A) 2
B) 4
C) 6
D) 8
Correct Answer: B
Solution :
\[C{{H}_{3}}-C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}\xrightarrow[{}]{C{{l}_{2}}}\] \[C{{H}_{3}}-C{{H}_{2}}-\underset{(I)}{\mathop{\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,-}}\,C{{H}_{2}}Cl\] \[\underset{(II)}{\mathop{+C{{H}_{3}}-C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{3}}}}\,\] \[\underset{(III)}{\mathop{C{{H}_{3}}-\underset{*}{\mathop{\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{C}}\,}}\,H-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}}}\,\] \[\underset{(IV)}{\mathop{+\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,-C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}}}\,\] I and III have chirality*. Thus, in all four (including enantiomers) optical isomers are obtained.
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