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.You need to login to perform this action.
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