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

WHAT to look at: where the atom chain touches nitrogen.

• (a) $C{H}_3-N{O}_2$: carbon is bonded straight to the nitrogen → nitro compound (nitromethane).
• (b) $C{H}_3-O-N=O$: carbon touches an oxygen first, and that oxygen holds the nitrogen → nitrite ester (methyl nitrite).

WHY it matters: they are isomers (same formula $C{H}_{3}N{O}_2$) but reduce/behave completely differently. Nitro → amine; nitrite ester is an entirely separate species.

The one rule we use: an oxidation state is found by pretending every bond is fully ionic — each shared pair is handed entirely to the more electronegative atom. Then we add up charges. Assign known atoms first, let the unknown be a letter, and force the total to match the real charge of the piece.

Proving N = +3 in $R-N{O}_2$ (step by step):

1. Oxygen is more electronegative than N, so each O takes its shared electrons → each O counts as $-2$. Two O's: $2\times (-2)=-4$.
2. Nitrogen is more electronegative than carbon, so at the C–N bond the electrons go to N. That C–N bond therefore does not oxidise N; we account for it by treating the $-N{O}_2$ fragment (N + 2 O, i.e. everything past the ring/chain) as carrying the ionic charge $-1$ — because when N keeps the C–N electrons, the fragment "$N{O}_2$" viewed on its own is the nitrite-type anion skeleton with net charge $-1$.
3. Let the oxidation state of N be $n$. Sum of oxidation numbers in the fragment must equal that fragment charge $-1$: $n+(-4)=-1\Rightarrow n=+3.$ So N is $+3$ in a nitro group. (This is the standard textbook value; the two N–O bonds are equivalent by resonance, but for counting we treat both O's as $-2$.)

Proving N in $-N{H}_2$ (same rule, same letter): 4. Hydrogen is less electronegative than N, so at each N–H bond the electrons go to N → each H counts as $+1$. Two H's: $+2$. 5. The C–N bond again sends electrons to N, so the $-N{H}_2$ fragment (N + 2 H) carries fragment charge $-1$, exactly as before. 6. Let the oxidation state of N here be $n$ again (same unknown symbol, new molecule): $n+(+2)=-1\Rightarrow n=-3.$ So N is $-3$ in an amine.

Electrons gained: from $+3$ down to $-3$ is a drop of $+3-(-3)=6$, i.e. the nitrogen gains 6 electrons (reduction = gain of electrons). $[H]$ count (balance O and H):

• The 2 oxygens leave as $2H_{2}O$ → that mops up $4H$.
• The nitrogen grows two H arms ($-N{H}_2$) → that adds $2H$.
• Total $=4+2=6$ hydrogen atoms $=\boxed{6[H]}$.

$Ar-N{O}_2+6[H]⟶Ar-N{H}_2+2H_{2}O$

Each $H_2$ molecule supplies $2$ hydrogen atoms. So $6[H]=\frac{6}{2}=3H_2$. $Ar-N{O}_2+3H_2\stackrel{Pt}{\to }Ar-N{H}_2+2H_{2}O$

Reagent: Fe / HCl (iron in hydrochloric acid). Why Fe/HCl and not Sn/HCl? Both give full reduction, but Fe/HCl is cheaper — iron is far less costly than tin. Where the hydrogen actually comes from: the acid supplies it. On the iron surface the reaction is $Fe+2HCl\to FeC{l}_2+2[H]$ (or, equivalently, $Fe$ reduces $H^{+}$ from HCl to give the nascent hydrogen). So it is the $H^{+}$ (protons) of HCl that are reduced to the active $[H]$ that then attacks the nitro group — HCl is the hydrogen/proton source, not water. (In the classic Béchamp variant a small amount of acid catalyses regeneration of $F{e}^{2+}/Fe$ cycling with added iron, but the reducing hydrogen still traces back to protons from the acid, never from water acting as reductant.) $C_6{H}_{5}N{O}_2+6[H]\stackrel{Fe/HCl}{\to }{C}_6{H}_{5}N{H}_2+2H_{2}O$ Product: aniline ($C_6{H}_{5}N{H}_2$).

Reagent: $LiAl{H}_4$ or $H_2/Ni$. Why $LiAl{H}_4$ is safe here: the substrate is aliphatic (open-chain, no aromatic $Ar$ ring), so there is no benzene ring to let the intermediates couple into an azo compound. $LiAl{H}_4$ cleanly delivers hydride. Product: $C{H}_{3}C{H}_{2}N{H}_2$ (ethylamine).

Reagent: ==$AgN{O}_2$== (silver nitrite). First, two words we need (defined):

• A nucleophile is an electron-rich atom that attacks a carbon; an ambident nucleophile has two different atoms that can each do the attacking — here the $N$ or either $O$ of $N{O}_2^{-}$ (see Ambident nucleophiles).
• "Hard" vs "soft": a hard site is small, and holds its negative charge tightly in one place (the $N$ end of $N{O}_2^{-}$, where the lone pair is concentrated); a soft site is larger and more polarisable, its charge more smeared out (the $O$ ends, where charge is spread by resonance). Hard prefers to react with hard partners; soft with soft.
• A protic solvent (water, ethanol) has O–H/N–H bonds and can hydrogen-bond to the ion; an aprotic / low-polarity solvent (ether) cannot.

Now the WHY, made concrete:

• In a protic solvent, the small, charge-dense $N$ end gets wrapped in a tight cage of hydrogen bonds (solvent H's cling to concentrated negative charge). That cage physically blocks the $N$ from reacting, leaving the poorly-solvated, more exposed $O$ end free to attack → nitrite ester $R-O-N=O$. This is why $NaN{O}_2$ in water/ethanol gives the O-product.
• In $AgN{O}_2$ + a low-polarity/aprotic solvent there are no H-bonds to cage the $N$; the reaction runs $S_{N}2$-like (a tight, direct back-side push on carbon), which the hard, localised $N$ lone pair does best because it is a concentrated, well-directed donor. Silver also yanks the halide out as insoluble $AgBr$, speeding the substitution. Result → C–N nitroalkane. $C{H}_{3}C{H}_{2}C{H}_{2}Br+AgN{O}_2\stackrel{\text{ether, mild}}{\to }C{H}_{3}C{H}_{2}C{H}_{2}N{O}_2+AgBr$ Product: 1-nitropropane. (Substrate is primary → clean $S_{N}2$, little elimination; keep the temperature modest to avoid $E2$.)

$Ar-N{O}_2\to \underset{\text{nitroso}}{\underbrace{Ar-NO}}\to \underset{\text{hydroxylamine}}{\underbrace{Ar-NHOH}}\to Ar-N{H}_2$ Nitroso → hydroxylamine → amine. Each step adds $2[H]$ and strips oxygen; acid keeps pushing all the way to the amine before the intermediates can wander off.

Figure — the reduction ladder. Alt-text / description: four coloured circles descend left-to-right like a staircase — $Ar-N{O}_2$ (orange, N at $+3$) at top-left, then $Ar-NO$ (blue), $Ar-NHOH$ (green), down to $Ar-N{H}_2$ (red, N at $-3$) at bottom-right. Each grey arrow between circles is labelled "$+2[H]$"; the y-axis runs from $+3$ (top) to $-3$ (bottom), so reduction reads as a literal downhill staircase in oxidation state.

WHAT forms (coupling of two rings through their nitrogens): $\text{azoxybenzene}\to \text{azobenzene}\to \text{hydrazobenzene}$ The mechanistic WHY (two competing rates): the key reactive pair is nitroso ($Ar-NO$) plus hydroxylamine ($Ar-NHOH$). They can do either of two things:

• Path A — keep reducing (add more $[H]$, go on to the amine). This step needs a proton: each reduction increment consumes $H^{+}$.
• Path B — couple (the $NHOH$ nitrogen condenses onto the $N=O$ of a nitroso molecule, expelling water) → azoxy → azo → hydrazo.

In acid: $H^{+}$ is abundant, so Path A is fast — every nitroso/hydroxylamine intermediate is protonated and reduced onward almost as soon as it forms. Its lifetime is too short to bump into another molecule and couple. Protonation also makes the intermediate a better electron acceptor (more electrophilic at N), so it grabs the next $[H]$ rather than acting as a nucleophile toward another ring. Coupling (Path B, which is bimolecular — it needs two molecules to meet) is starved out. Result: clean aniline. In neutral/base: protons are scarce, so Path A stalls. The nitroso and hydroxylamine now accumulate and live long enough to find each other; the electron-rich $NHOH$ nitrogen attacks the electrophilic nitroso — the bimolecular Path B wins. Result: azoxy/azo/hydrazo. Which coupled product you stop at then depends on how much reducing power is added: mild → azoxy; stronger alkaline reduction (e.g. Zn/NaOH, glucose) → azo → hydrazo. So acidity controls kinetics (which path), and reductant strength controls how far down Path B you travel.

Product: ==$m$-nitroaniline== (one $N{O}_2$ reduced, one untouched). Why only one? Ammonium/sodium sulfide is a mild, partial reductant — it lacks the driving power to reduce both groups. It reduces just one $N{O}_2\to N{H}_2$ and stops. Contrast: a full $6[H]$ reagent (Sn/HCl) would reduce both → $m$-phenylenediamine.

Figure — reagent strength controls the count. Alt-text / description: an orange starting circle "$m$-dinitrobenzene (2 × NO₂)" forks into two arrows. The upper green arrow, labelled "(NH₄)₂S mild — reduces ONE", leads to a green circle "$m$-nitroaniline (NO₂ + NH₂)". The lower blue arrow, labelled "Sn/HCl 6[H] — reduces BOTH", leads to a blue circle "$m$-phenylenediamine (2 × NH₂)". Same substrate, different reagent strength, different count of groups reduced.

Step 1 — Nitration: $C_6{H}_6+HN{O}_3\underset{55^{\circ }C}{\overset{conc.H_{2}S{O}_4}{\to }}{C}_6{H}_{5}N{O}_2+H_{2}O$ The real electrophile is the ==nitronium ion $N{O}_2^{+}$==, made by $H_{2}S{O}_4$ protonating $HN{O}_3$ then losing water. (See Electrophilic aromatic substitution.) Step 2 — Reduction: $C_6{H}_{5}N{O}_2+6[H]\stackrel{Sn/HCl}{\to }{C}_6{H}_{5}N{H}_2+2H_{2}O$ Net: benzene → nitrobenzene → aniline (here $Ar=C_6{H}_5$).

Aniline is diazotised (treat with $NaN{O}_2+HCl$ at $0$–$5^{\circ }C$) to give a diazonium salt $C_6{H}_5{N}_2^{+}$. See Diazonium salts — synthesis from aniline. That is the whole payoff: cheap benzene → nitrobenzene → aniline → diazonium → dyes, aryl halides, phenols. The nitro step is just the cheap on-ramp.

Reagent: Fe/HCl (or Sn/HCl) — a dissolving-metal / $[H]$-generator reduction. Why NOT $H_2/Pt$: catalytic hydrogenation would also saturate the vinyl $C=C$ into $-C{H}_{2}C{H}_3$, destroying the styrene. The metal/acid method reduces the nitro group but leaves the alkene intact. Product: $p$-aminostyrene ($H_{2}N$–$C_6{H}_4$–$CH=C{H}_2$).

(a) Nitroethane (aliphatic): $LiAl{H}_4$ reduces it cleanly to the amine → ethylamine $C{H}_{3}C{H}_{2}N{H}_2$. There is no aromatic ring, so the reduction runs straight to $-N{H}_2$ with no side pathway. (b) Nitrobenzene (aromatic, $Ar-N{O}_2$): $LiAl{H}_4$ does NOT give aniline. It stops at coupled azobenzene $C_6{H}_5-N=N-C_6{H}_5$. Why: $LiAl{H}_4$ generates aromatic nitroso/hydroxylamine intermediates but supplies no acidic protons to drive them onward — so, exactly as in L3-Q2, the long-lived intermediates couple (two rings joined through $N=N$) instead of finishing to the amine. The takeaway rule: $LiAl{H}_4$ → amine for aliphatic nitro; → azo compound for aromatic nitro. To make aniline from nitrobenzene, use Sn/HCl, Fe/HCl, or $H_2/Pt$ instead.

Step 1 — Dinitrate: benzene $\to$ $m$-dinitrobenzene using conc. $HN{O}_3$ / conc. $H_{2}S{O}_4$ (excess, hotter). The first $-N{O}_2$ is a meta-director and deactivator, so the second nitro enters meta → $m$-dinitrobenzene. Step 2 — Selective partial reduction: treat with ==$(N{H}_4{)}_{2}S$== (mild sulfide) → reduces one nitro group only → $m$-nitroaniline. Why not Sn/HCl in step 2? It would reduce both groups → $m$-phenylenediamine. The mild sulfide is what preserves one nitro.

Each $Sn$ delivers $2[H]$. We need $6[H]$, so moles of $Sn=6/2=\boxed{3}$. (Real prep uses excess $Sn$, but the stoichiometric minimum is $3Sn$ per nitro group.)

Each $N{O}_2\to N{H}_2$ needs $6[H]$. Two groups → $2\times 6=12[H]$. As gas: $12[H]=12/2=6H_2$.

What does "$Ar$" stand for? ::: an aryl (benzene-type) ring attached through a ring carbon, e.g. $C_6{H}_5-$. What does "$[H]$" stand for? ::: one hydrogen-atom equivalent of reducing power (1 e⁻ + 1 H⁺); a counting token, not a bottleable species; $2[H]=1H_2$. How many $[H]$ for one $Ar-N{O}_2\to Ar-N{H}_2$? ::: $6[H]$ ($=3H_2$); N goes $+3\to -3$ (6 e⁻), 2 O leave as $2H_{2}O$ (4H) + 2H onto N. In Fe/HCl reduction, where do the reducing hydrogens come from? ::: from the protons ($H^{+}$) of HCl, reduced on the iron surface to nascent $[H]$ — not from water. Reagent for nitroalkane from R–X, and solvent? ::: $AgN{O}_2$ in a low-polarity/aprotic solvent (N-bonded product), not $NaN{O}_2$ in polar protic (O-bonded). Three acidic intermediates in order? ::: nitroso $\to$ hydroxylamine $\to$ amine. Why does acid prevent azo coupling? ::: abundant $H^{+}$ reduces the intermediates onward too fast for the bimolecular coupling to compete. Reagent that reduces only one of two nitro groups? ::: $(N{H}_4{)}_{2}S$ / $N{a}_{2}S$. $LiAl{H}_4$ on aromatic nitro gives? ::: azobenzene, not aniline (no acidic protons → intermediates couple). Moles of Sn for $6[H]$? ::: 3 (each Sn gives $2[H]$).