A) \[{{H}_{3}}C-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}CH{{l}_{3}}CH-C{{H}_{2}}-C{{H}_{3}}\]
B) \[{{H}_{2}}C=\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{2}}OHC{{H}_{3}}l{{H}_{3}}C-C{{H}_{2}}-C{{H}_{2}}-OH\]
C) \[{{H}_{3}}C-C{{H}_{2}}-CHO\,\,\,\,\,CH{{l}_{3}}\,\,\,\,\,{{H}_{3}}C-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{3}}\]
D) \[{{H}_{3}}C-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}\,\,\,\,\,CH{{l}_{3}}\,\,\,\,\,\,{{H}_{3}}C-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{3}}\]
Correct Answer: A
Solution :
The compound reacts with \[{{l}_{2}}/NaOH\], so it must be a methyl ketone and it should give iodoform \[(CH{{l}_{3}})\] in this reaction. \[\therefore A\]is \[C{{H}_{3}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,C{{H}_{3}}\](acetone) \[C{{H}_{3}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{\underset{A}{\mathop{C}}\,}}\,-C{{H}_{3}}\xrightarrow{{{l}_{2}}/NaOH}\underset{\begin{smallmatrix} B \\ yellow\text{ }ppt. \end{smallmatrix}}{\mathop{C{{H}_{3}}}}\,\] \[+C{{H}_{3}}COONa+Nal+{{H}_{2}}O\] \[C{{H}_{3}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{\underset{A}{\mathop{C}}\,}}\,C{{H}_{3}}\xrightarrow[\begin{smallmatrix} (Clemmensens \\ reduction) \end{smallmatrix}]{Zn-Hg/HCl}\underset{\begin{smallmatrix} propane \\ C \end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{3}}}}\,\]
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