question_answer

                 chlorination of n-butane, will be:                                                          

A)                 meso form

B)                 racemic mixture

C)                 d-form

D)                 -form

Correct Answer: B

Solution :

                Chlorination of n-butane takes place by free radical mechanism as -                 Step I: \[C{{l}_{2}}\xrightarrow{\,}\,\overset{\centerdot }{\mathop{Cl}}\,+\overset{\centerdot }{\mathop{Cl}}\,\]                 Step II: \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}+\,\overset{\centerdot }{\mathop{Cl}}\,\xrightarrow{\,}\]                 Step III: \[C{{H}_{3}}\overset{\centerdot }{\mathop{CH}}\,C{{H}_{2}}C{{H}_{3}}+C{{l}_{2}}\to \]