Question

Propionic acid with Br₂/P yields a dibromo product. Its structure would be:

विकल्प

• CH₂Br,CHBr,COOH

• \[\begin{array}{cc}
  \phantom{}\ce{Br}\\
  \phantom{}|\\
  \phantom{...............}\ce{\underset{}{H - C - CH2COOH}}\\
  \phantom{}|\\
  \phantom{}\ce{Br}\\
  \end{array}\]

• CH₂Br,CH₂,COBr

• \[\begin{array}{cc}
  \phantom{}\ce{Br}\\
  \phantom{}|\\
  \phantom{....}\ce{\underset{}{CH3 - C - COOH}}\\
  \phantom{}|\\
  \phantom{}\ce{Br}\\
  \end{array}\]

Answer 1

\[\begin{array}{cc}
\phantom{}\ce{Br}\\
\phantom{}|\\
\phantom{....}\ce{\underset{}{CH3 - C - COOH}}\\
\phantom{}|\\
\phantom{}\ce{Br}\\
\end{array}\]

Explanation:

\[\begin{array}{cc}
\ce{H\phantom{.......................................}Br}\phantom{..}\\
\phantom{.......}|\phantom{.........................................}{}|\phantom{...........}\\
\ce{CH3 - C - COOH->[Br2/P][-2HBr]CH3 - C - COOH}\\
\phantom{.......}|\phantom{.........................................}{}|\phantom{...........}\\
\ce{H\phantom{.......................................}Br}\phantom{..}\\
\end{array}\]

This reaction is called as Hell-Volhard-Zelinsky (HVZ) reduction.