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

Step 1 — What "oxidation state" even means (the counting rule)

WHAT. Before we can say nitrogen goes "from +3 to −3", we must agree on what those numbers count. An oxidation state is a bookkeeping number: pretend every bond's electrons go entirely to the more electron-greedy atom, then count how many electrons the atom has gained (−) or lost (+) compared to when it's alone.

WHY this tool. We need a single number that tells us "how loaded up is this nitrogen?" so that we can measure the size of the job the reducing agent must do. Oxidation state is exactly that gauge — it converts a messy molecule into one integer.

PICTURE. Look at the figure. Each bond is a tug-of-war rope. Oxygen (cyan, greedy) pulls the shared pair fully to itself; nitrogen (amber) pulls harder than hydrogen. Count the arrows leaving nitrogen minus arrows coming in — that is its oxidation number.

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

Step 2 — The starting point: nitrogen at +3

WHAT. In the nitro group $- N O_{2}$ , nitrogen shares three electron pairs with oxygen in total, and is joined to one carbon.

WHY. We apply the Step-1 rule to get the before number — the top of the hill we must roll down.

PICTURE. In the figure, three ropes' worth of shared pairs leave the amber N toward the two equivalent greedy oxygens (each shared pair scores $+ 1$ ), and one rope goes to carbon (a tie, $0$ ).

$three N-O shared pairs (+ 1) + (+ 1) + (+ 1) + N-C (0) = + 3$

each $(+ 1)$ = one N–O shared pair, oxygen wins the tug → nitrogen loses one electron's worth.
$(0)$ = the N–C bond, near-tie → contributes nothing.
$+ 3$ = the total: nitrogen is electron-poor, "loaded up" with three lost electrons.

Step 3 — The finish line: nitrogen at −3

WHAT. In the amine group $- N H_{2}$ , nitrogen is bonded to two hydrogens and one carbon.

WHY. Same rule, applied to the after molecule — the bottom of the hill.

PICTURE. Now the two ropes go to hydrogens, and nitrogen wins each ( $- 1$ apiece); the carbon rope is still a tie.

$N-H (- 1) + N-H (- 1) + N-C (0) = - 3$

$(- 1)$ = an N–H bond, nitrogen wins → it gains one electron's worth.
$- 3$ = the total: nitrogen is now electron-rich, "fully reduced", relaxed.

Step 4 — The height of the hill: a 6-electron drop

WHAT. Subtract finish from start.

WHY. The word reduction means gain of electrons. The gap between the two oxidation numbers is literally the number of electrons nitrogen must gain.

$Δ = start, Step 2 (+ 3) - finish, Step 3 (- 3) = 6 electrons gained$

$+ 3$ = the loaded nitro nitrogen.
$- 3$ = the relaxed amine nitrogen.
$6$ = electrons that must flow into nitrogen. This is the "6" in "6[H]".

PICTURE. A vertical energy-ladder: N climbs down six rungs. Each rung is one electron delivered by the reducing agent.

Step 5 — Where the atoms actually go (the mass balance)

WHAT. Electron-counting told us "six". Now we confirm it by a completely independent method: balancing atoms. If the two methods agree, we trust the six.

WHY this second tool. Oxidation states are bookkeeping; a mass balance is physical reality (atoms in = atoms out). Two roads to the same "6" is a proof, not a coincidence.

PICTURE. The figure splits the six incoming H atoms into two destinations:

$A r - 2 oxygens here N O_{2} + 6 [H] ⟶ A r - 2 H land here N H_{2} + 2 oxygens leave here 2 H_{2} O$

Term-by-term:

2 oxygens on the left must exit. The only clean exit is as water $H_{2} O$ . Two waters need $2 \times 2 = 4$ hydrogen atoms.
2 hydrogens must appear on nitrogen to build $- N H_{2}$ : that is $2$ more H atoms.
$4 + 2 = 6$ hydrogen atoms — matching Step 4 exactly. ✔

Step 6 — Where the six [H] come from (three delivery trucks)

WHAT. The equation demands 6[H]. Three lab methods supply exactly that; only the delivery differs.

WHY. Same destination (aniline), but the truck you choose changes cost and selectivity — that's why the parent lists three.

PICTURE. Three parallel pipes all feeding the same "6[H]" tank.

(a) Tin + acid. The honest half-reaction is that tin gives up two electrons, and two protons ride along; the bookkeeping shorthand is: $S n + 2 H C l ⟶ S n C l_{2} + 2 [H]_{(to substrate)}$

$2 [H]$ per tin event ⇒ three events deliver the needed six electron-proton pairs.

(b) Iron + acid. The same electron-donor idea, iron in place of tin: $F e + 2 H C l ⟶ F e C l_{2} + 2 [H]_{(to substrate)}$

Iron is chosen because it is simply cheaper than tin, which is why industry prefers it. (Note: the HCl here is genuinely consumed stoichiometrically — each event uses up $2 H C l$ ; iron does not hand the acid back.)

(c) Hydrogen gas on platinum. Here the electrons and protons arrive already packaged as $H_{2}$ molecules, and the metal splits them: $3 H_{2} P t 6 [H]_{on surface} = 6 electron-proton pairs$

$3 H_{2}$ = three hydrogen molecules; the Pt surface cleaves each $H_{2}$ into two surface-bound H that add to the substrate → $3 \times 2 = 6$ . Identical electron count, delivered as gas rather than from acid.

Step 7 — Edge case A: what if you under-supply the hydrogens?

WHAT. If the reductant is weak (e.g. ammonium sulfide $(N H_{4})_{2} S$ ), it cannot deliver all six to every nitro group at once. The reaction stalls at partial reduction.

WHY it matters. This is not a failure — it's a tool. A molecule with two nitro groups can have exactly one reduced, leaving the other intact.

PICTURE. The energy-ladder from Step 4, but the sulfide truck runs out of fuel partway: one $N O_{2}$ reaches $- N H_{2}$ , the neighbouring one stays put.

$m - C_{6} H_{4} (N O_{2})_{2} (N H_{4})_{2} S m - O_{2} N - C_{6} H_{4} - N H_{2} (m-nitroaniline)$

Left dinitrobenzene has two loaded nitrogens.
Weak reductant = not enough 6[H] packets for both.
Product keeps one $N O_{2}$ (still $+ 3$ ) and reduces one to $N H_{2}$ ( $- 3$ ).

Step 8 — Edge case B: the intermediate rungs (acid vs neutral)

WHAT. Nitrogen does not leap from $+ 3$ to $- 3$ in one bound. It descends through real, isolable species. In acid the ladder is: $+ 3 A r - N O_{2} \to + 1 A r - N O \to - 1 A r - N H O H \to - 3 A r - N H_{2}$

$A r - N O$ (nitroso), N at $+ 1$ — one oxygen already gone (departed as the first water).
$A r - N H O H$ (hydroxylamine), N at $- 1$ — the last oxygen hangs on as $- O H$ (departs as the second water on the next step).
$A r - N H_{2}$ , N at $- 3$ — home.

WHY conditions matter. In acid, protons are everywhere, so each intermediate is grabbed and pushed onward → clean amine. In neutral/basic medium the half-reduced pieces linger and bump into each other, coupling two nitrogens through a chain: azoxy $\to$ azo $\to$ hydrazo compounds. The medium chooses the product.

PICTURE. Two forked paths: the acid path plunges straight down to $- N H_{2}$ ; the neutral path detours sideways, stepping azoxyarene → azoarene ( $A r - N = N - A r$ ) → hydrazoarene ( $A r - N H - N H - A r$ ).

Recall Why does

$L i A l H_{4}$ give azo with aromatic nitro? $L i A l H_{4}$ works in a non-acidic (aprotic) medium, so the nitroso/hydroxylamine intermediates are free to couple — giving azobenzene $P h - N = N - P h$ rather than aniline. For aromatic amines, use Sn/HCl, Fe/HCl, or H₂/Pt. ::: The acidic path pushes each intermediate onward before it can couple; the non-acidic path lets them meet and bond N=N.

The one-picture summary

Everything above compressed into a single diagram: nitrogen slides down the six-electron hill; the two oxygens peel off as two waters during the descent; six electron-proton pairs — from Sn/HCl, Fe/HCl, or $3 H_{2}$ /Pt — do all the work; and the side-exits (partial reduction, and the full neutral coupling chain azoxy → azo → hydrazo) branch off where the fuel or the acid runs short.

Recall Feynman retelling of the whole walkthrough

Picture nitrogen wearing a backpack of two oxygens. We first weigh how loaded it is with a bookkeeping trick: greedy oxygens pull electrons away, so nitrogen sits high at +3. In the finished amine, two friendly hydrogens instead give electrons to nitrogen, so it relaxes down to −3. The gap is six — six electrons must flow in. In acid every electron comes riding a proton, so that's six electron-proton pairs, which we write as six [H] (a bookkeeping symbol — never actual free H atoms in a bottle). Where do they go? A quick atom-count settles it: four of them carry the two backpack-oxygens away as two waters while the reduction happens, and the last two become nitrogen's two new H arms. Now, who delivers the six? Tin or iron in acid are electron-donors that feed protons onto the molecule two pairs at a time (three batches — and each batch really uses up the acid), or you blow three H₂ molecules over a platinum sponge that snaps each into two — same six either way. Finally, two warnings: give too little fuel (mild sulfide) and only one of two nitro groups gets reduced (useful!); run it in non-acid conditions and the half-done pieces bump together, walking the coupling chain azoxy → azo → hydrazo instead of reaching the amine. Same hill, but the fuel level and the acid decide where you land.

Connections

Oxidation states of nitrogen (the +3 → −3 counting used in Steps 1–4)
Reducing agents in organic chemistry (Sn/HCl, Fe/HCl, H₂/Pt, LiAlH₄, sulfide)
Amines — basicity and reactions (what the product does next)
Diazonium salts — synthesis from aniline (why we wanted aniline at all)
Electrophilic aromatic substitution (how the nitro group got there)
Ambident nucleophiles (aliphatic nitro preparation, from the parent)