MCQ
${{C}_{6}}{{H}_{5}}-CH=CHCHO\xrightarrow{X}{{C}_{6}}{{H}_{5}}CH=CHC{{H}_{2}}OH$. In the above sequence $ X$ can be
- A${H_2}/Ni$
- ✓$NaB{H_4}$
- C${K_2}C{r_2}{O_7}/{H^ + }$
- DBoth $(a)$ and $(b)$
✓
Answer
Correct option: B.
$NaB{H_4}$
b
(b) $NaB{H_4}$ and $LiAl{H_4}$ attacks only carbonyl group and reduce it into alcohol group. They do not attack on double bond.
(b) $NaB{H_4}$ and $LiAl{H_4}$ attacks only carbonyl group and reduce it into alcohol group. They do not attack on double bond.
$\underset{\text{cinnamic aldehyde}}{\mathop{{{C}_{6}}{{H}_{5}}-CH=CHCHO}}\,\xrightarrow{NaB{{H}_{4}}}$ $\underset{\text{cinnamic alcohol}}{\mathop{{{C}_{6}}{{H}_{5}}-CH=CH.C{{H}_{2}}OH}}\,$
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