Read the passage carefully and answer the Questions.
Amines are classified as primary (1°), secondary (2°) and tertiary (3°) depending upon the number of hydrogen atoms replaced by alkyl or aryl groups in ammonia molecules. Aliphatic amines are a stronger bases due to +I effect of an alkyl groups. Intermolecular hydrogen bonding in primary and secondary amines affects the boiling points of amines. The intermolecular association is more in primary amines than in secondary amines as there are more number of hydrogen atoms available for hydrogen bond formation. Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum and also by reduction with metals in an acidic medium. Gabriel synthesis is used for the preparation of primary amines. Primary amines react with nitrous acid to give unstable diazonium-salts which can be used to synthesize a variety of compounds.
Identify the amine possessing least basic strength
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → p-Nitroaniline
An amine is a base because of the lone pair on nitrogen; the more available this lone pair, the stronger the base. In aniline the lone pair is partly delocalised into the benzene ring by resonance, so aniline is a much weaker base than ammonia. A group at the para position changes this further:
An electron-donating group (\( \ce{-OCH3} \) by resonance, \( \ce{-CH3} \) by hyperconjugation and +I effect) pushes electron density towards nitrogen and makes the amine a stronger base than aniline.
An electron-withdrawing group such as \( \ce{-NO2} \) pulls the lone pair further into the ring by its −R and −I effects and makes the amine a weaker base than aniline.
The order of basic strength is therefore:
\( \displaystyle p\text{-methoxyaniline} \gt p\text{-toluidine} \gt \text{aniline} \gt p\text{-nitroaniline} \)
So the least basic amine is p-nitroaniline.
Why the other options are incorrect
Option (1) Aniline — Incorrect. Its lone pair is delocalised into the ring, but no nitro group withdraws it further, so it is more basic than p-nitroaniline.
Option (3) p-Toluidine — Incorrect. The methyl group releases electrons, so p-toluidine is slightly more basic than aniline.
Option (4) p-Methoxyaniline — Incorrect. The methoxy group donates electrons by resonance; this is the strongest base of the four.
NCERT (2026-27 edition):
Basic Character of Amines – “… We know that greater the number of resonating structures, greater is the stability. Thus you can infer that aniline (five resonating structures) is more stable than anilinium ion. Hence, the proton acceptability or the basic nature of aniline or other aromatic amines would be less than that of ammonia. In case of substituted aniline, it is observed that electron releasing groups like –OCH3, –CH3 increase basic strength whereas electron withdrawing groups like –NO2, –SO3H, –COOH, –X decrease it.”