question_answer

Which of the following sequence of reactions (reagents) can be used for the conversion of \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}C{{H}_{3}}\] into \[{{C}_{6}}{{H}_{5}}CH=C{{H}_{2}}\]?

A)  \[SOC{{l}_{2}};{{H}_{2}}O\]

B)  \[SOC{{l}_{2}};alc.KOH\]

C)  \[C{{l}_{2}}/hv;{{H}_{2}}O\]

D)  \[{{C}_{6}}{{H}_{5}}OH\]

Correct Answer: B

Solution :

In this reaction \[S{{O}_{2}}C{{l}_{2}}\] causes free radical substitution at benzylic position and ale. \[KOH\] causes elimination.