A) \[{{H}_{2}}O,{{H}_{2}}S{{O}_{4}}\]
B) Aqueous KOH
C) \[MgS{{O}_{4}},NaB{{H}_{4}}/{{H}_{2}}O\]
D) \[{{B}_{2}}{{H}_{6}},{{H}_{2}}{{O}_{2}},O{{H}^{-}}\]
Correct Answer: B
Solution :
\[\underset{\operatorname{Propene}-1}{\mathop{C{{H}_{3}}-CH=C{{H}_{2}}+aq.KOH}}\,=\xrightarrow{{}}\] \[\underset{\operatorname{Propanol}-1}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}OH}}\,\]
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