Answer to question in chapter 2, paragraph A.

The "improper" torsion angle Cl-C-H-Br with fixed value zero keeps the four atoms in one plane.
This helps (not guarantees) to maintain the mirror plane present in 2-bromopropane during the reaction, and to prevent elimination to occur.
More symmetry can be added, by forcing the C4 and C5 angles to be equal.
In such a case symmetry is inserted in between the z-matrix and the reaction path data, separated by a blank or zero line:
...SYMMETRY in the keyword line
.
.
.
H	1.1     1  	110.0   1   300.0 1  4  2  5
0 
4 2 5        (set angle for C5 equal to angle C4)
0
3.0 2.9  2.8  2.7  2.6  2.5  2.4  2.3  2.2  2.1  2.0  1.9  1.8  1.7
Actually, the Cl - C - Br angle should better not be used in the z-matrix, as the danger exists that it comes close to 180 degrees. Maybe not in this compound, but in case of three identical substituents to the central carbon, it will. This problem can be solved either by finding a different path for the definition of Cl ( e.g. Cl - C - C(sub) - Br), or by placing a dummy atom in a strategic position.
The latter strategy is followed by Jeffrey Gosper in his examples.
Make a drawing to see where the dummy atom is placed!

AM1 SYMMETRY T=3600 NOINTER CHARGE=-1
Sn2 reaction at a methyl carbon
C-Cl distance decreased
Cl      0.0     0       0.0     0       0.0     0       0       0       0
C       4.0    -1       0.0     0       0.0     0       1       0       0
XX      1.0     0       90.0    0       0.0     0       2       1       0
Br      1.9     1       90.0    0       180.0   0       2       3       1
H       1.0     1       71.0    1       0.0     0       2       1       3
H       1.0     0       71.0    0       120.0   0       2       1       3
H       1.0     0       71.0    0       240.0   0       2       1       3
0       0.0     0       0.0     0       0.0     0       0       0       0
5 1 6 7
5 2 6 7

3.5 3.0 2.9 2.8 2.7 2.6 2.5 2.4 2.3 2.2 2.15 2.1 2.05 2.0 1.95 1.9 1.85 1.8 
1.75

Back to text of 2A.