Nitro compounds — preparation, reduction to amines (Sn - HCl, Fe - HCl, H₂ - Pt): Worked examples

This page is a drill. The parent note built the theory; here we walk every kind of problem the topic can throw at you, so no exam question surprises you.

The scenario matrix

Every nitro-reduction problem is really a choice of two knobs: how much reducing power and what else is in the molecule that could get hurt. The table below lists every case class. The worked examples that follow each cover one (or more) cells.

Cell	Case class	What decides the answer
A	Single aromatic $- N O_{2}$ , full reduction	Any 6[H] reagent → amine
B	Two aromatic $- N O_{2}$ , full reduction	Strong reagent → both → diamine
C	Two aromatic $- N O_{2}$ , partial reduction	Mild sulfide → one → nitro-amine
D	Aliphatic $- N O_{2}$	$H_{2}$ /Ni or $L i A l H_{4}$ safe → amine
E	Aromatic $- N O_{2}$ + $L i A l H_{4}$ (the trap)	Gives azo, not amine
F	Molecule with a $C = C$ or $C = O$ to protect	Chemoselectivity: avoid $H_{2}$ /Pt
G	Electron-counting / stoichiometry	Balance $[H]$ and $H_{2}$
H	Word problem (industry choice)	Cheap, abundant metal → Fe/HCl
I	Degenerate / limiting: neutral medium, no acid drive	Coupling → azoxy/azo/hydrazo

Prerequisites you may want open: Oxidation states of nitrogen, Reducing agents in organic chemistry, Ambident nucleophiles.

Example 1 — Cell A: the plain case

Worked example Reduce nitrobenzene to aniline and state the reagent.

Forecast: guess the product and how many $[H]$ before reading.

Step 1. Identify the change: $C_{6} H_{5} - N O_{2} \to C_{6} H_{5} - N H_{2}$ . Why this step? You must know the target before choosing a reagent — the target is the fully reduced amine.

Step 2. Count electrons on nitrogen. In $- N O_{2}$ nitrogen is $+ 3$ ; in $- N H_{2}$ it is $- 3$ . Why this step? The oxidation-state jump tells you the reducing power needed: $+ 3 \to - 3$ is a 6-electron gain.

Step 3. Translate to $[H]$ : $6$ electrons + protons $= 6 [H]$ . Balance atoms — two O atoms leave as $2 H_{2} O$ (uses $4 H$ ), two H add to N (uses $2 H$ ), total $4 + 2 = 6$ . ✔

Step 4. Pick a 6[H] reagent: $S n / H C l$ . $C_{6} H_{5} N O_{2} + 6 [H] S n / H C l C_{6} H_{5} N H_{2} + 2 H_{2} O$

Verify: Count H on the reagent side: $6 [H] = 6$ H atoms $\to$ $4$ go to $2 H_{2} O$ , $2$ to $- N H_{2}$ . Nitrogen count 1 = 1, oxygen 2 = 2 (both in water). Balanced. ✔

Example 2 — Cell G: molecular-hydrogen stoichiometry

Example 3 — Cell C: partial reduction with a mild sulfide

Worked example Give the product of

m-dinitrobenzene with $(N H_{4})_{2} S$ . Forecast: does one nitro survive, or do both fall?

Step 1. Count nitro groups: two ( $- N O_{2}$ at positions 1 and 3). Why this step? With more than one reducible group, the strength of the reagent chooses how many fall.

Step 2. $(N H_{4})_{2} S$ is a mild, partial reductant (see the definition above) — enough driving power for one $- N O_{2}$ only. Why this step? Sulfide is a weak electron source; once one group is reduced, the newly formed $- N H_{2}$ is electron-donating and deactivates the ring toward further reduction, so it halts. → m-nitroaniline (one $- N H_{2}$ , one $- N O_{2}$ ).

Verify: Electron bookkeeping. One group reduced $= 6$ electrons used; the other still $+ 3$ . Product $=$ m-nitroaniline $C_{6} H_{6} N_{2} O_{2}$ . Nitrogen conserved (2 = 2), only one O-pair removed. ✔

Example 3b — Cell B: full reduction of both nitro groups

Worked example Now reduce the

same m-dinitrobenzene with $S n / H C l$ (excess). What is the product, and how does the decision differ from Example 3? Forecast: guess whether one or both groups fall this time.

Step 1. Identify the reagent's power: $S n / H C l$ delivers unlimited $6 [H]$ per group. Why this step? Cell B's whole decision is "is the reagent strong enough to finish every nitro group?" — and a metal/acid system is.

Step 2. Apply $6 [H]$ to each of the two nitro groups → both become $- N H_{2}$ . Why this step? Unlike the sulfide of Example 3, there is no power limit forcing a stop after one group. → m-phenylenediamine (benzene-1,3-diamine). $C_{6} H_{4} (N O_{2})_{2} + 12 [H] S n / H C l C_{6} H_{4} (N H_{2})_{2} + 4 H_{2} O$

Verify: Two groups reduced $= 12$ electrons $= 12 [H] = 6 H_{2}$ . Reducing the second $- N O_{2}$ consumes an extra $6 [H]$ and produces two more $H_{2} O$ . So m-nitroaniline $C_{6} H_{6} N_{2} O_{2}$ and m-phenylenediamine $C_{6} H_{8} N_{2}$ differ by exactly $6 [H]$ minus $2 H_{2} O$ : $H$ change $= 8 - 6 = 2$ (net H gained on N after 2 waters leave), O change $= 2 - 0 = 2$ (removed as the two waters). ✔

Example 4 — Cell D: aliphatic nitro, the safe zone

Worked example Reduce nitroethane

$C H_{3} C H_{2} N O_{2}$ to its amine. Which reagents are safe? Forecast: name the amine and two reagents.

Step 1. Target: $C H_{3} C H_{2} N O_{2} \to C H_{3} C H_{2} N H_{2}$ (ethylamine). Still a $+ 3 \to - 3$ change → $6 [H]$ . Why this step? Oxidation-state logic is identical for aliphatic and aromatic nitro; only the azo-coupling risk differs.

Step 2. Because the nitro is aliphatic, the dreaded azo-coupling side reaction (which needs two aromatic rings to conjugate) cannot happen. Why this step? This is what unlocks $L i A l H_{4}$ here — see Example 5 for why it fails on aromatics.

Step 3. Safe reagents: $H_{2} / N i$ or $L i A l H_{4}$ . $C H_{3} C H_{2} N O_{2} L i A l H_{4} or H_{2} / N i C H_{3} C H_{2} N H_{2}$

Verify: Formula check: $C_{2} H_{5} N O_{2} + 6 [H] \to C_{2} H_{5} N H_{2} + 2 H_{2} O$ . LHS: C2 H(5+6)=11 N1 O2. RHS: amine $C_{2} H_{7} N$ has H7, two waters H4 → H11, O2. Balanced. ✔

Example 5 — Cell E: the $L i A l H_{4}$ trap on an aromatic

Worked example A student treats

nitrobenzene with $L i A l H_{4}$ and writes "aniline." Correct them. Forecast: what actually forms?

Step 1. $L i A l H_{4}$ on an aromatic nitro group does not stop at the amine. Why this step? Under $L i A l H_{4}$ 's aprotic, non-acidic conditions the partly-reduced nitroso/hydroxylamine intermediates on two rings couple together before full reduction.

Step 2. The coupled product is azobenzene, $C_{6} H_{5} - N = N - C_{6} H_{5}$ . Why this step? This is the same coupling pathway the parent note flags for neutral/basic media (Cell I) — $L i A l H_{4}$ recreates that non-acidic environment.

Step 3. To actually get aniline, use $S n / H C l$ , $F e / H C l$ , or $H_{2} / P t$ (acidic or catalytic conditions that drive all the way through).

Verify: Nitrogen conservation. Two nitrobenzenes ( $2 \times N$ ) → one azobenzene has $2$ N in $- N = N -$ . So two molecules of substrate give one of product — the tell-tale sign of a coupling (not a simple reduction). ✔

Example 6 — Cell F: protect a $C = C$ (chemoselectivity)

Worked example

4-nitrostyrene ( $O_{2} N - C_{6} H_{4} - C H = C H_{2}$ ) must become the amine without touching the vinyl $C = C$ . Choose the reagent. Forecast: is $H_{2} / P t$ allowed here?

Step 1. List reducible groups: the $- N O_{2}$ (target) and the $C = C$ (must survive). Why this step? Chemoselectivity problems are won by listing everything a reagent could attack.

Step 2. Reject $H_{2} / P t$ (or Pd/Ni): it is not chemoselective — it hydrogenates $C = C$ to $- C H_{2} C H_{3}$ as well. Why this step? Pt splits $H_{2}$ indiscriminately; any $π$ bond on the surface gets saturated.

Step 3. Choose a metal/acid ( $F e / H C l$ or $S n / H C l$ ) or a sulfide — these deliver $[H]$ chemically to the nitro group and leave isolated $C = C$ alone. $O_{2} N - C_{6} H_{4} - C H = C H_{2} F e / H C l H_{2} N - C_{6} H_{4} - C H = C H_{2}$

Verify: Only the nitro nitrogen changes oxidation state ( $+ 3 \to - 3$ , a 6-electron reduction); the two vinyl carbons keep the same oxidation state. Product degree of unsaturation still = ring (4) + one $C = C$ (1) = 5, unchanged at the vinyl group. ✔

Example 6b — Cell F: protect a $C = O$ (chemoselectivity)

Worked example

4-nitroacetophenone ( $O_{2} N - C_{6} H_{4} - C O - C H_{3}$ ) must become the amino-ketone without touching the $C = O$ . Choose the reagent. Forecast: which of $H_{2} / P t$ , $F e / H C l$ , $L i A l H_{4}$ survives the carbonyl?

Step 1. List reducible groups: the $- N O_{2}$ (target) and the ketone $C = O$ (must survive). Why this step? Same chemoselectivity discipline as Example 6 — enumerate every vulnerable group first.

Step 2. Reject $L i A l H_{4}$ : it reduces the $C = O$ to $- C H (O H) -$ (and on an aromatic nitro it gives azo anyway — Cell E). Reject $H_{2} / P t$ : prolonged hydrogenation can also attack the carbonyl and it is not chemoselective. Why this step? Both are indiscriminate on carbonyls; the ketone would be destroyed.

Step 3. Choose $F e / H C l$ (or $S n / H C l$ ): metal/acid delivers $[H]$ to the nitro group but leaves an isolated ketone $C = O$ intact. $O_{2} N - C_{6} H_{4} - C O - C H_{3} F e / H C l H_{2} N - C_{6} H_{4} - C O - C H_{3}$

Verify: Only the nitro nitrogen changes ( $+ 3 \to - 3$ , 6-electron reduction); the carbonyl carbon keeps its oxidation state. The product p-aminoacetophenone still contains one $C = O$ — degree of unsaturation = ring (4) + $C = O$ (1) = 5, unchanged. ✔

Example 7 — Cell H: the industrial word problem

Worked example A dye factory reduces

tonnes of nitrobenzene to aniline daily and wants the cheapest full reduction. Which of $S n / H C l$ , $F e / H C l$ , $H_{2} / P t$ ? Forecast: pick one and justify on cost.

Step 1. All three give complete reduction (all are effectively 6[H]). So the tie-breaker is economics. Why this step? When every option reaches the target, choose on the secondary criterion the question emphasises — here, cost at scale.

Step 2. $S n$ is expensive; $P t$ is a precious-metal catalyst (costly, and not chemoselective). Iron is cheap and abundant, so $F e / H C l$ is the least-cost bulk reagent. Why this step? At tonne scale the price of the consumed metal dominates. (Note: the HCl is genuinely consumed stoichiometrically — $F e + 2 H C l \to F e C l_{2} + 2 [H]$ — so it is a reactant, not a catalyst; iron's low price, not any acid recycling, is what wins.)

Step 3. Choose $F e / H C l$ . $C_{6} H_{5} N O_{2} + 6 [H] F e / H C l C_{6} H_{5} N H_{2} + 2 H_{2} O$

Verify: Electron count still 6[H] = 3 $H_{2}$ equivalents (Examples 1–2) — same chemistry, cheaper metal. ✔ (This is why nitrobenzene is made industrially: it is a cheap on-ramp to aniline and hence diazonium chemistry — see Diazonium salts — synthesis from aniline.)

Example 8 — Cell I: the degenerate case (no acid drive)

Worked example Nitrobenzene is reduced in

neutral / weakly alkaline medium (e.g. Zn + $N H_{4} C l$ , or $A s_{2} O_{3}$ /NaOH). What class of product forms, and why not aniline? Forecast: guess whether you get an amine or something coupled.

Step 1. Recall the acidic pathway intermediates: nitroso ( $A r - N O$ ) → hydroxylamine ( $A r - N H O H$ ) → amine. Why this step? In acid, protons push each intermediate forward fast, so you reach the amine. Remove the acid drive and the intermediates linger.

Step 2. In neutral/basic medium the lingering $A r - N O$ and $A r - N H O H$ condense with each other instead of being reduced onward. Why this step? Coupling is a bimolecular condensation that only wins when full reduction is slowed — exactly the non-acidic case.

Step 3. The controlled ladder of coupled products: $A r - N O_{2} \to azoxy A r - N (O) = N - A r \to azo A r - N = N - A r \to hydrazo A r - N H - N H - A r$ Milder/less reduction → azoxy; more → azo; strong basic reduction → hydrazo.

Verify: Nitrogen accounting: azoxy, azo, and hydrazo all pair two aryl-N units, so two nitrobenzenes give one product — the coupling fingerprint (compare Example 5, $L i A l H_{4}$ → azobenzene, the same azo cell). ✔

Recall

Recall Which cell does each reagent land you in?

$S n / H C l$ on nitrobenzene ::: Cell A — full reduction to aniline. $(N H_{4})_{2} S$ on m-dinitrobenzene ::: Cell C — one group only → m-nitroaniline. $S n / H C l$ on m-dinitrobenzene ::: Cell B — both groups → m-phenylenediamine. $L i A l H_{4}$ on nitrobenzene ::: Cell E — azobenzene, NOT aniline. $H_{2} / P t$ on 4-nitrostyrene or 4-nitroacetophenone ::: Cell F trap — also attacks the $C = C$ / $C = O$ ; wrong choice. Neutral medium reduction ::: Cell I — azoxy/azo/hydrazo coupling. Cheapest industrial full reduction ::: Cell H — Fe/HCl (cheap, abundant iron).

Connections

Reducing agents in organic chemistry
Oxidation states of nitrogen
Ambident nucleophiles
Amines — basicity and reactions
Diazonium salts — synthesis from aniline
Electrophilic aromatic substitution