question_answer

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

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

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

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

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

Correct Answer: A

Solution :

(i) \[{{C}_{5}}{{H}_{10}}O\] must be aldehyde or ketone because it reacts with \[N{{H}_{2}}OH\]. (ii) It must be aldehyde as it does not give Fehling test. (iii) It must have \[C{{H}_{2}}CO\] group because it gives iodoform test. \[{{H}_{3}}C-\overset{\begin{smallmatrix}  Cl \\  | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\mathop{C}}\,}}\,-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}\xrightarrow{hydrolysis}\]            \[\underset{(A)}{\mathop{{{C}_{5}}{{H}_{10}}C{{l}_{2}}}}\,\]