A) LAH
B) \[HI+P\]
C) \[NaAl{{H}_{4}}\]
D) \[{{B}_{2}}{{H}_{6}}/{{H}_{2}}O\]
Correct Answer: D
Solution :
[d] 1. LAH reduces both ester and acid to alcohols. 2. \[HI+P\]reduces both ester and acid to alkane. 3. \[NaAl{{H}_{4}}\] reduces only ester to aldehyde and alcohol. 4. \[{{B}_{2}}{{H}_{6}}/{{H}_{2}}O\] selectively reduces acid to alcohol (although it can also reduce ester to alcohols) but acid group is reduced in the presence of ester.You need to login to perform this action.
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