$CH_3-CH_2-CH(Cl)-CH_3$ obtained by chlorination of n-butane, will be : -
A
Meso form
B
Racemic mixture
C
d-form
D
l-form
The answer is B.
Solution
Chlorination of n-butane at C-2 goes by a free radical mechanism: $CH_3CH_2CH_2CH_3 \rightarrow CH_3CH_2\dot{C}HCH_3 \rightarrow CH_3CH_2CHClCH_3$ The intermediate secondary free radical is $sp^2$ hybridised and planar. The chlorine molecule can attack this planar radical equally from both faces, giving the two enantiomers of 2-chlorobutane in equal amounts. So the product is a racemic mixture (optically inactive). It cannot be a meso form, since 2-chlorobutane has only one chiral centre.
ConceptRacemic product through planar radical
Haloalkanes and Haloarenes in past papers
55 questions from this chapter have appeared across 25 exam years.
