Question

The percentage of n-propyl chloride obtained in the chlorination of propane is about 56%.

• A. True
• B. False

Answer 1

The correct option is B False

The chlorination of propane gives two isomers products 1- chloropropane and 2-chloropropane. Their relative ratio would be given as :
$\frac{1-Chloropropane}{2-Chloropropane}=\frac{No.of1H}{No.of2H}\times \frac{Reactivityof1H}{Reactivityof2H}=\frac{6}{2}\times \frac{1.0}{3.8}=\frac{6.0}{7.6}$
Percentage of $n-$ isomers = $\frac{6.0}{6.0+7.6}\times 100=$ 44%