Question

$$ \mu $$ observed = $$ \sum \mu_iX_i $$
where $$ \mu_i $$ is the dipole moment of the stable conformer and $$ X_i $$ is the mole fraction of that conformer.
Write the stable conformer for $$ (Z - CH_2 - CH - Z) $$ in Newman's projection.
If $$ \mu_{solution} = 1.0\,D $$ and mole fraction of the antiform $$ = 0.82 $$ , find the $$ \mu $$ gauche form .

Answer 1

Mole fraction of anti-form $$ = 0.82 $$
Mole fraction of gauche form $$ = (1 - 0.82) = 0.18 $$
$$ \mu_{obs} = 1.0 $$
$$ \mu_{obs} = \mu_1 X_1 + \mu_2X_2 $$
$$ 1.0 = \mu (anti) \times 0.82 + \mu(gauche) \times 0.18 $$
$$ \mu(anti) = 0 $$ (from Newman's projection , two Z vectors cancel and all H vectors cancel)
$$ 1.0 = 0 \times 0.82 + \mu (gauche) \times 0.18 $$
$$ \mu(gauche) = \dfrac{1.0}{0.18} = 5.55\,D $$