Q27 / 120 · VITEEE 2019 · Chemistry

Starting from propanoic acid, the following reactions were carried out Propanoic acid $\stackrel{\mathrm{SOCl}_{2}}{\longrightarrow} \mathrm{X} \stackrel{\mathrm{NH}_{3}}{\longrightarrow} \mathrm{Y} \stackrel{\mathrm{Br}_{2}+\mathrm{KOH}}{\longrightarrow} \mathrm{Z}$ What is the compound $Z$ ?

  • A
    $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{Br}$
  • Correct answer
    B
    $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$
  • C
    Figure
  • D
    $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$

Solution

$$ \begin{array} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \stackrel{\mathrm{SOCl}_{2}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCl} \\ +\mathrm{SO}_{2}+\mathrm{HCl} \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCl}+\mathrm{NH}_{3} \rightarrow \mathrm{CH}_{3} \underset{(\mathrm{Y})}{\mathrm{CH}_{2} \mathrm{CONH}_{2}}+\mathrm{HCl} \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CONH}_{2}+\mathrm{Br}_{2} / \mathrm{NaOH} \rightarrow \mathrm{CH}_{3} \underset{\mathrm{Ethylamine, (Z)}}{\mathrm{CH}_{2} \mathrm{NH}_{2}+\mathrm{CO}_{2}} \end{array} $$

Hence, the correct answer is option: 'B'.

Relevant Questions