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Computational Chemistry Exercises First Exercise: Building and measuring

• Build each molecule, select, measure & record the bond length and the bond angle; • Do a energy minimisation calculation (or geometry optimisation) & re-record length & angle. • Use just 1 significant digit after the decimal point for bond angle & none for bond length (see example) Molecule Bond length Model MMFF AM1 PM3 Exptal. / pm H3CCH2CH3 C–C 153 152 151 151 153 H3COCH3 C–O 142 H3CSeCH3 C–Se 194 SF6 S–F 156 S2F10 S–S 221 H2O O–H 96 H2Se Se–H 146 PF3 P–F 156 PCl3 P–Cl 204

Molecule Bond angle Model MMFF AM1 PM3 Exptal. / °

H3CCH2CH3 C–C–C 109.5 111.6 111.8 111.7 109.5 H3COCH3 C–O–C 112.0 H3CSeCH3 C–Se–C 96.2 SF6 F–S–F 90.0 S2F10 F–S–F 90.0 H2O H–O–H 104.5 H2Se H–Se–H 91.0 PF3 F–P–F 109.5 PCl3 Cl–P–Cl 112.0

• Building: After sketching the molecule, click on ‘View’ icon, measure bond length & angles, then click on ‘Energy minimise’ icon and re-measure; then do a calculation & re-measure. • Calculation: o Click on Setup/Calculations o Check ‘Equilibrium geometry’ at ‘ground state’ and ‘semi-empirical’, ‘AM1’; then click “Submit” – you can provide a filename to save your work in or re-use an existing one • Measuring: Depress one of the question mark buttons and click on bond(s)

View Energy minimise Measure

- 1 - Name: Second Exercise: Structure vs energy

Hexasilabenzene can exist in several isomeric forms, see sketches below. Sax et al. [J. Comp. Chem. 1988, 9:564] found that the prismane, isomer #2, is the most stable, do you agree?

Compute the various heats of formation, ∆HF, using the AM1 semi-empirical method.

H H Si H Si Si Si Si H Si H

H 1

H H Si Si Si H Si Si H Si H 2 H H Si Si Si Si H H Si Si H H 3 H H Si Si Si H Si Si H H Si H 4

[Hint: for #4 start with a cyclohexane ring]

∆HF (1) = kcal/mol

∆HF (2) = kcal/mol

∆HF (3) = kcal/mol

∆HF (4) = kcal/mol

- 2 - Name: Third Exercise: Potential fuels and

Cubanes are an interesting family of carbon compounds; the parent , C8H8, is shown below.

It is likely that tetranitrocubane, symmetrical C8H4(NO2)4, may be ; the object of this exercise is to compute the heat released per gram of tetranitrocubane by first computing the heat of formation of the compound and then the heat of combustion by using the equations:

C8H4(NO2)4 + 5 O2 = 8 CO2 + 2 H2O + 2 N2

In general, the heat or enthalpy of reaction is given by:

∆H(reaction) = ∆HF(products) – ∆HF(reactants) for this particular reaction we have:

∆HF(combustion) = 8 × ∆HF(CO2) + 2 × ∆HF(H2O) – ∆HF[C8H4(NO2)4]

Compute the heat of reaction, ∆H(combustion), per gram of compound for each of the four potential fuels or explosives listed in the Table below.

Compound Relative molar mass ∆HF (PM3) ∆H (comb.) ∆H (comb.) per gram kcal mol–1 kcal mol–1 kcal g–1 Octane, C8H18 114 – 50.1 – 1,223 10.7 Cubane 104 Tetranitrocubane 287 Octanitrocubane 471 2,4,6–trinitrotoluene 227 – – 94.1 – – Water – – 57.8 – –

- 3 - Name: Fourth Exercise: Vibrational analysis

• Build the molecule acetone, (H3C)2C=O • Using a semi-empirical method such as PM3 do a geometry optimisation or energy minimisation (make sure that before you “Submit” the job that you have checked the Compute/Frequencies (IR) box first. • Examine the output and compare the calculated heat of formation with experiment, see Table 1. • Compare the computed dipole moment with the experimental value, Table 1; in which direction is the dipole moment oriented?

O

• Examine the vibrational modes via Display/Vibrations, identify some of the most important vibrations (and thereby the resulting infrared spectrum); animate and measure the C=O stretching vibration, comparing it with the experimental value in Table 1. • How many vibrational modes should acetone have altogether? [Hint: acetone has 10 atoms; you learnt a formula in 2nd year which relates the two.]

Table 1: Acetone data

Computed Experimental

–1 Heat of formation / kcal mol – 51.9 Dipole moment / Debyes 2.88 –1 C=O stretch / cm 1,737

Typically MO calculations overestimate the vibrational wavenumbers by about 10–12 %.

Project: Benzyne

The benzyne molecule, C6H4, is believed to be an intermediate in nucleophilic aromatic substitution reactions; there is a controversial report that benzyne can be trapped at very low temperature in a matrix and that an IR band at 2,085 cm–1 is associated with the C–C triple bond stretching vibration.

Show that the computed triple bond vibration for benzyne is not inconsistent with the report; use the ab initio method Hartree-Fock with a 3-21G(*) basis set. You may assume a scaling factor of 10 % so that your computed frequency should be compared to ≈ (2,085 × 1.1) ≈ 2,294 cm–1.

- 4 - Name:

Fifth Exercise: Wavefunctions

Build fluoride, H–F, run a semi-empirical AM1 geometry optimisation including frequency calculation, inspect the detailed output and compare it to that below.

PC SPARTAN PRO Semi-Empirical Program: (PC/x86) Release 6.0.4 Hydrogen fluoride, HF Run type: Geometry optimization (Analytical Gradient in FREQ) Model: RHF/AM1 restricted Hartree-Fock/Austin Method 1 Number of shells: 3 2 S shells 1 P shells Number of basis functions: 5 Number of electrons: 8 eight electrons Use of molecular symmetry enabled Molecular charge: 0 neutral species Spin multiplicity: 1 all spin paired; multiplicity=2S+1 If 1 unpaired e S=1/2; 2 unpaired e’s S=1; etc. Point Group = CIV Order = 1 Nsymop = 5 This system has 1 degrees of freedom

Initial Hessian option Hessian from AM1 calculation used. Max. Max. Neg. Cycle Energy Grad. Dist. Eigen 1 -74.2762 0.00003 0.00000

Heat of Formation: -74.276 kcal/mol key property

Estimating Force Constant matrix by central-differences Hessian Estimation Complete Memory Used: 44.37 Kb

Semi-empirical CPU Time : 000:00:00.1

Closed-Shell Molecular Orbital Coefficients HOMO LUMO ↓ ↓ MO: 1 2 3 4 5 Eigenvalues: -1.82822 -0.63289 -0.51768 -0.51768 0.24632 (ev): -49.74838 -17.22191 -14.08675 -14.08675 6.70263

A1 A1 ??? ??? A1 1 H1 S -0.37583 -0.46287 0.00000 0.00000 -0.80281 2 F1 S -0.91940 0.29466 0.00000 0.00000 0.26052 3 F1 PX 0.00000 0.00000 -0.53905 0.84228 0.00000 4 F1 PY 0.00000 0.00000 -0.84228 -0.53905 0.00000 5 F1 PZ 0.11597 0.83602 0.00000 0.00000 -0.53631

OK, so the 1st MO is built according to:

Ψ1 = 0.92φ()F1s − 0.12φ(F 2pz) + 0.38φ(H1s) Mulliken Population Analysis - 5 - Name:

AO atom type Occupancy Fraction of the total no. of electrons in each AO ------Mulliken analysis probably not very reliable; 1 H1 S 0.710991 natural population analysis is preferred nowadays 2 F1 S 1.864258 3 F1 PX 2.000000 4 F1 PY 2.000000 5 F1 PZ 1.424751

Atom Occupancy Charge ------H1 0.710991 0.289009 F1 7.289009 -0.289009

Total Charge = 0.000000

Q-minus(NAO) = -0.2890 Q-plus(NAO) = 0.2890

Normal Modes and Vibrational Frequencies (cm-1)

4459.18 Compare experimental value of 3,958 cm–1 A1 just the one vibration of course X Y Z Cartesian coordinates (x,y,z) of each atom H1 0.000 0.000 0.971 F1 0.000 0.000 -0.051

Zero-point vibrational energy is 6.375 kcal/mol

Standard Thermodynamic quantities at 298.15 K and 1.00 atm

Translational Enthalpy: 0.889 kcal/mol Rotational Enthalpy: 0.592 kcal/mol Vibrational Enthalpy: 6.375 kcal/mol

Translational Entropy: 34.921 cal/mol.K Rotational Entropy: 6.128 cal/mol.K Vibrational Entropy: 0.000 cal/mol.K

Reason for exit: Successful completion Properties Program CPU Time : 000:00:00.1 Finished

Project: Build the water molecule, do an energy minimisation at the AM1 level, calculate the MOs, plot and describe each orbital in your own words. View the orbitals via Setup/Surfaces/Add, from ‘Surfaces’ choose HOMO, then HOMO{-} picking 1, 2 and 3 in turn; when you have added all the MOs go to Setup/Submit. When calculation has finished, clicking on appropriate MO will display it (a second click will delete it). If you left click on the surface, you can choose ‘Mesh’ or ‘Transparent’ from the bottom right to get a better view of the MO.

- 6 - Name: Sixth Exercise: Walsh diagrams

Walsh diagrams are useful in predicting or understanding molecular geometry. They correlate energy changes of molecular orbitals between a reference geometry, frequently of high symmetry, and a deformed structure of lower symmetry.

Distance and angle ‘padlocks’

Sketch the water molecule, aligning it on screen; set a restraint or constraint (by clicking on ‘angle padlock’ icon, see above; go to bottom right, click on padlock and type 90.0 followed by return; a pink bow appears and the HOH angle is now held at 90 degrees during any subsequent energy calculation). Click on minimise icon (E ↓), now do a geometry optimisation with AM1 but click the “Subject to Constraints” box first. Record the values of all the four occupied molecular orbital energies (in electron-volts or eV; use just 3 significant figures, eg, –18.277 eV) and that of the bond angle in the Table below

Angle MO #1 MO #2 MO #3 MO #4 ( º ) A1 B1 A1 B2 90 -36.544 -17.381 -15.694 -12.527 100 -36.462 -18.002 -15.152 -12.486 120 140 160 180

-10 B2 -15 A1 Your results should show B1 the same trends as in the -20 Figure. Describe each orbital and

gy -25 explain why the orbital r energies change as the H– Ene O–H angle goes from 90º -30 to 180º. Note that the actual water molecule has -35 a bond angle of 103.5º; a A1 compromise solution.

-40 80 100 120 140 160 180 HOH bond angle

- 7 - Name: Seventh Exercise: Determination of transition state

Heating ethyl formate (pyrolysis) results in the elimination of ethene and the formation of formic acid:

HCOOC2H5 Æ H2C=CH2 + HCOOH

O OH + O H O H

Build ethyl formate, HCOOC2H5, choosing the correct geometry, minimise with AM1, save one copy ‘Ethyl_formate’ for later use & another as ‘Ethyl_formate_TS’ for use now. Select Reaction from Build menu (or curved arrow icon); click on bond ‘a’ then on ‘b’; then on ‘c’ & ‘d’ and finally on ‘e’ followed by a shift click on the methyl H to be transferred and on the O-atom to receive it. In this case it is relatively easy to guess what the transition state is going to look like – a 6-membered cyclic structure; this is not always the case.

e H O d a

c O b H

With all three arrows in place, click on equilibrium icon (twin arrows) at the bottom right of screen Enter Calculations dialogue, specify TS geometry, semi-empirical & AM1. Click on frequencies. Submit job, when finished, save file and examine geometry. What is the wavenumber of the imaginary frequency (in cm–1)?

Animate this imaginary (or negative) frequency or vibration. Question: Is vibrational motion consistent with reaction? - 8 - Name: Turn off animation by re-entering Vibrations & clicking on imaginary frequency. • Bring up Surfaces dialogue, click on Add … select density (bond) & none from Property menu & click on OK. • Click on Add … select ‘density (bond) & potential’ from Property menu & click on OK.

• Submit job; when it finishes. • Enter Surfaces & click on ‘density completed 0.08’ To answer the questions below you will want to change the default opaque view to either a mesh or a transparent view; do this by clicking anywhere on graphic, and then select either Transparent or Mesh from the menu to the right of Style (bottom right) to replace the opaque solid density surface by a mesh or transparent solid view. Question: Is the single C–O bond in ethyl formate nearly fully cleaved? Question: Is the migrating H midway between C & O? Click off ‘density Completed 0.08’ then click on ‘density/potential’ to display the electrostatic potential. Question: Check migrating H colour code; is it ? H+ (blue) H• (green) or H– (red)

Calculation of activation energy

Open ‘ethyl_formate’; enter Calculations dialogue, specify a single-point computation using the density functional theory or DFT method pBP/DN**, click OK & submit job. Record the energy in atomic units. Close the molecule and open the file ‘ethyl_formate_TS’. Enter Calculations, and as before use the DFT/pBP/DN** computation. Submit job and record the final energy. When both calculations are complete, compute the activation energy from the difference between the total energies of transition state and ethyl formate (use molecular properties from the Display menu) N. B. : 1 atomic unit (au) = 627.51 kcal/mol

Computed activation energy (in kcal / mol) Question: How does your value compare with the experimental values which range from 40–44 kcal / mol?

- 9 - Name: Eight Exercise: Multiple molecules

Molecular recognition or the ability of molecules to orient themselves with respect to another is a very important topic. Hydrogen-bonding is usually the dominant force in these interactions; here we shall compute the orientation of two molecules of acetic/ethanoic acid and attempt to measure the strength of the hydrogen-bonds formed. It is well known that acetic acid is dimeric in the vapour phase with the approximate structure:

O..... H O

H3C CH3 O H..... O

• Build acetic acid, minimise it with PM3, record the O–H bond length and the heat of formation. • Insert a second acetic acid molecule on-screen. • Set a constraint on both OH to O= distances of about 200 pm and minimise the resulting structure. • Now remove the constraints and do an energy minimisation at the PM3 level. • Record the new O–H bond length and the heat of formation of the dimer.

–1 r (O–H) / pm ∆HF / kcal mol

Monomer

Dimer

∆r (O–H)

Now you can calculate the hydrogen-bond strengths from:

Strength of hydrogen-bond = ∆HF (dimer) – 2 × ∆HF (monomer) = kcal mol–1

What is the hydrogen-bond length, that is, r (O·····H)? pm What structure do you think would the dimer of 9-methyladenine adopt?

NH2 N N

N N

CH3

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