The treatment of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}$ with chlorine in the presence of phosphorus…

The treatment of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}$ with chlorine in the presence of phosphorus gives
  1. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCl}$
  2. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}$
  3. $\mathrm{CH}_{3} \mathrm{CH}(\mathrm{Cl}) \mathrm{COOH}$
  4. $\mathrm{CH}_{2}(\mathrm{Cl}) \mathrm{CH}_{2} \mathrm{COOH}$

Solution

Treatment with chlorine and a catalytic amount of phosphorus leads to the selective $\alpha$-chlorination of carboxylic acids ($\mathrm{HVZ}$ reaction).
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}+\mathrm{Cl}_{2} \stackrel{\mathrm{RedP}}{ightarrow} \mathrm{CH}_{3} \mathrm{CHCl}-\mathrm{COOH}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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