Isoniazid is used in the treatment of tubercu | vedclass

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(B) In Isoniazid,there are three nitrogen atoms labeled as $(I)$,$(II)$,and $(III)$.$(I)$ is a pyridine nitrogen,which is $sp^2$ hybridized and its lo

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$II$

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(B) In Isoniazid,there are three nitrogen atoms labeled as $(I)$,$(II)$,and $(III)$.$(I)$ is a pyridine nitrogen,which is $sp^2$ hybridized and its lone pair is not involved in resonance,making it moderately basic.$(II)$ is an amide nitrogen $(-CONH-)$. Its lone pair is involved in resonance with the adjacent carbonyl group $(C=O)$,which significantly reduces its basicity.$(III)$ is an amine nitrogen $(-NH_2)$ attached to the amide nitrogen. It is more basic than the amide nitrogen $(II)$ due to the lack of direct resonance with the carbonyl group.Comparing these,the amide nitrogen $(II)$ is the least basic because its lone pair is delocalized into the carbonyl group,making it the least available for protonation. Therefore,the correct option is $B$.

Isoniazid is used in the treatment of tuberculosis. Identify the least basic $N$ atom present in it.

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