A) \[{{H}_{2}}O/{{H}_{2}}S{{O}_{4}}\]
B) \[Hg{{(OAc)}_{2}}/{{H}_{2}}O\]followed by \[NaB{{H}_{4}}\]
C) \[{{B}_{2}}{{H}_{6}}\]followed by \[{{H}_{2}}{{O}_{2}}\]
D) \[C{{H}_{3}}C{{O}_{2}}H/{{H}_{2}}S{{O}_{4}}\]
Correct Answer: C
Solution :
\[6C{{H}_{3}}-CH=C{{H}_{2}}+{{B}_{2}}{{H}_{6}}\xrightarrow{{{H}_{2}}{{O}_{2}}}\]\[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}OH\]
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