A) \[\underset{\begin{smallmatrix} \ \ \ \ \ \ \ \ \ \,| \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{C}_{2}}{{H}_{5}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \end{smallmatrix}}{\mathop{C{{H}_{3}}-C=O\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\,\]
B) \[\overset{\,\,\,\,\,\,\,\overset{\,\,\,\,\,\,\,\,\,{{C}_{2}}{{H}_{5}}}{\mathop{|}}\,}{\mathop{\underset{\underset{\,\,\,\,\,\,\,\,\,\,\,\,{{C}_{2}}{{H}_{5}}}{\mathop{\,\,\,\,\,\,\,|}}\,}{\mathop{C{{H}_{3}}-C-OH}}\,}}\,\]
C) \[\underset{\underset{\,\,\,\,\,\,\,\,\,\,C{{H}_{3}}}{\mathop{\,\,\,\,\,\,\,|}}\,}{\mathop{C{{H}_{3}}-C=O}}\,\]
D) \[\overset{\overset{\,\,\,\,\,\,\,\,\,\,\,\,\,{{C}_{2}}{{H}_{5}}}{\mathop{\,\,\,\,\,\,\,\,\,|}}\,}{\mathop{\underset{\underset{\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}}}{\mathop{\,\,\,\,\,\,\,|}}\,}{\mathop{C{{H}_{3}}-C=O}}\,}}\,\]
Correct Answer: B
Solution :
\[C{{H}_{3}}-\overset{\overset{O}{\mathop{|\ |}}\,}{\mathop{\underset{Ester}{\mathop{C}}\,}}\,-O\,{{C}_{2}}{{H}_{5}}+{{C}_{2}}{{H}_{5}}MgBr\to \]\[\underset{\,\underset{{{C}_{2}}{{H}_{5}}}{\mathop{|\,\,\,\,\,\,\,\,\,}}\,}{\mathop{\overset{\overset{\,\,\,\,\,\,O\,{{C}_{2}}{{H}_{5}}}{\mathop{|\,\,\,\,\,\,\,\,\,}}\,}{\mathop{C{{H}_{3}}-C-O\,MgBr}}\,}}\,\xrightarrow{-Mg(O{{C}_{2}}{{H}_{5}})Br}\underset{\underset{\,\,\,\,\,\,\,\,\,\,\,\,\,{{C}_{2}}{{H}_{5}}}{\mathop{\,\,\,\,\,\,\,\,|}}\,}{\mathop{C{{H}_{3}}-C=O}}\,\] \[\xrightarrow{{{C}_{2}}{{H}_{5}}Mg\,Br}\underset{\underset{{{C}_{2}}{{H}_{5}}}{\mathop{|\,\,\,\,\,\,\,\,\,\,\,}}\,}{\mathop{\overset{\overset{{{C}_{2}}{{H}_{5}}}{\mathop{|\,\,\,\,\,\,\,\,\,\,\,}}\,}{\mathop{C{{H}_{3}}-C-O\,Mg\,Br}}\,}}\,\underset{H.OH}{\mathop{\xrightarrow{-Mg(OH)Br}}}\,\]\[\underset{\underset{\,\,\,\,\,\,\,\,\,\,{{C}_{2}}{{H}_{5}}}{\mathop{\,\,\,|}}\,}{\mathop{\overset{\overset{\,\,\,\,\,\,\,\,\,\,{{C}_{2}}{{H}_{5}}}{\mathop{\,\,|}}\,}{\mathop{{{H}_{3}}C-C-OH}}\,}}\,\] 3° alcohol
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