Q4 / 180 · COMEDK 2025 (10 May Shift 2) · Chemistry
Choose the correct statement:
- Correct answerAAnilinium ion is less resonance stabilised than Aniline.
- BAliphatic amines are weaker bases than $\mathrm{NH}_3$ while aromatic amines are stronger bases than $\mathrm{NH}_3$.
- CSec-butylamine is optically inactive because Nitrogen atom of the $-\mathrm{NH}_2$ group is achiral.
- DGabriel phthalimide synthesis is used for preparing both Ethyl amine and Aniline.
Solution
(1) Correct: In aniline ($C_6H_5NH_2$), the lone pair on N is delocalized into the benzene ring, forming multiple resonance structures. In the anilinium ion ($C_6H_5NH_3^+$), the lone pair is donated to a proton and can no longer participate in resonance. Hence anilinium ion is less resonance stabilised than aniline.
(2) Incorrect: Aliphatic amines are stronger bases than $NH_3$ due to the $+I$ effect of alkyl groups. Aromatic amines are weaker bases than $NH_3$ because their lone pair is delocalized into the benzene ring.
(3) Incorrect: sec-Butylamine ($CH_3CH_2CH(NH_2)CH_3$) has a chiral carbon bonded to four different groups ($-H$, $-CH_3$, $-CH_2CH_3$, $-NH_2$), making it optically active.
(4) Incorrect: Gabriel phthalimide synthesis involves $S_N2$ reaction with alkyl halides. Aryl halides do not undergo $S_N2$, so aniline cannot be prepared by this method.
The correct statement is (1).
(2) Incorrect: Aliphatic amines are stronger bases than $NH_3$ due to the $+I$ effect of alkyl groups. Aromatic amines are weaker bases than $NH_3$ because their lone pair is delocalized into the benzene ring.
(3) Incorrect: sec-Butylamine ($CH_3CH_2CH(NH_2)CH_3$) has a chiral carbon bonded to four different groups ($-H$, $-CH_3$, $-CH_2CH_3$, $-NH_2$), making it optically active.
(4) Incorrect: Gabriel phthalimide synthesis involves $S_N2$ reaction with alkyl halides. Aryl halides do not undergo $S_N2$, so aniline cannot be prepared by this method.
The correct statement is (1).
Hence, the correct answer is option: 'A'.