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

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(A) Statement-$I$: Oxidation of $p$-nitrotoluene with acidic $KMnO_4$ yields $p$-nitrobenzoic acid. The $-NO_2$ group is a strong electron-withdrawing

Vedclass · Accepted Answer

Both Statement-$I$ and Statement-$II$ are correct

Vedclass · Answer

(A) Statement-$I$: Oxidation of $p$-nitrotoluene with acidic $KMnO_4$ yields $p$-nitrobenzoic acid. The $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect). The presence of this group on the benzene ring increases the acidity of the carboxylic acid group compared to benzoic acid. Thus, Statement-$I$ is correct.Statement-$II$: Reduction of $p$-nitrotoluene with $Sn/HCl$ followed by neutralization yields $p$-toluidine ($4$-methylaniline). The $-CH_3$ group is an electron-donating group ($+I$ and hyperconjugation effect). Electron-donating groups increase the basicity of the amine by increasing electron density on the nitrogen atom. Therefore, $p$-toluidine is more basic than aniline. Thus, Statement-$II$ is correct.

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