Computational Chemistry

Winter Quarter 1996

( Due 2 / 9 / 96 )

You will need to do the following in order for your Cray account to work. You can begin the problem set at that point.

A. Telnet to sk.msc.edu (Macs in the microlab should have an NCSA Telnet macro that will get you there, or you can simply enter sk.msc.edu at the prompt after Open Connection in the File menu).

B. Provide your correct uid (e.g., me106**, whatever ** is for you).

C. The initial password on your account is ch#m8003. Enter these characters exactly.

D. You should now be logged in successfully, and you will be prompted to enter a terminal type. Unless you know you have a different kind of terminal, enter vt100 (this will probably be listed as the default). If you are using MacX to communicate, enter xterm (which will almost certainly be listed as the default).

E. To keep the news items from showing up every time you enter, type news, and scroll through using the space bar. Type q to end. You can read them if you are interested.

F. In order to change your password, enter the command passwd. You will first be asked for your old password (i.e., ch#m8003) and then be prompted for a new one. Enter a password of at least six characters, preferably with a number or special character in it (e.g., my@acct). You will be asked to re-enter it so as to verify you did not type it incorrectly the first time. If the two passwords match, your new password will be active next time-go on to step F. If you make a mistake at any point you will be dropped and ch#m8003 will remain the password. Repeat this step and type more carefully . . .

G. At the next prompt, type the following exactly: cp ../me10601/.cshrc . and hit return (that last "dot" is not a period to my sentence, it needs to be typed). You have now copied the script that will set up your shell when you log in from now on. Type csh to execute the script (this is the only time you will have to execute it yourself).

H. You're ready to roll! To run an AMSOL job you will need to have an input file named myfile.dat (where myfile is whatever you want, but .dat is critical). Remember, to create this you need to use the vi editor. You can either create a file directly while in vi, or go into input mode and paste something from the Mac. Once your input file is complete, type amsolb myfile (don't include the .dat suffix) and it will be submitted to the batch queue to run. You can check the queue by typing qsr-that will show you everything that is running presently (for more info about what everything means, you could type man qstat; qsr is aliased in your .cshrc to mean qstat -R.) When you don't see your job anymore, it is complete (although that doesn't necessarily mean it was successful!) If everything worked, you will have a myfile.out and myfile.arc file with the results of the run found therein (the .arc file is very condensed compared to the .out file). Note that the amsolb command limits your time in the queue to only about 3 minutes. None of the problems assigned should take that long, so if you run out of time, there is probably something wrong (a very bad geometry, for instance).

I. You can look at the output files by using more myfile.out, in which case it will scroll one page at a time, or by using vi, in which case you can use all of the editor commands to move around (be careful not to change the file, though).

J. When you log out, be careful to clean up old files you don't need (for instance, running AMSOL in batch creates files myfile.com*, myfile.qe, and myfile.qo-these can all be deleted). To delete a file, type rm filename. Be careful, though, once you rm a file, it's gone forever!

K. Your account has about 30 minutes of allocation (you can view your allocation with the msiquery command). Don't waste time! You need it to last you for the next problem set too! On any day, you get 5 minutes of interactive time (for editing, using Xmol, etc.), so don't try staying logged in all day, as you'll eventually get booted.

The Problems: (note that there is an appendix to simplify the compilation of some answers-written comments should still be provided separately in several instances)

1. Use AM1 and PM3 to calculate the "heats of formation" of the following molecules: F2, Cl2, ClF, and ClF3. (i) Experimental heats of formation are available for the first three molecules; locate the data and decide which Hamiltonian is better for these calculations. Explain your answer. (ii) An alternative method to evaluate the utility of the methods is to calculate the enthalpy of reaction for F2 + Cl2 t 2 ClF. Based on this approach, which Hamiltonian is better and why? (iii) So, armed with this analysis and any other data you can find, comment on your ClF3 calculations.

2. DNA replication must proceed with extremely high fidelity in order to minimize the possibly disastrous effects of random base-pair mismatches (i.e., mutations). At left below is the standard Watson-Crick CG base pair. One possible source of mutations during DNA replication is matching to tautomeric forms of the bases that are present at equilibrium. For instance, the imino tautomer of cytosine can form a strongly hydrogen-bonded reverse-Watson-Crick pair with "normal" C (at right below). The frequency of such point mutations is presumably related to the relative population of imino-C compared to C during replication. Using AM1, calculate the relative energies of these two tautomers of cytosine (take R as a methyl group in order to keep things simple-you do not need to include the other base in this calculation). At 298 K, what is the ratio of normal tautomer to imino tautomer (assuming AM1 energies to be free energies)? To be viable, organisms need roughly part per million to part per billion fidelity in their DNA replication. Should you be panicking? Why or why not? Please attach your .arc file for each tautomer to your completed problem set.


                   Normal C_G base pair                           Abnormal Im C_C Base Pair

3. The Diels Alder reaction of chlorocyclopentadiene (ccp) and acrylonitrile (an) is shown below. The cycloaddition may proceed to give two regioisomeric products (for simplicity, we're going to consider only these products of endo cycloaddition, and ignore the two exo possibilities).


                                                            "ortho"                  "meta"
                                                           cycloadduct             cycloadduct

Assuming the reaction were allowed to proceed to equilibrium in the products (so-called "thermodynamic control"), what would be the ratio of ortho and meta cycloadducts? (Again, assume AM1 energies to be free energies.) If the reaction is stopped prior to any equilibration of cycloadducts (so-called "kinetic control") what would be the corresponding ratio? [Hopefully you are immediately aware that you will need to know the transition state energies for the two possible cycloadditions. These TS structures are not entirely trivial to locate, since they involve the simultaneous making of two bonds and there is no symmetry to take advantage of. To assist you, I have provided below an input file with Z-matrix for the optimized TS structure leading to meta cycloadduct. It is trivial to turn this into an excellent starting guess for the ortho TS geometry. Note the use of a dummy atom in the representation of the nitrile functional group. This is done because the program will stop if a bond angle of greater than 180° is reached, which can easily happen when a geometry optimization on a molecule with a nearly collinear three-atom system is proceeding. You will probably want to use this trick for acrylonitrile too (vide infra).]




pm3 tstate cycles=200
orthots.dat

  C     0.000000  0    0.000000  0    0.000000  0   0  0  0      -0.1579
  C     1.407420  1    0.000000  0    0.000000  0   1  0  0      -0.1165
  C     1.407388  1  108.222809  1    0.000000  0   2  1  0      -0.1337
  C     1.401980  1  108.878497  1    0.655392  1   3  2  1      -0.0794
  C     1.516806  1  107.421165  1  -19.325044  1   1  2  3      -0.0679
  C     2.111649  1   98.126062  1   74.935865  1   1  2  3      -0.1074
  C     2.192134  1   97.219388  1  -73.671052  1   4  3  2      -0.0540
  H     1.090130  1  125.771391  1  176.630449  1   4  3  2       0.1145
  C     1.424320  1  100.983452  1  -55.362859  1   7  4  3      -0.1074
  H     1.107449  1  110.935221  1  -87.949048  1   5  1  2       0.0900
  Cl    1.672221  1  124.240841  1 -171.008255  1   1  2  3       0.0721
  H     1.089270  1  125.781035  1  165.511688  1   2  1  5       0.1304
  H     1.103272  1  114.899884  1  151.455100  1   5  1  2       0.0835
  H     1.091466  1   91.984616  1  168.985907  1   6  1  2       0.0878
  H     1.089692  1  125.128040  1 -174.154754  1   3  2  1       0.1296
  H     1.089675  1   97.550695  1   55.006343  1   6  1  2       0.0910
  H     1.100839  1   91.146960  1 -170.046145  1   7  4  3       0.1149
  Xx    5.000000  0   90.000000  0  180.000000  0   9  7  4
  N     1.161237  1   89.492365  1 -179.267975  1   9 18  7      -0.0897
   0    0.000000  0    0.000000  0    0.000000  0   0  0  0

Oversimplified, but nevertheless often useful, models have been developed to predict regioselectivity in Diels Alder reactions like this one. In particular, one begins by examining the highest occupied and lowest unoccupied orbitals on the diene and dienophile (i.e., HOMOs and LUMOs, which are nearly always the relevant pi orbitals in these simple cases). One interaction occurs with a smaller energy separation than the other, and is hence the most relevant for a frontier orbital analysis. For a "standard" Diels Alder reaction, it is the HOMO-diene/LUMO-dienophile interaction (thus standard Diels Alder reactions are accelerated by pi donors on the diene (raises HOMO energy) and pi acceptors on the dienophile (lowers LUMO energy)). For an "inverse-electron-demand" Diels Alder, it is the HOMO-dienophile/LUMO-diene separation that is smallest. What are the HOMO and LUMO energies of ccp and an? (You will need to include the VECTORS keyword in your input file in order to have the molecular orbitals and their energies printed.) Is this a standard Diels Alder reaction or an inverse-electron-demand? Once you have the relevant HOMO and LUMO, look at the orbital coefficients. Since the molecules are planar, and the pi orbitals use only the out of plane p basis functions, all of the coefficients in the HOMO and LUMO should be zero (or very close to zero) except for the coefficients for one kind of p function (probably pz). List those coefficients for each atom in ccp and an. The simplified models suggest that regioselectivity can be predicted by comparing coefficients and matching the largest coefficient for a diene terminus to the largest for the dienophile vinyl unit (and, obviously, the smallest to the smallest). Does this work in this case?

Appendix:

1.

Molecule
AM1 DHf (kcal/mol)
PM3 DHf (kcal/mol)
Expt. DHf (kcal/mol)
F2    
Cl2    
ClF    
Reaction
AM1 DH (kcal/mol)
PM3 DH (kcal/mol)
Expt. DH (kcal/mol)
F2 + Cl2 --> 2 ClF    
Molecule
AM1 DHf (kcal/mol)
PM3 DHf (kcal/mol)
Expt. DHf (kcal/mol)
ClF3   
N/A

Structure of CF3:

2.

Tautomer
AM1 DHf (kcal/mol)
Rel. energy (kcal/mol)
% of equilibrium at 298 K (assuming no other tautomers)
Cytosine    
Iminocytosine    


3.

Structure
AM1 DHf (kcal/mol)
HOMO energy (eV)
LUMO energy (eV)
acrylonitrile    
chlorocyclopentadiene    
Structure
AM1 DHf (kcal/mol)
Rel. energy (kcal/mol)
% of equilibrium at 298 K (assuming no other cycloadducts)
ortho cycloadduct    
meta cycloadduct    
Structure
AM1 DHf (kcal/mol)
Activation barrier (kcal/mol)
% of corresponding product at 298 K (assuming kinetic control)
ortho TS structure    
meta TS structure    
     


For the smallest HOMO-LUMO gap (label which orbital corresponds to which molecule) list the pi coefficients (note that absolute sign is arbitrary-only relative sign matters. That is, it doesn't matter if your largest coefficient is negative on the acrylonitrile vinyl unit but positive on a chlorocyclopentadiene terminus, those are still the two centers that would be bonded to predict regioselectivity using simple models).

Reaction is standard/inverse-electron-demand Diels Alder (circle one).