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
${C_6}{H_5}CONHC{H_3} \,\mathop {\xrightarrow{{Zn - Hg/}}}\limits_{HCl} \,{C_6}{H_5}C{H_2}NHC{H_3}$
${C_6}{H_5}CONHC{H_3} \,\mathop {\xrightarrow{{Zn - Hg/}}}\limits_{HCl} \,{C_6}{H_5}C{H_2}NHC{H_3}$
This reaction is known as Clemmenson reduction.
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