Question

Which of the following alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature?

Options

• \[\ce{CH3CH2 - CH2 - OH}\]

• \[\begin{array}{cc}
  \phantom{}\ce{CH3CH2 - CH - OH}\\ \phantom{...}\phantom{}|\\
  \phantom{......}\ce{CH3}
  \end{array}\]

• \[\begin{array}{cc}
  \ce{CH3CH2-CH-CH2OH}\\
  |\phantom{..}\\
  \phantom{.}\ce{CH3}
  \end{array}\]

• \[\begin{array}{cc}
  \phantom{........}\ce{CH3}\\
  \phantom{.....}\phantom{}|\\
  \phantom{}\ce{CH3CH2 - C - OH}\\
  \phantom{.....}\phantom{}|\\ \phantom{........}\ce{CH3}
  \end{array}\]

Answer 1

\[\begin{array}{cc}
\phantom{........}\ce{CH3}\\
\phantom{.....}\phantom{}|\\
\phantom{}\ce{CH3CH2 - C - OH}\\
\phantom{.....}\phantom{}|\\ \phantom{........}\ce{CH3}
\end{array}\]

Explanation:

As tertiary carbocation is the most stable, so tertiary alcohols are most reactive towards cone. HCl. Hence, the reaction can be conducted at room temperature only, while primary and secondary alcohols require the presence of a catalyst ZnCl₂.