Question

What would be the product when ethene is oxidised with cold dil. KMnO₄ solution?

Options

• \[\begin{array}{cc}
  \ce{CH2-CH2}\\
  \phantom{}|\phantom{......}|\phantom{...}\\
  \phantom{}\ce{OH}\phantom{...}\ce{OH}\phantom{.}
  \end{array}\]

• \[\begin{array}{cc}
  \ce{H - C - H}\\
  ||\\
  \ce{O}
  \end{array}\]

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

• \[\ce{CO2 + H2O}\]

Answer 1

\[\begin{array}{cc}
\ce{CH2-CH2}\\
\phantom{}|\phantom{......}|\phantom{...}\\
\phantom{}\ce{OH}\phantom{...}\ce{OH}\phantom{.}
\end{array}\]

Explanation:

\[\begin{array}{cc}
\phantom{................}\ce{CH2-CH2}\phantom{}\\
\ce{CH2 = CH2 ->[cold dil][KMnO4] \phantom{}|\phantom{......}|\phantom{}}\phantom{.......}\\
\phantom{................}\ce{OH}\phantom{...}\ce{OH}\phantom{.}
\end{array}\]