Question

(i) ${\mu }_{obs}=\sum _i{\mu }_i{x}_i$, where ${\mu }_i$ is the dipole moment of a stable conformer of the molecule $Z-C{H}_2-C{H}_2-Z$ and $x_i$ is the mole fraction of the stable conformer
Given: ${\mu }_{obs}=1.0D$ and $x\left(Anti\right)$ $=0.82$
Draw all the stable conformers of $Z-C{H}_2-C{H}_2-Z$ and calculate the value of $\mu \left(Gauche\right)$.

(ii) Draw the stable conformer of $Y-CHD-CHD-Y$ (mesoform).
When $Y=C{H}_3$ (rotation about $C_2-C_3$) $Y=OH$ (rotation about $C_1-C_2$) in Newmann projection.

Answer 1

The compound $Z-C{H}_2-C{H}_2-Z$ will have just two stable conformers i.e. Gauche and anti forms.
(Gauche form) (Anti form)
Mole fraction of anti conformer of $Z-C{H}_2-C{H}_2-Z$, x(Anti) $=0.82$
Therefore, mole fraction of Gauche form, x(Gauche) $=1-0.82=0.18$
The antiform of $Z-C{H}_2-C{H}_2-Z$ will have zero dipole moment as its individual bond dipole moments will be cancelled out by one another.
${\mu }_{obs}=\mu \left(Gauche\right).x\left(Gauche\right)+\mu \left(Anti\right).x\left(Anti\right)$
$1=\mu \left(Gauche\right)\times 0.18+0\times 0.82$
$\mu \left(Gauche\right)=\frac{1}{0.18}=5.56D$
The stable conformer of $Y-CHD--CHD-Y$ (mesoform) when $Y-C{H}_3$ and rotated about $C_2-C_3$.
The stable conformer of $Y-CHD--CHD-Y$ (mesoform) when $Y=OH$ and rotated about $C_1-C_2$.