A) \[{{H}_{2}}/pt\]
B) \[Zn-Hg/HCl\]
C) \[Li/N{{H}_{3}}\]
D) \[NaB{{H}_{4}}\]
Correct Answer: D
Solution :
\[NaB{{H}_{4}}\] is a selective reducing agent, it reduces \[C=O\] group to alcohol. It does not reduce olefin (alkene) double bond. |
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