JEE Mains · Chemistry · STD 12 - 8.2 Carboxylic acids and their derivative
But \(-2-\) ene on reaction with alkaline \(KMnO_4\) at elevated temperature followed by acidification will give
- A One molecule of \(CH_3CHO\) and one molecule of \(CH_3COOH\)
- B \(2\) molecules of \(CH_3CHO\)
- C \(2\) molecules of \(CH_3COOH\)
- D \(\begin{array}{*{20}{c}}
{C{H_3} - CH - CH - C{H_3}}\\
{\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,}\\
{OH\,\,\,\,\,\,\,\,OH}
\end{array}\)
Answer & Solution
Correct Answer
(C) \(2\) molecules of \(CH_3COOH\)
Step-by-step Solution
Detailed explanation
\(C{H_3} - CH = CH - C{H_3}\xrightarrow[\Delta ]{{KMn{O_4}}}\) \(Images\) \( \to \,2C{H_3} - CHO\,\xrightarrow{{[O]}}2C{H_3} - COOH\)
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