Cyclohexane z-matrix

The chair conformation of cyclohexane contains a symmetry axis (S6), and perpendicular to it, a number of C2 axes. These axes lie in the 'mean' plane of the carbon skeleton, which disects the carbon-carbon bonds.
The main difference between cyclohexane and cyclopropane is the fact that this plane is not a mirror plane. Therefore, there are more independent degrees of freedom.

One way to place dummy atoms is to take the points where the S6 axis crosses the two planes, parallel to the 'mean' plane, defined by two sets of three C atoms:
the C1-C3-C5 plane and the C2-C4-C6 plane.
The distance between XX1 and XX2 is the variable which determines the puckering of the ring, i.e. the distance of each C to the 'mean' plane.

The degrees of freedom (variables) are:

In the input file below, one such a scheme is implemented.
There are six variables, of which two deserve an explanation: the hydrogen angles.
For the axial ones: H9-C3-XX1 is close to 90 degrees, but not exactly 90 by definition.
The equatorial H15 angle is the second variable. In combination these two angles determine the H-C-H angle and the H-C-C-H torsion angles.
All torsion angles in the z-matrix are constants!
 
 AM1 T=3600 SYMMETRY NOINTER NOXYZ
 z-matrix for cyclohexane
 with symmetry
XX   0.00 0   0.00 0   0.00 0  0  0  0		two dummy atoms
XX   1.00 1   0.00 0   0.00 0  1  0  0
 C   1.40 1  90.00 0   0.00 0  1  2  0		the carbon skeleton
 C   1.40 0  90.00 0 120.00 0  1  2  3		  three around XX1
 C   1.40 0  90.00 0 240.00 0  1  2  3
 C   1.40 0  90.00 0  60.00 0  2  1  5		  three around XX2
 C   1.40 0  90.00 0 120.00 0  2  1  6
 C   1.40 0  90.00 0 240.00 0  2  1  6
 H   1.10 1  90.00 1 180.00 0  3  1  2		three axial H's
 H   1.10 0  90.00 0 180.00 0  4  1  2		  on XX1 side
 H   1.10 0  90.00 0 180.00 0  5  1  2
 H   1.10 0  90.00 0 180.00 0  6  2  1		  three on XX2 side
 H   1.10 0  90.00 0 180.00 0  7  2  1
 H   1.10 0  90.00 0 180.00 0  8  2  1
 H   1.10 1 160.00 1   0.00 0  3  1  2		six equatorial H's
 H   1.10 0 160.00 0   0.00 0  4  1  2
 H   1.10 0 160.00 0   0.00 0  5  1  2
 H   1.10 0 160.00 0   0.00 0  6  2  1
 H   1.10 0 160.00 0   0.00 0  7  2  1 
 H   1.10 0 160.00 0   0.00 0  8  2  1
 0   0.00
 3  1  4  5  6  7  8				symmetry lines
 9  1  10  11  12  13  14 
 9  2  10  11  12  13  14 
15  1  16  17  18  19  20
15  2  16  17  18  19  20
The result, together with a discussion, can be found on a separate page.

If you have developed an alternative way of applying symmetry in the z-matrix, please let me know!

You can to me at borkent@cmbi.kun.nl