Explanation

(d)
[image:10_image102.emf]
Its monochloro derivatives are follows
[image:10_image103.emf]
It will exist as enantiomers pair $d$ and $l$-forms
[image:10_image104.emf]
no asymmetric C atom
Cl
|
(iii) $CH_{3} - CH - CH - CH_{3}$
|
$CH_{3}$
It will exist as enantiomeric pair ($d$-and $l$- forms)
[image:10_image105.emf]
No asymmetric carbon atom
Hence, only two enantiomeric pairs will be obtained by the monochlorination of 2-methylbutane.