Question

Dipole moment of a compound C- CH2 - CH2 - Cl is 1.0 D. If dipole moment of its Gauche form is 5.55 D. What will be
mole fraction of the anti form ? (Hanti = 0) (take 1/5.55 = 0.18)

Answer 1

Explanation:

The compound Z−CH

2

−CH

2

−Z will have just two stable conformers i.e. Gauche and anti forms.

(Gauche form) (Anti form)

Mole fraction of anti conformer of Z−CH

2

−CH

2

−Z, x(Anti) =0.82

Therefore, mole fraction of Gauche form, x(Gauche) =1−0.82=0.18

The antiform of Z−CH

2

−CH

2

−Z will have zero dipole moment as its individual bond dipole moments will be cancelled out by one another.

μ

obs

=μ(Gauche).x(Gauche)+μ(Anti).x(Anti)

1=μ(Gauche)×0.18+0×0.82

μ(Gauche)=

0.18

1

=5.56D

The stable conformer of Y−CHD−−CHD−Y (mesoform) when Y−CH

3

and rotated about C

2

−C

3

.

The stable conformer of Y−CHD−−CHD−Y (mesoform) when Y=OH and rotated about C

1

−C

2

.