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