question_answer

\[{{H}_{2}}C=CH-C{{H}_{2}}-C{{H}_{2}}-\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}\] \[\xrightarrow[Pyridine]{SOC{{l}_{2}}}(A)\xrightarrow[({{H}_{2}}O)]{{{O}_{3}}/Zn}\underset{{{C}_{5}}{{H}_{9}}ClO}{\mathop{(B)}}\,\xrightarrow{NaB{{H}_{4}}}(C)\] Compound (C) is

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

B) \[HOC{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-Cl\]

C) \[HO-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-\overset{Cl}{\mathop{\overset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}\]

D) \[HO-C{{H}_{2}}-C{{H}_{2}}-\underset{Cl}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{2}}-C{{H}_{3}}\]

Correct Answer: C

Solution :

[c] \[{{H}_{2}}C=CH-C{{H}_{2}}-C{{H}_{2}}-\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}\xrightarrow[Pyridine]{SOC{{l}_{2}}}\]