Aniline is a weaker base than ammonia because the lone pair on its nitrogen is pulled into the benzene ring (resonance). So the lone pair is "busy" and less free to grab a proton (H+). Memory hook: in aniline the lone pair is SHARED with the ring, so it is not FREE to act as a base.
In ammonia the nitrogen lone pair is free, so it easily grabs a proton (strong base). In aniline the lone pair is pulled into the benzene ring by resonance, so it is less available and aniline is the weaker base.
Your doubts, answered
What exactly makes a base strong or weak here?
Basicity means how easily nitrogen can donate its lone pair to grab a proton (H+). If the lone pair is free and sits fully on nitrogen, the base is strong. If the lone pair is pulled away or spread out, the base is weak. This one idea decides every basicity question in NEET Amines.
Why does the benzene ring make aniline weaker than ammonia?
In ammonia (NH3) the lone pair sits fully on nitrogen, so it is 100% free to take a proton. In aniline (C6H5NH2) the nitrogen is joined to a benzene ring. The lone pair moves into the ring by resonance (delocalisation). So the lone pair is partly spread over the ring and less available on nitrogen. Less available lone pair means weaker base.
How do resonance structures prove aniline is a weak base?
NCERT says aniline is a resonance hybrid of FIVE structures, because the lone pair spreads into the ring. But the anilinium ion (aniline after taking H+) has only TWO structures. More resonance structures means more stable. Since neutral aniline is MORE stabilised by resonance than its protonated form, aniline 'does not want' to take a proton. That is why it is a weak base.
Is aniline weaker than ammonia AND weaker than methylamine?
Yes, both. Aromatic amines like aniline are weaker than ammonia because the aryl ring withdraws the lone pair. Aliphatic amines like methylamine (CH3NH2) are stronger than ammonia because the +I effect of the alkyl group pushes electron density onto nitrogen. So the order is: methylamine > ammonia > aniline.
Does aniline being a weak base change its reactions in acid?
Yes. Because aniline is still a base, in strong acid it gets protonated to the anilinium ion (C6H5NH3+). The -NH3+ group has no lone pair to donate, is electron withdrawing, and becomes meta directing. This is why nitration of aniline in strong acid also gives some m-nitroaniline. NEET tests this directly.
Is the nitrogen in aniline sp or sp2 or sp3?
The nitrogen in aniline is basically sp3, but slightly flattened toward sp2 so its lone pair can overlap better with the ring. It is NOT sp hybridised. A common trap option says aniline is more basic because N is sp hybridised, which is wrong.
⚠️ The NEET trap ✗ Aryl amines are more basic than alkyl amines because the nitrogen lone pair is NOT delocalized into the ring. ✓ Aryl amines are LESS basic than alkyl amines because the nitrogen lone pair IS delocalized into the aromatic ring pi system, lowering electron density on nitrogen. 🧠 If the lone pair enters the ring, it becomes LESS free, so the base gets WEAKER, never stronger. Delocalisation always lowers aniline's basicity.
Real NEET questions
NEET 2016 Phase 1
The correct statement regarding the basicity of aryl amines is:
A · Aryl amines are generally less basic than alkyl amines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring pi electron system. ✓
B · Aryl amines are generally more basic than alkyl amines because the nitrogen lone-pair electrons are not delocalized by interaction with the aromatic ring pi electron system.
C · Aryl amines are generally more basic than alkyl amines because of aryl group.
D · Aryl amines are generally more basic than alkyl amines, because the nitrogen atom in aryl amines is sp-hybridized.
Solution: In aniline (C6H5NH2) the nitrogen lone pair is in conjugation with the benzene ring and is delocalized over the ortho and para positions. This lowers the electron density on nitrogen, so the lone pair is less free to bind a proton. Hence aryl amines are weaker bases than alkyl amines (where the +I effect of the alkyl group raises electron density on N). The nitrogen in aniline is sp3 (slightly flattened), not sp, so option D is also wrong. Correct answer: A.
NEET 2017
The correct increasing order of basic strength for the following compounds is: (I) Aniline C6H5NH2, (II) p-Nitroaniline (NO2 at para), (III) p-Toluidine (CH3 at para).
A · II < III < I
B · III < I < II
C · III < II < I
D · II < I < III ✓
Solution: Basicity depends on how available the nitrogen lone pair is. In p-nitroaniline the -NO2 group is strongly electron withdrawing (-I and -R); it pulls the lone pair even more, so it is the WEAKEST base (II). Plain aniline (I) already has its lone pair pulled into the ring, so it is in the middle. In p-toluidine the -CH3 group is electron donating (+I), which pushes electron density back to nitrogen, making it the STRONGEST of the three (III). So the increasing order is II < I < III. Correct answer: D.
NEET 2018
Nitration of aniline in strong acidic medium also gives m-nitroaniline because
A · in absence of substituents the nitro group always goes to the m-position.
B · in electrophilic substitution reactions the amino group is meta directing.
C · inspite of substituents the nitro group always goes to only the m-position.
D · in strong acidic medium aniline is present as the anilinium ion. ✓
Solution: Because aniline is still a base, in strong acid it gets protonated to the anilinium ion (C6H5NH3+). The -NH3+ group carries a full positive charge, has no lone pair to donate, is electron withdrawing (deactivating), and is therefore meta directing. So a significant part of the nitration product is m-nitroaniline. Correct answer: D. This question links directly to aniline's basicity: it acts as a base and gets protonated in acid.
Solved Amines NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Ammonia is more basic. In ammonia the nitrogen lone pair is fully available on nitrogen. In aniline the lone pair is pulled into the benzene ring, so it is less available and aniline is the weaker base.
What is the pKb of ammonia and how does aniline compare?
The pKb of ammonia is 4.75. Aliphatic amines have pKb about 3 to 4.22 (stronger than ammonia). Aromatic amines like aniline have higher pKb (weaker), because the aryl group withdraws the lone pair. Higher pKb means weaker base.
How many resonance structures does aniline have?
Aniline is a resonance hybrid of five structures because the lone pair delocalises into the ring. The anilinium ion (protonated aniline) has only two. More resonance stabilisation of the neutral form makes aniline reluctant to take a proton, so it is a weak base.
Is the effect on aniline due to +I effect or resonance?
It is mainly resonance (delocalisation of the lone pair into the ring, also called the -R effect of the phenyl group). There is also a small -I effect of the sp2 ring carbon. Both lower electron density on nitrogen, making aniline a weaker base than ammonia.
Why is this concept important for NEET?
NEET asks basicity order questions almost every year (2016, 2017, 2019, 2025). Understanding why aniline is weak (lone pair goes into the ring) lets you rank any set of amines and answer trap statements about delocalisation correctly.
