A) 3
B) 1
C) 4
D) 2
Correct Answer: D
Solution :
\[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{2}}C{{H}_{3}}\xrightarrow{C{{l}_{2}}/hv}\underset{*\,\,2\,\,enantiomeric\,\,pairs}{\mathop{\underset{C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-\overset{*}{\mathop{\underset{Cl}{\mathop{\underset{|}{\mathop{C}}\,}}\,}}\,H-C{{H}_{3}}\,\,(d,\ell )}{\mathop{\underset{+}{\mathop{C{{H}_{3}}-\underset{C{{H}_{2}}-Cl}{\mathop{\underset{|}{\mathop{\overset{*}{\mathop{C}}\,}}\,}}\,H-C{{H}_{2}}C{{H}_{3}}\,\,(d,\ell )}}\,}}\,}}\,\]
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