Question

Write Clemmenson's reduction reaction of propanone.

Answer 1

(i) Reducing the carbonyl group \[\begin{array}{cc}\backslash\phantom{........}\\\ce{C = O}\\/\phantom{........}\end{array}\] with zinc amalgam (Zn – Hg) in concentrated hydrochloric acid results in the formation of the methylene group (–CH₂–).

\[\begin{array}{cc}
\backslash\phantom{........................................}\\
\ce{C = O + 4[H] ->[Zn - Hg][conc. HCl] -CH2- + H2O}\\
/\phantom{.........................................}
\end{array}\]

(ii) Acetaldehyde on reduction with Zn – Hg in concentrated HCl forms ethane.

\[\begin{array}{cc}
\ce{O}\phantom{................}\\
//\phantom{...................}\\
\ce{CH3 - C + 4[H] ->[Zn - Hg][conc. HCl] \underset{Ethane}{C2H6} + H2O}\\
\backslash\phantom{....................}\\
\ce{H}\phantom{..................}
\end{array}\]

(iii) Acetone on reduction with Zn – Hg in concentrated HCl forms propane.

\[\begin{array}{cc}
\ce{O}\phantom{..........................................}\\
||\phantom{..........................................}\\
\ce{\underset{Acetone}{CH3 - C - CH3} + 4[H] ->[Zn - Hg][conc. HCl] \underset{Propane}{CH3 - CH2 - CH3} + H2O}
\end{array}\]