Question

An alkyl iodide on refluxing with aqueous KOH solution gave isopropyl alcohol. The structure of alkyl iodide could be:

Options

• \[\ce{CH3 - CH2 - CH2 - I}\]

• \[\begin{array}{cc}
  \ce{CH3 - CH - CH3}\\
  |\phantom{..}\\
  \ce{I}\phantom{..}
  \end{array}\]

• \[\ce{CH3 - CH2 - CH2 - CH2 - I}\]

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

Answer 1

\[\begin{array}{cc}
\ce{CH3 - CH - CH3}\\
|\phantom{..}\\
\ce{I}\phantom{..}
\end{array}\]

Explanation:

Hydrolysis of an alkyl halide with aqueous KOH yields the corresponding alcohol. A primary alkyl halide becomes primary alcohol, while a secondary alkyl halide becomes secondary alcohol. As a result, when isopropyl alcohol is produced by alkaline hydrolysis of an alkyl iodide, the alkyl iodide must be isopropyl iodide.