Question

obtained by chlorination of n-butane, will be

• A. meso form
• B. racemic mixture
• C. d-form
• D. I-form

Answer 1

Answer: (B)

Chlorination of $n$-butane takes place via free radical formation i.e., $Cl _{2} \xrightarrow{ h v} Cl ^{\bullet}+ Cl ^{\bullet}$
$CH _{3}- CH _{2}- CH _{2}- CH _{3} \xrightarrow{ Cl _{2} / hv }$

$Cl ^{\bullet}$ may attack on either side and give a racemic mixture of $2$-chlorobutane which contain $50 \% d$ form and $50 \% l$-form.