Some Basic Principles Of Organic Chemistry Question 153
Question: How many enantiomer pairs are obtained by monochlorination of 2, 3-dimethylbutane [Kerala PMT 2004]
Options:
A) Nil
B) Four
C) Two
D) Three
E) One
Show Answer
Answer:
Correct Answer: E
Solution:
$ \underset{\text{2, 3-dimethyl butane}}{\mathop{CH_3-\underset{CH_3\ }{\mathop{\underset{|}{\mathop{C}},H-}},\underset{CH_3\ }{\mathop{\underset{|}{\mathop{C}},H-}},CH_3+Cl_2}},\to $
$ \underset{\text{2, 3-dimethyl chloro butane}}{\mathop{CH_3-\underset{CH_3\ }{\mathop{\underset{|}{\mathop{C}},H-}},\underset{CH_3\ }{\mathop{\underset{|}{\overset{\bullet }{\mathop{C}}},H-}},CH_2Cl}}, $ Due to the presence of chiral carbon it shows the optical activity and its mirror image are non-superimposable so it shows one enantiomer pair.
$ \begin{matrix} \underset{(I)}{\mathop{CH_3-\underset{CH_3\ }{\mathop{\underset{|}{\mathop{C}},H-}}\underset{CH_3\ }{\mathop{\underset{|}{\mathop{C}},H-}},CH_2Cl}}, & \underset{(II)}{\mathop{ClCH_2-\underset{CH_3\ }{\mathop{\underset{|}{\mathop{C}},H-}},\underset{CH_3\ }{\mathop{\underset{|}{\mathop{C}},H-}},CH_3}}, \\ \end{matrix} $
