Given below are two statements:Statement I: H | vedclass

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(D) Statement $I$ is false: The reaction of benzamide with $Br_2$ and $NaOH$ (Hofmann bromamide degradation) produces aniline,not benzylamine.Statemen

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Statement $I$ is false but Statement $II$ is true

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(D) Statement $I$ is false: The reaction of benzamide with $Br_2$ and $NaOH$ (Hofmann bromamide degradation) produces aniline,not benzylamine.Statement $II$ is true: In the strong acidic medium of nitration $(H_2SO_4)$,aniline gets protonated to form anilinium ion $(-NH_3^+)$,which is meta-directing. Hence,$m$-nitroaniline is formed in a significant amount due to the deactivating and meta-directing nature of the $-NH_3^+$ group.

Given below are two statements:
Statement $I$: Heating benzamide with bromine in an ethanolic solution of sodium hydroxide will give benzylamine.
Statement $II$: Nitration of aniline with $HNO_3/H_2SO_4$ at $288 \ K$ produces $m$-nitroaniline in higher amount than $o$-nitroaniline (pH adjusted).

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Given below are two statements:Statement $I$: Heating benzamide with bromine in an ethanolic solution of sodium hydroxide will give benzylamine.Statement $II$: Nitration of aniline with $HNO_3/H_2SO_4$ at $288 \ K$ produces $m$-nitroaniline in higher amount than $o$-nitroaniline (pH adjusted).