In the reaction $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{CS}_{2}…

In the reaction $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{CS}_{2} \frac{\mathrm{HgCl}_{2}}{\Delta} \longrightarrow$ the product obtained is
  1. phenyl isocyanide
  2. phenyl cyanide
  3. $\mathrm{p}$ -amino benzene
  4. phenyl isothiocyanate sulphonic acid

Solution

$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{CS}_{2} ightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH} . \mathrm{CS} \cdot \mathrm{SH} \stackrel{\mathrm{HgCl}_{2}}{\longrightarrow}$
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}=\mathrm{C}=\mathrm{S}+\mathrm{HgS}+2 \mathrm{HCl}$
The reaction is called mustard oil reaction.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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