highlight these differences.

1. Build benzene. Click on . Select Benzene from the Rings menuand click on screen. Click on .

2. Build pyridine. In order to put both benzene and pyridine into thesame “document”, select New Molecule (not New) from the Filemenu. Benzene (Rings menu) is still selected. Click anywhere onscreen. Select aromatic nitrogen from the model kit and doubleclick on one of the carbon atoms (not a free valence). Click on and then click on . To go between the two molecules in yourdocument, use the and keys at the bottom left of the screen(or use the slider bar).

3. Select Calculations... (Setup menu) and specify calculation ofequilibrium geometry using the HF/3-21G model. Click on OKto dismiss the dialog. Select Surfaces (Setup menu). Click onAdd... at the bottom of the Surfaces dialog to bring up the AddSurface dialog. Select density from the Surface menu andpotential from the Property menu. Click on OK. Leave theSurfaces dialog on screen. Select Submit (Setup menu) andsupply the name “benzene and pyridine”.

Tutorial 5W 7/1/04, 3:19 PM31

32

4. When completed, bring up the spreadsheet (Spreadsheet underthe Display menu) and check the box to the right of the molecule“Label” (leftmost column) for both entries. This allows benzeneand pyridine to be displayed simultaneously. However, the motionsof the two molecules will be “coupled” (they will move together).Select (uncheck) Coupled from the Model to allow the twomolecules to be manipulated independently. In turn, select (clickon) each and orient such that the two are side-by-side.

5. Double click on the line “density potential...” inside the Surfacesdialog. Electrostatic potential maps for both benzene and pyridinewill be displayed. Change the scale so that the “neutral” color is“green”. Select Properties (Display menu) and click on one ofthe maps to bring up the Surface Properties dialog.

Type “-35” and “35” inside the boxes underneath “Property Range”(press the Enter key following each data entry). The “red” regionsin benzene, which are most attractive to an electrophile, correspondto the molecule’s π system, while in pyridine they correspond tothe σ system in the vicinity of the nitrogen. Note that the π systemin benzene is “more red” than the π system in pyridine (indicatingthat it is more susceptible to electrophilic attack here), but thatthe nitrogen in pyridine is “more red” than the π system in benzene(indicating that pyridine is overall more susceptible to attack byan electrophile).

Further discussion of the use of such maps is provided in theessay “Electrostatic Potential Maps: Charge Distributions”.

6. Remove “benzene and pyridine” and any dialogs from the screen.

Tutorial 5W 7/1/04, 3:19 PM32

33

6WWeak vs. Strong Acids

This tutorial shows how electrostatic potential maps may be used todistinguish between weak and strong acids, and quantify subtledifferences in the strengths of closely-related acids. It also showshow information can be retrieved from Spartan’s database.

Chemists know that nitric acid and sulfuric acids are strong acids,acetic acid is a weak acid, and that ethanol is a very weak acid. Whatthese compounds have in common is their ability to undergoheterolytic bond fracture, leading to a stable anion and a “proton”.What distinguishes a strong acid from a weak acid is the stability ofthe anion. NO3

– and HOSO3– are very stable anions, CH3CO2

– issomewhat less stable and CH3CH2O

– is even less so.

One way to reveal differences in acidity is to calculate the energy ofdeprotonation for different acids, e.g., for nitric acid.

HONO2 H+ + NO3–

This involves calculations on both the neutral acid and on the resultinganion (the energy of a proton is zero). An alternative approach,illustrated in this tutorial, involves comparison of electrostatic potentialmaps for different acids, with particular focus on the potential in thevicinity of the “acidic hydrogen”. The more positive the potential,the more likely will dissociation occur, and the stronger the acid.

1. Build nitric acid. Click on to bring up the organic model kit.Select Nitro from the Groups menu and click anywhere on screen.Add sp3 oxygen to the free valence on nitrogen. Click on .Build sulfuric acid. Select New Molecule (not New) from theFile menu. Select Sulfone from the Groups menu and clickanywhere on screen. Add sp3 oxygen to both free valences onsulfur. Click on . Build acetic acid. Again select New Molecule.Select Carboxylic Acid from the Groups menu and click

Tutorial 6W 7/1/04, 3:23 PM33

34

anywhere on screen. Add sp3 carbon to the free valence atcarbon. Click on . Finally, build ethanol. Select New Moleculeand construct from two sp3 carbons and an sp3 oxygen. Click on

, and then on .

2. Bring up the Calculations dialog and specify calculation ofequilibrium geometry using the HF/6-31G* model. Click on OK.Bring up the Surfaces dialog and click on Add... (at the bottomof the dialog). Select density from the Surface menu and potentialfrom the Property menu in the Add Surface dialog which appears.Click on OK. Leave the Surfaces dialog on screen. Submit forcalculation with the name “acids”.

3. When completed, bring up the spreadsheet and check the boximmediately to the right of the molecule label for all four entries.The four molecules will now be displayed simultaneously onscreen. Select (uncheck) Coupled from the Model menu so thatthey may be independently manipulated, and arrange on screensuch that the “acidic” hydrogens are visible.

Manipulations normally refer only to the selected molecule. To rotateor translate all molecules together, hold down the Ctrl (Control)key in addition to the left or right buttons, respectively, while movingthe mouse.

4. Double click on the line “density potential ...” inside the Surfacesdialog. Electrostatic potential maps for all four acids will bedisplayed. Examine the potential in the vicinity of the acidichydrogen (one of the two equivalent acidic hydrogens for sulfuricacid). Change the scale (color) to highlight differences in thisregion. Select Properties (Display menu) and click on one of themaps. Type “0” and “90” inside the boxes underneath “PropertyRange” in the Surface Properties dialog. Press the Enter keyfollowing each data entry.

“Blue” regions identify acidic sites, the more blue the greater theacidity. On this basis, rank the acid strength of the four compounds.

5. Remove “acids” and any open dialogs from the screen.

Tutorial 6W 7/1/04, 3:23 PM34

35

6. One after the other, build trichloroacetic, dichloroacetic,chloroacetic, formic, benzoic, acetic and pivalic acids (structuralformulae are provided in the table below). Put all into the samedocument (New Molecule instead of New following the firstmolecule). Click on when you are finished.

acid pKa acid pKa

trichloroacetic (Cl3CCO2H) 0.7 benzoic (C6H5CO2H) 4.19dichloroacetic (Cl2CHCO2H) 1.48 acetic (CH3CO2H) 4.75chloroacetic (ClCH2CO2H) 2.85 pivalic ((CH3)3CCO2H) 5.03formic (HCO2H) 3.75

7. Note that the name of the presently selected molecule in thedocument appears at the bottom of the screen. This indicates thata HF/3-21G calculation is available in Spartan’s database. Clickon to the left of the name, and then click on Replace All in thedialog which results. Structures obtained from HF/3-21Gcalculations will replace those you have built.

8. Enter the Calculations dialog and specify a single-point-energyHF/3-21G calculation. Click on OK. Enter the Surfaces dialog.Click on Add..., select density from the Surface menu andpotential from the Property menu in the Add Surface dialogwhich appears and then click on OK. Leave the Surfaces dialogon screen. Submit for calculation. Name it “carboxylic acids”.

9. Bring up the spreadsheet. Expand it so that you can see all sevenmolecules, and that three data columns are available. Click insidethe header cell for a blank column. Click on Add... at the bottomof the spreadsheet, select Name from the list of entries and clickon OK. The name of each molecule will appear. Next, click insidethe header cell of an available data column, type “pKa” and pressthe Enter key. Enter the experimental pKa’s (see above) into theappropriate cells under this column. Press the Enter key followingeach entry. Finally, click inside the header cell of the next availabledata column and type “potential”. Press the Enter key.

Tutorial 6W 7/1/04, 3:23 PM35

36

10.After all calculations have completed, arrange the molecules suchthat the “acidic hydrogen” is visible. You need to check the box tothe right of the “Label” column in the spreadsheet for each entry,and select (uncheck) Coupled from the Model menu.

11.Double click on the line “density . . .” inside the Surfaces dialogto turn on the electrostatic potential map for each molecule. Bringup the Properties dialog, remove the checkmark from GlobalSurfaces, and click on the Reset button at the top of the dialog.The property range will now apply to the individual molecules.Enter the maximum value (most positive electrostatic potential)into the appropriate cell of the spreadsheet (under “potential”),and press the Enter key.

12.Plot experimental pKa vs. potential. Bring up the Plots dialog(Display menu), select pKa under the X Axis menu and potentialfrom the Y Axes list, and click on OK. The data points areconnected by a cubic spline. For a least squares fit, selectProperties from the Display menu, click on the curve, and selectLinear LSQ from the Fit menu in the Curve Properties dialog.

13.Close “carboxylic acids” and remove any remaining dialogs fromthe screen.

Tutorial 6W 7/1/04, 3:23 PM36

37

7WInternal Rotation in n-Butane

This tutorial illustrates the steps required to calculate the energy ofa molecule as a function of the torsion angle about one of its bonds,and to produce a conformational energy diagram.

Rotation by 1800 about the central carbon-carbon bond in n-butanegives rise to two distinct “staggered” structures, anti and gauche.

HHCH3

H HCH3

CH3CH3

CH3

H H

anti gauche

H

Both of these should be energy minima (conformers), and the correctdescription of the properties of n-butane is in terms of a Boltzmannaverage of the properties of both conformers (for discussion see theessay “Total Energies and Thermodynamic and Kinetic Data”).

This tutorial shows you how to calculate the change in energy as afunction of the torsion angle in n-butane, place your data in aspreadsheet and make a conformational energy diagram.

1. Click on to bring up the organic model kit. Make n-butanefrom four sp3 carbons. Click on to dismiss the model kit.

2. Set the CCCC dihedral angle to 00 (syn conformation). Click on then, one after the other, click on the four carbon atoms in

sequence. Type “0” (00) into the text box to the right of “dihedral...”at the bottom right of the screen and press the Enter key.

3. Select Constrain Dihedral from the Geometry menu. Click againon the same four carbons you used to define the dihedral angle,and then click on at the bottom right of the screen. The iconwill change to indicating that a dihedral constraint is to be

Tutorial 7W 7/1/04, 3:24 PM37

38

imposed. Select Properties (Display menu) and click on theconstraint marker on the model on screen. This leads to theConstraint Properties dialog.

4. Check Dynamic inside the dialog. This leads to an extended formof the Constraint Properties dialog which allows the single(dihedral angle) constraint value to be replaced by a “range” ofconstraint values.

Leave the value of “0” (0°) in the box to the right of Value as it is,but change the contents of the box to the right of to to “180”(1800). Be sure to press the Enter key after you type in the value.The box to the right of Steps should contain the value “10”. (If itdoes not, type “10” in this box and press the Enter key.) Whatyou have specified is that the dihedral angle will be constrainedfirst to 0°, then to 20°*, etc. and finally to 180°. Click on todismiss the dialog.

5. Bring up the Calculations dialog and select Energy Profile fromthe menu to the right of “Calculate”, and Semi-Empirical fromthe menu to the right of “with”. Click on Submit at the bottom ofthe dialog and provide the name “n-butane”.

* The difference between constraint values is given by: (final-initial)/(steps-1).

Tutorial 7W 7/1/04, 3:24 PM38

39

6. When the calculations on all conformers have completed, theywill go into a document named “n-butane.Profile1”. Open thisdocument ( ). (You might wish to first close “n-butane” to avoidconfusion.) Align the conformers to get a clearer view of therotation. Select Align Molecules from the Geometry menu and,one after the other, click on either the first three carbons or thelast three carbons. Then click on the Align button at the bottomright of the screen, and finally click on . Bring up the spreadsheet(Display menu), and enter both the energies relative to the 180°or anti conformer, and the CCCC dihedral angles. First, click onthe label (“M010”) for the bottom entry in the spreadsheet (thisshould be the anti conformer), then click on the header cell for theleft most blank column, and finally, click on Add... at the bottomof the spreadsheet. Select rel. E from among the selections in thedialog which results, kJ/mol from the Energy menu and click onOK. To enter the dihedral angle constraints, click on , click onthe constraint marker and click on at the bottom of the screen(to the right of the value of the dihedral angle constraint). Finally,click on .

7. Select Plots... (Display menu). Select Dihedral (Con1) from theitems in the X Axis menu and rel. E(kJ/mol) from the Y Axeslist. Click on OK to dismiss the dialog and display a plot.

The curve (a so-called “cubic spline”) smoothly connects the datapoints. You can see that it contains two minima, one at 180° (the

Tutorial 7W 7/1/04, 3:24 PM39

40

anti form) and one around 60° (the gauche form). The former islower in energy.

Further discussion of the potential energy surface for n-butaneamong other systems is provided in the essay “Potential EnergySurfaces”.

8. Remove any molecules and any remaining dialogs from the screen.

Tutorial 7W 7/1/04, 3:24 PM40

41

8WEne Reaction

This tutorial illustrates the steps involved in first guessing and thenobtaining a transition state for a simple chemical reaction. Followingthis, it shows how to produce a “reaction energy diagram”.

The ene reaction involves addition of an electrophilic double bond toan alkene with an allylic hydrogen. The (allylic) hydrogen istransferred and a new carbon-carbon bond is formed, e.g.

HH1

2

3

5

4

The ene reaction belongs to the class of pericyclic reactions whichincludes such important processes as the Diels-Alder reaction andthe Cope and Claisen rearrangements.

Spartan may be used to locate the transition-state for the ene reactionof ethylene and propene and then show the detailed motions which theatoms undergo during the course of reaction. It is easier to start from1-pentene, the product of the ene reaction, rather than from the reactants.

1. Build 1-pentene. Click on to bring up the organic model kit.Click on the Groups button, select Alkene from the menu andclick anywhere on screen. Select sp3 carbon ( ) and build a three-carbon chain onto one of the free valences on ethylene. Adjust theconformation of the molecule such that all five carbons and oneof the “hydrogens” (free valences) on the terminal methyl groupform a “6-membered ring.” You can rotate about a bond by firstclicking on it (a red torsion marker appears on the bond) and thenmoving the mouse “up and down” while holding down on boththe left button and the Alt key. Do not minimize. Click on .

Tutorial 8Ws 7/1/04, 3:27 PM41

42

2. Select Transition States from the Search menu (or click on the icon in the toolbar). Orient the molecule such that both the CH

bond involving the hydrogen which will “migrate” and the C4-C5

bond are visible (see figure on previous page for numbering). Clickon the CH bond and then on the C4-C5 bond. An “arrow” will bedrawn.* Orient the molecule such that both the C3-C4 and the C2-C3 bonds are visible. Click on the C3-C4 bond and then on the C2-C3 bond. A second arrow will be drawn. Orient the molecule suchthat the C1=C2 bond, C1 and the hydrogen (on C5) which willmigrate are all visible. Click on the C1=C2 bond and, while holdingdown on the Shift key, click on C1 and then on the hydrogen. Athird arrow will be drawn.

If you make a mistake, you can delete an arrow by clicking on itwhile holding down on the Delete key. Alternatively, select Deletefrom the Build menu (or click on the icon in the toolbar), thenclick on the arrow and finally click on . When all is in order,click on at the bottom right of the screen. Your structure willbe replaced by a guess at the transition state.

3. Select Calculations... (Setup menu). Select Transition StateGeometry from the menu to the right of “Calculate” and Semi-Empirical from the menu to the right of “with”. Check IR to theright of “Compute” and click on Submit with the name “enetransition state”.

4. When the calculation has completed animate the motion of atomsalong the reaction coordinate. Select Spectra under the Displaymenu. Click on the top entry of the list in the Spectra dialog. Itcorresponds to an imaginary frequency, and will be designated

* Formally, this corresponds to migration of a pair of electrons from the tail of the arrow tothe head. However, Spartan treats arrows simply as a convenient and familiar nomenclaturewith which to search its database of transition states.


43

with an “i” in front of the number*. Make certain that the associatedvibrational motion is consistent with the reaction of interest andnot with some other process.

5. Controls at the bottom of the Spectra dialog allow for changingboth the amplitude of Vibration (Amp) and the number of stepswhich make up the motion (Steps). The latter serves as a “speedcontrol”. Change the amplitude to “0.3” (type “0.3” in the box tothe right of Amp and press the Enter key). Next, click on MakeList at the bottom of the dialog. This will give rise to a documentcontaining a series of structures which follow the reactioncoordinate down from the transition state both toward reactantand product. To avoid confusion, it might be better to removeoriginal molecule from the screen. Click on “ene transition state”(the vibrating molecule) and close it. Also remove the Spectradialog by clicking on .

6. Enter the Calculations dialog (Setup menu) and specify a single-point energy calculation using the semi-empirical model (the sametheoretical model used to obtain the transition state and calculatethe frequencies). Make certain that Global Calculations at thebottom of the dialog is checked before you exit the dialog. Next,enter the Surfaces dialog and specify evaluation of two surfaces:a bond density surface and a bond density surface onto which theelectrostatic potential has been mapped. Click on Add . . ., selectdensity (bond) for Surface and none for Property and click onOK. Click on Add . . . again, select density (bond) for surfaceand potential for Property and click on OK. Make certain thatGlobal Surfaces at the bottom of the dialog is checked. Leavethe Surfaces dialog on screen.

7. Submit for calculation (Submit from the Setup menu). Name it“ene reaction”. Once the calculations have completed, enter the

* Recall from the tutorial “Infrared Spectrum of Acetone”, that frequency is proportional tothe square root of the force constant divided by the reduced mass. The force constantassociated with the reaction coordinate is negative because the energy “curves downward”at the transition state (see the essay “Potential Energy Surfaces”). Since the reduced massis positive, the ratio of force constant to reduced mass is a negative number, meaning that itssquare root is an imaginary number.


44

Surfaces dialog and, one after the other, select the surfaces whichyou have calculated. For each, step through the sequence ofstructures ( and ) keys at the bottom of the screen) or animatethe reaction ( ). Note, in particular, the changes in bondingrevealed by the bond density surface. Also pay attention to the“charge” on the migrating atom as revealed by the sequence ofelectrostatic potential maps. Is it best described as a “proton” (bluecolor), hydrogen atom (green color) or “hydride anion” (red color)?

Further discussion of the use of electrostatic potential mapsto investigate charge distributions is provided in the essay“Electrostatic Potential Maps: Charge Distributions”.

9. Close “ene reaction” as well as any remaining dialogs.


45

9WProteins and Nucleotides

This tutorial illustrates models appropriate to large biomolecules, inparticular, ribbon displays of protein and nucleotide “backbones”and display of hydrogen bonds. Biomolecule building is not illustratedand no calculations are performed.

Treatment of very large molecules, proteins and nucleotides(“biopolymers”) most important among them, requires models whichare simpler than those which are appropriate for small organic andinorganic molecules. This refers both to display and manipulation,where much detail needs to be eliminated, and to the calculation ofstructure and properties, where molecular mechanics needs to replacequantum mechanics.

This tutorial uses an unusual “protein-RNA complex” to illustrate avariety of models for the display of biopolymers, including ribbondisplays to elucidate the “backbone” and hydrogen-bond displays toshow how the structure is “held together”.

1. Open “tutorial 9” in the “tutorials” directory. This contains a PDB(Protein Data Bank) file* of a protein-RNA complex. Note that asimple ribbon display, demarking the backbones of the proteinand RNA chains has replaced the usual structure models (“tube”,“ball-and-spoke”, etc.). To see why this is necessary, turn “on”(select) one of these models from the Model menu. The detail hascompletely obliterated the most important structure feature, thatthe molecule is made up of two “intertwined” segments. Note,however, that a space-filling model (Space Filling under the Modelmenu) does provide indication of overall size and shape of thecomplex. When you are done, select Hide from the Model menu.

* PDB designation 1A1T. R.N. de Guzman, Z.R. Wu, C.C. Stalling, L. Pappalardo, P.N. Borerand M.F. Summers, Science, 279, 384 (1998).

Tutorial 9W 7/1/04, 3:28 PM45

46

2. The structure of this complex has been determined by NMRspectroscopy which gives several alternative conformers. Stepthrough them ( and at the bottom left of the screen) to seewhere they are similar and where they differ. In particular, noticethat a large portion of the overall structure remains basically thesame from one conformer to another.

3. Select Configure... from the Model menu and click on theRibbons tab in the dialog which results. Select By Residue under“Coloring” in the Configure Ribbons dialog and click on OK.The model is now colored according to amino acid/nucleotidebase. Click on the various “color bands” to see what they are (labelsare provided at the bottom right of the screen).

Hydrogen bonding is known to be a decisive factor in determiningthe three-dimensional structures of biopolymers. The base pairs incomplementary strands which make up DNA are “held together” byhydrogen bonds. Helical structures in proteins are also maintainedby hydrogen bonds as are neighboring strands in so-called β sheets.

4. Select Hydrogen Bonds (Model menu). Single “dotted lines”represent hydrogen bonds throughout the “protein part” of thecomplex, and “sets” of two or three dotted lines in the “RNA”part. The latter form the connections between nucleotide basesand the number of lines in each set actually allows you to identifywhat the bases are.

5. Select Tube (Model menu). Also, select (uncheck) Hydrogensfrom this menu. You can now see in greater detail the structure ofthe complex and the positions of the hydrogen bonds.

6. Close “tutorial 9”.

Tutorial 9W 7/1/04, 3:28 PM46

123

7Atomic and Molecular OrbitalsChemists have developed a variety of methods for describing electronsin molecules. Lewis structures are the most familiar. These drawingsassign pairs of electrons either to single atoms (lone pairs) or pairs ofatoms (bonds)*. The quantum mechanical “equivalents” are atomicand molecular orbitals which arise from solution of (approximate)Schrödinger equations for atoms and molecules, respectively.Molecular orbitals are spread throughout the entire molecule, that is,they are “delocalized”. Because of this, they are typically moredifficult to interpret than Lewis structures.

Orbital Surfaces

Molecular orbitals provide important clues about chemical reactivity,but before we can use this information we first need to understandwhat molecular orbitals look like. The following figure shows tworepresentations, a “hand” drawing and a Spartan-generated image ofan unoccupied molecular orbital of hydrogen molecule, H2.

H H

unoccupied molecular orbital in hydrogen

Open “hydrogen empty”. Note that except for the colors (sign of theorbital) the two sides of the graphic are identical. The junction between“red” and “blue” regions is where the value of the orbital is zero.Close “hydrogen empty” when you are finished.

* Treatment here is limited to systems in which all electrons are paired. Spartan can also treatsystems with one or more unpaired electrons (“radicals”, triplet states, etc.).

Essay 7 7/2/04, 9:13 AM123

124

The hand drawing shows the orbital as two circles and a dashed line.The circles identify regions where the orbital takes on a significantvalue, either positive (shaded) or negative (unshaded). The dashedline identifies locations where the orbital’s value is exactly zero (a“node”). The drawing is useful, but it is also limited. We only obtaininformation about the orbital in two dimensions, and we only learnthe location of “significant” regions and not how the orbital buildsand decays inside and outside of these regions.

The Spartan-generated image depicts the same orbital as a “surface”of constant value. The surface is “accurate” in that it is derived froman authentic (but approximate) calculated solution to the quantummechanical equations of electron motion. Equally important, theimage is three-dimensional. It can be manipulated using Spartan, andcan be looked at from a variety of different perspectives. Note thatwhat we call an “orbital surface” actually consists of two distinctsurfaces represented by different colors. The two surfaces have thesame meaning as the two circles in the orbital drawing. They identifyregions where the orbital takes on a significant value, either positive(blue) or negative (red). The orbital node is not shown, but we canguess that it lies midway between the two surfaces (this follows fromthe fact that the orbital’s value can only change from positive tonegative by passing through zero).

Atomic Orbitals

Atomic orbitals, “descriptions of atoms”, are the fundamental buildingblocks from which molecular orbitals “descriptions of molecules”are assembled. The familiar atomic orbitals for the hydrogen atomare in fact exact solutions of the Schrödinger equation for this oneelectron system. They form an infinite collection (a “complete set”),the lowest-energy member representing the “best” location for theelectron, and higher-energy members representing alternativelocations. Orbitals for “real” many-electron atoms are normallyassumed to be similar in form to those of hydrogen atom, the onlydifference being that, unlike hydrogen, more than the lowest-energyatomic orbital is utilized. In practical quantum chemical calculations,

Essay 7 7/2/04, 9:13 AM124

125

atomic orbitals for many-electron atoms are made up of sums anddifferences of a finite collection of hydrogen-like orbitals (see theessay “Theoretical Models”).

It is common practice to divide the full set of atomic orbitals intocore and valence orbitals, and further to “ignore” the former. Valenceorbitals for an element in the first long row of the Periodic Table are2s, 2px, 2py and 2pz, and for the second long row are 3s, 3px, 3py and3pz. In the case of first-row elements, a single orbital, 1s, liesunderneath (is a core orbital) while in the case of second-row elements,a set of five orbitals, 1s, 2s, 2px, 2py and 2pz, lie underneath.

Open “fluoride and chloride”. On the top row are the four valence orbitalsof fluoride anion and on the bottom row the four valence orbitals ofchloride anion. You can select among them by clicking (left mouse button)on each in turn. First note that the three 2p orbitals in fluoride are identicalexcept for the direction in which they point. The same is true for thethree 3p orbitals in chloride. Next, note that the valence orbitals in chlorideare larger than those in fluoride. Atoms further down in the PeriodicTable are generally larger than analogous atoms further up. Close “fluorideand chloride” when you are finished.

Orbitals and Chemical Bonds

Although molecular orbitals and Lewis structures are both used todescribe electron distributions in molecules, they are used for differentpurposes. Lewis structures are used to count the number of bondingand non-bonding electrons around each atom. Molecular orbitals arenot useful as counting tools, but orbitals and orbital energies are usefultools for describing chemical bonding and reactivity. This sectiondescribes a few common orbital shapes and illustrates their use.

Molecular orbital surfaces can extend over varying numbers of atoms.If the orbital surface (or surfaces) is confined to a single atom or toatoms which are not “close together”, the orbital is regarded as non-bonding. If the orbital contains a surface that extends continuouslyover two neighboring atoms, the orbital is regarded as “bonding”with respect to these atoms. Adding electrons to such an orbital willstrengthen the bond between these atoms and cause them to draw

Essay 7 7/2/04, 9:13 AM125

126

closer together, while removing electrons will have the opposite effect.Two different kinds of bonding orbitals are depicted below. Thedrawing and surface on the left correspond to a σ bond while thedrawing and surface on the right correspond to a π bond.

A B A B

σ bonding π bonding

Open “nitrogen bonding”. The image on the left corresponds to the σbonding orbital of N2, while that on the right corresponds to one of twoequivalent π bonding orbitals. Switch to a transparent or mesh model tosee the underlying molecular skeleton. For Windows, click on the graphicand select the appropriate entry from the menu which appears at the bottomright of the screen. For Mac, position the cursor over the graphic, holddown either the left or right button and select from the menu which appearsalongside. Note that the σ orbitals is drawn in a single color (insofar asNN bonding is concerned) while the π orbital is made up of “red” and“blue” parts. This indicates a “node” or a break in the latter, although notinvolving the NN bond. Close “nitrogen bonding” when you are finished.

It is also possible for an orbital to contain a node that divides theregion between two neighboring atoms into separate “atomic” regions.Such an orbital is regarded as “antibonding” with respect to theseatoms. Adding electrons to an antibonding orbital weakens the bondand “pushes” the atoms apart, while removing electrons from suchan orbital has the opposite effect. The following pictures showdrawings and orbital surfaces for two different kinds of antibondingorbitals. As above, the left and right-hand sides correspond to σ andπ type arrangements, respectively.

node

A B

node

A B

σ antibonding π antibonding

Essay 7 7/2/04, 9:13 AM126

127

Open “nitrogen antibonding”. The image on the left corresponds to theσ antibonding (σ*) orbital of N2 while that on the right corresponds toone of the two equivalent π antibonding (π*) orbitals. Switch to a meshor transparent surface to see the underlying molecular skeleton. Notethat the σ* orbital has a single node (change in color from “red” to “blue”)in the middle of the NN bond, while the π* orbital has two nodes (one inthe middle of the NN bond and to the other along the bond). Close“nitrogen antibonding” when you are finished.

Bonds can be strengthened in two different ways, by adding electronsto bonding orbitals, or by removing electrons from antibondingorbitals. The converse also holds. Bonds can be weakened either byremoving electrons from bonding orbitals or by adding electrons toantibonding orbitals. An example of the consequences is provided inthe activity “Removing and Adding Electrons”.

Singlet Methylene

Molecular orbitals in molecules which contain many atoms aretypically spread throughout the molecule (they are “delocalized”).Delocalized orbitals have complicated shapes and contain multipleinteractions that may be bonding, non-bonding, antibonding, or anymixture of all three. Nevertheless, these shapes can still be brokendown into two-atom interactions and analyzed using the principlesoutlined earlier. This process is illustrated for a triatomic molecule,“singlet” methylene, CH2. (“Singlet” refers to the fact that the eightelectrons in this highly reactive molecule are organized into four pairs,and that each pair of electrons occupies a different molecular orbital.The lowest-energy state of methylene is actually a triplet with threeelectron pairs and two unpaired electrons.)

The lowest energy molecular orbital of singlet methylene is not veryinteresting in that it looks like a 1s atomic orbital on carbon. Theelectrons occupying this orbital restrict their motion to the immediateregion of the carbon nucleus and do not significantly affect bonding.Because of this restriction, and because the orbital’s energy is verylow, this orbital is referred to as a “core” orbital and its electrons arereferred to as “core” electrons.

Essay 7 7/2/04, 9:13 AM127

128

core orbital for methylene

The next orbital is much higher in energy. It consists of a single surfacewhich is delocalized over all three atoms. This means that it issimultaneously (σ) bonding with respect to each CH atom pair.

CH bonding orbital for methylene

The next higher energy orbital is described by two surfaces, a positive(blue) surface that encloses one CH bonding region and a negative(red) surface that encloses the other CH bonding region*. Since eachsurface encloses a bonding region, this orbital is also (σ) bondingwith respect to each CH atom pair. This reinforces the bondingcharacter of the previous orbital. The node that separates the twosurfaces passes through the carbon nucleus, but not through either ofthe CH bonding regions, so it does not affect bonding.

CH bonding orbital for methylene

Thus, the two CH bonds in the Lewis structure for singlet methyleneare “replaced” by two “bonding” molecular orbitals.

The highest-occupied molecular orbital (the HOMO) is also describedby two orbital surfaces. One surface extends into carbon’s “non-bonding” region opposite the two hydrogens. The other surfaceencompasses the two CH bonding regions. Although it is hard to

* Note that the absolute signs (colors) of a molecular orbital are arbitrary, but that the relativesigns (colors) indicate bonding and antibonding character.

Essay 7 7/2/04, 9:13 AM128

129

track the exact path of the orbital node in this picture, it happens topass almost exactly through the carbon. This means that this particularorbital possesses only weak CH bonding character (it is H---Hbonding). It turns out that the non-bonding character of the orbital ismuch more important than the bonding character, in that it leads tothe fact that singlet methylene is able to behave as an electron-pairdonor (a “nucleophile”).

HOMO of methylene

The above analysis shows that while the occupied orbitals of singletmethylene are spread over all three atoms, they are comprehensible.The orbitals divide into two groups, a single low-energy “core” orbitaland three higher-energy “valence” orbitals. The latter consist of twoCH bonding orbitals and a non-bonding orbital on carbon. There isno one-to-one correspondence between these orbitals and the Lewisstructure. The bonding orbitals are not associated with particular bonds,and the non-bonding orbital contains bonding interactions as well.

Open “methylene bonding”. Four images appear corresponding to thecore and three valence orbitals of singlet methylene. Switch to a mesh ortransparent surface to see the underlying molecular skeleton. Close“methylene bonding” when you are finished.

Singlet methylene also possesses unoccupied molecular orbitals. Theunoccupied orbitals have higher (more positive) energies than theoccupied orbitals, and these orbitals, because they are unoccupied,do not describe the electron distribution in singlet methylene.*

Nevertheless, the shapes of unoccupied orbitals, in particular, thelowest-unoccupied orbital (LUMO), is worth considering because itprovides valuable insight into the methylene’s chemical reactivity.

* There is no direct analogy between unoccupied molecular orbitals and Lewis structures(which attempt to describe electron pair bonds and non-bonding electron pairs).

Essay 7 7/2/04, 9:13 AM129

130

The LUMO in methylene has non-bonding character, and looks like a2p atomic orbital on carbon. This suggests that singlet methylene shouldbe able to behave as an electron-pair acceptor (an “electrophile”). Note,however, that were the molecule to accept electrons, these would gointo non-bonding orbital; carbon would become more electron-rich,but the CH bonds would not be much affected.

LUMO of methylene

Open “methylene LUMO” and switch to a mesh or transparent surfaceto see the underlying skeleton. Close “methylene LUMO” when you arefinished.

Frontier Orbitals and Chemical Reactivity

Chemical reactions typically involve movement of electrons from an“electron donor” (base, nucleophile, reducing agent) to an “electronacceptor” (acid, electrophile, oxidizing agent). This electronmovement between molecules can also be thought of as electronmovement between molecular orbitals, and the properties of these“electron donor” and “electron acceptor” orbitals provide considerableinsight into chemical reactivity.

The first step in constructing a molecular orbital picture of a chemicalreaction is to decide which orbitals are most likely to serve as theelectron-donor and electron-acceptor orbitals. It is obvious that theelectron-donor orbital must be drawn from the set of occupied orbitals,and the electron-acceptor orbital must be an unoccupied orbital, butthere are many orbitals in each set to choose from.

Orbital energy is usually the deciding factor. The highest-energyoccupied orbital (the HOMO) is most commonly the relevant electron-donor orbital and the lowest-energy unoccupied orbital (the LUMO)is most commonly the relevant electron-acceptor orbital. For example,

Essay 7 7/2/04, 9:13 AM130

131

the HOMO and LUMO of singlet methylene (σ and π non-bondingorbitals, respectively) would serve as the donor and acceptor orbitals.The HOMO and LUMO are collectively referred to as the “frontierorbitals”, and most chemical reactions involve electron movementbetween them. In this way, the energy input required for electronmovement is kept to a minimum.

Energy

LUMO

HOMO

frontier molecular orbitals

unoccupiedmolecularorbitals

occupiedmolecularorbitals

One very important question for chemists concerns chemicalselectivity. Where more than one combination of reagents can react,which combination will react first? The answer can often be foundby examining frontier orbital energies. Consider a set of electron-donor reagents, where chemical reaction requires electron donationfrom the donor’s HOMO. It is reasonable to expect that the donorwith the highest energy HOMO will give up its electrons most easilyand be the most reactive. Electron-acceptor reagents should followthe opposite pattern. The reagent with the lowest energy LUMOshould be able to accept electrons most easily and be the most reactive.For a mixture of several donor and acceptor reagents, the fastestchemical reaction would be expected to involve the reagentcombination that yields the smallest HOMO-LUMO energy gap.

Another kind of selectivity question arises when a molecule hasmultiple reactive sites. In this case, then the orbital’s energy is uselessas a guide to “site selectivity”, and only the shape of the relevantorbital is important. For example, the enolate of ethyl acetate shownbelow might react with an electrophile (E+) at two different sites.

Essay 7 7/2/04, 9:13 AM131

132

H2C

E

OEtC

H2C OEtC

EOO

E+ E+

H2C OEtC

O–

H2C OEtC

O

–

Because the anion acts as an electron donor, we can find clues to itsreactivity preferences by examining the shape of its HOMO. Eventhough the HOMO is delocalized over several sites, the largestcontribution clearly comes from the terminal carbon atom. Therefore,we expect electron movement and bond formation to occur at thiscarbon, and lead to the product shown on the left.

HOMO for enolate of ethyl acetate

Open “ethyl acetate enolate HOMO”, and switch to a mesh or transparentsurface to see the underlying skeleton. Note that the orbital is most heavilyconcentrated on carbon, meaning that electrophilic attack will likely occurhere. Close “ethyl acetate enolate HOMO” when you are finished.

The Woodward-Hoffmann Rules

In certain cases, multiple frontier orbital interactions must beconsidered. This is particularly true of so-called “cycloadditionreactions”, such as the Diels-Alder reaction between 1,3-butadieneand ethylene.

+

The key feature of this reaction is that the reactants combine in a waythat allows two bonds to form simultaneously. This implies two

Essay 7 7/2/04, 9:13 AM132

133

different sites of satisfactory frontier orbital interaction (the two newbonds that form are sufficiently far apart that they do not interactwith each other during the reaction). If we focus exclusively on theinteractions of the terminal carbons in each molecule, then threedifferent frontier orbital combinations can be imagined.

individual interactions reinforce

individual interactions cancel

In all combinations, the “upper” orbital components are the samesign, and their overlap is positive. In the two cases on the left, thelower orbital components also lead to positive overlap. Thus, thetwo interactions reinforce, and the total frontier orbital interaction isnon zero. Electron movement (chemical reaction) can occur. The right-most case is different. Here the lower orbital components lead tonegative overlap (the orbitals have opposite signs at the interactingsites), and the total overlap is zero. No electron movement and nochemical reaction can occur in this case.

As it happens, the frontier orbital interactions in the Diels-Aldercycloaddition shown above are like those found in the middle drawing,i.e., the upper and lower interactions reinforce and the reaction proceeds.

Open “1,3-butadiene+ethylene”. The image on top is the LUMO ofethylene while that on the bottom is the HOMO of 1,3-butadiene. Theyare properly “poised” to interact, but you can manipulate themindependently. Close “1,3-butadiene+ethylene” when you are finished.

Cycloaddition of two ethylene molecules (shown below), however,involves a frontier orbital interaction like that found in the rightdrawing, so this reaction does not occur.

+ X

Essay 7 7/2/04, 9:13 AM133

134

Open “ethylene+ethylene”. The image on top corresponds to the LUMOof one ethylene while that on the bottom corresponds to the HOMO ofthe other ethylene. You can manipulate them independently or in concert(hold down on the Ctrl key while you carry out rotation and translation).Note, that in this case, the two individual “atom-atom” interactions cancel.Close “ethylene+ethylene” when you are finished.

The importance of orbital overlap in determining why certain chemicalreactions proceed easily while other “similar reactions” do not go atall was first advanced by Woodward and Hoffmann, and collectivelytheir ideas are now known as the Woodward-Hoffmann rules.

Essay 7 7/2/04, 9:13 AM134

135

21Electron Densities:

Sizes and Shapes of MoleculesHow “big” is an atom or a molecule? Atoms and molecules require acertain amount of space, but how much? A gas can be compressedinto a smaller volume but only so far. Liquids and solids cannot beeasily compressed. While the individual atoms or molecules in a gasare widely separated and can be pushed into a much smaller volume,the atoms or molecules in a liquid or a solid are already close togetherand cannot be “squeezed” much further.

Space-Filling Models

Chemists have long tried to answer the “size” question by using aspecial set of molecular models known as “space-filling” or “CPK”models. The space-filling model for an atom is simply a sphere offixed radius. A different radius is used for each element, and thisradius has been chosen to reproduce certain experimentalobservations, such as the compressibility of a gas, or the spacingbetween atoms in a crystal. Space-filling models for molecules consistof a set of interpenetrating atomic spheres. This reflects the idea thatthe chemical bonds that hold the molecule together cause the atomsto move very close together. “Interpenetration” can be used as acriterion for chemical bonding. If two atomic spheres in a space-filling model strongly interpenetrate then the atoms must be bonded.

space-filling models for ammonia (left), trimethylamine (center) and 1-azaadamantane (right)

Essay 8 7/2/04, 9:26 AM135

136

Space-filling models for ammonia, trimethylamine and1-azaadamantane show how “big” these molecules are, and also showthat the nitrogen in ammonia is more “exposed” than the correspondingnitrogen atoms in trimethylamine and 1-azaadamantane.

Open “amines space filling”. Space-filling models for ammonia,trimethylamine and 1-azaadamantane all appear on screen. Carbon atomsare colored dark grey, hydrogen atoms white and nitrogen blue. Notethat the models clearly reveal the extent to which the nitrogen is“exposed”. Close “amines space filling” when you are finished.

Electron Density Surfaces

An alternative technique for portraying molecular size and shape relieson the molecule’s own electron cloud. Atoms and molecules are madeup of positively-charged nuclei surrounded by a negatively-chargedelectron cloud, and it is the size and shape of the electron cloud thatdefines the size and shape of an atom or molecule. The size and shapeof an electron cloud is described by the “electron density” (the numberof electrons per unit volume). Consider a graph of electron density inthe hydrogen atom as a function of distance from the nucleus.

distance from nucleus

electrondensity

The graph brings up a problem for chemists seeking to define atomicand molecular size, in that the electron cloud lacks a clear boundary.While electron density decays rapidly with distance from the nucleus,nowhere does it fall to zero. Therefore, when atoms and molecules“rub up against each other”, their electron clouds overlap and mergeto a small extent.

Essay 8 7/2/04, 9:26 AM136

137

electron densityfor second molecule

molecular "boundary"

electron densityfor first molecule

What this means is that it is really not possible to say how big amolecule is. The best that can be done is to pick a value of the electrondensity, and to connect together all the points which have this value.The criteria for selecting this value is the same as that for selectingatomic radii in space-filling models, the only difference being thatonly a single “parameter” (the value of the electron density) is involved.The result is an electron density surface which, just like a space-fillingmodel, is intended to depict overall molecular size and shape.

electron density surfaces for ammonia (left), trimethylamine (center)and 1-azaadamantane (right)

Open “amines electron density”. Electron density surfaces for ammonia,trimethylamine and 1-azaadamantane all appear on screen. Switch to amesh or transparent surface in order to see the underlying skeletal model.On Windows, click on one of the surfaces, and select Mesh orTransparent from the menu which appears at the bottom right of thescreen. On the Mac, position the cursor over the graphic, hold downeither the left or right button and select from the menu which appearsalongside. With “mesh” selected, change the model to Space Filling(Model menu). This allows you to see how similar the electron densityrepresentation is to that offered by a simple space-filling model. Close“amines electron density” when you are finished.

Both space-filling and electron density models yield similar molecularvolumes, and both show differences in overall size among molecules.Because the electron density surfaces provide no discernible boundaries

Essay 8 7/2/04, 9:26 AM137

138

between atoms, the surfaces may appear to be less informative thanspace-filling models in helping to decide to what extent a particularatom is “exposed”. This “weakness” raises an important point,however. Electrons are associated with a molecule as a whole and notwith individual atoms. (As a consequence, it is actually not possibleto define the charge on an atom; see the activity “Charges on Atomsin Molecules”.) The space-filling representation of a molecule withits discernible atoms does not reflect reality.

Bond Density Surfaces

Another useful surface, termed the “bond density surface”, is onethat marks points corresponding to a much higher value of the electrondensity*. Since points of high electron density are located much closerto the atomic nuclei, bond density surfaces enclose relatively smallvolumes, and do not give a correct impression of molecular size. Onthe other hand, bond density surfaces identify regions correspondingto bonding electron density, and the volume of these surfaces may beroughly correlated with the number of electrons that participate inbonding. In this sense, bond density surfaces are analogous toconventional line drawings and skeletal models.

The bond density surface for hex-5-ene-1-yne clearly shows whichatoms are connected, although it does not clearly distinguish single,double and triple carbon-carbon bonds.

C CH C

C

HH

CHH

H

H

H

bond density surface for hex-5-ene-1-yne

* An even higher value of the electron density leads to a surface in which only “spheres” ofelectrons around the non-hydrogen atoms are portrayed. This serves to locate the positionsof these atoms and is the basis of the X-ray diffraction experiment.

Essay 8 7/2/04, 9:26 AM138

139

Open “hex-5-ene-1-yne bond density”, and switch to a mesh ortransparent surface to see the connection between the chemical bonds ina conventional model and the electron density. Close “hex-5-ene-1-ynebond density” when you are finished.

The usefulness of the bond density surface is more apparent in thefollowing model of diborane. The surface clearly shows that there isrelatively little electron density between the two borons. Apparentlythere is no boron-boron bond in this molecule. This is informationextracted from the bond density surface model, and has been obtainedwithout reference to any preconceived ideas about the bonding indiborane. (See the activity “Too Few Electrons”.)

HB

HB

H

H

H

H

bond density surface for diborane

Open “diborane bond density”, and switch to a mesh or transparent surfaceto see how few electrons are actually collected in the region between thetwo borons. Close “diborane bond density” when you are finished.

Bond density surfaces can also be informative in describing changesin bonding in moving from reactants to products through a transitionstate in a chemical reaction. For example, heating ethyl formate causesthe molecule to fragment into two new molecules, formic acid andethylene. A line drawing can show which bonds are affected by theoverall reaction, but it cannot tell us if these changes occur all atonce, sequentially, or in some other fashion.

O OH+

O

H

O O

H

O

On the other hand, the bond density surface is able to providequantitative information.

Essay 8 7/2/04, 9:26 AM139

140

bond density surfaces for the reactant, ethyl formate (left), pyrolysis transition state (center)and for the products, formic acid and ethylene (right)

Compare the bond density surface in the pyrolysis transition state tothose of the reactant and the products. The CO single bond of thereactant is clearly broken in the transition state. Also, the migratinghydrogen seems more tightly bound to oxygen (as in the product)than to carbon (as in the reactant). It can be concluded that thetransition state more closely resembles the products than the reactants,and this provides an example of what chemists call a “late” or“product-like” transition state.

To see the smooth change in electron density throughout the course of theethyl formate pyrolysis reaction, open “pyrolysis bond density”. Click on

at the bottom left of the screen to animate the graphic (click on tostop the animation). Switch to a mesh or transparent surface to follow thechange in bonding. Close “pyrolysis bond density” when you are finished.

Additional information about the ethyl formate pyrolysis reaction isfound in the activity “What Do Transition States Look Like?”, anda closely related example is provided in the tutorial “Ene Reaction”.

Essay 8 7/2/04, 9:26 AM140

141

9Electrostatic Potential Maps:

Charge DistributionsThe charge distribution in a molecule can provide critical insightinto its physical and chemical properties. For example, moleculesthat are charged, or highly polar, tend to be water-soluble, and polarmolecules may “stick together” in specific geometries, such as the“double helix” in DNA. Chemical reactions are also associated withcharged sites, and the most highly-charged molecule, or the mosthighly-charged site in a molecule, is often the most reactive. The“sign” of the charge is also important. Positively-charged sites in amolecule invite attack by bases and nucleophiles, while negatively-charged sites are usually targeted by acids and electrophiles.

One way to describe a molecule’s charge distribution is to give anumerical “atomic charge” for each atom. A particularly simple andfamiliar recipe yields so-called “formal charges” directly from Lewisstructures. (For more detail, see the activity “Charges on Atoms inMolecules”.) Unfortunately, formal charges are arbitrary. In fact, allmethods for assigning charge are arbitrary and necessarily bias thecalculated charges in one way or another. This includes methods basedon quantum mechanics.

An attractive alternative for describing molecular charge distributionsmakes use of a quantity termed the “electrostatic potential”. This isthe energy of interaction of a point positive charge with the nucleiand electrons of a molecule. The value of the electrostatic potentialdepends on the location of the point positive charge. If the point chargeis placed in a region of excess positive charge (an electron-poorregion), the point charge-molecule interaction is repulsive and theelectrostatic potential will be positive. Conversely, if the point chargeis placed in a region of excess negative charge (an electron-rich

Essay 9 7/2/04, 9:28 AM141

142

region), the interaction is attractive and the electrostatic potentialwill be negative. Thus, by moving the point charge around themolecule, the molecular charge distribution can be surveyed.

Electrostatic potentials can be depicted in various ways. For example,it is possible to make an electrostatic potential “surface” by findingall of the points in space where the electrostatic potential matchessome particular value. A much more useful way to show molecularcharge distribution is to construct an electrostatic potential “map”.This is done first by constructing an electron density surfacecorresponding to a space-filling model (see the essay “ElectronDensities: Sizes and Shapes of Molecules”). The electrostaticpotential is then “mapped” onto this surface using different colors torepresent the different values of the electrostatic potential. Mappingrequires an arbitrary choice for a color scale. Spartan uses the“rainbow”. Red, the low energy end of the spectrum, depicts regionsof most negative (least positive) electrostatic potential, and blue depictsthe regions of most positive (least negative) electrostatic potential.Intermediate colors represent intermediate values of the electrostaticpotential, so that potential increases in the order: red < orange < yellow< green < blue.

The connection between a molecule’s electron density surface, itselectrostatic potential surface, and an electrostatic potential map isillustrated below for benzene. The electron density surface definesmolecular shape and size. It performs the same function as aconventional space-filling model by indicating how close twobenzenes can get in a liquid or in a crystal.

space-filling model for benzene electron density surface for benzene

Essay 9 7/2/04, 9:28 AM142

143

Open “benzene electron density”. Two different images, a space-fillingmodel and an electron density surface, appear side by side. You can switchbetween the two models (in order to manipulate them) by clicking on eachin turn. Notice how similar they are. To get an even clearer impression,switch to a mesh surface. On Windows, click on the graphic and selectMesh from the menu which appears at the bottom right of the screen. Forthe Mac, position the cursor over the graphic, hold down either the left orright button and select from the menu which appears alongside. Switch toa space-filling model (Space Filling from the Model menu). The two arenow superimposed. Close “benzene electron density” when you are finished.

An electrostatic potential surface corresponding to points where thepotential is negative shows two different surfaces, one above the faceof the ring and the other below. Since the molecule’s π electrons lieclosest to these surfaces, we conclude that these electrons areresponsible for the attraction of a point positive charge (or anelectrophile) to the molecule. An electrostatic potential surfacecorresponding to points where the potential is positive has acompletely different shape. It is disk-shaped and wrapped fairly tightlyaround the nuclei. The shape and location of this surface indicatesthat a point positive charge is repelled by this region, or that a pointnegative charge (a nucleophile) would be attracted here.

negative (left) and positive (right) electrostatic potential surfaces for benzene

Open “benzene electrostatic potential”. Two different images appear sideby side. The one on the left depicts a surface of constant negative potential,while the one on the right depicts a surface of equal positive potential.You can switch between the two models (in order to manipulate them)by clicking on each in turn. Close “benzene electrostatic potential” whenyou are finished.

Essay 9 7/2/04, 9:28 AM143

144

Next, combine the electron density and electrostatic potential surfacesto produce a so-called electrostatic potential map. The map for benzeneconveys both the molecule’s size and shape as well as its chargedistribution in a compact and easily interpretable manner. The size andshape of the map are, of course, identical to that of the electron densitysurface, and indicate what part of the molecule is easily accessible toother molecules (the “outside world”). The colors reveal the overallcharge distribution. The faces of the ring, the π system, are “red”(electron rich), while the plane of the molecule and especially thehydrogens are “blue” (electron poor).

electrostatic potential map for benzene

Open “benzene electrostatic potential map”. Manipulate the image toconvince yourself that the “red” regions are on the π faces and the “blue”regions are around the edges. Close “benzene electrostatic potential map”when you are finished.

A comparison of electrostatic potential maps for benzene and pyridineis provided in the tutorial “Electrophilic Reactivity of Benzene andPyridine”. This clearly points out the value of the models in revealingdifferences in chemical properties and reactivity.

Electrostatic potential maps have now made their way into mainstreamgeneral and (especially) organic chemistry textbooks as a means ofdisplaying charge distributions in molecules. In addition, they havefound application as a “natural” step beyond “steric models” forinterpreting and predicting the way in which molecules fit together.A good example of this follows from the electrostatic potential mapfor benzene, which recall is “negative” on the π faces and “positive”around the periphery. The benzene dimer would, therefore, beexpected to exhibit a “perpendicular” geometry, to best accomodate

Essay 9 7/2/04, 9:28 AM144

145

Coulombic interactions, instead of a parallel arrangement which mightbe expected to be favored on steric grounds.

electrostatic potential maps for parallel (left) and perpendicular (right) benzene dimers

Open “benzene dimer electrostatic potential map”. Note that the parallelarrangement forces the negative region of one benzene onto the negativeregion of the other, while the perpendicular structure associates thenegative region of one benzene with a positive region of the other. Close“benzene dimer electrostatic potential map” when you are finished.

Of greater interest is the structure of benzene in solid state. “Intuition”suggests a parallel stack. Afterall, benzene is “flat” and flat things(plates, pancakes, hamburgers) make stacks. However, Coulombs lawfavors a perpendicular arrangement.

perpendicularparallel

Essay 9 7/2/04, 9:28 AM145

146

The experimental X-ray crystal structure shows a perpendiculararrangement, but in three dimensions. There are two “lessons” here.Intermolecular interactions go beyond steric interactions and sometimesour simple “one-dimensional” view of the world will lead us astray.

X-ray crystal structure of benzene

Open “benzene crystal”. Manipulate in order to see the packing of benzenemolecules. Close “benzene crystal” when you are finished.

Electrostatic potential maps may also be used to describe in greatdetail the “workings” of chemical reactions. For example, a map maybe used to show the transfer of negative charge during the Sn2 reactionof cyanide with methyl iodide.

N C N C CH3 + ICH3 I– –

Open “Sn2 cyanide+methyl iodide”. One “frame” of a sequence ofelectrostatic potential maps for the Sn2 reaction will appear. Animate byclicking on at the bottom left of the screen (stop the animation byclicking on ). Note that the negative charge (red color) flows smoothlyfrom cyanide to iodide during the reaction. Note also, that cyanide (asthe reactant) is “more red” than iodide (as the product). Iodide is betterable to carry negative charge, i.e., it is the better leaving group. Switch tomesh or transparent map to “see” the making and breaking of “bonds”.Close “Sn2 cyanide+methyl iodide” when you are finished.

Essay 9 7/2/04, 9:28 AM146

147

10Local Ionization Potential

Maps and LUMO Maps:Electrophilic and

Nucleophilic ReactivitiesThe Hammond Postulate states that the transition state in a one-stepreaction will more closely resemble the side of the reaction that ishigher in energy. Thus, the transition state of an endothermic reactionwill more closely resemble the products. Conversely, the transitionstate of an exothermic reaction will resemble the reactants. One wayto rationalize the Hammond postulate is to suggest that similarity inenergy implies similarity in structure. That is, the transition state willresemble whichever reactants or products to which it is closer in energy.As seen in the reaction coordinate diagrams below this is the productin an endothermic reaction and the reactants in an exothermic reaction.

Energy

reaction coordinate reaction coordinate

reactant

product reactant

product

transition state transition state

Energyendothermic reaction exothermic reaction

The Hammond Postulate provides a conceptual basis both for theWoodward-Hoffmann rules (see the essay “Atomic and MolecularOrbitals”) and for the use of graphical models. Both consider theproperties of reactants as an alternative to direct calculations oftransition states and reaction pathways as a way to assess chemical

Essay 10 7/2/04, 9:43 AM147

148

reactivity and selectivity. In this context, two models stand out as beingparticularly useful: the local ionization potential map for electrophilicreactions and the LUMO map for nucleophilic reactions.

Local Ionization Potential Maps and Electrophilic Reactivity

The local ionization potential provides a measure of the relative easeof electron removal (“ionization”) at any location around a molecule.For example, a surface of “low” local ionization potential for sulfurtetrafluoride demarks the areas which are most easily ionized, isclearly recognizable as a lone pair on sulfur.

local ionization potential for sulfur tetrafluoride

The local ionization potential by itself is not generally a useful model.However, a map of the local ionization potential onto an electrondensity surface is a useful model, in that it reveals those regions fromwhich electrons are most easily ionized.

"electron density"

Local ionization potential maps may be employed to reveal sites whichare susceptible to electrophilic attack. For example, they show boththe positional selectivity in electrophilic aromatic substitution (NH2

directs ortho/para, and NO2 directs meta), and the fact that π-donorgroups (NH2) activate benzene while electron-withdrawing groups(NO2) deactivate benzene.

Essay 10 7/2/04, 9:43 AM148

149

NH2 NO2

local ionization potential maps for benzene (left), aniline (center) and nitrobenzene (right)

Open “benzene, aniline, nitrobenzene local ionization potential maps”.Here, the color red corresponds to regions of lowest ionization potential(most accessible to electrophiles). Note, that the π system in aniline ismore accessible than the π system in benzene (the standard) and that theortho and para positions are more accessible than meta positions. Notealso, that the π system in nitrobenzene is less accesible than the π systemin benzene and that here the meta positions are more accessible than theortho and para positions, in accord with observation.

LUMO Maps and Nucleophilic Reactivity

As elaborated in an earlier essay (“Atomic and Molecular Orbitals”),the energies and shapes of the so-called frontier molecular orbitalsoften fortell the “chemistry” which a particular molecule mightundergo. Maps of “key” molecular orbitals may also lead to informativemodels. The most common of these is the so-called “LUMO map”, inwhich the (absolute value) of the lowest-unoccupied molecular orbital(the LUMO) is mapped onto an electron density surface. This fortellswhere an electron pair (a nucleophile) might attack.

"electron density"

LUMO

electron pair

Essay 10 7/2/04, 9:43 AM149

150

A good example is provided by the LUMO map for cyclohexenone.

O

α

β

LUMO map for cyclohexenone showing positions for nucleophilic attack in blue

The LUMO shows which regions of a molecule are most electrondeficient, and hence most subject to nucleophilic attack. In this case,one such region is over the carbonyl carbon, consistent with theobservation that carbonyl compounds undergo nucleophilic additionat the carbonyl carbon. Another region is over the β carbon, againconsistent with the known chemistry of α,β-unsaturated carbonylcompounds, in this case conjugate or Michael addition.

OCH3HO

CH3Li (CH3)2CuLi

CH3

carbonyl addition Michael addition

O

The buildup of positive charge on the β carbon leading to possibilityof Michael addition could have been anticipated from resonancearguments, although the LUMO map, like an experiment, has theadvantage of “showing” the result (“you don’t have to ask”).

Open “cyclohexenone LUMO map” and click on the resulting surface.Switch to a mesh or transparent surface in order to see the underlyingskeletal model. On Windows, click on the graphic and select Mesh orTransparent from the menu which appears at the bottom right of the screen.On the Mac, position the cursor over the graphic, hold down on either theleft or right button and select from the menu which appears alongside.Orient the molecule such that the two “blue regions” are positioned overthe appropriate carbons and then switch back to a solid surface.

Essay 10 7/2/04, 9:43 AM150

153

1How Big are Atoms

and Molecules?The “sizes” of atoms and molecules depend not on the nuclei butrather on the distribution of electrons. Molecules with more electronswill tend to be larger (require more space) than comparable moleculeswith fewer electrons. Molecules in which electrons are “looselybound” will tend to be larger than molecules with the same numberof “more tightly bound” electrons.

Electron distribution, or electron density as it is commonly referredto, is obtained as part of a quantum chemical calculation. The electrondensity is a “molecular property” which can actually be measuredexperimentally (using X-ray diffraction). However, X-ray diffractionis almost exclusively used only to locate the regions in a molecule ofhighest electron density, which in turn locates the nuclei. As detailedin the essay “Electron Densities: Sizes and Shapes of Molecules”, asurface of electron density with a value of 0.002 electrons/au3 providesa measure of overall molecular size and in most cases closelyresembles a conventional space-filling (CPK) model.

In this activity, you will explore relationships between number ofelectrons and atomic/molecular size and between the “looseness” ofthe electron distribution and size.

1. “Build” lithium cation, atom and anion, and put all into a singledocument. Specify calculation of a single-point energy using theHF/6-31G* model.

Inside the inorganic model kit, select the appropriate element fromthe Periodic Table and one coordinate from the available hybrids,and click on screen. Then select Delete from the Build menu andclick on the free valence. To group the atoms, select New Molecule

Section D1 7/2/04, 9:51 AM153

154

instead of New after you have built the first. Set Total Charge insidethe Calculations dialog to Cation (+1) for lithium cation and to Anion(-1) for lithium anion, and Multiplicity to Doublet for lithium atom.Also, remove the checkmark beside Global Calculations (ApplyGlobally for the Mac) at the bottom of the dialog.

Finally, specify calculation of a density surface for all three species.When the calculations have completed, place all three densitysurfaces side-by-side on screen.

Bring up the spreadsheet (Spreadsheet from the Display menu). ForWindows, check the box to the right of “Label” (first column) for allthree atoms. For Mac, click on at the bottom left of the screen andcheck the box to the left of the atom name (in the spreadsheet) for allthree atoms. Also, select (uncheck) Coupled from the Model menu(to turn it “off”), so that the three atoms can be moved independently.

Compare the three surfaces. Which is smallest? Which is largest?How does “size” relate to the number of electrons? Which surfacemost closely resembles a conventional space-filling model? Whatdoes this tell you about the kinds of molecules which were usedto establish the space-filling radius of lithium?

2. Build methyl anion, CH3–, ammonia, NH3, and hydronium cation,

H3O+, and put all three molecules into a single document.

To build hydronium cation, start from ammonia, move to the inorganicmodel kit, select oxygen from the Periodic Table and double click onnitrogen.

Specify calculation of equilibrium geometry using the HF/6-31G*model. Adjust the total charge for methyl anion and hydroniumcation. Finally, request calculation of a density surface for all threemolecules. When the calculations have completed, place all threedensity surfaces side-by-side on screen. Which is smallest? Whichis largest? How does size relate to the total number of electrons?How does it relate to the total nuclear charge?

Section D1 7/2/04, 9:51 AM154

155

2The Changing Nature

of HydrogenWhat is the charge on hydrogen in a molecule? We tend to think ofhydrogen as “neutral” or nearly so in hydrocarbons, while we give ita partial positive charge in a molecule like hydrogen fluoride and apartial negative charge in a molecule like sodium hydride. This isconveyed in our chemical nomenclature “sodium hydride”, meaninglike H–.

Is hydrogen unique among the elements in that it is able (and willing)to change its “personality” to reflect its chemical environment? Arethe changes as large as the nomenclature suggests “protic to hydridic”or are they much more subtle? Can we detect and “measure” anychanges with the tools available to us? These are questions whichyou will explore in this activity.

1. One after the other, build hydrogen molecule and the one-heavy-atom hydrides of first-row elements: LiH, BeH2 (linear), BH3

(trigonal planar), CH4 (tetrahedral), NH3 (trigonal pyramidal), H2O(bent) and HF. Put them into a single document, and obtainequilibrium geometries using the HF/6-31G* model.

You need to use the inorganic model kit to construct LiH, BeH2 and BH3.

Is there a correlation between the calculated charges on hydrogenin the first-row hydrides and the difference in electronegativitybetween hydrogen and the first-row element? Use theelectronegativities tabulated below.

H 2.2 B 2.0 O 3.4Li 1.0 C 2.6 F 4.0Be 1.6 N 3.0

Section D2 7/2/04, 9:52 AM155

156

2. Request an electron density surface for each of the molecules.This reflects the overall “size” of the molecule as well as the sizeof hydrogen inside the molecule. Display the surfaces side-by-side on screen to allow visual comparison. Do you see a trend inthe size of hydrogen as you move from lithium hydride to hydrogenfluoride? If so, for which molecule is hydrogen the smallest? Forwhich is it the largest? Are your results consistent with the previouscharacterization of hydrogen as taking on “protic” or “hydridic”identity depending on the environment? Elaborate.

While you cannot calculate the electron density for proton (thereare no electrons), you can for the hydride anion, H–. “Build” H–,calculate its energy using the HF/6-31G* model as well as surfaceof electron density. How does the size of hydride anion comparewith the “largest hydrogen” in your compounds? What does thissay regarding the extent to which the bond in this compound hasdissociated into ions?

3. Request an electrostatic potential map for each of the molecules.This reflects the distribution of charge, the color red meaningexcess negative charge and the color blue meaning excess positivecharge. Display the maps side-by-side on screen to allow visualcomparison. “Measure” the maximum (or minimum) value of theelectrostatic potential at hydrogen in each of the compounds.

Bring up the Surface Properties dialog by selecting Properties fromthe Display menu and clicking on a surface. Remove the checkmarkfrom Global Surfaces (Apply Globally on the Mac) and click on theReset button. The property range given in the dialog will now apply tothe selected molecule. Repeat for all molecules.

Is there a correlation between maximum (minimum) potential athydrogen and difference in electronegativity between the twoatoms which make up the bond?

Section D2 7/2/04, 9:52 AM156

157

3Too Few Electrons

At first glance, the structure of diborane, B2H6, would seem unusual.Why shouldn’t the molecule assume the same geometry as ethane,C2H6, which after all has the same number of heavy atoms and thesame number of hydrogens?

HB

HB

HHH

HC C

diborane

H

H

H

HH

H

ethane

The important difference between the two molecules is that diboranehas two fewer electrons than ethane and is not able to make the samenumber of bonds. In fact, it is ethylene which shares the same numberof electrons to which diborane is structurally related.

C C

ethyleneH H

HH

This activity explores isoelectronic (equal electron) relationships, andshows how they can be employed to anticipate the structures ofmolecules which have too few electrons to make the “required” bonds.We will begin with the relationship between ethylene and diborane(both of which structures are well known) and then try to predict thegeometry of the (unknown) borane analogue of acetylene.

1. Build ethylene and diborane, and put them into the same document.

Construct diborane using the inorganic model kit. Select boron fromthe Periodic Table and the five coordinate trigonal-bipyramid structurefrom the list of atomic hybrids. Notice that an icon of fragment isdisplayed at the top of the model kit. Identify the axial and equatorialpositions in this fragment,

Section D3 7/2/04, 9:53 AM157

158

B

axial

equatorial

axial

equatorial

and click on one of the equatorial positions. In response a yellowcircle will move to this position. Next click anywhere on screen, andorient the fragment such that you can clearly identify axial andequatorial positions. Click on an equatorial position to make a two-boron fragment.

B B

Next select hydrogen from the Periodic Table and two coordinatelinear from the list of hybrids. Click on the upper axial free valenceon one boron and then on the lower axial free valence on the otherboron. You are left with the structure.

B

H

B

H

Select Make Bond from the Build menu, and click on the free valenceof the upper hydrogen and then on the axial free valence on theadjacent boron. Repeat for the lower hydrogen.

B

H

B

H

Select Minimize from the Build menu to produce a “reasonable”geometry for diborane.

2. Obtain equilibrium geometries for ethylene and diborane usingthe HF/3-21G model. Also request that all six valence molecularorbitals for the two molecules be drawn. After the calculationshave completed, sketch the two sets of orbitals. As best you can,associate each valence orbital in ethylene with its counterpart indiborane. Focus on similarities in the structure of the orbitals andnot on their “position” in the lists of orbitals. To what orbital in

Section D3 7/2/04, 9:53 AM158

159

diborane does the π orbital in ethylene (the HOMO) best relate?How would you describe this orbital in diborane? Is it BB bonding,BH bonding or both?

3. Build acetylene, C2H2, and optimize its geometry using the HF/3-21G model. Request all valence molecular orbitals be drawn.On the basis of the orbital shapes (for acetylene) and on yourexperience with diborane, suggest at least two “plausible”geometries for the hypothetical molecule B2H4. Build each, (usetechniques similar to those you used to build diborane) calculateits equilibrium geometry using the HF/3-21G model and requestall valence molecular orbitals. Also request calculation of theinfrared spectrum.

Of the possible geometries you suggested, which is favoredenergetically? Is this structure actually a minimum on the energysurface? To tell, examine the infrared spectrum for the existenceof imaginary frequencies (see the essay “Potential EnergySurfaces”). Are any or all of your other selections also energyminima? Sketch the molecular orbitals of your preferred B2H4

geometry alongside those of acetylene and, as best you can, pairthem up. Pay particular attention to the orbitals relating to the twoπ bonds in acetylene.

Section D3 7/2/04, 9:53 AM159

161

4Too Many Electrons

What happens to electron pairs which are “left over” after all bondshave been formed? Is each electron pair primarily associated with asingle atom and directed tetrahedrally from this atom as implied byLewis structures, or is it “spread out”? Are these extra pairs (“lonepairs”) just “hangers on”, or do they contribute to (or even dominate)the “chemistry”.

In this activity you will examine molecules with “too many electrons”(the exact opposite you did in the previous activity “Too FewElectrons”). You will use electrostatic potential surfaces to probewhere these extra (non-bonded) electrons reside and try to associatewhat you find with conventional (Lewis) structures.

1. Build ammonia, water and hydrogen fluoride, and put them in thesame document. Request equilibrium geometries using the HF/6-31G* model. Before you submit for calculation, also requestcalculation of electrostatic potential surfaces for the threemolecules. The requested isosurfaces correspond to an electrostaticpotential isovalue of -80 kJ/mol. These will demark the highlyelectron-rich regions for the three molecules.

After the calculations have completed, display all three potentialsurfaces side-by-side on screen.

Bring up the spreadsheet. For Windows, check the box to the right ofthe first (“Label”) column for each molecule. For Mac, click on atthe bottom left of the screen and check the box to the left of themolecule name (in the spreadsheet) for each molecule. You also needto select (uncheck) Coupled from the Model menu in order to“uncouple” the motions of the three molecules so that you can movethem independently.

Section D4 7/2/04, 9:57 AM161

162

Describe the three surfaces and relate them to the conventionalLewis structures.

NO

HHH HH

F

H

In this context, rationalize the unusual shape of the potential forwater. Also, elaborate and rationalize the difference in the shapesof the ammonia and hydrogen fluoride potentials (which at firstglance appear to be nearly identical).

2. Revisit the isoelectronic series, methyl anion, ammonia andhydronium cation (see the activity “How Big are Atoms andMolecules?”).

If you have performed this earlier activity, you don’t need to buildand optimize these molecules a second time, but only computeelectrostatic potential surfaces.

Display the three potential surfaces side-by-side on screen. Forwhich does the potential extend furthest away from the nuclei?For which is the extension the least? Does this correlate with thesizes of the electron densities (see previous activity)? What dothe relative sizes (extensions) of the potential tell you about therelative “likelihood” of these three molecules to act as electronsources (“nucleophiles”)?

3. Hydrazine, N2H4, is a very “high-energy” molecule and is used asa “rocket fuel”. Like ethane, it should prefer a staggeredarrangement of hydrogens (see the essay “Potential EnergySurfaces”). The difference is that two of the CH bonds in ethanehave been replaced by electron pairs in hydrazine. A consequenceof this is that there are two staggered arrangements, one with theelectron pairs anti and the other with the electron pairs gauche.

N NH

HH

N NH

H

H

H

anti hydrazine gauche hydrazine

H

Section D4 7/2/04, 9:57 AM162

163

On the basis of the same arguments made in VSEPR theory(“electron pairs take up more space than bonds”) you might expectthat anti hydrazine would be the preferred structure. Here, youwill use quantum chemical calculations to test your “intuition”.

Build both anti and gauche hydrazine and put into the samedocument. Request equilibrium geometries using the HF/6-31G*model. Also request calculation of electrostatic potential surfaces.

With hydrazine on screen (in the builder), you can rotate about theNN bond by first clicking on it (a red arrow will ring the bond; if thering disappears, click again), and then moving the mouse “up anddown” with both the Alt key (space bar for the Mac) and left buttondepressed. Rotate into the proper (anti or gauche) structure and thenselect Minimize from the Build menu to give a better geometry. Theanti structure will have C2h symmetry and the gauche structure willhave C2 symmetry.

After the calculations have completed, examine the energies forthe two molecules. Which is favored? By how much? Is this resultin accord with what you would expect from VSEPR theory?

The energy will be given in atomic units (au). To convert to kJ/mol,multiply by 2625. The HOMO energy (in eV) which will be neededbelow is also available in this dialog. To convert from eV (electronvolts) to kJ/mol, multiply by 96.49.

Examine the two potential surfaces side-by-side on screen. Is therea noticeable difference in the extent to which the two electronpairs interact (“delocalize”) between the two conformers? If so,is the “more delocalized” conformer lower or higher in energythan the “less delocalized” conformer?

You can rationalize your result by recognizing that when electronpairs interact they form combinations, one of which is stabilized(relative to the original electron pairs) and one of which is destabilized.Destabilization is greater than stabilization, meaning that overallinteraction of two electron pairs is unfavorable energetically.

Section D4 7/2/04, 9:57 AM163

164

HOMO

stabilized combination

destabilized combination

In terms of molecular orbital theory, the higher-energy(destabilized) combination of electron pairs is the HOMO. Youcan judge the extent to which the electron pairs interact (and theoverall destabilization) by measuring the energy of the HOMO.Which hydrazine conformer has the higher HOMO energy? Isthis the higher-energy conformer? If so, is the difference in HOMOenergies similar to the energy difference between conformers?

Section D4 7/2/04, 9:57 AM164

165

5Removing and

Adding ElectronsMolecular orbital theory leads to a description of a molecule in whichelectrons are assigned (in pairs) to functions called molecular orbitals,which in turn are made up of combinations of atom-centered functionscalled atomic orbitals (see the essay “Atomic and MolecularOrbitals”). Except where “core” electrons are involved, molecularorbitals are typically delocalized throughout the molecule and showdistinct bonding or antibonding character. Loss of an electron from aspecific molecular orbital (from excitation by light or by ionization)would, therefore, be expected to lead to distinct changes in bondingand changes in molecular geometry.

Not all molecular orbitals are occupied. This is because there aremore than enough atomic orbitals to make the number of molecularorbitals needed to hold all the electrons. “Left-over combinations”or unoccupied molecular orbitals are also delocalized and also showdistinct bonding or antibonding character. Normally, this is of noconsequence. However, were these orbitals to become occupied (fromexcitation or from capture of an electron), then changes in moleculargeometry would also be expected.

In this activity, you will examine the highest-occupied molecularorbital (the HOMO) and the lowest-unoccupied molecular orbital(the LUMO) for a number of simple molecules and try to “guess”what changes in geometry would occur from electron removal fromthe former and electron addition to the latter. You will then obtaingeometries for the radical cation (electron removal) and radical anion(electron addition) for each of the molecules to see how successfulyour guesses were. Finally, you will attempt to rationalize the observedgeometry of the first excited state of a simple molecule.

Section D5 7/2/04, 9:58 AM165

166

1. Build ethylene, H2C=CH2, formaldimine, H2C=NH, andformaldehyde, H2C=O, and put all in a single document. Obtainequilibrium geometries using the HF/ 6-31G* model and followingthis request HOMO and LUMO surfaces.

2. Examine the HOMO for each of the three molecules, withparticular focus on bonding or antibonding character among theatoms involved. “Guess” what would happen to the geometryaround carbon (remain planar vs. pyramidalize), to the C=X bondlength and (for formaldimine) to the C=NH bond angle were anelectron to be removed from this orbital. Choose from thepossibilities listed below.

remove electron add electronfrom HOMO to LUMO

H2C=CH2 geometry remain planar remain planararound carbon pyramidalize pyramidalize

C=C lengthen lengthenshorten shortenremain the same remain the same

H2C=NH geometry remain planar remain planararound carbon pyramidalize pyramidalize

C=N lengthen lengthenshorten shortenremain the same remain the same

<CNH increase increasedecrease decreaseremain the same remain the same

H2C=O geometry remain planar remain planararound carbon pyramidalize pyramidalizeC=O lengthen lengthen

shorten shortenremain the same remain the same

On the basis of your “guesses”, build radical cations for ethylene,formaldimine and formaldehyde, and then obtain their actualequilibrium geometries using the HF/6-31G* model.

Section D5 7/2/04, 9:58 AM166

167

Inside the Calculations dialog, set Total Charge to Cation (+1) andMultiplicity to Doublet.

Are the calculated structures in line with what you expect? Elaborate.

3. Examine the LUMO for each of the three molecules. Guess whatwould happen to the geometry around carbon, C=X bond lengthand (for formaldimine) the C=NH bond angle were an electron tobe added to this orbital. Choose from the possibilities listed onthe previous page.

On the basis of your “guesses”, build radical anions for ethylene,formaldimine and formaldehyde, and then obtain their actualequilibrium geometries using the HF/6-31G* model.

Inside the Calculations dialog, set Total Charge to Anion (-1) andMultiplicity to Doublet.

Are the calculated structures in line with what you expect? Elaborate.

4. The first excited state of formaldehyde (the so-called n→π* state)can be thought of as arising from the “promotion” of one electronfrom the HOMO (in the ground-state of formaldehyde) to theLUMO. The experimental equilibrium geometry of the moleculeshows lengthening of the CO bond and a pyramidal carbongeometry (ground-state values in parentheses).

C OH

H1.32Å (1.21Å)

154° (180°)

Rationalize this experimental result on the basis of what you knowabout the HOMO and LUMO in formaldehyde and yourexperience with calculations on the radical cation and radical anionof formaldehyde.

Section D5 7/2/04, 9:58 AM167

169

6Water

Water is unusual (unique) in that it incorporates an equal number(two) of electron pairs (electron-donor sites) and “acidic hydrogens”(electron-acceptor sites) in such a “small package”.

two electron pairs OHH

two acidic hydrogens

Water molecules use these two “complementary resources” fully byforming a network of electron-donor/electron-acceptor interactions(hydrogen bonds), with each water molecule participating in up tofour hydrogen bonds. Other “similar” molecules like ammonia andhydrogen fluoride, are no match, simply because they incorporate anunequal number of electron-donor and electron-acceptor sites

one electron pair N three acidic hydrogensH

HH

three electron pairs F one acidic hydrogenH

The ability of water to establish a network of hydrogen bonds accountsfor its unusually high boiling point (other molecules of similarmolecular weight are gases at temperatures where water is a liquid).

This activity lets you “see” the network of hydrogen bonds in a tinysample of water (a so-called water cluster) and then lets you see theeffect of introducing different molecules.

1. Build a cluster of water molecules, and display hydrogen bonds.

Select sp3 oxygen in the organic model kit, hold down the Insert key(option key for the Mac) and click repeatedly at different locationson the screen. Turn the cluster every few molecules to obtain a three-dimensional structure. Continue until your cluster has at least 30-40

Section D6 7/2/04, 10:01 AM169

170

water molecules in it. Select Minimize from the Build menu. Whencomplete, select Hydrogen Bonds from the Model menu.

Focus your attention on one or a few water molecules in “themiddle” of your cluster. Are these involved in the “maximum”number of hydrogen bonds? Elaborate. Measure the lengths ofthe hydrogen bonds in your cluster. Do they all fit into a narrowrange (± 0.05Å) as would be the case for normal (covalent) bonds,or do they show more widely ranging behavior? Is what youobserve consistent with the fact that hydrogen bonds are muchweaker than covalent bonds? Elaborate.

Display your water cluster as a space-filling model. This providesan alternative view of hydrogen bonding (spheres representinghydrogens “interpreting” into spheres representing oxygens). It alsogives an “impression” of how much “space” is left in a sample ofliquid water. Describe what you see (with regard to these two issues).

2. Add a molecule of ammonia into (the center of) your cluster andminimize the energy. How many hydrogen bonds are there to theammonia molecule which you added? On average, how manyhydrogen bonds are there to the water molecules immediatelysurrounding the ammonia molecule? Does the situation appear tobe similar or different from that for the “pure” water cluster? Hasyour cluster noticeably expanded or contracted in the vicinity ofammonia? (Look at a space-filling model.) Would you expectwater to dissolve ammonia? Elaborate.

3. Replace ammonia by methane and minimize the energy.

Select sp3 carbon from the organic model kit and double click on thenitrogen of ammonia in your cluster.

Has the cluster noticeably expanded or contracted in the vicinityof methane? Rationalize your result in terms of changes inhydrogen bonding (relative to the “pure” water cluster). Wouldyou expect water to dissolve methane?

Section D6 7/2/04, 10:01 AM170

171

7Beyond VSEPR Theory

Valence State Electron Pair Repulsion (VSEPR) theory allowschemists to anticipate the geometry about an atom in a molecule usingtwo simple rules.

1. Start with a coordination geometry which depends only on thenumber of electron pairs associated with an atom (a bondcontributes one electron pair).

number of electron pairs coordination

2 linear3 trigonal planar4 tetrahedral5 trigonal bipyramidal6 octahedral

2. Given this coordination geometry, position the bonds such thatbond angles between lone pairs are as large as possible and,following this, that bond angles involving lone pairs and bondsare as large as possible.

In effect, VSEPR theory tells us to keep lone pairs as far apart fromeach other as possible, and then as far away from bonds as possible,and finally to keep bonds as far away from other bonds as possible.

VSEPR theory will generally provide a “unique” structure, but willnot tell whether other structures are “possible” and, if they are, howthey relate in stability to that of the “best” structure. Quantum chemicalmodels are able to do this. This activity explores their utility as aviable alternative to VSEPR theory.

1. The sulfur atom in sulfur tetrafluoride is “surrounded” by fivepairs of electrons. According to VSEPR theory what is thegeometry of SF4? Build this structure. Perform a HF/3-21G

Section D7 7/2/04, 10:02 AM171

172

calculation to obtain an equilibrium geometry and energy for themolecule, and request an infrared spectrum.

While you are waiting for the calculation to complete, propose anadditional structure for SF4 which satisfies the “first rule” (goodcoordination geometry), but which does not fully satisfy the“second rule”. Note that the alternative structure dealt with in thetutorial “Sulfur Tetrafluoride: Building an Inorganic Molecule”does not even satisfy the first rule. Build your alternative structureand obtain its geometry, energy and infrared spectrum using theHF/3-21G model.

Examine the energies of both your preferred and alternative SF4

structures. Is the geometry of SF4 preferred by VSEPR theorythat with the lower (more negative) total energy according to thecalculations? If not, what is the HF/3-21G structure of sulfurtetrafluoride? If VSEPR and HF/3-21G models are in agreementwith regard to the “best” structure for SF4, is the higher-energyalternative structure actually an energy minimum? Elaborate. Hint:look at the calculated infrared spectrum for imaginary frequencies(see the essay “Potential Energy Surfaces” for a discussion). Ifthe second structure is an energy minimum, what would be thecomposition of an equilibrium mixture of the two forms at roomtemperature? Use a special case of the Boltzmann equation foronly two molecules “in equilibrium” (see the essay “Total Energiesand Thermodynamic and Kinetic Data” for a discussion).

A B

[A]/[B] = e–1060(E – E )A B

where [A]/[B] is the ratio of molecules A and B in equilibriumand EA and EB are their energies in atomic units.

2. The chlorine atom in chlorine trifluoride, like the sulfur in sulfurtetrafluoride, is “surrounded” by five pairs of electrons. Repeatthe analysis you performed on SF4 for this molecule. Start withthe structure that VSEPR theory suggests and look for stablealternatives. Do the VSEPR and quantum chemical models reach

Section D7 7/2/04, 10:02 AM172

173

the same conclusion with regard to preferred structure? Does astable alternative structure exist?

3. The xenon atom in xenon tetrafluoride is surrounded by six pairsof electrons. Repeat the analysis you performed on SF4 for thismolecule. Start with the structure that VSEPR theory suggestsand look for a stable alternative. Do the two models reach thesame conclusion with regard to preferred structure? Is there a stablealternative structure?

4. According to VSEPR theory, xenon hexafluoride, XeF6, is notoctahedral. Why not? How many pairs of electrons are associatedwith xenon in XeF6? Does VSEPR theory tell you what thestructure should actually be? “Ask” the quantum chemicalcalculations by starting out with a six-coordinate structure whichhas been distorted from octahedral.

Build octahedral XeF6 using the inorganic model kit. Set several of theFXeF bond angles to “non 90º” values. For each, select Measure Anglefrom the Geometry menu, select a FXeF bond angle, replace the 90°value which appears at the bottom of the screen by a different valueand press the Enter key (return key for the Mac). Do not minimize inthe builder or your structure will revert back to octahedral symmetry.

Obtain an equilibrium geometry using the HF/3-21G model. Thiswill require several tens of minutes as your “guess” is likely notvery close to the real equilibrium structure. Is the resulting XeF6

geometry distorted from octahedral in accord with the predictionof VSEPR theory?

To get an estimate of the energy gained by distorting XeF6, buildoctahedral XeF6 and obtain its equilibrium geometry using theHF/3-21G model. Also request an infrared spectrum. Thiscalculation will also require several tens of minutes. Whencompleted, compare the energy of this structure with that of thenon-octahedral form above. How much stabilization has beengained as a result of distortion? How does this energy differencecompare with those you obtained earlier in examining differentstructures of SF4, ClF3 and XeF4?

Section D7 7/2/04, 10:02 AM173

174

Examine the infrared spectrum of octahedral XeF6. Does it showany imaginary frequencies? What does this tell you about whetheroctahedral XeF6 is an energy minimum on the overall energysurface? If there are imaginary frequencies, animate their motions.Is the direction of the vibrational motion (away from an octahedralgeometry) consistent with the non-octahedral structure you located?

Section D7 7/2/04, 10:02 AM174

175

8Bond Angles in

Main-Group HydridesThe HNH bond angles in ammonia are 106.7°, somewhat less thanthe tetrahedral values (109.5°). So too is the HOH bond angle inwater (104.5°). These data are usually rationalized by suggesting thatthe lone pair on nitrogen and the two lone pairs on oxygen “take upmore space” than NH and OH bonds, respectively (see the activity“Too Many Electrons”). As seen from the experimental data in thetable below, HXH bond angles in second-row and heavier main-groupanalogies of ammonia and water deviate even more from “ideal”tetrahedral values.

NH3 106.7 AsH3 92.1 H2O 104.5PH3 93.3 SbH3 91.6 H2S 92.1H2Se 90.6 H2Te 90.3

Is this further reduction in bond angle due to increased size of lonepairs on the heavy elements or are other factors involved? You willexplore the question in this activity.

1. Build the eight hydrides listed above and obtain the equilibriumgeometry of each using the HF/3-21G model. Put them all in onedocument. Do the calculations show the same trend in bond anglesas seen in the experimental data? Point out any significantexceptions.

2. Calculate and display electrostatic potential surfaces for all thehydrides. Set the isovalue to -10. What is the ordering of sizes oflone pairs (as indicated by the electrostatic potential surfaces) inthe series NH3, PH3, AsH3, SbH3? In the series H2O, H2S, H2Se,H2Te?

Section D8 7/2/04, 10:03 AM175

176

You will have noticed that the sizes of the lone pairs appear todecrease, not increase, in going to the heavier analogues of ammoniaand water. Other factors must be at work. You will examine twopossibilities, electrostatics and orbital hybridization.

Coulomb’s law states that “like charges repel” and will seek to moveas far apart as possible. Do charges on hydrogen increase in movingdown the Periodic Table?

3. Examine hydrogen charges in ammonia and its analogues. Dothey increase (hydrogen becoming more positive), decrease orremain about the same in moving to heavier analogues? Rationalizeyour result in terms of what you know about the electronegativitiesof nitrogen and its heavier analogues (relative to theelectronegativity of hydrogen). Given Coulomb’s law and ignoringany other factors, predict the trend in HXH bond angles in theseries NH3, PH3, AsH3, SbH3.

Repeat your analysis for water and its analogues.

Both the bonds and lone pair(s) in ammonia, water and their heavieranalogues are commonly viewed as made up of sp3 hybrids. It isreasonable to expect that “p contribution” to the bonds (which arelower in energy than the lone pair(s), will increase as the energy ofthe (atomic) p orbitals move closer to the energy of the s orbital.

4. In order to get a measure of relative valence s and p orbitals (2s,2p for first-row elements, 3s, 3p for second-row elements, etc.)perform calculations on the Noble gas atoms from each row (Ne,Ar, Kr and Xe).

Use the inorganic model kit and delete any free valences. Check Orbitalsinside the Calculations dialog. The orbital energies will be written tothe text output (Output under the Display menu).

Do valence s and p orbitals move closer, move further apart ordon’t significantly alter their relative positions in going from Neto Xe? If they do change their relative positions, how would youexpect the HXH bond angles to change in moving from NH3 toSbH3 and from H2O to H2Te? Elaborate.

Section D8 7/2/04, 10:03 AM176

177

9Bond Lengths

and HybridizationEach of the carbons in ethane is surrounded by four atoms in a roughlytetrahedral geometry, while each carbon in ethylene is surroundedby three atoms in a trigonal-planar geometry and each carbon inacetylene by two atoms in a linear geometry. These structures can berationalized by suggesting that the valence 2s and 2p orbitals of carbonare able to “combine” either to produce four equivalent “sp3 hybrids”directed toward the four corners of a tetrahedron, or three equivalent“sp2 hybrids” directed toward the corners of an equilateral triangle,or two equivalent “sp hybrids” directed along a line.

C C C

four sp3 hybrids three sp2 hybrids two sp hybrids

In the first instance, no atomic orbitals remain, while in the secondinstance, a 2p atomic orbital, directed perpendicular to the plane madeby the three sp2 hybrids, remains and in the third instance, a pair of2p atomic orbitals, directed perpendicular to the line of the two sphybrids and perpendicular to each other, remain. Thus, the “bonding”in ethane is described by four sp3 hybrids, that in ethylene by threesp2 hybrids and a p orbital and that in acetylene by two sp hybridsand two p orbitals.

2p atomic orbitals are higher in energy and extend further from carbonthan the 2s orbital. The higher the “fraction of 2p” in the hybrid, themore it will extend. Therefore, sp3 hybrids will extend further thansp2 hybrids, which in turn will extend further than sp hybrids. As aconsequence, bonds made with sp3 hybrids should be longer thanthose made with sp2 hybrids, which should in turn be longer than

Section D9 7/2/04, 10:03 AM177

178

bonds made with sp hybrids. In this activity, you will first test suchan hypothesis and explore its generality. Finally, you will search fora molecule with a very short carbon-carbon single bond.

1. Build ethane, ethylene, and acetylene and put into a singledocument. Obtain HF/3-21G equilibrium geometries for all threemolecules. Is the ordering in CH bond lengths what you expecton the basis of the hybridization arguments presented above?Using the CH bond length in ethane as a standard, what is the %reduction in CH bond lengths in ethylene? In acetylene? Is therea rough correlation between % reduction in bond length and % of2p in the hybrid?

2. Obtain the HF/3-21G equilibrium geometry for cyclopropane andmeasure the CH bond length. Based on your experience in theprevious part, would you say that the carbons are sp3 hybridized?Elaborate.

3. Choose one or more of the following sets of molecules: propane,propene and propyne or fluoroethane, fluoroethylene andfluoroacetylene or chloroethane, chloroethylene andchloroacetylene. Build all molecules in the set and again obtainHF/3-21G equilibrium geometries. When completed, measure theC–C (C–F or C–Cl) bond lengths.

Is the ordering of bond lengths the same as that observed for theCH bond lengths in ethane, ethylene and acetylene? Are the %reductions in bond lengths from the appropriate standards(propane, fluorethane and chloroethane) similar (±10%) to thoseseen for ethylene and acetylene (relative to ethane)?

4. How short can a carbon-carbon single bond be? Propose one ormore “shortest bond” candidates and perform HF/3-21Gcalculations. (You might find several of your “candidates” in thedatabase of HF/3-21G calculations supplied with Spartan. Lookfor the molecule name at the bottom of the screen.) Is your shortestcarbon-carbon single bond shorter or longer than the typicalcarbon-carbon double bond?

Section D9 7/2/04, 10:03 AM178

179

10Dipole Moments

The dipole moment provides a measure of the extent to which chargeis non uniformly distributed in a molecule. In a molecule like H2,where both “sides” are the same and the charge on both atoms isnecessarily equal, the dipole moment is zero. Increasing the differencein charge increases the dipole moment. The magnitude of the dipolemoment also depends on the extent to which charge is separated. Thelarger the separation of charge, the larger the dipole moment. Theoverall situation is particularly simple for a diatomic molecule wherethe dipole moment is proportional to the product of the absolutedifference in charge between the two atoms, |qA – qB|, and the bondlength, rAB. The greater the difference in charge and the greater thebond length, the greater the dipole moment.

dipole moment α qA–qB rAB

In this activity you will explore this relationship. You will alsoexamine to what extent the difference in atomic electronegativitiesbetween the atoms in a diatomic molecule anticipate the differencein atomic charge, and so can be used to estimate dipole moments.

1. Build hydrogen molecule, hydrogen fluoride, hydrogen chloride,hydrogen bromide, and hydrogen iodide. Put all into a singledocument. Obtain HF/3-21G equilibrium geometries for all fivemolecules. Is there a “reasonable” correlation between calculateddipole moments and the product of bond lengths andelectronegativity differences? Use the electronegativities tabulatedbelow.

H 2.2 F 4.0Li 1.0 Cl 3.2Na 0.9 Br 3.0

I 2.7

Section D10 7/2/04, 10:04 AM179

180

If so, does the correlation properly reproduce the fact that thedipole moment in hydrogen is zero?

2. Repeat your analyses for the series: lithium hydride, lithiumfluoride, lithium chloride, lithium bromide and lithium iodide andfor the series: sodium hydride, sodium fluoride, sodium chloride,sodium bromide and sodium iodide.

You will need to use the inorganic model kit to build these molecules.

Section D10 7/2/04, 10:04 AM180

181

11Charges on Atoms

in MoleculesWhat is the charge on an individual atom in a molecule? Surprisingly,this is not a simple question. While the total charge on a molecule iswell defined, being given as the sum of the nuclear charges (atomicnumbers) minus the total number of electrons, defining charges onindividual atoms requires accounting both for the nuclear charge andfor the charge of any electrons uniquely “associated” with theparticular atom. It is certainly reasonable to expect that the nuclearcontribution to the total charge on an atom is simply the atomicnumber, but it is not at all obvious how to partition the total electrondistribution by atoms. Consider, for example, the electron distributionfor a simple diatomic molecule like hydrogen fluoride.

H F

Here, the surrounding “line” is a particular “isodensity surface” (seethe essay “Electron Densities: Sizes and Shapes of Molecules”, saythat corresponding to a van der Waals surface and enclosing a largefraction of the total electron density. In this picture, the surface hasbeen drawn to suggest that more electrons are associated with fluorinethan with hydrogen. This is entirely reasonable, given the knownpolarity of the molecule, i.e., δ+H-Fδ-, as evidenced experimentallyby the direction of its dipole moment.

H F

It is, however, not at all apparent how to divide this surface between thetwo nuclei. Are any of the divisions shown below better than the rest?

Section D11 7/2/04, 10:04 AM181

182

H F H F H F

Clearly not! In fact, there is no “best” division. It is possible tocalculate (and measure) the number of electrons in a particular volumeof space, but it is not possible to say how many electrons belong toeach atom.

Despite their ambiguity, charges are part of the everyday vocabularyof chemists. Most common are so-called “formal charges” whichfollow from a “back of the envelope” recipe:

formalcharge

number ofvalence electrons

number oflone-pair electrons

number of bonds(single bond equivalents)= – –

In general, it will be possible to write a Lewis structure (for anuncharged molecule) which will lead to formal charges for all atomsof zero. There is no guarantee, however, that this Lewis structurewill properly account for the actual geometry and/or chemicalbehavior of the molecule. Also, formal charges are not able to revealsubtle differences among atoms in the same or in different molecules.They are really nothing more than a “bookkeeping” device.

Several different approaches are available for assigning atomic chargesbased on quantum chemical calculations. The approach implementedin your edition of Spartan fits the energy which a point charge “feels”as it approaches a molecule (the so-called “electrostatic potential”;see the essay “Electrostatic Potential Maps: Charge Distributions”),by a model in which the nuclei and electron distribution are replacedby “atomic charges”. Such an approach would be expected to lead tocharges which are much more “realistic” than formal charges andwhich vary with subtle changes in environment.

The purpose of this activity is both to point out differences betweenquantum chemical charges and formal charges, and to show the utilityof the former in rationalizing molecular structure and properties.

1. Sulfur difluoride is bent, sulfur tetrafluoride is a trigonal bipyramidmissing one (equatorial) “arm” and sulfur hexafluoride is

Section D11 7/2/04, 10:04 AM182

183

octahedral. Draw “proper” Lewis structures for each and assignformal charges at sulfur and at fluorine. Does the (formal) chargeon sulfur change from one molecule to another?

Build all three molecules and put into a single document. Obtainequilibrium geometries using the HF/3-21G model. Also requestelectrostatic potential maps. When completed, measure the chargeat sulfur in each of the three compounds.

The charges (in units of electrons) are available under the AtomProperties dialog (Properties under the Display menu and click onan atom).

Does the calculated charge on sulfur change from one moleculeto another? If so, for which molecule is sulfur the least charged?The most charged? Also compare the three electrostatic potentialmaps recalling that colors near “red” depict excess negative chargeand that colors near “blue” depict excess positive charge. On thebasis of the calculated charges and electrostatic potential maps,draw alternative (to those you provided above) Lewis structures(or sets of Lewis structures) for the three molecules.

2. Draw a Lewis structure for cyanide anion (CN–), and assign formalcharges. Does your structure incorporate a double bond likeformaldimine (H2C=NH) or a triple bond like hydrogen cyanide(HC≡N)? On which atom does the negative charge reside?

To see if your Lewis structure presents a “realistic” picture, obtainequilibrium geometries for cyanide anion, formaldimine andhydrogen cyanide using the HF/6-31G* model. According to yourcalculations, is the CN bond in cyanide anion closer to double ortriple (compare it to bond lengths in formaldimine and hydrogencyanide)? Which atom bears the negative charge, or is it distributedover both carbon and nitrogen?

Further discussion of cyanide anion is provided in the activity“SN2 Reaction of Cyanide and Methyl Iodide”.

Section D11 7/2/04, 10:04 AM183

184

3. Draw two different “reasonable” geometries for ozone, O3, provideLewis structures for each and assign formal charges to the oxygenatoms in each. Build both (put into the same document) and obtaintheir equilibrium geometry using the HF/6-31G* model. Whichof your structures is lower in energy? Is it in accord with theexperimentally known equilibrium geometry? (Because ozone isso important in the production of “smog”, it is well studied andits structure has been determined.) If the preferred structure hastwo (or three) distinct oxygen atoms, which is most positivelycharged? Most negatively charged? Is your result consistent withformal charges?

Section D11 7/2/04, 10:04 AM184

185

12What Makes A Strong Acid?

HF is a much stronger acid than H2O, which in turn is a stronger acidthan NH3. This parallels a decrease in the electronegativity of theatom bonded to hydrogen (F > O > N) and presumably to a decreasein bond polarity. In other words, the hydrogen in HF is more positivethan the hydrogens in H2O, which are in turn more positive than thehydrogens in NH3. It might be expected, therefore, that acid strengthwould decrease in moving from HF to HI, paralleling the decrease inelectronegativity of the halogen.

F > Cl > Br > I4.0 3.2 3.0 2.7

In fact the opposite is true, and HI is the strongest acid in the seriesand HF is the weakest. Clearly, factors other than differences in bondpolarity caused by differences in electronegativity are at work.

The key is recognizing that acid strength directly relates to the energyof bond fracture into separated positive and negative ions, the so-called heterocyclic bond dissociation energy.

HX → H+ + X–

The present activity relates only to acidity in the gas phase. Gas phaseheterolytic bond dissociation energies are much larger than thecorresponding energies in a solvent such as water. This is because thesolvent acts to stabilize the charged dissociation products much morethan it does the uncharged reactants. See the next activity “Is a StrongBase Always a Strong Base?” for a discussion of solvent effects on acid/base properties.

In this activity, you will first compute heterolytic bond dissociationenergies for HF, HCl, HBr and HI to establish whether or not thesereflect the observed ordering of acidities.

Section D12 7/2/04, 10:05 AM185

186

1. Build HI, HCl, HBr and HI and calculate their equilibriumgeometries using the HF/3-21G model. Also “build” F–, Cl–, Br–

and I– and perform single-point energy calculations on each. Putthese in a single document.

To build an atom, first build the associated hydride and delete the freevalence. Make certain that you set Total Charge to Anion in theCalculations dialog.

Compute heterolytic bond dissociation energies for the fourmolecules. (The energy of the proton is 0).

Is the ordering of calculated bond dissociation energies the sameas the ordering of acidities observed for these compounds?

Heterolytic bond dissociation in these compounds leads toseparated ions, one of which, H+, is common to all. Is it reasonableto expect that bond dissociation energy will follow the ability ofthe anion to stabilize the negative charge. One measure is providedby an electrostatic potential map.

2. Calculate electrostatic potential maps for the four anions anddisplay side-by-side on screen in the same color scale.

Which ion, F–, Cl–, Br– or I–, best accommodates the negativecharge? Which most poorly accommodates the charge? Elaborate.Is there a correlation between the “size” of the ion and its abilityto accommodate charge? Elaborate. Overall, is the ability toaccommodate charge in the atomic anion reflect the heterolyticbond dissociation energy of the corresponding hydride?

Section D12 7/2/04, 10:05 AM186

187

13Is a Strong Base

Always a Strong Base?What makes a strong base? In the absence of solvent, the mostimportant factor is stabilizing (delocalizing) the positive charge. Ingeneral, “bigger” groups should be more effective “delocalizers” thansmaller groups. For example, it is to be expected that a methyl groupis superior to hydrogen, meaning that methylamine is a stronger basethan ammonia, dimethylamine stronger than methylamine andtrimethylamine stronger than dimethylamine.

NH3 < MeNH2 < Me2NH < Me3N

increasing base strength

This situation is less obvious where a solvent is present. Here, thesolvent might be expected to stabilize a localized positive charge morethan it would a delocalized charge. In the case of the methylamines,solvent stabilization of (protonated) ammonia should be greater thanthat of (protonated) methylamine, which in turn should be greaterthan stabilization of (protonated) dimethylamine, and so forth.

In this activity, you will first apply the HF/6-31G* model to investigatethe relative “gas phase” basicities of the methylamines. You will then“correct” your data for the effects of aqueous solvent using anapproximate quantum chemical model. While the latter cannotrealistically be expected to provide a quantitative account of relativeaqueous-phase basicities, it should be sufficient to allow you to “see”the effects that solvent has in altering gas-phase basicities.

1. Build ammonia, methylamine, dimethylamine and trimethylamineand obtain equilibrium geometries using the HF/6-31G* model.Next, build protonated forms for the four amines and obtainequilibrium geometries using the HF/6-31G* model.

Section D13 7/2/04, 10:15 AM187

188

To build the protonated amines, build the analogous hydrocarbons (methane,ethane, propane and isobutane), bring up the inorganic model kit, select Nfrom the Periodic Table and double click on the appropriate carbon.

Work out the energies of the following reactions:

MeNH2 + NH4+ MeNH3

+ + NH3

Me2NH + NH4+ Me2NH2

+ + NH3

Me3N + NH4+ Me3NH+ + NH3

What is the ordering of methylamine basicities in the gas phase? Isthe effect of methyl substitution in altering basicity consistentthroughout the series?

2. Repeat your analysis for the “aqueous calculations”.

“Aqueous phase” energies (Eaq) and relative aqueous phase energies(RelEaq) are available from the spreadsheet.

What is the ordering of basicities in water? Is the range of basicitysmaller, greater or about the same as the range you observed inthe gas phase? Rationalize your result. Compare the changes inaqueous basicity in moving from ammonia to trimethylamine withthe analogous changes in gas-phase basicity.

3. Experimental gas and aqueous-phase basicities (in kJ/mol) forthe methylamines (relative to ammonia) are tabulated below.

∆Hgas ∆Haq

ammonia 0 0methylamine 38 8dimethylamine 67 8trimethylamine 79 3

How well do these compare with your results both insofar asabsolute numbers and with regard to “trends”?

Section D13 7/2/04, 10:15 AM188

189

14Which Lewis Structure

is Correct?Some molecules cannot be adequately represented in terms of a singleLewis structure, but require a series of Lewis structures, which takenas a whole, provide an adequate representation. Such a picture isunambiguous where the individual Lewis structures are all “the same”(different but equivalent arrangements of bonds) and are, therefore,equally important. For example, taken together the two (equivalent)Lewis structures for benzene lead to the experimental result that thesix carbon-carbon bonds are identical and midway in length between“normal” single and double bonds.

The situation is less clear where the Lewis structures are not allequivalent. For example, two of the three Lewis structures whichcan be written for naphthalene are equivalent, but the third is different.

equivalent

In this case, any conclusions regarding molecular geometry dependon the relative importance (“weight”) given to the individual Lewisstructures. Choosing all three Lewis structures to have equal weightleads to the result that four of the bonds in naphthalene (which aredouble bonds in two of the three Lewis structures) should be shorterthan the remaining ring bonds, (which are double bonds in only oneof the three Lewis structures). This is exactly what is observedexperimentally.

Section D14 7/2/04, 10:16 AM189

190

1.42Å 1.43Å

1.42Å

1.38Å

In the first part of this activity, you will attempt to guess the trends inbond lengths in anthracene and in phenanthrene by assuming that allLewis structures are equally important (as done for naphthaleneabove), and compare your guesses with actual geometries obtainedfrom HF/3-21G calculations.

1. Draw the complete set of Lewis structures for anthracene andphenanthrene.

anthracene phenanthrene

Assuming that each Lewis structure contributes equally, assignwhich if any of the carbon-carbon bonds should be especially shortor especially long. Next, obtain equilibrium geometries for thetwo molecules using the HF/3-21G model.

Are your assignments consistent with the results of thecalculations? If not, suggest which Lewis structures need to beweighed more heavily (or which need to be weighed less heavily)in order to bring the two sets of data into accord.

Pyridine and pyridazine are each represented by a pair of Lewisstructures.

pyridine

NN

pyridazine

NN

NN

While the two structures are the same for pyridine, they are markedlydifferent for pyridazine. In this part of the activity, you will compare

Section D14 7/2/04, 10:16 AM190

191

calculated bond distances in pyridazine (using those in pyridine as a“reference”) to decide whether or not the two Lewis structures shouldbe given equal weight and if not, which is the more important.

2. Obtain equilibrium geometries for pyridine and pyridazine usingthe Hartree-Fock 3-21G model.

To build pyridine, start with benzene, select aromatic nitrogen and doubleclick on one carbon. To build pyrazine, start with benzene, select aromaticnitrogen and double click on two adjacent carbons.

Using the calculated geometry of pyridine as a reference, wouldyou conclude that the two Lewis structures for pyridazine areequally important? If not, which should be given more weight?

Finally, perform the same analyses of a pair of more complexheterocyclic compounds.

3. Draw all Lewis structures for quinoline and for isoquinoline.

quinoline isoquinoline

NN

Given what you know about the geometry of pyridine (see previouspart) and assuming that each of the Lewis structures contributesequally, assign which if any of the carbon-carbon bonds in thetwo molecules should be especially short or especially long. ObtainHF/3-21G geometries for quinoline and isoquinoline to supportor refute your conclusions.

Section D14 7/2/04, 10:16 AM191

193

15Is Azulene Aromatic?

Aromatic molecules are (thermodynamically) more stable than mighthave been anticipated. The famous case is benzene. Here, the firststep in its complete hydrogenation (to cyclohexane) is endothermic,while both of the remaining steps are exothermic.

+H2

∆H = 25 kJ/mol

+H2

∆H = –109 kJ/mol

+H2

∆H = –117 kJ/mol

The difference in the hydrogenation energy between the first stepand either the second or third steps (134 kJ/mol and 142 kJ/mol,respectively) provides a measure of the aromatic stabilization.

Aromatic molecules may also be distinguished in that they incorporatebonds intermediate in length between normal (single and double)linkages. For example, all carbon-carbon bond lengths in benzeneare 1.39Å, which is longer than a double bond (1.30 to 1.34Å) butshorter than a single bond (1.48 to 1.55Å).

Is azulene, known for its intense blue color and the basis of numerousdyes, aromatic as is its isomer naphthalene? Both moleculesincorporate 10 π electrons in a planar fused-ring skeleton.

naphthaleneazulene

In this activity, you will compare energies, geometries and electrostaticpotential maps for azulene and naphthalene in an effort to decide.

1. Build azulene and naphthalene and obtain their equilibriumgeometries using the HF/3-21G model. Is azulene more stable(lower in energy), less stable or about as stable as naphthalene? If

Section D15 7/2/04, 10:16 AM193

194

it is less stable, is the energy difference between the two isomersmuch less, much greater or about the same as the “aromaticstabilization” of benzene? On the basis of energy, would youconclude that azulene is or is not aromatic?

Calculate an “average” carbon-carbon bond length in azulene. Isthis average similar to the carbon-carbon bond length in benzene?Next, calculate the mean absolute deviation from the average toprovide a measure of the uniformity of bond lengths. Is thisdeviation similar to the corresponding quantity for naphthalene?On the basis of uniformity in bond lengths alone, would youconclude that azulene is or is not aromatic?

It is common to suggest that azulene is made up of the “fusion” oftwo aromatic ions, both with 6 π electrons, cycloheptatrienyl(tropylium) cation and cyclopentadienyl anion.

+ +

cycloheptatrienylcation

cycloheptadienylanion azulene

+ ––

This being the case, the “cycloheptatrienyl side” of azulene shouldbe “positively charged” (relative to naphthalene) while the“cyclopentadienyl side” should be “negatively charged”.

2. Request electrostatic potential maps for naphthalene and azulene,and display them in the same scale and side-by-side on screen.

Set the “color scale” for both molecules to be the same and centered at“0”. For each molecule, select Properties from the Display menu andclick on the electrostatic potential map. Inside the Surface Propertiesdialog which results, change the property range to be the same for bothmolecules (-30 to 30 is a good range).

Do you see evidence of charge separation in azulene? Is it in theexpected direction? What effect would you expect charge separationto have on the energy of azulene? Elaborate.

Section D15 7/2/04, 10:16 AM194

195

16Why is Pyrrole a Weak Base?

Pyrrole and indole are known to be very weak bases, in strikingcontrast to the strong basicity exhibited by “related” aromatic aminessuch as pyridine, quinoline and isoquinoline.

NH N

H N NN

pyrrole indole pyridine quinoline isoquinoline

Assuming that protonation occurs on nitrogen in all compounds, thereason for the difference in basicity is clear. The nitrogen in pyridine(quinoline, isoquinoline) incorporates a non-bonded pair of electronsin the plane of the ring. Protonation does not directly affect the πsystem and the aromaticity of the ring. On the other hand, the“available” electrons on nitrogen in pyrrole (indole) are part of thering’s π system. Protonation “removes” electrons (or at least localizesthem in NH bonds) leading to loss of aromaticity.

In this activity, you will first employ electrostatic potential maps toestablish that the nitrogen in pyrrole (indole) is indeed less susceptibleto protonation than the nitrogen in pyridine (quinoline, isoquinoline).You will then perform calculations on “isomers” of protonated pyrrole(indole) to establish where protonation is actually likely to occur.

1. Build pyrrole and pyridine. (Optionally, build indole, quinolineand/or isoquinoline.) Obtain equilibrium geometries andelectrostatic potential maps for all molecules using the HF/3-21Gmodel. Display the maps side-by-side on screen.

Are there significant differences in the electrostatic potential atnitrogen in pyridine (quinoline, isoquinoline) and in pyrrole(indole), both in terms of magnitude and direction of maximum

Section D16 7/2/04, 10:17 AM195

196

potential (most negative)? Are any differences consistent with theobserved difference in basicities of pyridine and pyrrole and withthe qualitative rationale used to explain these differences? Elaborate.On the basis of its electrostatic potential map, would you expectthe nitrogen in pyridine (quinoline, isoquinoline) to be the mostbasic site? If not, where is the most basic site? Is the nitrogen inpyrrole (indole) the most basic site? If not, where is that site?

2. Build nitrogen-protonated pyrrole and obtain its equilibriumgeometry using the HF/3-21G model. Also obtain equilibriumgeometries for the two alternative (carbon-protonated) forms. (Ifyou also examine protonated indole, consider only isomersresulting from protonation of the five-membered ring.)

N

nitrogen protonated

H H

+ NH

+

NH

+

carbon protonated

HH

H H

Which of the three “isomers” is lowest in energy? Is your resultconsistent with what you expected based on examination ofelectrostatic potential maps? Elaborate.

3. Obtain and compare proton affinities of pyridine (quinoline,isoquinoline) and pyrrole (indole). This is simply the differencein energy between the neutral molecule and protonated form (theenergy of the proton is zero). You already have all the data forpyrrole, but you will need to calculate the geometry of protonatedpyridine using the HF/3-21G model. Which is the stronger base(larger proton affinity), pyridine or pyrrole? Is your resultconsistent with experiment?

Section D16 7/2/04, 10:17 AM196

197

17Not the Sum of the Parts

N,N-dimethylaniline is more basic than pyridine, which leads to theexpectation that the “aniline nitrogen” in 4-(dimethylamino)pyridinewill be more basic than the “pyridine nitrogen”.

NMe2

N,N-dimethylaniline pyridine

N

N

N(CH3)2

4-(dimethylamino)pyridine

While the site of protonation (the more basic site) in 4-(dimethylamino)pyridine is unknown, there is evidence to suggest that the reverse istrue and that the “pyridine nitrogen” is actually the more basic of thetwo. Specifically, addition of methyl iodide to 4-(dimethylamino)pyridine leads exclusively to the “pyridine adduct”.

N

N(CH3)2

CH3Inot

N

N(CH3)2

N

N(CH3)3

CH3

+

I – I –

+

In this activity, you will first confirm (or refute) that the preferredsite of protonation in 4-(dimethylamino)pyridine is the pyridinenitrogen. If it is, then you will examine the interaction of thedimethylamino substituent with pyridine in both neutral andprotonated 4-(dimethylamino)pyridine for clues to its behavior.

1. Build both forms of protonated 4-(dimethylamino)pyridine andobtain the geometry of each using the HF/3-21G model. Whichprotonated form is the more stable? Is your result consistent with

Section D17 7/2/04, 10:17 AM197

198

the product observed upon addition of methyl iodide? Is the energyof the other protonated form close enough so that one might expectto see both methyl cation adducts? Elaborate.

Assuming that you find 4-(dimethylamino)pyridine to favorprotonation on the pyridine nitrogen, the next step is to establish whathas caused the reversal from that noted in the “parent compounds”(N,N-dimethyl-aniline and pyridine). Either this is due to stabilizationof protonated pyridine by the dimethylamino substituent or todestabilization of protonated dimethylaniline by the change in thearomatic ring from benzene to pyridine, or both. To tell, you willexamine reactions 1 and 2 which separate the two protonated formsinto their respective components.

NH+

NMe2

+

NMe2

+

NH+

(1)

N

NMe2

+

NMe2

+

N

(2)

+H H

+

2. Build all the molecules required for reactions 1 and 2 and obtainequilibrium geometries using the HF/3-21G model. (You alreadyhave data for the two protonated forms of 4-(dimethylamino)pyridine. Does the dimethylamino substituent stabilize ordestabilize protonated pyridine? Does the change from carbon tonitrogen in the aromatic ring stabilize or destabilize protonatedN,N-dimethylaniline? Which, if either, is the dominant factor behindthe preference for protonation in 4-(dimethylamino) pyridine?

Section D17 7/2/04, 10:17 AM198

199

18Stereoisomers vs. Conformers.

A Matter of DegreeStereoisomers are molecules with the same molecular formula inwhich the constituent atoms are connected to each other (bonded) inthe same way but differ in their three-dimensional arrangement. Forexample, cis and trans-2-butene are stereoisomers but 2-methylpropene and cis (or trans) 2-butene are not.

C C

CH3

HH

CH3

C C

H

CH3H

CH3

C C

H

HCH3

CH3

cis-2-butene trans-2-butene 2-methylpropene

C C

CH3

CH3

anti-n-butane gauche-n-butane

HH

HH

C C

H

CH3

HH

HCH3

anti and gauche-n-butane like cis and trans-2-butene have the samemolecular formula, the same arrangement of bonds but different three-dimensional geometry, but are not considered to be stereoisomers.Rather, they are referred to as conformers (see the tutorial “InternalRotation in n-Butane”).

The difference is one of degree. Interconversion of cis and trans-2-butene is “difficult” because it requires fracture of a π bond, whereasinterconversion of anti and gauche n-butane is “easy” as it onlyinvolves rotation about a carbon-carbon single bond. In morequantitative terms, cis-2-butene needs to surmount an activationbarrier of roughly 210 kJ/mol in order to isomerize to trans-2-butene,while gauche n-butane needs only to climb a 10 kJ/mol “hill” inorder to yield the anti conformer. Isomerization of cis-2-butene totrans-2-butene will be very slow, while rotation of gauche-n-butaneto anti-n-butane will be fast.

This activity explores a situation where it is not clear whether theterm “isomer” and “conformer” is the more appropriate.

Section D18 7/2/04, 10:18 AM199

200

While amides, such as formamide, may be represented in terms of asingle “uncharged” Lewis structure, both spectroscopic and chemicalevidence suggests that such a picture is inappropriate, and that theCN bond may exhibit characteristics of a double bond. This suggestsa significant contribution of the “charged” Lewis structure to theoverall description.

N C

O

HH

H

N C

O–

HH

H+

1. Build formamide, H2NCHO. Obtain an equilibrium geometry usingthe HF/3-21G model. For comparison, obtain equilibriumgeometries for both methylamine, CH3NH2 and for methyleneimine,H2C=NH. Is the CN bond length in formamide shorter than that inmethylamine? Is it closer in length to the double bond inmethyleneimine or to the single bond in methylamine?

2. Next, build a guess at the “transition state” for rotation about theCN bond in formamide. Start with formamide and twist the CNbond such that the NH2 and CHO groups are approximatelyperpendicular. Specify calculation of a transition-state geometryusing the HF/3-21G model and also request an infrared spectrum.When the calculations have completed, verify that your structurecorresponds to a transition state, and that the motion associatedwith the imaginary frequency is consistent with rotation aboutthe CN bond.

Compare the energy of the transition state to that of formamide.Is it in the range of a “normal” single-bond rotation or closer tothat cis-trans isomerization of a double bond? Compare the CNbond in the transition state to that in formamide. Is it shorter, longeror about the same length? Is this result consistent with theenergetics of the process?

Section D18 7/2/04, 10:18 AM200

201

19Enantiomers.

The Same and Not the SameEnantiomers are non-superimposible mirror images. While theynecessarily have identical physical properties, “under the rightconditions” they may exhibit entirely different chemical behavior.The usual analogy is the human hand. Left and right hands are non-superimposible mirror images (they are enantiomers) and are identicalin all respects. However, a right hand “shaking” another right handprovides an entirely different “experience” than the same right handshaking a left hand.

1. One of the enantiomers of carvone occurs naturally in carawaywhile the other is found in spearmint oil. These enantiomers areresponsible for the characteristic odors of these materials.Ibuprofen is an analgesic sold under various names, includingAdvil, Motrin, and Nuprin. The material is sold as a mixture, butonly one enantiomer acts as an analgesic. The other enantiomer isinactive. This means that 800 mg of ibuprofen contains only 400mg of analgesic. The two enantiomers of limonene havecompletely different tastes. One has the taste of lemon (as thename implies) and the other tastes of orange.

Ocarvone ibuprofen

CO2H

limonene

Each of these molecules incorporates a single chiral center. Identifyit, and draw R and S forms of each compound.

Section D19 7/2/04, 10:21 AM201

202

2. Select one (or more) molecule and bring it onto the screen.

Select carvone, ibuprofen and/or limonene from the files in the“activities” directory. Both R and S forms will be placed in a singledocument.

Add R/S labels to your model to confirm that your assignments inthe previous step are correct.

Configure... from the Model menu and check R/S in the dialog whichresults.

3. Compare total energies and dipole moments for the twoenantiomers of the compound you selected. Are the energies anddipole moments for the two enantiomers of carvone (ibuprofenand limonene) the same or are they different?

You could have performed this activity by building and calculatingthe enantiomers of carvone (ibuprofen or limonene) instead of retrievingthem from the “activities” directory. In this case, you would need toexamine the different possible conformers available for each, whichwould entail performing a series of different equilibrium geometrycalculations and, following that, selecting the lowest-energy conformer.

Section D19 7/2/04, 10:21 AM202

203

20Diastereomers

and Meso CompoundsWe have seen in the previous activity that molecules with a singlechiral center exist as a pair of enantiomers, the properties of whichare identical. The situation is different where there are two chiralcenters. In the case where the two chiral centers are different, as forexample in 2-chloro-3-fluorobutane, there are four different chiralityassignments; RR, RS, SR and SS, leading to four distinct molecules.However, there are two distinct kinds of relationships between thefour molecules, enantiomeric relationships (as in the previous activity)and diasteriomeric relationships.

1. Draw all four forms of 2-chloro-3-fluorobutane and assign R/Schirality for each center.

2. Bring 2-chloro-3-fluorobutane onto the screen.

Select 2-chloro-3-fluorobutane from the files in the “activities”directory. All four forms will be placed in a single document.

Attach R/S labels to your models to confirm that your assignmentsin the previous step are correct.

Configure... from the Model menu and check R/S in the dialog whichresults.

3. Compare total energies and dipole moments among the fourmolecules. How many different sets of energies and dipolemoments are there? Are molecules with the same energy and dipolemoment enantiomers (non-superimposible mirror images) or dothey bear a different relationship to each other? Try to superimpose

Section D20 7/2/04, 10:22 AM203

204

to find out what is the relationship between molecules withdifferent charges and dipole moments.

In the case where the two chiral centers are the same, as for examplein 2,3-difluorobutane, there are also four different chiralityassignments: RR, RS, SR and SS. Two of these lead to moleculeswhich are enantiomers. The other two are the same (a mesocompound) but are different from the first two.

4. Draw all four forms of 2,3-difluorobutane and assign R/S chiralityto each center.

5. Bring 2,3-difluorobutane onto the screen (all four forms will beplaced in a single group).

Select 2,3-difluorobutane from the files in the “activities directory.

Attach R/S label to your models to confirm that your assignmentsin the previous step are correct.

6. Compare total energies and dipole moments among the fourmolecules. How many different sets of energies and dipolemoments are there? Are molecules with the same energy and dipolemoment enatiomers or do they bear a different relationship. Tryto superimpose to find out. What is the relationship betweenmolecules with different energies and dipole moments?

You could have performed this activity by building and calculatingthe different stereoisomers of 2-chloro-3-fluorobutane and of 2,3-difluorobutane instead of retrieving them from the “activities”directory. In this case, you would need to examine the differentpossible conformers available for each, which would entail performinga series of different equilibrium geometry calculations and, followingthat, selecting the lowest-energy conformer.

Section D20 7/2/04, 10:22 AM204

205

21Are Reactive Intermediates

“Normal” Molecules?Bromine, Br2, adds to alkenes stereospecifically in a stepwise fashion.The first step involves formation of a “cyclic” bromonium ionintermediate that then undergoes backside attack by Br– (or anothernucleophile) to give only trans products.

Br2 BrorNu

H BrBr H

or

H BrNu H

Br+

What is the structure of the reactive intermediate, a so-calledbromonium ion? Does it take the form of a “saturated” three-membered ring (like cyclopropane or oxirane) as drawn above, or isit better represented in terms of a weak complex between the cationof bromine atom and an olefin, or is bromine only bonded to onecarbon leaving the positive charge on the other carbon?

Br+

or

Br+

C C C Cor

C C

Br

+

The geometries of the three alternatives should be sufficiently differentto allow you to tell. In particular, the ring structure should exhibit aCC length typical of a single bond (1.48 - 1.55 Å) while the CC bondin a complex should resemble that in a free olefin (1.30 - 1.34Å). Inboth of these the bromine will be equidistant from the two carbons,in contrast to the situation for the open structure.

In this activity, you will calculate the geometry of “cyclic” bromoniumion to see which description (three-membered ring or complex) is a

Section D21 7/2/04, 10:22 AM205

206

better fit. You will also obtain a structure for the open form ofbromonium ion to see whether it is more or less stable than the cyclicform. You will then obtain a LUMO map for your best structure to seewhere a nucleophile would most likely attack. Finally, you will examine“ring” and “open” structures for analogous reactive intermediates inwhich bromine cation “attaches” to benzene rather than to the alkene.

1. Build ethylene bromonium ion, both as a cyclic structure and astwo different “open” forms, and put all three in the same document.

Br

C CH HH H

C C

H

Br H

H

+H

+

C C

H

Br

H

H

H

+

To build cyclic ethylene bromonium ion, start with oxirane, O

H2C CH2, bring

up the inorganic model kit, select Br from the Periodic Table and doubleclick on oxygen. To build the open structures, start with methyl bromide,bring up the inorganic model kit, select C from the Periodic Table andplanar trigonal from the list of hybrids and click on a free valence. UseMeasure Dihedral from the Geometry menu to set the dihedral anglein one conformer to 90° and in the other conformer to 0°.

Obtain equilibrium geometries for all three structures using theHF/3-21G model. Do all three forms appear to be energy minimaor do one or more “collapse” to another? Elaborate. Whichstructure is the lowest in energy? Is the cyclic structure betterrepresented as a three-membered ring or as a complex? Elaborate.

2. Obtain a LUMO map for your lowest-energy structure (only).

A LUMO map, which indicates the extent to which the lowest-unoccupiedmolecular orbital (LUMO) “can be seen” at the “accessible surface” of amolecule, results from displaying the (absolute) value of the LUMO,indicating the “most likely” regions for electrons to be added, i.e., fornucleophilic attack to occur, on top of a surface of electron density,delineating the space taken up by a molecule. See the essay “LocalIonization Potential Maps and LUMO Maps: Electrophilic andNucleophilic Reactivity”.

Section D21 7/2/04, 10:22 AM206

207

Where is the LUMO most concentrated? Given that nucleophilicattack should occur here, is this consistent with the observedstereochemistry of Br2 addition? Elaborate.

Bromine also reacts with arenes but leads to substitution rather thanaddition. The overall process is believed to involve an ionicintermediate analogous to ethylene bromonium ion.

Br2

Br

Br

+ HBr

+

As with ethylene bromonium ion, both cyclic and open structures forthe intermediate are plausible.

+

Br+Br

3. Obtain HF/3-21G equilibrium geometries for both cyclic and openintermediates.

To build the cyclic structure, start with cyclic bromonium ion and to addfour sp2 carbons to make the ring. Use the inorganic model kit to build theopen intermediate. Form a six-carbon ring from five trigonal planar andone tetrahedral hybrid and change four of the bonds involving two trigonalplanar carbons from (single) to (partial double).

Which structure is lower in energy? Is this the “same” structurepredicted for ethylene bromonium ion?

Section D21 7/2/04, 10:22 AM207

209

22Molecular Shapes I.

To Stagger or Not to StaggerOne of the first “rules” dictating molecular shape that organicchemistry students learn is that “single bonds stagger”. Ethane isdiscussed and a plot presented showing that the staggered form is anenergy minimum while the eclipsed form is an energy maximum.

C C

H

H

staggered eclipsed

HH

HH

C C

H H

HHH

H

The next example, is inevitably n-butane where more than one staggeredform (and more than one eclipsed form) are possible. As with ethane,the staggered forms (so-called anti and gauche conformers) are energyminima while the eclipsed forms (syn and skew) are energy maxima.

C C

H3C

CH3

anti gauche

HH

HH

C C

H3C

HH

C C

H3C

syn skew

HH

C C

H3C H

CH3

HHH

CH3

HH

H

CH3H

Does the “staggered rule” extend to bonds involving sp2 hybridizedelements, most important, sp2 hybridized carbon? In this activity, youwill examine the shapes of molecules incorporating bonds betweensp2 and sp3 carbons to see if it does.

1. Build 1-butene and set and “lock” the C=CCC dihedral angle tobe 0°. Next, define a range of values for this dihedral angle startingfrom 0° and going to 180° in 20° steps.

Section D22 7/2/04, 10:23 AM209

210

After you have built 1-butene, select Measure Dihedral, click on thefour carbons in order, type 0.0 into the box to the right of Dihedral (C1,C2, C3, C4) at the bottom right of the screen and press the Enter key(return key on the Mac). For Windows, select Constrain Dihedral ( ),click on the same four atoms and click on the icon at the bottom rightof the screen. The icon will change to (locked). For the Mac, click onthe icon at the bottom right of the screen. The icon will change to .Next, select Properties from the Display menu. Click on the magentacolored constraint marker on your model to bring up the ConstraintProperties dialog. For Windows, check Dynamic inside the dialog andtype “180” into the second text box to the right of Value and press theEnter key. For Mac, type “180” in the To box and “10” in the Steps box.

Obtain the energy of 1-butene as a function of the CCCC dihedralangle. Use the HF/3-21G model. Plot the energy of 1-butene as afunction of the C=CCC dihedral angle. How many energy minimaare there? How many energy minima would there be if you hadvaried the dihedral angle from 0° to 360° instead of from 0° to180°? Elaborate. Characterize the structures of the energy minimaas “staggered” or “eclipsed” relative to the CC double bond.Characterize the structures of the energy maxima. Formulate a“rule” covering what you observe.

Next, consider the conformational preference in cis-2-butene, amolecule where “eclipsing” should result in strong unfavorable stericinteractions.

2. Build cis-2-butene. Lock both HCC=C dihedral angles to 0°(eclipsed). Next, define a range of values for only one of thesedihedral angles from 0° to 180° in 20° steps. As with 1-butene,obtain the energy of cis-2-butene as a function of this dihedralangle using the HF/3-21G model, and construct a plot. Characterizethe structure of the energy minima as “staggered” or “eclipsed”relative to the CC double bond. Do you see any evidence thatother structural parameters, that is, bond lengths and/or bondangles, have significantly altered in order to accommodate yourresult? Elaborate.

Section D22 7/2/04, 10:23 AM210

211

23Molecular Shapes II.

cis 1,3-DienesIn order for 1,3-butadiene to undergo Diels-Alder cycloaddition, itneeds to be in a cis (or nearly cis) conformation.

O+ O

O

OO

O

Is this a minimum energy shape for the diene? Certainly it benefitsfrom having the two double bonds coplanar. However, it also placesthe pair of “inside” hydrogens in close proximity presumably resultingin unfavorable steric repulsion.

H

H

H

H steric repulsion

In this activity, you will first examine the energy profile for rotationaround the central carbon-carbon (single) bond in 1,3-butadiene tosee if the syn form is an energy minimum and if not what the “closest”minimum-energy form actually is.

1. Build 1,3-butadiene and “lock” the C=CC=C dihedral angle to be0°. Next, define a range of values for this dihedral angle startingfrom 0° and going to 180° in 20° steps. Obtain and plot the energyof 1,3-butadiene as a function of the C=CC=C dihedral angle usingthe HF/3-21G model.

Section D23 7/2/04, 10:23 AM211

212

Instructions for carrying out these operations with Spartan have beenprovided in the previous activity “Molecular Shapes I. To Stagger orNot to Stagger”.

Describe the lowest energy minima. Are the double bondscoplanar? Is it suitable for Diels-Alder cycloaddition? If not, isthere a second energy minima? Are the double bonds coplanar inthis structure? If not, what is the difference in energy betweenthis structure and the “closest” structure in which the double bondsare coplanar?

2. Suggest one or more 1,3-dienes which have C=CC=C dihedralangles close to 0°. Test your suggestions by structures obtainedfrom the HF/3-21G model.

Section D23 7/2/04, 10:23 AM212

213

24Molecular Shapes III.When is Axial Better?

Substituents on cyclohexane, or “cyclohexane-like” rings, may eitherbe equatorial or axial, for example, methylcyclohexane.

CH3

equatorial axial

H

H

CH3

The equatorial arrangement is favored in the majority of situations,but the difference in energy between the two is often small enough(4 - 12 kJ/mol) for the axial arrangement to be detected.

There is one very important exception to the “equatorial rule”, notfor cyclohexane itself but for derivatives of tetrahydropyran, a closelyrelated molecule.

tetrahydropyran

O

Here, electronegative substituents on the carbon adjacent to oxygentypically prefer an axial arrangement. The so-called anomeric effectis particularly important in carbohydrate chemistry. An example isprovided in the activity “Molecular Shapes V. Which ConformerLeads to Product?”.

The usual explanation for the equatorial cyclohexanes is that asubstituent in the axial position will “run into” the pair of axialhydrogens.

Section D24 7/2/04, 10:24 AM213

214

CH3H

H

This is a “steric” (crowding) argument. May non-steric considerationsalso play a role? In this activity, you will look for substitutedcyclohexanes that prefer to be axial. Specifically, you will draw onCoulomb’s Law “charge separation requires energy” as a means tooverride (or at least reduce) unfavorable sterics. Dipole moment willbe employed as a measure of charge separation. A reasonable startingpoint is fluorocyclohexane. The carbon fluorine bond is highly polar(C+–F–), giving rise to the possibility of a large Coulombiccontribution, while fluorine is normally viewed as a “small”substituent, thereby minimizing steric factors.

1. Build both equatorial and axial fluorocyclohexane andobtain equilibrium geometries using the HF/3-21G model. As areference, also perform HF/3-21G calculations on equatorial andaxial methylcyclohexane.

Which arrangement, equatorial or axial is predicted to be lower inenergy? How does this result compare with that found formethylcyclohexane? Is the favored fluorocyclohexane structure alsothe one with the lower dipole moment? If so, what is the differencein dipole moments between the two structures? How does thisdifference compare with the difference in dipole moments betweenthe two methylcyclohexane structures? Given what you observe,for which system, fluorocyclohexane or methylcyclohexane, wouldyou expect “charge separation effects” to be more significant?

2. trans-1,2-difluorocyclohexane can exist as either a diaxial or adiequatorial structure. Build both and examine the relativeorientation of the two CF bonds. For which would you expect thedipole moment to be smaller? Elaborate. Given your prediction aboutthe relative magnitudes of the dipole moments in the two structuresand your results from the first part of this activity, would you expecttrans-1,2-difluorocyclohexane to be diaxial or diequatorial?

Section D24 7/2/04, 10:24 AM214

215

25Molecular Shapes IV.

The “Other” CyclohexaneCyclohexane plays a central role in organic chemistry. Not only is itincorporated into a wide variety of important compounds, but it alsoserves as one of the pillars on which the rules of organicstereochemistry have been built. Cyclohexane is drawn as a “chair-like” structure in which all bonds are staggered (see the activity“Molecular Shapes I. To Stagger or Not to Stagger”) As discussedin the previous exercise “Molecular Shapes III. When Axial isBetter”, this leads to two sets of hydrogens, so-called equatorial andaxial hydrogens, and to the possibility that a substituted cyclohexanewill exist in two different shapes.

X

X

equatorially-substitutedcyclohexane

axially-substitutedcyclohexane

There is an additional shape available to cyclohexane (and substitutedcyclohexanes) which also satisfies the “staggered rule”. It is generallydescribed as a “twist-boat” or “skew-boat” structure, the boatdesignation, meaning that opposite methylene groups in the ring pointtoward each other rather than away from each other as in the chairstructure. (See also the essay “Potential Energy Surfaces”.)

H

H

H

H

H

H

H

H

awaytoward

twist-boat chair

Section D25 7/2/04, 10:24 AM215

216

In this activity, you will locate the twist-boat form of cyclohexaneand then attempt to rationalize why it is seldom given notice.

1. Build a form of cyclohexane which “looks like” a twisted boat.The easiest way to do this is start with a chain of six sp3 carbonsand to rotate around individual carbon-carbon bonds. Make certainthat minimization in the builder does not lead either to the chairstructure or to a “non-twisted” boat structure.* When you aresatisfied, obtain an equilibrium geometry using the HF/3-21Gmodel. Also perform a HF/3-21G geometry optimization on“normal” (chair) cyclohexane. Examine your alternativecyclohexane structure. Does it appear to satisfy the requirementthat single bonds stagger each other? Examine the energy of thealternative relative to that of chair cyclohexane. Is it about thesame (within 1-2 kJ/mol) or significantly higher? What wouldyou expect the relative equilibrium populations of the two formsto be at room temperature (use the Boltzmann equation)?

2. Identify that site in your alternative cyclohexane structure whichyou believe to be the least “crowded”. One after the other,substitute this site with methyl, fluoro and cyano groups andcalculate the equilibrium geometry of each. Also obtainequilibrium geometries for the corresponding equatorially-substituted chair cyclohexanes. Are any of the alternativesubstituted cyclohexanes close enough in energy to the “normal”chair structures to be detectable in an equilibrium mixture at roomtemperature (> 1%)?

* Even though the non-twisted boat is an energy maximum, if you start with a C2v symmetrystructure, it will be maintained in the optimization procedure. A futher example of this isgiven in the activity “Transition States are Molecules Too”.

Section D25 7/2/04, 10:24 AM216

217

26Molecular Shapes V. Which

Conformer Leads to Product?Successful application of molecular modeling to the description ofreactivity and product selectivity assumes knowledge of the structureof the reactant. As many, indeed most, molecules will have morethan one conformation, this means knowledge of the “best” (lowestenergy) conformation. A simple example is provided by reaction ofsodium borohydride with the spiroketal 1 which proceeds with highstereospecificity.*

O O

O

1

O O

OH

13

O O

OH

1:

NaBH4CeCl3

In order to model the process (which “face” of the carbonyl reactsfastest) it is necessary to know which of the four possible confomersof 1, which differ in whether the oxygen in each ring is equatoriallyor axially disposed relative to the other ring, is most abundant.

In this activity, you will use HF/3-21G calculations to determine whichof the four different conformers of 1 is lowest in energy and thenmodel the selectivity of borohydride addition to this conformer usinga LUMO map.

1. One after the other, build all four conformers of 1 and obtain anequilibrium geometry for each using the HF/3-21G model. Whichconformer is the lowest in energy? Are any other conformers closeenough in energy to contribute significantly (>1%) to an equilibriummixture at room temperature? Can you offer any precedents to yourassignment of favored conformer? Can you offer an explanation?

* De Shong et al. J. Org. Chem. 1991, 56, 3207.

Section D26 7/2/04, 10:30 AM217

218

2. Obtain a LUMO map for the lowest-energy conformer (only).

A LUMO map, which indicates the extent to which the lowest-unoccupiedmolecular orbital (LUMO) “can be seen” at the “accessible surface” of amolecule, results from displaying the (absolute) value of the LUMO,indicating the “most likely” regions for electrons to be added, i.e., fornucleophilic attack to occur, on top of a surface of electron density,delineating the space taken up by a molecule. See the essay “LocalIonization Potential Maps and LUMO Maps: Electrophilic andNucleophilic Reactivity” for additional insight.

3. When the calculation has completed, display the LUMO map. Inthis particular case, the LUMO will be localized on the carbonylcarbon, and the question of interest will be at which “face” of thecarbonyl group is the LUMO more visible, that is at which facenucleophilic attack is likely to occur. At which face of the carbonylcarbon is the LUMO more visible? Is this result consistent withthe experimental stereochemistry for nucleophilic addition?Elaborate.

4. Obtain LUMO maps for the remaining three conformers of 1.Which (if any) give the same preference for nucleophilic additionas the lowest-energy conformer? Which (if any) give the oppositepreference?

Section D26 7/2/04, 10:30 AM218

219

27SN2 Reaction of Cyanide

and Methyl IodideSN2 is usually the first reaction encountered by a beginning studentof organic chemistry. The reaction of cyanide with methyl iodide,leading to acetonitrile and iodide is typical.

N C– + CH3I CH3CN + I–

It proceeds via an “inversion” mechanism in which the nucleophile(cyanide) approaches the substrate (methyl iodide) “under theumbrella” made by carbon and its three hydrogens. In response, theumbrella opens (flattens out), leading to a five-coordinate carboncenter (the transition state) with partially-formed bonds involvingboth the nucleophile (cyanide) and the leaving group (iodide).

H

C IC

HH

N

–

Inversion continues and finally leads back to a four-coordinatetetrahedral carbon in which the cyanide has replaced iodide. (For an“animation” of this reaction showing migration of charge, see theessay “Electrostatic Potential Maps: Charge Distributions”.) Theimportance of the SN2 reaction, aside from the fact that it substitutesone group on carbon for another, is that the inversion of chiral carboncenter leads to change of chirality at this center.

A great deal of effort is expended “talking about” SN2 as it relates tothe inversion of carbon. While this is, for the most part, warranted(synthesis of chiral molecules is a challenging enterprise), there areother important questions which could be . . . and should be . . . askedin order to truly understand what is “going on” in a simple SN2reaction. Two questions form the basis of this activity.

Section D27 7/2/04, 10:31 AM219

220

Why does cyanide react at carbon and not at nitrogen?

Doesn’t the fact that nitrogen is more electronegative than carbon(3.0 vs. 2.6) imply that the “extra electrons” (the negative charge)should reside primarily on nitrogen and not carbon, and that nitrogenshould be the source of the “attacking” electron pair? The key here isasking the right question. It isn’t so much an issue of where the extraelectrons are, but rather where the electrons which are most“available” and hence most likely to react are. According to molecularorbital theory, the most available electrons reside in the molecularorbital with the highest energy, the so-called highest-occupiedmolecular orbital or HOMO.

1. Build cyanide anion. Obtain its equilibrium geometry using theHF/3-21G model and request calculation of the HOMO. Displaythe HOMO. Is it bonding, antibonding or non-bonding? Does ithave significant concentration on both carbon and nitrogen? Is itmore concentrated on carbon or on nitrogen? How do yourobservations relate to cyanide serving as a “carbon nucleophile”,a “nitrogen nucleophile” or both? Elaborate.

Why does iodide leave following nucleophilic attack onto carbon?

The real question here is whether or not we can “explain” whatactually takes place (loss of iodide). The key is asking what happensto the electrons when they are put into methyl iodide. The obviousanswer is that they go into a molecular orbital which is both emptyand is as low an energy as is available. This is the so-called lowest-unoccupied molecular orbital or LUMO.

2. Build methyl iodide. Obtain its equilibrium geometry using theHF/3-21G model and request calculation of the LUMO. Displaythe LUMO. Is it bonding, antibonding or non-bonding? If it isbonding or antibonding, which bond(s) is (are) likely to be affectedby the addition of electrons (from the nucleophile)? What changesin bond length(s) would you expect? How does your observationrelate to what actually happens in the SN2 reaction?

Section D27 7/2/04, 10:31 AM220

221

28Transition States

are Molecules TooHow can you tell a transition state from a stable molecule? The“reaction coordinate diagram” discussed in the essay “PotentialEnergy Surfaces” gives you the answer: “a transition state is an energymaximum along the reaction coordinate while a stable molecule isan energy minimum”, but how exactly are you to put this informationto use in classifying a particular molecule? The key is detailedknowledge of the way molecules vibrate as a result of their absorbinglow-energy (infrared) light.

A diatomic molecule exhibits a single vibrational motion correspondingto expansion and contraction of the bond away from its equilibriumposition. The frequency (energy) of vibration is proportional to thesquare root of the ratio of the “force constant” and the “reduced mass”(see also the tutorial “Infrared Spectrum of Acetone”).

frequency α force constant

reduced mass

The force constant corresponds to the curvature of the energy surfacein the vicinity of the minimum (it is the second derivative of the energywith respect to change in distance away from the equilibrium value).In effect, the magnitude of the force constant tells us whether themotion is “easy” (shallow energy surface meaning a low force constant)or “difficult” (steep energy surface meaning a high force constant).

Analysis of vibrational motions and energies (frequencies) inpolyatomic molecules is more complicated, but follows from the samegeneral principles. The main difference is that the vibrational motionsin polyatomic molecules seldom correspond to changes in individual

Section D28 7/2/04, 10:37 AM221

222

bond lengths, bond angles, etc., but rather to combinations of thesemotions. These combinations are called “vibrational modes” or“normal modes”. A particularly simple example of this has alreadybeen provided for water molecule in the tutorial “Basic Operations”.

This activity is intended to help you draw connections between the“formalism” and the motions which polyatomic molecules actuallyundergo when they vibrate. It is also intended to show you howknowledge of a molecule’s vibrational frequencies will allow you tosay with confidence that a molecule is or is not a minimum energyspecies, and (if it is not) to say whether it could or could not be atransition state.

1. Build ammonia, NH3, and specify calculation of equilibriumgeometry and infrared spectrum using the HF/6-31G* model.Display the vibrational frequencies and one after the other, animatethe vibrational motions. Describe the motion associated with eachfrequency, and characterize each as being primarily bondstretching, angle bending or a combination of the two. Is bondstretching or angle bending “easier”? Do the stretching motionseach involve a single NH bond or do they involve combinationsof all three bonds?

All but a few elements occur naturally as a mixture of isotopes, whichshare the same number of protons and electrons but differ in thenumber of neutrons and so differ in overall mass. You are no doubtfamiliar with the isotopes of uranium. The common “stable” isotope,238U, has 92 protons and 92 electrons in addition to 146 neutrons,while the “radioactive” isotope 235U has the same number of protonsand electrons but only 143 neutrons.

Aside from their difference in mass, isotopes have virtually identicalphysical and chemical properties. Except for the lightest elementsthey are very difficult to distinguish and very difficult to separate. Inaddition, the electronic Schrödinger equation which the quantumchemical models in Spartan are based does not contain nuclear mass,meaning that potential energy surfaces for molecules with differentisotopes are identical. However, nuclear mass does figure into a variety

Section D28 7/2/04, 10:37 AM222

223

of physical properties, most important among them being vibrationalfrequencies (see equation at the beginning of this activity and alsothe essay “Potential Energy Surfaces”), and thermodynamicquantities such as entropy which depend on the vibrational frequencies.

2. Replace the three hydrogens in ammonia by three deuteriums,the isotope of hydrogen which contains one neutron (the “normal”isotope of hydrogen has no neutrons).

Select Properties from the Display menu and click on a hydrogen.Change Mass Number in the Atom Properties dialog which appearsfrom “1” to “2 (deuterium)”. Repeat for the other hydrogens.

Repeat the calculations and compare the resulting vibrationalfrequencies (for ND3) with those obtained above for NH3.Rationalize any differences in terms of the expression providedearlier for the vibrational frequency of a diatomic molecule.

3. Next, build ammonia as a planar molecule (as opposed to a“pyramidal”) molecule.

Use trigonal planar nitrogen instead of tetrahedral nitrogen in theorganic model kit

Calculate its equilibrium geometry and infrared spectrum using theHF/6-31G* model just as you did for pyramidal ammonia.

The first frequency listed is preceded by an “i”. This indicatesthat it is an “imaginary” (as opposed to a “real”) number. Givenwhat you know from diatomic molecules, and given that reducedmass is necessarily a positive quantity, what does this tell you thesign of the force constant for this particular vibrational motion?What does this tell you about the position of planar ammonia onthe potential energy surface? Describe the motion. Identify theanalogous motion in pyramidal ammonia. Is it a low or highfrequency (energy) motion?

Section D28 7/2/04, 10:37 AM223

224

A transition state, like a “stable molecule”, is a well-defined point onthe overall potential energy surface. It differs from a stable moleculein that not all of its “coordinates” are at minimum energy positionson the surface. Rather, one and only one coordinate is at an energymaximum on the surface. Liken the situation to a mountain pass,which is both a minimum (look to the left and right to see the mountainpeaks towering above) and a maximum (look forward and backwardto see the valleys below).

It is the “easiest way” to cross over a mountain range, just like atransition state in a chemical reaction is the easiest (lowest-energy)way to go between reactants and products.

Not all transition states are as simple as planar ammonia, connectingthe two equivalent forms of pyramidal ammonia, but all arecharacterized by a single imaginary vibrational frequency.

4. Bring the boat form of cyclohexane onto the screen. Thiscorresponds to a possible transition state connecting the chair andtwist boat forms of cyclohexane (see the essay “Potential EnergySurfaces” as well as the activity “Molecular Shapes IV. The“Other” Cyclohexane”).

Select “boat cyclohexane” from the files in the “activities” directory.

Locate the imaginary frequency and describe the motion of atomsas best as you can.

Section D28 7/2/04, 10:37 AM224

225

29What Do Transition States

Look Like?There is an enormous body of experimental knowledge aboutmolecular geometry. The structures of upwards of 400,000 crystallinesolids have been determined, primarily through X-ray crystallography.The diversity is enormous, ranging from small inorganic and organicmolecules to proteins, polymers and materials. In addition, thegeometries of more than 3000 small molecules have been determined,either in the gas phase or in solution. While not as diverse a collection,included are structures for a variety of highly-reactive species. Takenall together, this information has given chemists a clear picture ofwhat is “normal” and what is not, and enabled them to accurately“guess” the geometries of molecules that are not yet known.

What is completely missing from this picture is experimentalinformation about transition states. The reason is simple. A transitionstate is not a minimum on a potential energy surface and therefore cannotserve as a “trap” (see the previous activity “Transition States areMolecules Too” as well as the essay “Potential Energy Surfaces”). Inother words, transition states “do not exist” in the sense of being able toput a collection of them “into a bottle”. This is not a problem for quantumchemical calculations. Any molecule, real or imaginary, in fact, anycollection of nuclei and electrons, may be “calculated”, and the resultsof the calculations be used to show that the molecule could “exist” orcould be a “transition state”.

The purpose of this activity is to show you how to calculate transitionstates for simple chemical reactions, and to have you relate theirstructures to those of “normal” (stable molecules). You will firstexamine the rearrangement of methyl isocyanide, CH3NC, toacetonitrile, CH3CN. This is an example of a “unimolecular” process

Section D29 7/2/04, 10:38 AM225

226

in which a molecule remains “intact”, but reorganizes to give rise toa lower-energy geometry. You will confirm that the process isenergetically “downhill”, and then to identify and characterize thetransition state. Following this, you will then look at a somewhatmore complicated “bimolecular” reaction involving the splitting ofethyl formate into formic acid and ethylene.

1. Build both acetonitrile, CH3C≡N and methyl isocyanide, CH3N≡C.

To construct methyl isocyanide, first build propyne, CH3C≡CH anddelete the alkyne hydrogen (Delete from the Build menu). Next, bringup the inorganic model kit (click on the Inorganic tab at the top ofthe organic model kit), select N from the Periodic Table and doubleclick on the central carbon.

Obtain HF/3-21G equilibrium geometries for both molecules. Whichmolecule, methyl isocyanide or acetonitrile is more stable (lowerin energy) according to your calculations? How does the calculatedenergy difference between the two compare with the experimentalenthalpy difference of 86 kJ/mol in favor of acetonitrile?

2. Build a guess at the transition state for geometrical rearrangement.

Start with acetonitrile. Select Transition States from the Search menu( ), click on the CC bond and, while holding down the Shift key,click on the methyl carbon and on the nitrogen. Finally, click on atthe bottom of the screen ( on the Mac) to produce a guess at thetransition state.

Specify calculation of transition-state geometry using the HF/3-21G model and also request an infrared spectrum. Thecalculations will require several minutes.

When complete, first examine the vibrational (infrared)frequencies. Is there one imaginary frequency? If so, animate thevibrational motion associated with this frequency to convinceyourself that it corresponds to a “reasonable” reaction coordinate(connecting methyl isocyanide and acetonitrile).

Section D29 7/2/04, 10:38 AM226

227

Examine the geometry of the transition state. Does it incorporatea “full” triple bond (as does both reactant and product)? Is themigrating methyl group midway between reactant and product oris it “closer” to either the reactant or product? Elaborate. Giventhe thermodynamics of the reaction, is this result consistent withthe “Hammond Postulate”?

The Hammond Postulate states that the transition state for anexothermic reaction will more closely resemble reactants than products.

A somewhat more complicated reaction is that involving “splitting’of ethyl formate into formic acid and ethylene in response to heat (aso-called “pyrolysis” reaction). This reaction is quite similar to the“ene” reaction of 1-pentene (leading to propene and ethylene) alreadydiscussed in the tutorial “Ene Reaction”. Both reactions involve singlebond cleavage and transfer of a hydrogen.

C

CO

C

O C

CH

HO

CHHH

H

ethyl formate ethylene formic acid

H

O

H

H

∆+

HH H

C

CO

C

O

HHH

HH H

3. Build ethyl formate, ethylene and formic acid. Obtain equilibriumgeometries for all three molecules using the HF/3-21G model. Isthe reaction as written above exothermic or endothermic (see theessay “Total Energies and Thermodynamic and Kinetic Data”)?Based on this and on the Hammond Postulate, would you expectthe transition state to more closely resemble reactants or products?

4. Build a guess at the pyrolysis transition state.

Section D29 7/2/04, 10:38 AM227

228

Start with ethyl formate in a conformation in which one of the CH3

hydrogens is close to the (carbonyl) oxygen (as in the figure above).Click on . Click on bond “a” in the figure below and then click onbond “b”.

C

CO

C

O

HHH

a

bc

d

eHH

H

A curved arrow from bond “a” to bond “b” will be drawn (as shownabove). Next, click on bond “c” and then on bond “d”. A secondcurved arrow from bonds “c” to “d” will be drawn. Finally, click onbond “e” and, while holding down the Shift key, click on the (methyl)hydrogen to be transferred and on the (carbonyl) oxygen to receivethis hydrogen. A third curved arrow from bond “e” into the “space”between the hydrogen and oxygen will be drawn. If you make amistake, you can remove one or more “arrows” using Delete fromthe Build menu. With all three arrows in place, click on ( onthe Mac) at the bottom right of the screen. Your initial structure willbe replaced by a guess at the pyrolysis transition state.

Specify calculation of transition-state geometry using the HF/3-21G model and also request an infrared spectrum. Thecalculations will require several minutes to complete. Whencomplete, first examine the vibrational (infrared) frequencies. Isthere one imaginary frequency? If so, convince yourself that thiscorresponds to the reaction of interest.

5. Compare the geometry of the transition state with both ethylformate (the reactant) and formic acid and ethylene (the products).Is the CC bond at the transition state less or more than “halfway”between a single and double bond? Are the two CO bonds in thetransition state less or more than halfway to their lengths in theproduct? Overall, does the transition state appear to be more“reactant like” or more “product like”? Given the energetics ofthe reaction, is your result consistent with the Hammond postulate?

Section D29 7/2/04, 10:38 AM228

229

30Reactions that

“Twist and Turn”Carbenes are reactive intermediates in which one carbon atom issurrounded by only six valence electrons instead of the normalcomplement of eight. Triplet carbenes, for example, methylene (CH2),have two half-filled molecular orbitals, one in the plane of themolecule and the other perpendicular to the plane.

CH

H

methylene

Their chemistry closely resembles that of radicals. Singlet carbenes,for example, difluorocarbene (CF2) have two electrons in an in-planemolecular orbital, leaving the out-of-plane molecular orbital vacant.*

CF

F

difluorocarbene

This suggests that singlet carbenes should, in principle, either be ableto act as nucleophiles by “donating” their electron pair, or aselectrophiles, by “accepting” an electron pair.

In the first part of this activity, you will compare the HOMO and LUMOof difluorocarbene with the qualitative descriptions provided above.

* The molecular orbitals of singlet methylene have previously been described in the essay“Atomic and Molecular Orbitals”.

Section D30 7/2/04, 10:39 AM229

230

1. Build difluorocarbene.

Build methylene fluoride and delete the two free valences on carbon.

Obtain its equilibrium geometry using the HF/3-21G model andrequest HOMO and LUMO surfaces.

One after the other, display the HOMO and LUMO fordifluorocarbene, and point out any “significant” difference withthe qualitative descriptions provided above.

Among the “textbook reactions” of singlet carbenes is their additionto alkenes to yield cyclopropanes, e.g.

C C + CF2••

C

C C

FF

Here, a π bond is destroyed but two new σ bonds are formed (netgain of one bond). This reaction presents an interesting dilemma, inthat the “obvious” approach of the two molecules; that is, the approachleading directly to a cyclopropane with the “correct” geometry, hasthe in-plane orbital on the carbene pointing directly at the π orbitalon the alkene.

C C

C

FF

filled

filled

Both orbitals are filled and the resulting interaction is repulsive. Abetter approach is for the carbene to twist 90°. In this case, the emptyout-of-plane molecular orbital on the carbene points toward the πorbital leading to stabilizing interaction.

Section D30 7/2/04, 10:39 AM230

231

CF

F

C C

empty

filled

However, the cyclopropane product is now in the “wrong” geometry!

C C

C

FF

In the second part of this activity, you will again use the HF/3-21Gmodel to obtain the transition state for addition of difluorocarbeneto ethylene. You will examine the motion which the reagents take inorder to avoid unfavorable interaction between their filled molecularorbitals yet still land up with the proper geometry.

2. Build a guess at the transition state for difluorocarbene adding toethylene.

Start by building ethylene. Select sp3 carbon, click on the Insert key(option key on the Mac) at the bottom of the model kit and click onscreen. Add fluorines to two of the free valences on the sp3 carbon anddelete the remaining two free valences. You are left with two fragments,ethylene and difluorocarbene. Orient the two as to be poised for reaction.

C

C C

FF

Translations and rotations normally refer to the complete set of fragments,but that they can be made to refer to an individual fragment. For Windows,click on a fragment to identify it, and then hold down the Ctrl key whilemanipulations are being carried out. For Mac, click on one of thefragments to select it (the other fragments will be “dimmed”) and moveit around using the mouse. Click on the background to move bothfragments as a unit.

Section D30 7/2/04, 10:39 AM231

232

Click on . Click on the carbon on the CF2 fragment and then, whileholding down the Shift key, click first on the CF2 carbon and then on oneof the carbons on the ethylene fragment. A curved arrow will be drawn.

C

C C

FF

Next, click on the CC double bond and then, while holding down theShift key, click on the other ethylene carbon and then on the CF2 carbon.A second arrow is added to the structure.

C

C C

FF

With both arrows in place, click on ( on the Mac) at the bottomright of the screen.

Obtain a transition-state geometry using the HF/3-21G model andrequest an infrared spectrum. The calculations will require severalminutes.

Does the transition state more reflect the need to minimizeunfavorable interaction between the electron pairs ondifluorocarbene and ethylene, or does it more reflect the geometryof the product? Animate the vibrational mode (in the infraredspectrum) corresponding to the reaction coordinate, that is, themode described by an imaginary frequency. This allows you to“see” the motion of reactants as the approach and leave thetransition state. Account for this motion on the basis of the tworequirements noted above.

Section D30 7/2/04, 10:39 AM232

233

31Thermodynamic vs. Kinetic

Control of Chemical ReactionsOrganic chemists will easily recognize that cyclohexyl radical is morestable than cyclopentylmethyl radical, because they know that“6-membered rings are more stable than 5-membered rings”, and(more importantly) that "2° radicals are more stable than 1° radicals".It may come as a surprise then that loss of bromine from6-bromohexene leading initially to hex-5-enyl radical, resultsprimarily in product from cyclopentylmethyl radical, rather than fromthe (presumably) more stable cyclohexyl radical.

Br•

•

17%

Bu3SnHAlBN

2%

81%

∆

rearrangement•

hex-5-enyl radical

cyclopentylmethyl radical

cyclohexyl radical

There are two reasonable interpretations for this result: (i) that thereaction is thermochemically controlled but our understanding ofradical stability is “wrong”, and (ii) that the reaction is kineticallycontrolled.

Obtain relative energies for cyclohexyl and cyclopentylmethyl radicalsto determine the thermodynamic product.

Section D31 7/2/04, 10:39 AM233

234

1. Build cyclohexyl and cyclopentylmethyl radicals and obtain theirequilibrium geometries using the HF/3-21G model. Which is morestable? Is the energy difference large enough such that only one islikely to be observed? (Recall that, according to the Boltzmannequation, at room temperature an energy difference of 12 kJ/molcorresponds to a product ratio of >99:1.) Do you conclude thatring closure is under thermodynamic control?

Establish the kinetic product, i.e., which ring closure, to cyclohexylradical or to cyclopentylmethyl radical, is “easier”.

2. Build guesses for transition states for closure of hex-5-enyl radicalinto cyclohexyl radical and into methylcyclopentyl radical.

Build hex-5-enyl radical, click on , and draw reaction arrows as followsfor closure to cyclohexyl radical: from C1 to a new bond between C1 andC6; from the C5C6 bond to C5, and for closure to cyclopentylmethyl radical:from C1 to a new bond between C1 and C5; from the C5C6 bond to C6.Click on ( on the Mac).

5

4

32

1•

6

Which radical, cyclohexyl or cyclopentylmethyl, is more easilyformed? Given the relationship between transition-state energydifference, ∆E‡, and the ratio of major to minor (kinetic) products,

∆E‡ (kJ/mol) major: minor (at room temperature)

4 90:108 95:512 99:1

what is the approximate ratio of products suggested by thecalculations? How does this compare with what is observed? Doyou conclude that ring closure is under kinetic control?

Section D31 7/2/04, 10:39 AM234

235

32Anticipating Rates of

Chemical ReactionsThe rate of Diels-Alder reactions generally increases with π-donorability of the electron-donor group (EDG) on the diene, and withπ-acceptor ability of the electron-withdrawing group (EWG) on thedienophile.

EDG

EWG

Y

X

+ EDG = R, OR

EWG = CN, CHO, CO2H

The usual interpretation is that donors will “push up” the energy ofthe HOMO on the diene and acceptors will “push down” the energyof the LUMO on the dienophile. Any decrease in “HOMO-LUMOgap” should lead to stronger interaction between diene and dienophileand to a decrease in barrier.

HOMO

LUMO

dienophilediene

OrbitalEnergy

In this activity, you will first test such a hypothesis using the followingrate data for Diels-Alder cycloadditions involving cyclopentadieneas a diene and cyano-substituted alkenes as dienophiles (expressedin log units relative to the rate of cyclopentadiene and acrylonitrile).

acrylonitrile 0 1,1-dicyanoethylene 4.64trans-1,2-dicyanoethylene 1.89 tricyanoethylene 5.66cis-1,2-dicyanoethylene 1.94 tetracyanoethylene 7.61

Section D32 7/2/04, 10:40 AM235

236

You will then obtain transition states for reactions of cyclopentadieneand both acrylonitrile and tetracyanoethylene.

1. Build acrylonitrile, 1,1-dicyanoethylene, cis and trans-1,2-dicyanoethylene, tricyanoethylene and tetracyanoethylene andobtain their equilibrium geometries using the HF/3-21G model.Plot LUMO energy vs. the log of the relative rate. Whichdienophile has the smallest LUMO energy (smallest HOMO-LUMO gap)? Is this the dienophile which reacts most rapidly withcyclopentadiene? Which has the largest LUMO energy? Is thisthe compound which reacts most slowly? Does the HOMO-LUMOgap correlate with relative reaction rate?

2. Build guesses for transition states for Diels-Alder cycloadditionof cyclopentadiene and both acrylonitrile and tetracyanoethylene.

To build a guess at the transition state for cycloaddition of cyclopentadieneand acrylonitrile, first build cyclopentadiene, then press the Insert key(option key on the Mac) and build acrylonitrile on the same screen. Orientthe two molecules into an endo geometry.

NC1

2

12

3

4

Click on . Draw reaction arrows from the C1C2 bond in acrylonitrile toa new bond between C1 on acrylonitrile and C1 on cyclopentadiene; fromthe C1C2 to the C2C3 bonds in cyclopentadiene; from the C3C4 bond incyclopentadiene to a new bond between C4 on cyclopentadiene and C2

on acrylonitrile. Click on ( on the Mac).

Obtain transition states for the two reactions using the HF/3-21Gmodel. Also, build cyclopentadiene and obtain its geometry usingthe HF/3-21G model. Calculate activation energies for the twoDiels-Alder reactions, and then use the Arrhenius equation (seethe essay “Total Energies and Thermodynamic and Kinetic Data”)to calculate the difference in rates for the two reactions. How doesthe calculated difference compare with the experimental difference?

Section D32 7/2/04, 10:40 AM236

237

33Identifying Greenhouse Gases

The earth and all the other planets can be thought of as blackbodieswhich radiate into the universe. This provides a means to dissipatethe energy which falls on the planets due to the sun. Because of theirlow ambient temperatures, so-called blackbody radiation occursprimarily in the infrared.

The earth is actually warmer than such a picture would predict, dueto the fact that some of the blackbody radiation is absorbed by itsgaseous atmosphere. This is known as the “greenhouse effect”. Themagnitude of the effect, “greenhouse warming”, is obviously due toboth the extent and chemical makeup of the atmosphere.

To be an effective “greenhouse gas”, a molecule must absorb in theinfrared. Neither nitrogen nor oxygen, which together make up 99%of the earth’s atmosphere, satisfy this requirement. However, several“minor” atmospheric components, carbon dioxide, water and ozonemost important among them, absorb in the infrared and contributedirectly to greenhouse warming. These “subtract” from the earth’sblackbody radiation leaving actual radiation profile (in the range of500 to 1500 cm-1) that is given below.

1

2

3

0600 700 900 1100 1300 1500

Section D33 7/2/04, 10:41 AM237

238

This leads to a further requirement for an effective greenhouse gas,mainly that it absorbs in a frequency region where blackbody radiationis intense.

In this activity, you will calculate infrared spectra for molecules whichare already present in the atmosphere, or might be introduced throughhuman involvement. By matching any “strong” bands with theblackbody radiation profile given on the previous page, you shouldbe able to tell whether they are reasonable candidates for effectivegreenhouse gases.

This is an oversimplification. A more realistic model needs to accountnot only for intense infrared bands in the “proper place”, but also needsto establish the quantity of material likely to get into the atmosphere anda reasonable lifetime for this material in the atmosphere. These factorsare addressed in a beautiful paper proposing an experimental investigationsuitable for an undergraduate laboratory. (M.J. Elrod, J. Chem. Ed., 76,1702, 1999). The present activity is based on this paper.

Start by comparing calculated and experimental infrared spectra forcarbon dioxide. This will allow you to establish a “scaling factor”bringing calculated and measured frequencies into agreement.

1. Build carbon dioxide and calculate its equilibrium geometry andinfrared spectrum using the HF/6-31G* model.

How do the calculated infrared frequencies compare with theexperimental values given below? How do the calculated infraredintensities compare with the (qualitative) experimentaldesignations? Can you see evidence in the “experimental”blackbody profile for the presence of carbon dioxide in theatmosphere? Elaborate.

description of mode frequency cm-1 intensity

carbon dioxide bend 667 strongsymmetric stretch 1333 inactiveasymmetric stretch 2349 very strong

Section D33 7/2/04, 10:41 AM238

239

Propose a single multiplicative factor which, when applied to thecalculated frequencies, will bring them into best agreement withthe experimental values.

Methane and nitrous oxide (N2O) are both introduced into theatmosphere from agriculture. Methane is also released as a result ofoil recovery and fuel production. 1,1,1,2-tetrafluoroethane is nowwidely used in automobile air conditioners and is likely alsointroduced into the atmosphere in significant quantities.

2. Build methane, nitrous oxide and 1,1,1,2-tetrafluoroethane, andcalculate the equilibrium geometry and infrared spectrum of each.

Nitrous oxide is linear. To build it, start with Allene from the Groupsmenu and delete all four free valences. Then, bring up the inorganic modelkit, select N from the Periodic Table and double click on two adjacentcarbons. Finally, select O from the Periodic Table and double click onthe remaining carbon.

One after the other, display the infrared spectrum for each of thethree molecules. Using as a selection criterion the presence of a(modestly) strong infrared absorption in the range of 500 - 1500cm-1 (use scaled frequencies), comment as to whether each mightbe an effective greenhouse gas.

Section D33 7/2/04, 10:41 AM239

241

34“Unseen” Vibrations

The infrared spectrum of trans-1,2-dichloroethylene in the region of500 to 3500 cm-1 comprises only four lines, while nine lines can beseen in the spectrum of the corresponding cis isomer.

C C

H

Cl H

Cl

C C

H

Cl Cl

H

trans-dichloroethylene cis-dichloroethylene

Both molecules have six atoms and both will undergo twelve (3 timesnumber of atoms -6) different vibrational motions. It might, therefore,be expected that the infrared spectra of both would contain twelvelines. While a few of the “missing lines” are below the 500 cm-1

measurement range, the primary reason for the discrepancy is thatsome vibrational motions may give rise to absorptions which are tooweak to be observed or be “infrared inactive”, meaning that they willnot appear at all in the infrared spectrum.

While discussion of the theory underlying infrared spectroscopy isbeyond the present focus, it can be stated that in order for a particularvibrational motion to be “infrared active” (and hence “seen” in theinfrared spectrum), it needs to lead the change in the overall polarityof the molecule as reflected in a change in dipole moment. In fact,the intensity of absorption is proportional to the change in dipolemoment, meaning that vibrational motions which lead only to smallchanges in dipole moment, while “infrared active”, may be “tooweak” to actually be observed in the infrared spectrum.

In this activity, you will calculate the vibrational spectra of cis andtrans isomers of 1,2-dichloroethylene. Unlike the experimental(infrared) spectra, the calculations reveal all twelve vibrations foreach isomer as well as the expected intensity (including zero intensity).

Section D34 7/2/04, 10:41 AM241

242

A comparison with the experimental spectra will allow you to seewhich lines are unobserved because they are outside the measurementrange and which are unobserved because they are weak or inactive.

1. Build both trans and cis-dichloroethylene. Calculate theequilibrium geometry for each along with the infrared spectrumusing the HF/3-21G model.

Examine the infrared spectrum for cis-dichloroethylene. (It is betterto look at the table of frequencies and intensities rather than theactual spectrum.) Associate each vibration falling between 500 and3500 cm-1 and having an intensity 1% or greater of the maximumintensity with the experimental data provided below. Use the(experimentally assigned) “description of vibration” as well asthe labels describing the “symmetry of vibration” to assist you.

description of vibration symmetry of vibration frequency

CCCl deformation b1 571CH bend b2 697CCl stretch a1 711CCl stretch b1 857CH bend a1 1179CH bend b1 1303CC stretch a1 1587CH stretch b1 3072CH stretch a1 3077

Repeat your analysis with trans-1,2-dichloroethylene.

description of vibration symmetry of vibration frequency

CCl stretch bu 828CH bend au 900CH bend bu 1200CH stretch bu 3090

Do the calculations provide a quantitatively correct descriptionof the observed infrared spectra for cis and trans-dichloroethylene,

Section D34 7/2/04, 10:41 AM242

243

that is, properly assign those lines which are intense enough toactually be observed?

2. Identify the most intense line in the calculated infrared spectrumof trans-dichloroethylene. Make a list of structures correspondingto distortion of the molecule away from its equilibrium geometryalong this vibration. Calculate the dipole moment for each distortedstructure (using the HF/3-21G model) and plot them (vs. motionaway from the equilibrium geometry).

Click on the appropriate frequency inside the Spectra dialog. Next, changethe value inside the box to the right of Steps at the right of the dialog to“5” and click on Make List. A list of five structures “walking along” thevibrational coordinate will result. Enter the Calculations dialog andrequest a Single Point Energy calculation using the HF/3-21G model.Submit. When completed, enter the dipole moment into the spreadsheet.For Windows, click on the header cell for an empty column, then clickon the Add button at the bottom of the spreadsheet, select dipole fromthe menu and finally click on OK. For Mac, click on Add button at thebottom of the spreadsheet and drag dipole from the menu into thespreadsheet. (Alternatively, bring up the Molecule Properties dialog,click on Dipole at the bottom and drag it into the spreadsheet.) Bring upthe Plots dialog and select Molecule (the number in the list) under XAxis and Dipole under Y Axes.

What is the dipole moment of trans-dichloroethylene in itsequilibrium geometry? What, if anything, happens to the dipolemoment as the molecule is distorted along the selected vibrationalcoordinate?

3. Locate the vibrational frequency in the calculated infraredspectrum corresponding to the stretching of the CC bond. It shouldhave an intensity of zero. Perform the same dipole momentcalculations as above using this frequency. How is this situationdifferent from that in the previous step? How does your resulthere together with that from the previous step fit in with the factthat intensity depends on change in dipole moment?

Section D34 7/2/04, 10:41 AM243

245

35Benzyne

Benzyne has long been implicated as an intermediate in nucleophilicaromatic substitution, e.g.

–H2O–Cl–

Cl

OH– OH–

–H2O

benzyne

OH

While the geometry of benzyne has yet to be conclusively established,the results of a 13C labeling experiment leave little doubt that two(adjacent) positions on the ring are equivalent.

Cl

*KNH2

NH3

*

NH2

*

NH2

*+

1 : 1

* = 13C

The infrared spectrum of a species purported to be benzyne has beenrecorded and a line in the spectrum at 2085 cm-1 assigned to the C≡Cstretch.

In this activity, you will obtain an equilibrium geometry for benzyneusing the HF/3-21G model and, following this, obtain an infraredspectrum for the molecule. Comparison with the experimentalspectrum (specifically the line at 2085 cm-1 attributed to the C≡Cstretch) should allow you to comment one way or another about itsreported “sighting”.

Section D35 7/2/04, 10:42 AM245

246

1. Build benzyne.

Start with benzene and delete two adjacent free valences (Delete fromthe Build menu).

Also build 2-butyne and benzene. The structures of those moleculeswill help you to judge the bonding in benzyne. Additionally, thecalculated C≡C stretching frequency for 2-butyne, which is knownexperimentally, will serve to calibrate the calculations for benzyne.Obtain equilibrium geometries for all three molecules using theHF/3-21G model and following that, calculate infrared spectrafor benzyne and 2-butyne only.

2. After the calculations have completed, examine the geometry ofbenzyne, and compare it to the structures for 2-butyne and benzene.Does it incorporate a “real” triple bond (as does 2-butyne) or is thelength closer to that in benzene? Based on a comparison ofstructures among the three molecules, draw what you feel is anappropriate Lewis structure (or set of Lewis structures) for benzyne.

3. Display the infrared spectrum for 2-butyne. Locate the line in thespectrum corresponding to the C≡C stretch and record itsfrequency. You will use the ratio of the experimental C≡Cstretching frequency (2240 cm-1) and the calculated value to scalethe calculated vibrational frequencies for benzyne.

4. Display the infrared spectrum for benzyne. Is benzyne an energyminimum? How do you know? Locate the line in the spectrumcorresponding to the “C≡C” stretch. Is it “weak” or “intense”relative to the other lines in the spectrum? Would you expect thatthis line would be easy or hard to observe? Scale the calculatedfrequency by the factor you obtained (for 2-butyne) in the previousstep. Is your (scaled) stretching frequency in reasonable accordwith the reported experimental value of 2085 cm-1?

Section D35 7/2/04, 10:42 AM246

247

36Why are Silicon-CarbonDouble Bonds so Rare?

With the exception of so-called “phosphorous ylides”, compoundsincorporating a double bond between carbon and a second-rowelement are quite rare. Most curious perhaps is the dearth of stablecompounds incorporating a carbon-silicon double bond, in starkcontrast to the common occurrence of the carbon-carbon double bondin organic compounds.

In this activity, you will first examine the three-dimensional structuresof one or more of the silaolefins, 1-4, which have been synthesizedfor “hints” why such compounds have proven to be so elusive.

Si C

SiMe3

Me SiMe(Cme3)2

Me

Si C

OSiMe3

Me3Si

Me3Si

1 2(1-adamantyl)

Si CC

(CMe3)Me2SiC

(CMe3)Me2Si

3

CMe3

CMe3

Si C

(CMe3)Me2Si

Me3Si

3

You will then employ two different graphical models to compareboth electrophilic and nucleophilic reactivities of a simpler silaolefin,Me2Si=CMe2, with that of the analogous olefin, Me2C=CMe2.

1. Build one or more of the compounds, 1 - 4. Display the moleculeas a space-filling model. Given that the “chemistry” of olefins(and presumably of silaolefins as well) is associated with the πbond, do you anticipate any “problems” that the molecule mighthave reacting? Elaborate.

Section D36 7/2/04, 10:42 AM247

248

2. Build tetramethylsilaethylene, Me2Si=CMe2, as well as its carbonanalog 2,3-dimethyl-2-butene, Me2C=CMe2. Optimize thegeometries of both molecules using the HF/3-21G model andfollowing this, obtain both a local ionization potential map and aLUMO map. A local ionization potential map indicates the relativeease of electron removal (“ionization”) at the “accessible surface”of a molecule. This would be expected to correlate with the relativeease of addition of an electrophile. A LUMO map indicates theextent to which the lowest-unoccupied molecular orbital (theLUMO) can be “seen” at the “accessible surface” of a molecule.This indicates the “most likely” regions for electrons to be added,and would be expected to correlate with the likelihood ofnucleophilic attack. (Both types of maps are described andillustrated in the essay “Local Ionization Potential Maps andLUMO Maps: Electrophilic and Nucleophilic Reactivities”.)Place both molecules side-by-side on screen, and one after theother display the local ionization potential and LUMO maps.

First, bring up the spreadsheet. For Windows, check the boxes to theright of the first (“Labels”) column for each molecule. For Mac, click on

at the bottom left of the screen and then check the box to the left of themolecule name (in the spreadsheet) for each molecule. “Uncouple” themolecules so that they can be manipulated independently. Select (uncheck)Coupled from the Model menu.

On the basis of comparison of local ionization potential maps,would you conclude that the silaolefin is likely to be more or lessreactive than the olefin toward electrophiles? Elaborate. On thebasis of comparison of the LUMO maps, would you concludethat the silaolefin is likely to be more or less reactive than theolefin toward nucleophiles? Elaborate. How do your conclusionsfit with the known silaolefins? Elaborate.

Section D36 7/2/04, 10:42 AM248

249

37Carbon Monoxide andMetal-Ligand Bonding

Carbon monoxide is probably the single most common molecule toappear in organometallic compounds. Where a single metal is involved(or where two or more metals are involved but are widely separated),CO inevitably bonds “end on” from carbon, and contributes twoelectrons to the valence shell of the metal. Where two metals are“close”, carbon monoxide can alternatively bond to both, again fromcarbon (it can bridge). In this case, it contributes one electron to thevalence shell of each metal.

O C M:2 electrons

O CM

:1 electron

1 electron M

At first glance, both bonding modes use only the lone pair on carbonand neither should have much effect on the structure and propertiesof free CO. There is, however, evidence to suggest the contrary.Specifically, the infrared sketching frequency of CO complexed end-on to a metal is typically in the range of 1850 to 2100 cm-1, which issmaller than the frequency in free carbon monoxide (2143 cm-1).Changes are even greater in molecules where CO is a bridging group,typically falling in the range of 1700 - 1850 cm-1. It would appearthat the simple bonding models above involving only the lone pairneed to be modified. This is the subject of the present activity.

1. Build CO and optimize its geometry using the semi-empirical(PM3) model. Request the HOMO and LUMO. Display theHOMO. This corresponds to the molecular orbital in which thehighest-energy pair of electrons are held. Is it consistent with theusual Lewis structure for CO? Is the HOMO bonding, antibondingor essentially non-bonding between carbon and oxygen? What, ifanything, would you expect to happen to the CO bond strength as

Section D37 7/2/04, 10:43 AM249

250

electrons are donated from the HOMO to the metal? Elaborate. Isthis consistent with the changes seen in the infrared stretchingfrequency of carbon monoxide?

Reduction in bond strength will generally be accompanied by increasein bond length and decrease in stretching frequency.

Display the LUMO. This corresponds to the molecular orbital inwhich the next (pair of) electrons will go.

The LUMO in CO is one of a set of two equivalent “degenerate” orbitals.You can base any arguments either on the LUMO or the next orbital(LUMO+1).

Is the LUMO bonding, antibonding or essentially non-bondingbetween carbon and oxygen? What if anything would you expectto happen to the CO bond strength were electrons to be donated(from the metal) into this orbital? Elaborate. Is this consistentwith the changes seen in the infrared stretching frequency of carbonmonoxide?

To see if the metal center incorporates a high energy filled molecularorbital properly disposed to donate electrons into the LUMO of CO,you need to perform calculations on a simple organometallic fromwhich a carbon monoxide ligand has been removed. You will useFeCO4, arising from loss of CO from FeCO5.

2. Build FeCO5 and obtain its equilibrium geometry (a trigonalbipyramid) using the semi-empirical (PM3) model. Whencompleted, delete one of the equatorial CO ligands to make irontetracarbonyl and perform a single-point energy calculation.Request the HOMO.

Does the HOMO have significant amplitude in the location wherethe next carbon monoxide ligand (the one you removed) willattach? If so, are signs (colors) of the orbital components consistentwith the signs of the components for the LUMO in CO? Wouldyou expect electron donation to occur?

Section D37 7/2/04, 10:43 AM250

251

38Ethylene and

Metal-Ligand BondingTwo “limiting” structures can be drawn to represent ethylene “bonded”to a metal center. The first may be thought of as a “weak complex” inthat it maintains the CC double bond, while the second completelydestroys the double bond in order to form two new metal-carbon σbonds, leading to a three-membered ring (a so-called “metallacycle”).

M M

The difference between the two representations is one of degree and“real” metal-alkene complexes are expected to span the full range ofpossible structures. (A similar situation has already been describedin the activity “Are Reactive Intermediates “Normal” Molecules?”.)

In this activity, you will first examine the HOMO and LUMO ofethylene to see where electrons may be drawn from and where theymay be put back, and to understand the consequences which these“electron movements” will have on its geometry.

1. Obtain an equilibrium geometry for ethylene using the semi-empirical PM3 model and request the HOMO and LUMO. Displaythe HOMO. Is it bonding, antibonding or essentially non-bondingbetween the two carbons? What if anything should happen to theCC bond as electrons are donated from the HOMO to the metal?Specifically, do you expect the carbon-carbon bond length todecrease, increase or remain about the same? Elaborate. Displaythe LUMO of ethylene. This corresponds to the molecular orbitalwhere the next (pair of) electrons will go. Is this orbital bonding,

Section D38 7/2/04, 10:43 AM251

252

antibonding or essentially non-bonding between the two carbons?What, if anything, should happen to the CC bond as electrons aredonated (from the metal) into the LUMO? Elaborate. Is the expectedchange in the CC bond due to this interaction in the same directionor in the opposite direction as any change due to interaction of theHOMO with the metal? Elaborate.

To see if the metal center incorporates appropriate unfilled and filledmolecular orbitals to interact with the HOMO and LUMO of ethylene,respectively, perform calculations on FeCO4, arising from loss ofethylene from ethylene iron tetracarbonyl, CO4FeC2H4.

2. Build ethylene iron tetracarbonyl and obtain its equilibriumgeometry (a trigonal bipyramid with ethylene occupying anequatorial position with the CC bond in the equatorial plane)using the semi-empirical (PM3) model. When completed, deletethe ethylene ligand and perform a single-point energy calculation.Request both HOMO and LUMO.

Does the LUMO in FeCO4 have significant amplitude in theequatorial region where ethylene will fit to accept electrons fromthe (ethylene) HOMO? If so, are the signs (colors) of the LUMOorbital components consistent with the signs of the componentsof the ethylene HOMO? Would you expect electron donation fromethylene to be metal to occur? Does the HOMO in FeCO4 havesignificant amplitude in the equatorial region where ethylene willfit? If so, are the signs of its components consistent with the signsof the components of the ethylene LUMO? Would you expectelectron donation from the metal to ethylene to occur?

Section D38 7/2/04, 10:43 AM252

253

39The Chromium

Tricarbonyl “Substituent”Benzene and other aromatics (“arenes”) devoid of strong electronwithdrawing groups such as cyano or nitro, are normally immune tonucleophilic aromatic substitution, e.g., anisole is non-reactive, while4-cyanoanisole is reactive.

OMe

:Nu

OMe

Nu:

CN

substitutionno reaction

On the other hand, the analogous arene chromium tricarbonylcomplexes are typically highly reactive, giving rise primarily to metasubstitution products.

OMe

:Nu

Cr(CO)3

OMe

Nu

Does this imply that chromium tricarbonyl acts as an electron acceptor,having the same net effect on the arene ring as a directly bondedsubstituent such as cyano? This activity uses electrostatic potentialmaps to test such a hypothesis and further to see to what extent metalcomplexation is as effective as “conventional” ring substitution inenhancing and directing reactivity.

Section D39 7/2/04, 10:44 AM253

254

1. Build anisole, 4-cyanoanisole and anisole chromium tricarbonyl,and obtain PM3 (semi-empirical) equilibrium geometries for allthree molecules and following this, electrostatic potential maps.

To build anisole chromium tricarbonyl, bring up the inorganic model kit,select Cr from the Periodic Table and four-coordinate tetrahedral fromthe list of hybrids and click on screen. Add Carbon Monoxide (Ligandsmenu) to three of the free valences on chromium and Benzene (Ligandsmenu not Rings menu) to the remaining free valence. Bring up the organicmodel kit and introduce a methoxy group onto the (benzene) ring.

When the calculations have completed, put the molecules side-by-side on screen, and turn on the electrostatic potential maps.

Bring up the spreadsheet. For Windows, check the box to the right of“Label” (first column) for all three molecules. For Mac, first click on at the bottom left of the screen and then check the box to the left of themolecule name (in the spreadsheet) for each molecule. Also, select(uncheck) Coupled from the Model menu, so that the three moleculescan be moved independently.

The “blue” regions in the maps demark regions which are mostelectron deficient. Relative to anisole, is the (accessible) ring facein anisole chromium tricarbonyl more or less electron deficient?What does this suggest to you about the relative likelihood thatthe two molecules will undergo nucleophilic aromatic substitution?Is the complexed chromium tricarbonyl group more effective, lesseffective or about as effective as a para cyano group in increasingthe electron deficiency of the benzene ring? Is there any evidencein the electrostatic potential map for anisole chromium tricarbonylthat nucleophilic attack will occur meta to the methoxy substituent?Elaborate. Is there any evidence for this in calculated carboncharges? Elaborate.

To display the charge on an atom, select Properties from the Displaymenu and click on the atom. To “attach” charges to the model, selectConfigure... from the Model menu and check Charge.

Section D39 7/2/04, 10:44 AM254

255

40Vitamin E

Molecules with unpaired electrons (“radicals”) can cause biochemicaldamage through their reaction with the unsaturated fatty acids foundin cellular membranes. Vitamin E may play an active role in defendingcells from attack by reacting quickly with radicals to give stableproducts that can then be safely excreted. Such compounds are referedto as “antioxidants”.

HO

O

vitamin E

In order to be effective as a cellular antioxidant, vitamin E must beable to transfer a hydrogen atom to the offending radical, leading toa stable vitamin E radical, i.e.

R• + vitamin E RH + vitamin E•

Of course, vitamin E must also be soluble in the cellular membranewhich presumably is “hydrocarbon-like” as opposed to “water-like”.The fact that vitamin E is an effective antioxidant implies that it doesform a stable radical and that it is soluble in the membrane.

In this activity, you will first examine the electrostatic potential mapfor vitamin E for evidence that it should be soluble in a cellularmembrane. You will then look at the radical formed by hydrogenabstraction from vitamin E for evidence that it should be “stable”.Here, a map of spin density will be employed, revealing the extent towhich the radical site remains localized or is delocalized. You willthen examine an alternative to vitamin E and try to judge whether ittoo might be an effective antioxidant.

Section D40 7/2/04, 10:45 AM255

256

1. Bring vitamin E onto the screen.

Select “vitamin E” from the files in the “activities” directory.

Display the electrostatic potential map. Recall that the colors “red”and “blue” depict regions of excess negative and excess positivecharge, respectively, while the color “green” depicts regions whichare electrically neutral. The former would be expected to enhancesolubility in water while the latter would be expected to enhancesolubility in hydrocarbons. Would you expect vitamin E to besoluble in cellular membranes? Explain.

2. Bring vitamin E radical onto the screen.

Select “vitamin E radical” from the files in the “activities” directory.

From which atom has the hydrogen atom been abstracted? Examinethe calculated equilibrium geometry of the radical to decide whetherthe unpaired electron (the radical site) is localized on this atom orif it is spread over several atoms (delocalized)? If the latter, drawappropriate Lewis structures showing the delocalization. On thebasis of the calculated geometry, would you conclude that vitaminE radical should be especially stable? Elaborate.

In order to show the location of the unpaired electron, display thespin density. This depicts deviations from “perfect” electron pairingat different locations in the molecule. A spin density map limits thelocations to those on the accessible electron density surface.

3. Display the spin density map for vitamin E radical.

Colors near “blue” depict regions of maximum excess spin whilethose near “red” depict regions of least spin.

Is the map in accord with the calculated equilibrium geometry?Would you conclude that vitamin E radical should be especiallystable?

Section D40 7/2/04, 10:45 AM256

257

Other compounds have been used as antioxidants. These include3,5-di-tert-butyl-4-hydroxytoluene (butylated hydroxytoluene orBHT).

Me3C

OH

CMe3

BHT

Does this satisfy the same requirements demanded of vitamin E? Is itlikely to be as effective as vitamin E? Compare electrostatic potentialmaps and spin density maps to see.

3. Build BHT as well as the radical resulting from hydrogen removal.Obtain equilibrium geometries for both using the semi-empiricalmodel, and request on electrostatic potential map for BHT and aspin density map for BHT radical.

For BHT radical, you need to set Multiplicity inside the Calculationsdialog to Doublet. To request a spin density map, specify density forSurface and spin for Property inside the Surfaces dialog.

When both have completed, examine the electrostatic potentialmap for BHT. Relative to vitamin E, would you expect BHT to bemore or less soluble in a cellular membrane? Elaborate.

4. Finally, examine the spin density map for BHT radical. Relativeto vitamin E radical, is the spin more or less delocalized? Wouldyou expect BHT radical to be more or less easily formed (fromBHT) than vitamin E radical (from vitamin E)? Elaborate.

Section D40 7/2/04, 10:45 AM257

259

41Can DNA be Tricked?

Protons bound to heteroatoms in heterocyclic compounds are likelyto be very mobile in solution, and where there are two (or more)heteroatoms, different isomers related by shifts in protons among theheteroatoms may be present in equilibrium. The situation is so commonthat these isomers have been given a special name “tautomers”, andthe equilibrium commonly referred to as a “tautomeric equilibrium”.A simple and well known example involves 2-hydroxypyridine whichin protic media is in tautomeric equilibrium with 2-pyridone.

NNHHO O

2-hydroxypyridine 2-pyridone

The surprising fact is that the equilibrium actually lies in favor of2-pyridone despite the fact that 2-hydroxypyridine is an aromaticmolecule and would be expected to be very stable.

A much more important case concerns the four nucleotide bases usedin the construction of DNA, adenine, thymine, cytosine and guanine.Each base incorporates several heteroatoms and each can give rise toseveral tautomers. The structure of DNA assumes the predominanceof only one tautomer for each base (the one drawn in your textbook),which in turn hydrogen bonds to its “complementary base”. Were anyof the possible “alternative” tautomers present in significant amounts,the consequences could be catastrophic. In particular, the alternativemight very well select an alternative complementary base leading toerrors in replication.

In this activity, you will use HF/3-21G calculations to obtain energiesfor all “reasonable” tautomers (principal tautomer and all alternatives)for one (or more) of the nucleotide bases. This will allow you to say

Section D41 7/2/04, 10:45 AM259

260

with confidence whether tautomeric equilibrium presents a realdanger.

1. Build one of the nucleotide bases in its usual tautomeric form(see below). Use a methyl group to mimic the connection to thesugar-phosphate backbone in DNA.

N

N

N

NH

N

N

N

NHN

N

N

N

NH2

O

O

O

NH2 O

NH2

CH3 CH3CH3 CH3

cytosine thymine adenine guanine

2. One after the other, build all the alternative tautomers for yourselected nucleotide base (don’t worry about including more thanone stereoisomer for tautomers which incorporate iminefunctionality). Obtain equilibrium geometries for your full set ofstructures using the HF/3-21G model.

What is the energy of the alternative relative to the principaltautomer? What is its equilibrium abundance at room temperature(use the Boltzmann equation; see the essay “Total Energies andThermodynamic and Kinetic Data”)?

If you put both tautomers in the same document prior to calculation (NewMolecule instead of New to start building the second molecule), you canuse Spartan’s spreadsheet to provide both the relative energy and theBoltzmann weight. Bring up the spreadsheet and click on the entry forthe principal tautomer. For Windows, click on the header cell for an emptyspreadsheet column, then click on the Add button at the bottom of thespreadsheet, select both rel. E and Boltzmann Distribution from themenu of items which appears and finally, click on OK. For Mac, click onthe Add button at the bottom of the spreadsheet, and drag both Rel.Eand Boltzmann from the menu of items into the spreadsheet.

Is the alternative likely to be present in sufficient amount to affectbonding in DNA? Elaborate.

Section D41 7/2/04, 10:45 AM260

261

Section ECommon Terms and Acronyms

3-21G. A Split-Valence Basis Set in which each Core Basis Function is writtenin terms of three Gaussians, and each Valence Basis Function is split into twoparts, written in terms of two and one Gaussians, respectively.

6-31G*. A Polarization Basis Set in which each Core Basis Function is writtenin terms of six Gaussians, and each Valence Basis Function is split into twoparts, written in terms of three and one Gaussians, respectively. Non-hydrogenatoms are supplemented by a set of single Gaussian d-type functions.

Ab Initio Models. The general term used to describe methods seekingapproximate solutions to the Electronic Schrödinger Equation, but which donot involve empirical parameters.

Acid. A molecule which “desires” to give up a proton.

Acidity. The Energy (Enthalpy) of the reaction: AH → A– + H+. Typically givenrelative to a “standard” acid, AsH, that is ∆E(∆H) for AH + As

– → A– + AsH.

Activation Energy. The energy of a Transition State above that of reactants.Activation energy is related to reaction rate by way of the Arrhenius Equation.

Anion. An atom or molecule with a net charge of -1.

Antibonding Molecular Orbital. A Molecular Orbital which has a “break” ora “Node” between particular atomic centers. Adding electrons to such an orbitalwill weaken the bond while removing electrons will strengthen the bond. Theopposite is a Bonding Molecular Orbital.

Arrhenius Equation. An equation governing the rate of a chemical reaction asa function of the Activation Energy and the temperature.

Atomic Orbital. A Basis Function centered on an atom. Atomic orbitals typicallytake on the form of the solutions to the hydrogen atom (s, p, d, f... type orbitals).

Atomic Units. The set of units which remove all of the constants from inside theSchrödinger Equation. The Bohr is the atomic unit of length and the Hartreeis the atomic unit of energy.

Base. A molecule which “desires” to accept a proton.

Section E 7/2/04, 10:50 AM261

262

Basicity. The Energy (Enthalpy) of the reaction: B + H+ → BH+. Typicallygiven relative to a “standard” base, Bs, that is ∆E(∆H) for B + BsH+ → BH+ + Bs.

Basis Functions. Functions usually centered on atoms which are linearlycombined to make up the set of Molecular Orbitals. Except for Semi-EmpiricalModels where basis functions are Slater type, basis functions are Gaussian type.

Basis Set. The entire collection of Basis Functions.

Bohr. The Atomic Unit of length. 1 bohr = 0.529167Å.

Boltzmann Equation. The equation governing the distribution of products inThermodynamically-Controlled Reaction.

Bonding Molecular Orbital. A Molecular Orbital which has strong positiveoverlap between two particular atomic centers. Adding electrons to such an orbitalwill strengthen the bond, while removing electrons will weaken the bond. Theopposite is an Antibonding Molecular Orbital.

Bond Surface. An Isodensity Surface used to elucidate the bonding in molecules.The value of the density is typically taken as 0.1 electrons/bohr.3

Born-Oppenheimer Approximation. An approximation based on theassumption that nuclei are stationary. Applied to the Schrödinger Equation, itleads to the Electronic Schrödinger Equation.

Cation. An atom or molecule with a net charge of +1.

Chiral. A molecule with a non-superimposible mirror image.

Chiral Center. A tetrahedral atom (most commonly carbon) with four differentgroups attached. Two different arrangements (“stereoisomers”) are possible andthese are mirror images.

B

ADC

B

ADC

Chiral centers are labelled “R” or “S” depending on the arrangement of groups.Also known as a Stereocenter.

Closed Shell. An atom or molecule in which all electrons are paired.

Conformation. The arrangement about single bonds and of flexible rings.

Core. Electrons which are primarily associated with individual atoms and donot participate significantly in chemical bonding (1s electrons for first-rowelements, 1s, 2s, 2px, 2py, 2pz electrons for second-row elements, etc.).

Section E 7/2/04, 10:50 AM262

263

Coulombic Interactions. Charge-charge interactions which follow Coulomb’slaw. Stabilizing when charges are of opposite sign and destabilizing when theyare of the same sign.

Covalent Bond. A chemical bond which involves a significant sharing of a pairof electrons between the two atoms.

CPK Model. A molecular model in which atoms are represented by spheres,the radii of which correspond to van der Waals Radii. Intended to portraymolecular size and shape. Also known as a Space-Filling Model.

Diastereomers. Stereoisomers which differ in the stereochemistry (R or S) ofone or more (but not all) Chiral Centers.

Diffusion-Controlled Reactions. Chemical reactions without Transition States(or Activation Energies), the rates of which are determined by the speed inwhich molecules encounter each other and how likely these encounters are tolead to reaction. The combination of two radicals proceeds without ActivationEnergy and are examples of diffusion-controlled reactions.

Dipole Moment. A measure of the overall polarity of a molecule, taking intoaccount differences in nuclear charges and electron distribution.

Electron Density. The number of electrons per unit volume at a point in space.This is the quantity which is measured in an X-ray diffraction experiment.

Electronic Schrödinger Equation. The equation which results from incorporationof the Born-Oppenheimer Approximation to the Schrödinger Equation.

Electrophile. An electron pair acceptor. A molecule (or part of a molecule) whichdesires to interact (react) with an electron-rich reagent or Base.

Electrostatic Charges. Atomic charges chosen to best match the ElectrostaticPotential at points surrounding a molecule, subject to overall charge balance.

Electrostatic Potential. The energy of interaction of a positive point chargewith the nuclei and fixed electron distribution of a molecule.

Electrostatic Potential Map. A graph that shows the value of ElectrostaticPotential on an Electron Density Isosurface corresponding to a van der WaalsSurface.

Enantiomers. Stereoisomers which differ in the stereochemistry (R or S) of allChiral Centers. Enantiomers are non-superimposible mirror images.

Endothermic Reaction. A chemical reaction in which the Enthalpy is positive.

Energy(∆E). The heat given off (negative energy) or taken in (positive energy) by achemical reaction at constant volume. Quantum chemical calculations give the energy.

Section E 7/2/04, 10:50 AM263

264

Enthalpy (∆H). The heat given off (negative enthalpy) or taken in (positiveenthalpy) by a chemical reaction. Enthalpy is commonly equated to Energyfrom which it differs by a (small) pressure-volume (PV) term: ∆H = ∆E + P∆V.

Entropy (∆S). The extent of ordering (negative entropy) or disordering (positiveentropy) which occurs during a chemical reaction.

Equilibrium Geometry. A Local Minimum on a Potential Energy Surface.

Excited State. An electronic state for an atom or molecule which is not thelowest-energy or Ground State.

Exothermic Reaction. A chemical reaction in which the Enthalpy is negative.

Force Field. The set of rules underlying Molecular Mechanics Models.Comprises terms which account for distortions from ideal bond distances andangles and for Non-Bonded van der Waals and Coulombic Interactions.

Frontier Molecular Orbitals. The HOMO and LUMO.

Formal Charge. A “recipe” to assign charges to atoms: formal charge = numberof valence electrons - number of electrons in lone pairs - number of bonds (singlebond equivalents).

Free Energy; See Gibbs Energy

Gaussian. A function of the form xlymzn exp (αr2) where l, m, n are integers (0,1, 2 . . .) and α is a constant. Used in the construction of Basis Sets.

Gaussian Basis Set. A Basis Set made up of Gaussian Basis Functions.

Gibbs Energy (∆G). The combination of Enthalpy and Entropy which dictateswhether a reaction is favorable (spontaneous) or unfavorable at temperature T:∆G = ∆H - T∆S.

Global Minimum. The lowest energy Local Minimum on a Potential EnergySurface.

Ground State. The lowest energy electronic state for an atom or molecule.

Hammond Postulate. The idea that the Transition State for an exothermicreaction will more closely resemble reactants than products. This provides thebasis for “modeling” properties of Transition States in terms of the propertiesof reactants.

Hartree. The Atomic Unit of energy. 1 hartree = 627.47 kcal/mol.

Hartree-Fock Approximation. Separation of electron motions in many-electronsystems into a product form of the motions of the individual electrons.

Hartree-Fock Energy. The energy resulting from Hartree-Fock Models.

Section E 7/2/04, 10:50 AM264

265

Hartree-Fock Models. Methods in which the many-electron wavefunction inwritten terms of a product of one-electron wavefunctions. Electrons are assignedin pairs to functions called Molecular Orbitals.

Heterolytic Bond Dissociation. A process in which a bond is broken and acation and anion result. The number of electron pairs is conserved, but a non-bonding electron pair has been substituted for a bond.

HOMO. Highest Occupied Molecular Orbital. The highest-energy molecularorbital which has electrons in it.

Homolytic Bond Dissociation. A process in which a bond is broken and twoRadicals result. The number of electron pairs is not conserved.

Hybrid Orbital. A combination of Atomic Orbitals. For example, 2s and 2px,2py, 2pz orbitals may be combined to produce four equivalent sp3 hybrid orbitals,each pointing to different corners of a tetrahedron.

Hypervalent Molecule. A molecule containing one or more main-group elementsin which the normal valence of eight electrons has been exceeded. Hypervalentmolecules are common for second-row and heavier main-group elements butare uncommon for first-row elements.

Imaginary Frequency. A frequency which results from a negative curvature ofthe Potential Energy Surface. Equilibrium Geometries are characterized byall real frequencies while Transition States are characterized by one imaginaryfrequency.

Infrared Spectrum. The set of Energies corresponding to the vibrational motionswhich molecules undergo upon absorption of infrared light.

Ionic Bond. A chemical bond in which the pair of electrons is not significantlyshared by the two atoms.

Isodensity Surface. An Electron Density Isosurface. Bond Surfaces and SizeSurfaces may be used to elucidate bonding or to characterize overall molecularsize and shape, respectively.

Isopotential Surface. An Electrostatic Potential Isosurface. It may be used toelucidate regions in a molecule which are particularly electron rich and subjectto electrophilic attack and those which are particularly electron poor, subject tonucleophilic attack.

Isosurface. A three-dimensional surface defined by the set of points in spacewhere the value of the function is constant.

Isotope Effect. Dependence of molecular properties and chemical behavior onatomic masses.

Section E 7/2/04, 10:50 AM265

266

Kinetically-Controlled Reaction. A chemical reaction which has not gone allthe way to completion, and the ratio of products is not related to the relativeActivation Energies.

Kinetic Product. The product of a Kinetically-Controlled Reaction.

LCAO Approximation. Linear Combination of Atomic Orbitals approximation.Approximates the unknown Hartree-Fock Wavefunctions (MolecularOrbitals) by linear combinations of atom-centered functions (Atomic Orbitals).

Local Ionization Potential. A measure of the relative ease of electron removal(“ionization”) as a function of location.

Local Ionization Potential Map. A graph of the value of the Local IonizationPotential on an Isodensity Surface corresponding to a van der Waals Surface.

Local Minimum. Any Stationary Point on a Potential Energy Surface forwhich all coordinates are at energy minima.

Lone Pair. A Non-Bonded Molecular Orbital which is typically associatedwith a single atom.

LUMO. Lowest Unoccupied Molecular Orbital. The lowest-energy molecularorbital which does not have electrons in it.

LUMO Map. A graph of the absolute value of the LUMO on an IsodensitySurface corresponding to a van der Waals Surface.

Mechanism. The sequence of steps connecting reactants and products in anoverall chemical reaction. Each step starts from an equilibrium form (reactant orintermediate) and ends in an equilibrium form (intermediate or product).

Merck Molecular Force Field; See MMFF94.

Meso Compound. A molecule with two (or more) Chiral Centers with asuperimposible mirror image.

Minimal Basis Set. A Basis Set which contains the fewest functions needed tohold all the electrons on an atom and still maintain spherical symmetry.

MMFF94. Merck Molecular Force Field. A Molecular Mechanics Force Fieldfor organic molecules and biopolymers developed by Merck Pharmaceuticalsincorporated into Spartan.

Molecular Mechanics Models. Methods for structure, conformation and strainenergy calculation based on bond stretching, angle bending and torsionaldistortions, together with Non-Bonded Interactions, and parameterized to fitexperimental data.

Section E 7/2/04, 10:50 AM266

267

Molecular Orbital. A one-electron function made of contributions of BasisFunctions on individual atoms (Atomic Orbitals) and delocalized throughoutthe entire molecule.

Molecular Orbital Models. Methods based on writing the many-electronsolution of the Electronic Schrödinger Equation in terms of a product of one-electron solutions (Molecular Orbitals).

Multiplicity. The number of unpaired electrons (number of electrons with “down”spin) +1. 1=singlet; 2=doublet; 3=triplet, etc.

Node. A change in the sign of a Molecular Orbital. Nodes involving bondedatoms indicate that a particular Molecular Orbital is Antibonding with respectto the atoms.

Non-Bonded Interactions. Interactions between atoms which are not directlybonded. van der Waals Interactions and Coulombic Interactions are non-bonded interactions.

Non-Bonded Molecular Orbital. A molecular orbital which does not show anysignificant Bonding or Antibonding characteristics. A Lone Pair is a non-bondedmolecular orbital.

Non-Polar Bond. A Covalent Bond which involves equal or nearly equal sharingof electrons.

Nucleophile. An electron-pair donor. A molecule (or region of a molecule) which“desires” to interact (react) with an electron-poor reagent or Acid.

Octet Rule. The notion that main-group elements prefer to be “surrounded” byeight electrons (going into s, px, py, pz orbitals).

Orbital Symmetry Rules; See Woodward-Hoffmann Rules.

Open Shell. An atom or molecule in which one or more electrons are unpaired.Radicals are open-shell molecules.

PM3. Parameterization Method 3. A Semi-Empirical Model incorporated intoSpartan.

Point Group. A classification of the Symmetry Elements in a molecule.

Polar Bond. A Covalent Bond which involves unequal sharing of electrons.

Polarization Basis Set. A Basis Set which contains functions of higher angularquantum number (Polarization Functions) than required for the Ground Stateof the atom, in particular, d-type functions for non-hydrogen atoms. 6-31G* is apolarization basis set.

Section E 7/2/04, 10:50 AM267

268

Polarization Functions. Functions of higher angular quantum than required forthe Ground State atomic description. Added to a Basis Set to allow displacementof Valence Basis Functions away from atomic positions.

Potential Energy Surface. A function of the energy of a molecule in terms ofthe geometrical coordinates of the atoms.

Property Map. A representation or “map” of a “property” on top of an Isosurface,typically an Isodensity Surface. Electrostatic Potential Maps, LUMO Mapsand Spin Density Maps are useful property maps.

Pseudorotation. A mechanism for interconversion of equatorial and axial sitesaround trigonal bipyramidal centers, e.g., fluorines in phosphorous pentafluoride.

Quantum Mechanics. Methods based on approximate solution of theSchrödinger Equation.

Radical. A molecule with one or more unpaired electrons.

Rate Limiting Step. The step in an overall chemical reaction (Mechanism)which proceeds via the highest-energy Transition State.

Reaction Coordinate. The coordinate that connects the Local Minimacorresponding to the reactant and product, and which passes through a TransitionState.

Reaction Coordinate Diagram. A plot of energy vs. Reaction Coordinate.

Schrödinger Equation. The quantum mechanical equation which accounts forthe motions of nuclei and electrons in atomic and molecular systems.

Semi-Empirical Models. Quantum Mechanics methods that seek approximatesolutions to the Electronic Schrödinger Equation, but which involve empiricalparameters. PM3 is a semi-empirical model.

Size Surface. An Isodensity Surface used to establish overall molecular sizeand shape. The value of the density is typically taken as 0.002 electrons/bohr3.

Slater. A function of the form xlymzn exp (-ζr) where l, m, n are integers (0, 1, 2. . .) and ζ is a constant. Related to the exact solutions to the Schrödinger Equationfor the hydrogen atom. Used as Basis Functions in Semi-Empirical Models.

SOMO. Singly Occupied Molecular Orbital. An orbital which has only a singleelectron in it. The HOMO of a Radical.

Space-Filling Model; See CPK Model.

Spin Density. The difference in the number of electrons per unit volume of “up”spin and “down” spin at a point in space.

Section E 7/2/04, 10:50 AM268

269

Spin Density Map. A graph that shows the value of the Spin Density on anIsodensity Surface corresponding to a van der Waals Surface.

Split-Valence Basis Set. A Basis Set in which the Core is represented by asingle set of Basis Functions (a Minimal Basis Set) and the Valence is representedby two or more sets of Basis Functions. This allows for description of asphericalatomic environments in molecules. 3-21G is a split-valence basis set.

Stationary Point. A point on a Potential Energy Surface for which all energyfirst derivatives with respect to the coordinates are zero. Local Minima andTransition States are stationary points.

Stereocenter; See Chiral Center

Symmetry Elements. Elements which reflect the equivalence of different partsof a molecule. For example, a plane of symmetry reflects the fact that atoms onboth sides are equivalent.

Theoretical Model. A “recipe” leading from the Schrödinger Equation to ageneral computational scheme. A theoretical models needs to be unique andwell defined and, to the maximum extent possible, be unbiased by preconceivedideas. It should lead to Potential Energy Surfaces which are continuous.

Theoretical Model Chemistry. The set of results following from application ofa particular Theoretical Model.

Thermodynamically-Controlled Reaction. A chemical reaction which has goneall the way to completion, and the ratio of different possible products is relatedto their thermochemical stabilities according to the Boltzmann Equation.

Thermodynamic Product. The product of a reaction which is underThermodynamic Control.

Transition State. A Stationary Point on a Potential Energy Surface for whichall but one of the coordinates is at an energy minimum and one of the coordinatesis at an energy maximum. Corresponds to the highest-energy point on theReaction Coordinate.

Transition-State Geometry. The geometry (bond lengths and angles) of aTransition State.

Transition State Theory. The notion that all molecules react through a singlewell-defined Transition State.

Valence. Electrons which are delocalized throughout the molecule and participatein chemical bonding (2s, 2px, 2py, 2pz for first-row elements, 3s, 3px, 3py, 3pz forsecond-row elements, etc.).

Section E 7/2/04, 10:50 AM269

270

van der Waals Interactions. Interactions which account for short-range repulsionof non-bonded atoms as well as for weak long-range attraction.

van der Waals Radius. The radius of an atom (in a molecule), which is intendedto reflect its overall size.

van der Waals Surface. A surface formed by a set of interpreting spheres (atoms)with specific van der Waals radii, and which is intended to represent overallmolecular size and shape.

Vibrational Frequencies. The energies at which molecules vibrate. Vibrationalfrequencies correspond to the peaks in an Infrared and Raman Spectrum.

VSEPR Theory. Valence State Electron Pair Repulsion theory. A simpleempirical model used to predict the geometries of molecules given only the totalnumber of electrons associated with each center.

Wavefunction. The solutions of the Electronic Schrödinger Equation. In thecase of the hydrogen atom, a function of the coordinates which describes themotion of the electron as fully as possible. In the case of a many-electron systema function which describes the motion of the individual electrons.

Section E 7/2/04, 10:50 AM270

Documents

Modélisation moléculaire PPH-1003 Exercices, HIV-2012€¦ · Tutorial and User's Guide Wavefunction, Inc. 18401 Von Karman Avenue, Suite 370 Irvine, CA 92612 U.S.A. Wavefunction,