• # question_answer The number of possible enantiomeric pairs that can be produced during mono-chlorination of 2-methylbutane is [Pb. CET 2004] A) 3 B) 4 C) 1 D) 2

$\overset{1}{\mathop{C}}\,{{H}_{3}}-\underset{C{{H}_{3}}\ }{\mathop{\underset{|}{\overset{2}{\mathop{C}}}\,H-}}\,\overset{3}{\mathop{C}}\,{{H}_{2}}-\overset{4}{\mathop{C}}\,{{H}_{3}}$ Its monochloro derivatives are as follows : (i) $ClC{{H}_{2}}-\underset{C{{H}_{3}}\ \,}{\mathop{\underset{|}{\overset{\bullet }{\mathop{C}}}\,H-}}\,C{{H}_{2}}-C{{H}_{3}}$  or $C{{H}_{3}}-\underset{C{{H}_{2}}Cl\ \ \ \ \ \,}{\mathop{\underset{|}{\overset{\bullet }{\mathop{C}}}\,H-C{{H}_{2}}}}\,-C{{H}_{3}}$ It will exist as enantiomeric pair (d and l-forms) (ii) $C{{H}_{3}}-\underset{C{{H}_{3}}\ \ \ \ \,}{\mathop{\overset{Cl\ }{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,-}}\,C{{H}_{2}}}}\,-C{{H}_{3}}$ no asymmetric C atom (iii) $C{{H}_{3}}-\underset{C{{H}_{3}}\,}{\mathop{\underset{|}{\mathop{C}}\,H-}}\,\overset{Cl\ }{\mathop{\overset{|}{\mathop{C}}\,H}}\,-C{{H}_{3}}$ It will exist as enantiomeric pair (d- and l-forms) (iv) $C{{H}_{3}}-\underset{C{{H}_{3}}\ }{\mathop{\underset{|}{\mathop{C}}\,H-}}\,C{{H}_{2}}-C{{H}_{2}}-Cl$ No asymmetric carbon atom Hence, only two enantiomeric pairs will be obtained by the monochlorination of 2-methylbutane.