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"Nitrobenzene, Sn/HCl, gives aniline" is one of the first conversions students memorise in the amines chapter. It is also where marks leak: an unbalanced equation, a missing base step, or no answer to "why iron rather than tin?". This post works out the electron count, balances the reactions for tin and iron, explains NCERT's preference for iron, and ends with a yield calculation and a practice routine.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus open the nitrogen-compounds unit with "general methods of preparation, properties, reactions and uses". The CBSE Chemistry curriculum for 2026-27 lists "Preparation of Amines" in the Class 12 Amines unit. The method itself is the first one in section 9.4 of the NCERT Class 12 unit on amines (Reprint 2026-27): 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 acidic medium". Check this year's bulletin.
How many electrons does the nitro group need?
Work out the oxidation state of nitrogen.
In C6H5NO2: nitrogen is bonded to two more electronegative oxygens and one less electronegative carbon. Its oxidation state is +3.
In C6H5NH2: nitrogen is bonded to two hydrogens and one carbon, all less electronegative. Its oxidation state is −3.
The change is 6, so each nitro group takes six electrons. In hydrogen terms that is 6[H]: C6H5NO2 + 6[H] → C6H5NH2 + 2H2O. Two of the six hydrogens end up on nitrogen and four go into the two water molecules.
Worked example 1: balancing with tin
Tin is oxidised from 0 to +4, so each Sn atom gives 4 electrons. Two nitro groups need 12 electrons, which is 3 Sn atoms:
2C6H5NO2 + 3Sn + 12HCl → 2C6H5NH2 + 3SnCl4 + 4H2O
Check: 3 Sn and 12 Cl each side; O is 4 = 4; H is 10 + 12 = 22 on the left and 14 + 8 = 22 on the right.
In this much acid, aniline does not stay free. It is a base, so it takes up a proton to form the anilinium ion, C6H5NH3⁺. Adding sodium hydroxide at the end releases aniline:
C6H5NH3⁺ + OH⁻ → C6H5NH2 + H2O
Worked example 2: a yield calculation
Sample numbers for practice: 12.3 g of nitrobenzene is reduced with tin and HCl, and 7.0 g of aniline is collected after adding base. Find the mass of tin needed and the percentage yield.
Molar mass of C6H5NO2: 6(12.011) + 5(1.008) + 14.007 + 2(15.999) = 123.11 g/mol.
Moles of nitrobenzene: 12.3/123.11 = 0.0999 mol.
Tin needed: 3/2 × 0.0999 = 0.150 mol, which is 0.150 × 118.71 = 17.8 g.
Molar mass of C6H5NH2: 93.13 g/mol. Theoretical aniline: 0.0999 × 93.13 = 9.30 g.
Percentage yield: 7.0/9.30 × 100 = 75%.
Why does NCERT prefer iron scrap and HCl?
With iron going to Fe²⁺, each iron atom gives 2 electrons, so 3 Fe atoms reduce one nitro group:
C6H5NO2 + 3Fe + 6HCl → C6H5NH2 + 3FeCl2 + 2H2O
On paper that needs six HCl. NCERT's point is that the real reaction needs far less, because the "FeCl2 formed gets hydrolysed to release hydrochloric acid during the reaction. Thus, only a small amount of hydrochloric acid is required to initiate the reaction." That is the answer to write when asked why Fe/HCl is preferred.
When would a chemist choose something else?
OpenStax Organic Chemistry notes that arylamines are usually made by nitration followed by reduction. Catalytic hydrogenation over platinum "works well but is often incompatible" with other reducible groups such as C=C or C=O in the molecule. Iron, zinc, tin and tin(II) chloride in acidic aqueous solution also work, and SnCl2 "is particularly mild and is often used when other reducible functional groups are present". So the choice depends on what else the molecule carries.
Common mistakes and how to avoid them
Writing 3H2 without checking. 6[H] is the same as 3H2, but only if you also produce 2H2O. Balance oxygen first.
Forgetting the base step. In Sn/HCl the product is the anilinium salt until you add NaOH.
Wrong tin stoichiometry. Sn to Sn⁴⁺ gives 4 electrons, so the ratio is 3 Sn to 2 nitrobenzene.
Mixing up the reasons. "FeCl2 hydrolyses to give HCl" is NCERT's reason for iron. Do not attach it to tin.
Once you have aniline, the next steps usually protect or test the amine. See our posts on acetylating aniline before nitration and on the carbylamine test for primary amines. If oxidation states in organic molecules still feel shaky, the Fermi Concept Library covers every concept in chemistry, so you can revise them before balancing the next reduction.
A practice routine
Find the oxidation state of nitrogen in nitrobenzene, nitrosobenzene (C6H5NO) and aniline.
Balance the reduction of p-nitrotoluene with Sn/HCl, then with Fe/HCl.
Redo the yield example for 24.6 g of nitrobenzene and 15.0 g of aniline.
Write the NCERT reason for preferring iron in two sentences, from memory.
Tonight, write "+3 to −3, six electrons" next to the nitro group in your amines notes.
