A) \[NaB{{H}_{4}}\]
B) \[{{H}_{2}}-Pd/C\]
C) \[LiAl{{H}_{4}}\]
D) \[Zn-Hg/HCl\]
Correct Answer: D
Solution :
Amide can be converted to amine by reduction with \[Zn/Hg/HCl\]. \[{{C}_{6}}{{H}_{5}}CONHC{{H}_{3}}\xrightarrow{Zn-Hg/HCl}\] \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}NHC{{H}_{3}}\] \[Amide\xrightarrow[or\,LiAl{{H}_{4}}]{Na/{{C}_{2}}{{H}_{5}}OH}\text{ }amine.\] \[Amide+B{{r}_{2}}+KOH\xrightarrow{343K}p-amine\] (Hermanns bromamide reaction)You need to login to perform this action.
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