MCQ
Appropriate reducing agent for the following conversion is-

$\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\ 
  {C{H_2} = CH - C{H_2} - C - H} 
\end{array} \to C{H_3} - C{H_2} - C{H_2} - C{H_2}OH$

  • A
    $\operatorname{LiAlH}_{4 /} H _{2} O$
  • B
    $NaBH _{4} / H _{2} O$
  • C
    $Na + C _{2} H _{5} OH$
  • $B_{2} H_{6} / H ^{+}$

Answer

Correct option: D.
$B_{2} H_{6} / H ^{+}$
d
$\operatorname{LiAlH}_{4} / H _{2} O , NaBH _{4} / H _{2} O$ and $Na / C _{2} H _{5} OH$ will reduce only $- CHO ,$ and not $C \equiv C$ bond.

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