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(C) The iodoform test is given by compounds containing the $CH_{3}CO-$ group or the $CH_{3}CH(OH)-$ group. $CH_{3}CH(OH)CH_{2}CH_{3}$ is a secondary a

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$CH_{3}COOH$

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(C) The iodoform test is given by compounds containing the $CH_{3}CO-$ group or the $CH_{3}CH(OH)-$ group. $CH_{3}CH(OH)CH_{2}CH_{3}$ is a secondary alcohol with a $CH_{3}CH(OH)-$ group,so it responds to the test. $ICH_{2}COCH_{2}CH_{3}$ contains a ketone group,but it lacks the $CH_{3}CO-$ moiety; however,in the presence of base,it can undergo reactions,but $CH_{3}COOH$ (acetic acid) is a carboxylic acid. In acetic acid,the most acidic proton is the one attached to the oxygen atom $(COOH)$. The haloform reaction requires the deprotonation of an $\alpha$-hydrogen atom from a methyl ketone group $(CH_{3}CO-)$. Since $CH_{3}COOH$ does not contain a methyl ketone group,it does not undergo the iodoform reaction.

Amongst the following compounds,the one which would not respond to iodoform test is

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