A) \[{{H}_{2}}/Ni\]
B) \[NaB{{H}_{4}}\]
C) \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}^{+}}\]
D) both and
Correct Answer: B
Solution :
\[\text{NaB}{{\text{H}}_{\text{4}}}\]and \[\text{LiAl}{{\text{H}}_{\text{4}}}\]attacks only carbonyl group and reduce it into alcohol group. They do not attack on double bond. \[\underset{\text{Cinnamic}\,\text{aldehyde}}{\mathop{{{C}_{6}}{{H}_{5}}-CH=CHCHO}}\,\xrightarrow{\text{NaB}{{\text{H}}_{\text{4}}}}\] \[\underset{\text{Cinnamic}\,\text{alcohol}}{\mathop{{{C}_{6}}{{H}_{5}}=CH.C{{H}_{2}}OH}}\,\]
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