Exercise

Outline: in this exercise we will start a search for the Transition State (TS) in the Diels-Alder reaction between cyclopentadiene and ethene. Such a TS can serve as a basic skeleton, a starting point for a comparison of substituent effects in these reactants.
We use SYBYL to make the input file, and deliberately in such a way that the file has to be changed, for the sake of exercise.
See bottom remark for the faster way to handle this problem.

To find the TS, the most convenient starting point is the product. Norbornene contains the required connectivity.

Start SYBYL, Build/Edit menu, Sketch molecule and start drawing the skeleton in the following order:

	    C3
       C2	 C4

       C1	 C5
    	    C6
Then 'draw' C6 - C7 - C3 and an extra bond C1 - C2. Rotate the molecule (right mouse button) backwards and 'move' C7 from the center of the ring to a position above the plane.
ADDH hydrogens and leave the sketch menu (End select).
Minimize the structure (Compute, Minimize) using the default settings.
Under Compute, Interfaces, MOPAC, change the default settings to Optimization/Full and enter a name (nbn in the example); then Submit.
Very soon the 'job completed' message will appear in the Sybyl command window, and nine files with the name 'nbn' will be produced.
The .sta file is what MOPAC itself calls the .arc file. We continue with this file.

We can extract a new input file with z-matrix with the tail command:

tail -21 nbn.sta >nbnts.dat
The new input file, nbnts.dat, has to be prepared for a line search, in which we change the two bonds being broken simultaneously, using symmetry. In this examples it concerns the bonds C4 to C3 and C6 to C5. The file may look like this:
 AM1 T=3600
     ~nbnts.dat
 
  C    0.0000000  0      0.000000  0      0.000000  0    0    0    0     -0.1743
  C    1.3538998  1      0.000000  0      0.000000  0    1    0    0     -0.1743
  C    1.5248008  1    107.766748  1      0.000000  0    2    1    0     -0.0976
  C    1.5499905  1    105.449786  1     69.630685  1    3    2    1     -0.1580
  C    1.5427329  1    103.657767  1    -66.887830  1    4    3    2     -0.1581
  C    1.5222543  1    107.551308  1      0.143545  1    1    2    3     -0.0971
  C    1.5621438  1     99.733989  1    -33.852277  1    3    2    1     -0.1595
  H    1.0861020  1    129.482989  1   -178.708430  1    1    2    3      0.1410
  H    1.0862646  1    129.425064  1    178.685897  1    2    1    3      0.1404
  H    1.1029015  1    116.352041  1   -160.916774  1    3    2    1      0.1113
  H    1.1151317  1    110.249425  1    174.052113  1    4    3    2      0.0818
  H    1.1141312  1    111.384349  1     53.271707  1    4    3    2      0.0864
  H    1.1154219  1    111.089216  1   -118.954228  1    5    4    3      0.0818
  H    1.1147620  1    111.601982  1    119.993238  1    5    4    3      0.0857
  H    1.1028913  1    116.350181  1    160.601318  1    6    1    2      0.1111
  H    1.1101467  1    113.155065  1    167.540416  1    7    3    2      0.0875
  H    1.1100645  1    113.045659  1    -66.447383  1    7    3    2      0.0920

C4 - C3 is present in the file, and can be marked with a '-1', but the definition for C6 has to be changed, to (C6 -) C5 - C4 - C3.
In this case the values for the bond, angle and torsion can be maintained, they are more or less correct! Only the connectivity has to be changed to:
  C    1.5222543  0    107.551308  1      0.143545  1    5    4    3     -0.0971
At the end of the file first the symmetry relation is added (don't forget the keyword SYMMETRY in the first line!), then the reaction path:
 .
  H    1.1100645  1    113.045659  1    -66.447383  1    7    3    2      0.0920

 4  1  6

 1.6  1.7  1.8  1.9  2.0  2.1  2.2  2.3  2.4  2.5  3.0  4.0  5.0

This file can be submitted to MOPAC. In chapter 2 the result will be discussed.


There is a faster solution than the exercise described above.
Back to paragraph 1E.