question_answer

A compound\[A\to {{C}_{5}}{{H}_{10}}C{{l}_{2}}\]on hydrolysis gives \[{{C}_{5}}{{H}_{10}}O\]which reacts with\[NH{{  }_{2}}OH,\]forms iodoform, but does not give Fehling test. A is:

A)  \[C{{H}_{3}}-\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\overset{\begin{smallmatrix}  Cl \\  | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}\]

B)  \[C{{H}_{3}}-C{{H}_{2}}-\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\overset{\begin{smallmatrix}  Cl \\  | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}-C{{H}_{3}}\]

C)  \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-CH{{  }_{2}}-\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\overset{\begin{smallmatrix}  Cl \\  | \end{smallmatrix}}{\mathop{CH}}}\,\]

D)  \[C{{H}_{3}}-\overset{\begin{smallmatrix}  Cl \\  | \end{smallmatrix}}{\mathop{CH}}\,-\overset{\begin{smallmatrix}  Cl \\  | \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}-C{{H}_{3}}\]

Correct Answer: A

Solution :

                 A compound\[({{C}_{5}}{{H}_{10}}C{{l}_{2}})\]on hydrolysis gives \[{{C}_{5}}{{H}_{10}}O\]which reacts with hydroxylamine, so \[{{C}_{5}}{{H}_{10}}O\]must be aldehydic or ketonic group but it does not give Fehling test, so it must be a ketone and it form iodoform which is a characteristic test of\[C{{H}_{3}}CO-\]group. \[\underset{\begin{smallmatrix}  ({{C}_{5}}{{H}_{10}}C{{l}_{2}}) \\  \,\,\,\,\,\,\,\,(A) \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\overset{\begin{smallmatrix}  Cl \\  | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}}}\,-C{{H}_{2}}-C{{H}_{3}}\xrightarrow[{}]{~hydrolysis}\]                                \[\underset{(unstable)}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\overset{\begin{smallmatrix}  Cl \\  | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}}}\,-C{{H}_{2}}-C{{H}_{3}}\]