Reduction of Nitro Compounds to Amines (H2/Ni, Fe/HCl)
Chemistry · Amines · NEET
A nitro group (-NO2) is reduced to a primary amine (-NH2) by adding hydrogen. In the lab we use metal + acid (Fe/HCl, Sn/HCl, or Zn/HCl) and in industry we use H2 gas with a metal catalyst (Ni, Pd or Pt). So nitrobenzene (C6H5NO2) gives aniline (C6H5NH2). Memory hook: "Nitro takes 3 H2 (6 H) to reach amine" - the N goes from +3 down, gaining 6 hydrogen atoms and losing 2 oxygen as water.
Nitrobenzene is reduced to aniline by Fe/HCl, Sn/HCl, or catalytic H2 (Ni/Pd/Pt). The -NO2 group gains six hydrogen atoms and loses its two oxygens as water, giving a primary amine. NaBH4 cannot do this reduction - a frequent NEET trap.
Your doubts, answered
Does Fe/HCl really reduce a nitro group to an amine?
Yes. Iron with dilute HCl (Fe/HCl) reduces -NO2 to -NH2 and is the standard method in NCERT. The nitro compound gains 6 H atoms; the product is a primary amine plus water. For nitrobenzene: C6H5NO2 + 3H2 (from Fe/HCl) gives C6H5NH2 + 2H2O.
H2/Ni or Fe/HCl - which one should I use?
Both give the amine. H2/Ni (catalytic hydrogenation, also Pd or Pt) is fast and clean but the H2 will also reduce other reducible groups like C=C, -CHO or -CN in the same molecule. Fe/HCl (or Sn/HCl) is chemoselective - it reduces only the nitro group and leaves C=C alone, so it is preferred when the molecule has a double bond you want to keep.
Why does NCERT prefer Fe/HCl over Sn/HCl?
Both work, but Fe/HCl is cheaper and the HCl acts catalytically - it is regenerated during the reaction, so you need less of it. Sn/HCl is more expensive. That is why NCERT lists Fe/HCl as the preferred acidic-medium reagent for reducing nitro compounds.
Does NaBH4 reduce the aromatic nitro group?
No. NaBH4 is a mild reducing agent and does NOT touch an aromatic -NO2 group. This is a classic NEET trap: if a question offers NaBH4 on nitrobenzene expecting aniline, that step gives no reaction. You need Fe/HCl, Sn/HCl, or H2/catalyst.
What is the oxidation and hydrogen change when nitro becomes amine?
Nitrogen in -NO2 is highly oxidised. Reduction removes the two oxygen atoms (as water) and adds hydrogen to nitrogen, giving -NH2. Overall three molecules of H2 (six H atoms) are consumed. This is a 6-electron reduction, which is why so many equivalents of reducing agent are needed.
Can I use this reaction to make aniline in acidic medium and get it out?
In Fe/HCl the aniline first forms as its salt, anilinium chloride (C6H5NH3+Cl-), because the medium is acidic. You then add a base like NaOH to free the neutral aniline. Forgetting the base-liberation step is a common exam slip.
⚠️ The NEET trap ✗ Any reducing agent, including NaBH4 or Sn/HCl in every case, will reduce an aromatic nitro group to an amine. ✓ NaBH4 does NOT reduce an aromatic -NO2 group. Aromatic nitro reduction needs Fe/HCl, Sn/HCl, Zn/HCl, or H2 with Ni/Pd/Pt. In ReNEET 2026, C6H5NO2 with NaBH4 gave NO reaction, so it was NOT a source of aniline. 🧠 Nitro is stubborn - mild NaBH4 walks past it. You need metal+acid or H2/catalyst to knock it down to an amine.
Real NEET questions
2016
A given nitrogen-containing aromatic compound A reacts with Sn/HCl, followed by HNO2, to give an unstable compound B. B, on treatment with phenol, forms a coloured compound C with molecular formula C12H10N2O. The structure of compound A is:
A · Aniline, C6H5NH2
B · Nitrobenzene, C6H5NO2 ✓
C · Benzonitrile, C6H5CN
D · Benzamide, C6H5CONH2
Solution: Sn/HCl reduces the nitro group of nitrobenzene to aniline. Aniline is then diazotised with HNO2 to an unstable diazonium salt, which couples with phenol to give the azo dye C12H10N2O. So A must be nitrobenzene (the starting nitro compound that gets reduced).
2026
Identify the reactions which give aniline as the major product. (A) C6H5CN --LiAlH4-->; (B) C6H5CONH2 --KOH, Br2-->; (C) C6H5NO2 --NaBH4-->; (D) C6H5NHCOCH3 --HCl, H2O, heat-->.
A · A and B only
B · B and D only ✓
C · A and C only
D · C and D only
Solution: (C) is the key nitro-reduction trap: NaBH4 does NOT reduce the aromatic nitro group, so C6H5NO2 gives no reaction (no aniline). (A) LiAlH4 on benzonitrile gives benzylamine, not aniline. (B) Hofmann bromamide of benzamide gives aniline; (D) acid hydrolysis of acetanilide gives aniline. Correct: B and D.
2023
Choose the correct sequence of reagents in the conversion of 4-nitrotoluene to 2-bromotoluene.
A · Br2; Sn/HCl; NaNO2/HCl; H2O/H3PO2 ✓
B · Sn/HCl; Br2; NaNO2/HCl; H2O/H3PO2
C · NaNO2/HCl; Sn/HCl; Br2; H2O/H3PO2
D · Sn/HCl; NaNO2/HCl; Br2; H2O/H3PO2
Solution: Brominate first while -NO2 is still on the ring so Br goes to the required position, then reduce -NO2 to -NH2 with Sn/HCl, diazotise with NaNO2/HCl, and finally remove the diazonium group by deamination with H3PO2. The Sn/HCl step is exactly the reduction of a nitro group to an amine.
Solved Amines NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
