Amines MCQs for NEET — Chemistry Questions with Answers

The reaction sequence is as follows:

1. $CH_3CH_2CH_2NH_2$ (Propylamine) reacts with $HONO$ (Nitrous acid) to form $CH_3CH_2CH_2OH$ (n-Propanol).
2. n-Propanol reacts with $PCl_5$ (Phosphorus pentachloride) to form $CH_3CH_2CH_2Cl$ (n-Propyl chloride).
3. n-Propyl chloride reacts with $KCN$ (Potassium cyanide) to form $CH_3CH_2CH_2CN$ (n-Propyl cyanide).
4. n-Propyl cyanide is reduced by $Na/C_2H_5OH$ (Sodium in ethanol) to form $CH_3CH_2CH_2CH_2NH_2$ (n-Butylamine). So, the correct option is $[A] = CH_3CH_2CH_2OH, [B] = CH_3CH_2CH_2Cl, [D] = CH_3CH_2CH_2CH_2NH_2$.

Aniline ($C_6H_5NH_2$) on oxidation with $Na_2Cr_2O_7$ and $H_2SO_4$ forms p-benzoquinone ($C_6H_4O_2$). The reaction involves the oxidation of the amino group to a quinone structure. Hence, the correct answer is p-benzoquinone.

Hinsberg's reagent is benzene sulphonyl chloride ($C_6H_5SO_2Cl$). It is used to distinguish between primary, secondary, and tertiary amines. Primary amines form a soluble sulfonamide, secondary amines form an insoluble sulfonamide, and tertiary amines do not react. Hence, the correct answer is benzene sulphonyl chloride.

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.