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. \[\underset{Cinnamic\text{ }aldehyde}{\mathop{{{C}_{6}}{{H}_{5}}-CH=CHCHO}}\,\xrightarrow[{}]{NaB{{H}_{4}}}\] \[\underset{Cinnamic\text{ }alcohol}{\mathop{{{C}_{6}}{{H}_{5}}-CH=CH.C{{H}_{2}}OH}}\,\]
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