A) \[C{{H}_{3}}COCl\xrightarrow{LiAl{{H}_{4}}}\]
B) \[{{C}_{2}}{{H}_{5}}CHO\underset{{{H}^{+}}/{{H}_{2}}O}{\mathop{\xrightarrow{C{{H}_{3}}MgBr}}}\,\]
C) \[{{(C{{H}_{3}})}_{2}}CH{{C}_{2}}{{H}_{5}}\xrightarrow{Cu}\]
D)
\[\begin{matrix} H \\ {} \\ C{{H}_{3}} \\ \end{matrix}\,\,\,\,\,\,C=C\,\,\,\,\,\,\,\,\begin{matrix} C{{H}_{3}} \\ {} \\ C{{H}_{3}} \\ \end{matrix}\xrightarrow{C{{l}_{2}}}\]
Correct Answer: B
Solution :
\[{{C}_{2}}{{H}_{5}}CHO\underset{{{H}^{+}}/{{H}_{2}}O}{\mathop{\xrightarrow{C{{H}_{3}}MgBr}}}\,\,\underset{\,\,\,\,\,\,C{{H}_{3}}}{\mathop{\underset{|}{\mathop{\overset{\,H}{\mathop{\overset{|}{\mathop{{{C}_{2}}{{H}_{5}}-{{C}^{*}}-OH}}\,}}\,}}\,}}\,\] \[{{C}^{*}}\]-chiral carbon as all the four valencies are attached with different substituents or groups.
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