Chemistry MCQs for NEET — Practice Questions with Answers

The conversion of benzenamide to acetanilide involves two steps: first, the Hofmann bromamide reaction using $KOH/Br_2$ to convert benzenamide to aniline, and then acetylation of aniline using $CH_3COCl$ to form acetanilide.

It has chiral carbon So, it is optically active

$ 3 ^\circ $ amines do not react with acetyl chloride because they do not have replaceable H atom.

Benzylamine is stronger base because the lone pair on N atom is not
delocalised over the benzene ring

The given reaction sequence involves (1) nucleophilic substitution of chloroethane with $NaCN$ to form ethyl cyanide ($CH_3CH_2CN$), (2) hydrogenation with $Ni/H_2$ to form propylamine ($CH_3CH_2CH_2NH_2$), and (3) acetylation with $ (CH_3CO)_2O $ to form N-propylacetamide ($CH_3CH_2CH_2NHCOCH_3$).

Benzylamine is the strongest base among the given options. In benzylamine, the electron-donating nature of the benzyl group increases the electron density on the nitrogen atom, making it more basic. In contrast, the other compounds have electron-withdrawing groups or are less effective in increasing the electron density on the nitrogen atom.

Aromatic nitriles (ArCN) are not prepared by the reaction of ArX with KCN. This is because KCN is not efficient enough to substitute the halide group (X) in aromatic halides. The other options involve reactions that can effectively produce aromatic nitriles.

The presence of a nitro group (-NO2) on a benzene ring deactivates the ring towards electrophilic substitution. This is because the nitro group is an electron-withdrawing group due to its strong -I (inductive effect) and -M (mesomeric effect), which reduces the electron density on the benzene ring, making it less reactive towards electrophiles.