A) \[Cl-C{{H}_{2}}C{{H}_{2}}Cl\]
B) \[C{{H}_{3}}C{{H}_{2}}CHC{{l}_{2}}\]
C) \[C{{H}_{3}}C(C{{l}_{2}})C{{H}_{3}}\]
D) \[C{{H}_{2}}C{{l}_{2}}\]
Correct Answer: C
Solution :
\[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{3}}\xrightarrow[{}]{2{{H}_{2}}O}\left[ C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\overset{\begin{smallmatrix} OH \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{3}} \right]\] \[\xrightarrow[-{{H}_{2}}O]{Rearrangement}\underset{acetone}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} || \\ O \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}}}\,\] acetone forms sodium bisulphide addition product but does not give silver mirror with Tollen's reagent.
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