Question

On chlorination, an equimolar mixture of ethane and neopentane yields neopentyl chloride and ethyl chloride in the ratio $2.3:1$. The ratio of the reactivity of $1^{\circ }$ hydrogen in neopentane to that of a $1^{\circ }$ hydrogen in ethane is $y/100$. Calculate $z$ if $z=\frac{y}{57.5}$.

Answer 1

Number of primary hydrogen in ethane = 6

Number of primary hydrogen in neopentane = 12

$\frac{Yieldofneopentylchloride}{Yieldofethylchloride}=\frac{No.of{1}^{o}H\times Reactivityof{1}^{o}Hinneopentane}{No.of{1}^{o}H\times Reactivityof{1}^{o}Hinethane}$

$\frac{2.3}{1}=\frac{12\times Reactivityinneopentane}{6\times Reactivityinethane}$

$\therefore y=115$

So, $z=2$ .