Chemistry · Amines · NEET
You are right that free aniline is ortho/para-directing because the lone pair on nitrogen donates into the ring. But nitration needs a strongly acidic medium (conc. HNO3 + conc. H2SO4). In that acid, the lone pair is used up to hold a proton, so -NH2 becomes -NH3+. Now nitrogen has a positive charge, cannot donate electrons, and instead pulls electrons out of the ring. A positively charged, electron-withdrawing group is meta-directing, so a large amount of meta-nitroaniline forms.
In the acidic solution, aniline exists as a mixture: mostly protonated anilinium ion (meta-directing) but a small fraction stays as free aniline (very strongly ortho/para-directing). The free aniline reacts much faster, so it still contributes ortho and para product. That is why the real result is a mixture, roughly ortho 2%, meta 47%, para 51% (some books simplify it to about 50% meta). The point for NEET is that meta rises sharply compared to normal ortho/para directors.
Two problems. First, the nitronium ion (NO2+), the electrophile, is only made in a strongly acidic mixture, so you cannot avoid acid. Second, aniline is very easily oxidised, and conc. HNO3 is a strong oxidiser, so direct nitration also chars and destroys much of the aniline. To get a clean, mostly para product, chemists protect the -NH2 group by acetylation first.
The anilinium ion is C6H5-NH3+. It forms when the lone pair on the nitrogen of aniline accepts a proton (H+) from the acid. Once protonated, nitrogen has no free lone pair to share with the ring, so it stops being an activating ortho/para director and becomes a deactivating meta director.
No. Para-nitroaniline is usually the largest single fraction (around 51%), and meta is close behind (around 47%). The teaching point is not that meta is the majority, but that the meta fraction is unusually large (around 47%) compared to what a normal -NH2 director would give. NEET questions test the reason: acid protonates -NH2 to the meta-directing -NH3+.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Because the strongly acidic medium protonates the -NH2 group into the anilinium ion (-NH3+), which is electron-withdrawing and meta-directing, so meta product rises to about 47%.
Roughly ortho 2%, meta 47%, para 51%. Para is the largest single fraction, but the meta fraction is unusually high because of the protonated -NH3+ group.
The nitrogen in -NH3+ carries a positive charge and has no free lone pair, so it withdraws electron density from the ring. Electron-withdrawing groups direct the incoming electrophile to the meta position.
Protect the -NH2 group by acetylation (make acetanilide) before nitration. The -NHCOCH3 group stays ortho/para-directing and less easily protonated, giving mainly para product, then hydrolyse back to the amine.
Yes. Aniline is easily oxidised, and conc. HNO3 is a strong oxidiser, so direct nitration also destroys some aniline. This is another reason acetylation protection is preferred.
