Question

But-2-ene on reaction with alkaline KMnO₄ at elevated temperature followed by acidification will given:
1. \(\begin{array}{*{20}{c}} {{\rm{C}}{{\rm{H}}_3} - {\rm{CH}} - {\rm{CH}} - {\rm{C}}{{\rm{H}}_3}}\\ {{\rm{\;}}\left| {\rm{\;}} \right|{\rm{\;}}}\\ {{\rm{\;OH\;OH}}} \end{array}\)
2. One molecule of CH₃CHO and one molecule of CH₃COOH
3. 2 molecules of CH₃COOH
4. 2 molecules of CH₃CHO

Answer 1

Correct Answer - Option 3 : 2 molecules of CH₃COOH

Concept:

When 2-Butane reacts with the alkaline KMnO₄ (Potassium Manganate) at elevated temperature followed by acidification will give 2 moles of CH₃COOH (Acetic acid).

\({\rm{C}}{{\rm{H}}_3} - {\rm{CH}} = {\rm{CH}} - {\rm{C}}{{\rm{H}}_3}\mathop {\xrightarrow{{{\rm{\;KMN}}{{\rm{O}}_4}} {\rm{\;}}} }\limits_{\frac{{{{\rm{H}}^ + }}}{\Delta }} 2{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}\)

Thus, option (c) is the correct answer.