Some Basic Principles Of Organic Chemistry Question 364
Question: The number of possible enantiomeric pairs that can be produced during mono-chlorination of 2-methylbutane is [Pb. CET 2004]
Options:
A) 3
B) 4
C) 1
D) 2
Show Answer
Answer:
Correct Answer: D
Solution:
$ \overset{1}{\mathop{C}},H_3-\underset{CH_3\ }{\mathop{\underset{|}{\overset{2}{\mathop{C}}},H-}},\overset{3}{\mathop{C}},H_2-\overset{4}{\mathop{C}},H_3 $ Its monochloro derivatives are as follows : (i) $ ClCH_2-\underset{CH_3\ ,}{\mathop{\underset{|}{\overset{\bullet }{\mathop{C}}},H-}},CH_2-CH_3 $ or $ CH_3-\underset{CH_2Cl\ \ \ \ \ ,}{\mathop{\underset{|}{\overset{\bullet }{\mathop{C}}},H-CH_2}},-CH_3 $ It will exist as enantiomeric pair (d and l-forms) (ii) $ CH_3-\underset{CH_3\ \ \ \ ,}{\mathop{\overset{Cl\ }{\mathop{\underset{|}{\overset{|}{\mathop{C}}},-}},CH_2}},-CH_3 $ no asymmetric C atom (iii) $ CH_3-\underset{CH_3,}{\mathop{\underset{|}{\mathop{C}},H-}},\overset{Cl\ }{\mathop{\overset{|}{\mathop{C}},H}},-CH_3 $ It will exist as enantiomeric pair (d- and l-forms) (iv) $ CH_3-\underset{CH_3\ }{\mathop{\underset{|}{\mathop{C}},H-}},CH_2-CH_2-Cl $ No asymmetric carbon atom Hence, only two enantiomeric pairs will be obtained by the monochlorination of 2-methylbutane.
