A) \[{{H}_{2}}/Ni\]
B) \[NaE{{H}_{4}}\]
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. \[\underset{cinnamic\text{ }aldehyde}{\mathop{{{C}_{6}}{{H}_{5}}-CH=CHCHO}}\,\xrightarrow{NaB{{H}_{4}}}\] \[\underset{cinnamic\text{ }alcohol}{\mathop{{{C}_{6}}{{H}_{5}}-CH=CH.\,C{{H}_{2}}OH}}\,\]
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