MCQ
${C_6}{H_5}CONHC{H_3}$ can be converted into ${C_6}{H_5}C{H_2}NHC{H_3}$ by
  • A
    $NaB{H_4}$
  • B
    ${H_2} - Pd/C$
  • C
    $LiAl{H_4}$
  • $Zn - Hg/HCl$

Answer

Correct option: D.
$Zn - Hg/HCl$
d
${C_6}{H_5}CONHC{H_3} \,\mathop {\xrightarrow{{Zn - Hg/}}}\limits_{HCl} \,{C_6}{H_5}C{H_2}NHC{H_3}$

This reaction is known as Clemmenson reduction.

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