H Solutions to Problems
255
R B =−
πνμ 2
2
R 12 sin α
This result predicts a normal CD sign, with a lower negative branch (B-state)
and an upper positive branch (A-state) [16]. This is a typical right-handed helix,
corresponding to a rotation of the dipoles in the right-handed sense when going
from chromophore 1 to chromophore 2 along the inter-chromophore axis. In
the S-conformation the sign of α will change, and the CD spectrum will be
inverted.
6.5 The direct square of the e-irrep in D 2d yields four coupled states:
e × e = A 1 + A 2 + B 1 + B 2
(10)
The corresponding coupling coefficients are given in the table below. This table
is almost the same as the table for D 4 in Appendix F, but note that B 1 and
B 2 are interchanged. Such details are important, and therefore we draw again a
simple picture of the molecule in a Cartesian system. Both in D 4 and in D 2d ,
the B 1 and B 2 irreps are distinguished by their symmetry with respect to the ˆ
C ′
2
axes.
D 2d
E × EA 1
A 2
B 1
B 2
a 1
a 2
b 1
b 2
xx
1
√
2
00
−
1
√
2
yy
1
√
2
00
1
√
2
xy
0
1
√
2
1
√
2
0
yx
0
−
1
√
2
1
√
2
0
In the orientation of twisted ethylene, as indicated in the figure below, the directions of these axes are along the bisectors of x and y. In contrast, in the standard
orientation for D 4 they are along the x and y axes, while the bisector directions
coincide with the ˆ
C ′′
2 axes, and hence the interchange between B 1 and B 2 .
255
R B =−
πνμ 2
2
R 12 sin α
This result predicts a normal CD sign, with a lower negative branch (B-state)
and an upper positive branch (A-state) [16]. This is a typical right-handed helix,
corresponding to a rotation of the dipoles in the right-handed sense when going
from chromophore 1 to chromophore 2 along the inter-chromophore axis. In
the S-conformation the sign of α will change, and the CD spectrum will be
inverted.
6.5 The direct square of the e-irrep in D 2d yields four coupled states:
e × e = A 1 + A 2 + B 1 + B 2
(10)
The corresponding coupling coefficients are given in the table below. This table
is almost the same as the table for D 4 in Appendix F, but note that B 1 and
B 2 are interchanged. Such details are important, and therefore we draw again a
simple picture of the molecule in a Cartesian system. Both in D 4 and in D 2d ,
the B 1 and B 2 irreps are distinguished by their symmetry with respect to the ˆ
C ′
2
axes.
D 2d
E × EA 1
A 2
B 1
B 2
a 1
a 2
b 1
b 2
xx
1
√
2
00
−
1
√
2
yy
1
√
2
00
1
√
2
xy
0
1
√
2
1
√
2
0
yx
0
−
1
√
2
1
√
2
0
In the orientation of twisted ethylene, as indicated in the figure below, the directions of these axes are along the bisectors of x and y. In contrast, in the standard
orientation for D 4 they are along the x and y axes, while the bisector directions
coincide with the ˆ
C ′′
2 axes, and hence the interchange between B 1 and B 2 .