How many of the following compounds cannot decolourise $KMnO_{4}$ solution?
(a) $HC \equiv CH$ (b) $CH_{3} CH_{2} OH$ (c) $CH_{3} CHO$
(d) $CH_{3} COOH$ (e) $CH_{3} - O - CH_{3}$ (f) $CH_{3} - CH_{3}$
(g) $CH_{3} COOCH_{3}$ (h) $PhCH = CH_{2}$

Question

How many of the following compounds cannot decolourise $KMnO_{4}$ solution?

(a) $HC \equiv CH$ (b) $CH_{3} CH_{2} OH$ (c) $CH_{3} CHO$

(d) $CH_{3} COOH$ (e) $CH_{3} - O - CH_{3}$ (f) $CH_{3} - CH_{3}$

(g) $CH_{3} COOCH_{3}$ (h) $PhCH = CH_{2}$

TestHub · Accepted Answer

Question Explanation:

Identify the number of compounds from the list (a) to (h) that cannot decolorize KMnO₄ solution.

Concept:

This is based on KMnO₄ oxidation of Alkenes, Alkynes, Alcohols and Aldehydes.

Solution:

KMnO₄ is a strong oxidizing agent that is decolorized by compounds that are easily oxidized,

specifically: alkenes/alkynes (oxidized to diols/carboxylic acids), alcohols (oxidized to

aldehydes/ketones/carboxylic acids), and aldehydes (oxidized to carboxylic acids)

The compounds that cannot decolorize KMnO₄ are carboxylic acid (d), ether (e), alkane (f), and ester (g).

The total number is 4.

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