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

Intuition How to use this page

Every line below is a trap — a place where the obvious answer is wrong, or where a boundary case breaks a rule you thought was universal. Cover the answer, commit out loud, then reveal. If your reason doesn't match the reason given, that gap is the thing to learn. This page pairs with the parent topic.

One rule before we start, so no symbol is unearned:

$- N O_{2}$ = the nitro group: nitrogen bonded to carbon, carrying a $+$ formal charge, with two oxygens sharing a $-$ charge by resonance. Strongly electron-withdrawing.
$[H]$ = a single "hydrogen helper" = one proton + one electron delivered by a reducing system. " $6 [H]$ " just means six of these are consumed.
"Oxidation state of N" = a bookkeeping number for how electron-poor (high, $+$ ) or electron-rich (low, $-$ ) the nitrogen is. In $- N O_{2}$ it is $+ 3$ ; in $- N H_{2}$ it is $- 3$ .

The figures below are your visual reference bank: glance at them as you work the traps, so every abstract claim (equal bonds, nascent-hydrogen electron flow, protonated intermediates) has a concrete picture behind it.

Visual reference bank

The nitro group is a resonance hybrid — that's why its two N–O bonds are identical. Don't picture one single and one double bond; picture the average of the two drawings below, where the negative charge and the double-bond character are smeared equally over both oxygens.

Where does the "nascent hydrogen" $[H]$ come from? The metal surface hands electrons to protons from the acid. Follow the curved electron-flow arrows: tin atoms give up electrons (becoming $S n^{2 +}$ ), and $H^{+}$ ions from HCl grab them right at the metal surface to become reactive H atoms.

Acid vs. neutral medium — the fork that decides amine vs. azo. In acid, each intermediate is protonated and pushed onward before it can meet a neighbour. In neutral/basic medium, two intermediates (a nitroso and a hydroxylamine) couple and drift toward azoxy → azo → hydrazo products.

Chemoselectivity, drawn. $H_{2} / P t$ attacks every reducible bond it meets (it also eats the $C = C$ ). A metal/acid or sulfide system reduces the nitro group but leaves the alkene untouched — this is the picture behind the edge cases below.

True or false — justify

True or false: a nitro compound and a nitrite ester are the same substance written two ways.

False — they are isomers. Nitro (

R - N O_{2}

) bonds through nitrogen; the nitrite ester (

R - O - N = O

) bonds through oxygen, and they have entirely different chemistry.

True or false: the two N–O bonds in

- N O_{2}

are of different length (one single, one double).

False — resonance makes them equal (≈1.22 Å). Look at figure s01: the two drawings interconvert, so the real molecule is their average and neither oxygen "owns" the double bond. That shared negative charge is exactly why the group is so electron-withdrawing.

True or false: reducing

A r - N O_{2}

A r - N H_{2}

is a 3-electron change on nitrogen.

False — nitrogen goes from

+ 3

- 3

, a swing of six units, hence the

6 [H]

(or

3 H_{2}

True or false:

3 H_{2}

and

6 [H]

deliver the same reducing power.

True — each

H_{2}

molecule carries two hydrogen helpers, so

3 H_{2} = 6 [H]

; only the delivery vehicle differs (gas on metal vs. dissolving metal in acid).

True or false:

H_{2} / P t

is the most selective reagent for nitro → amine.

False — it is the least chemoselective. As figure s04 shows, it also hydrogenates

C = C

C = O

and

- C \equiv N

, so any such group elsewhere in the molecule is destroyed. See Reducing agents in organic chemistry.

True or false: in acidic reduction the nitro group jumps straight to the amine with no intermediates.

False — it passes through ==nitroso

(A r - N O)

→ hydroxylamine

(A r - N H O H)

→ amine== (figure s03). The intermediates matter because in non-acidic media they can couple instead.

True or false: nitration of benzene needs conc.

H_{2} S O_{4}

only as a solvent.

False — the sulfuric acid is the strongest acid present and its job is to generate the real electrophile, the ==nitronium ion

N O_{2}^{+}

== (

H N O_{3} + 2 H_{2} S O_{4} \to N O_{2}^{+} + H_{3} O^{+} + 2 H S O_{4}^{-}

). Without it almost no

N O_{2}^{+}

forms. See Electrophilic aromatic substitution.

True or false:

L i A l H_{4}

always turns nitrobenzene into aniline because it is a universal reducer.

False — under the standard textbook protocol (

L i A l H_{4}

in dry ether at or near reflux, then aqueous workup) aromatic nitro gives coupled ==azobenzene

(P h - N = N - P h)

==, because the ring-stabilised intermediates couple. Context matters: some forcing or modified conditions can push further, but for exam purposes and this parent note the answer is azobenzene, not the amine.

L i A l H_{4}

is reliable for aliphatic nitro → amine.

True or false:

(N H_{4})_{2} S

is too weak to reduce any nitro group.

False — it is weak but not useless; its weakness is a feature. In a dinitro compound it reduces exactly one

N O_{2}

, giving selectivity a strong reagent can't.

Spot the error

"To make a nitroalkane, react

R - X

with

N a N O_{2}

— the sodium salt is cheap and contains the nitrite ion."

Error:

N O_{2}^{-}

is ambident (attacks via N or O; see Ambident nucleophiles — the ion has lone pairs on both N and O ends). The sodium salt gives mostly the O-bonded nitrite ester

R - O - N = O

. Use ==

A g N O_{2}

== to get the true C–N nitroalkane.

"Use a tertiary alkyl halide with

A g N O_{2}

to get a high yield of nitroalkane."

Error: tertiary

R - X

favours elimination over substitution. Use a primary halide so the

S_{N} 2

-type displacement dominates and the nitroalkane forms.

S n + 2 H C l \to S n C l_{2} + C l_{2}

; the chlorine gas does the reducing."

Error: the metal–acid system generates ==nascent hydrogen

[H]

==, not

C l_{2}

S n + 2 H C l \to S n C l_{2} + 2 [H]

. As figure s02 shows, tin gives electrons to

H^{+}

at the surface; it is the hydrogen (electrons + protons) that reduces the nitro group.

"Fe/HCl needs a full stoichiometric amount of HCl per nitro group, so it's expensive."

Error: in Fe/HCl the HCl is effectively catalytic — iron regenerates the acid as the cycle turns — which is precisely why Fe/HCl is the industrial, cheap choice.

"m-dinitrobenzene +

(N H_{4})_{2} S

gives m-phenylenediamine."

Error: the mild sulfide reduces only one nitro group, giving m-nitroaniline. m-Phenylenediamine needs a full

6 [H]

-per-group reagent like Sn/HCl.

"Because

- N O_{2}

is electron-withdrawing, it makes benzene react faster in the next nitration."

Error: an electron-withdrawing group deactivates the ring toward further electrophilic attack and directs meta. It slows the next substitution, not speeds it.

Why questions

Why does the

- N O_{2} \to - N H_{2}

reduction need so many hydrogens rather than one or two?

Because two jobs stack: strip both oxygens (they leave as

2 H_{2} O

, needing

4 H

) and add two H's to nitrogen (

2 H

), totalling

6 [H]

— mirroring the six-unit oxidation-state drop.

Why does acidic medium drive nitrobenzene all the way to aniline, while neutral/basic medium can stop at azo/hydrazo products?

Follow figure s03. In acid the intermediates get protonated (e.g.

A r - N O \to A r - N (O H)^{+} \to

reduced onward, and

A r - N H O H \to A r - N H_{2}

) before they meet each other. In neutral/basic medium two neutral intermediate molecules couple (nitroso + hydroxylamine → azoxy → azo → hydrazo) before full reduction. Condition controls product.

Why choose Fe/HCl over Sn/HCl industrially when both give the amine?

Both fully reduce, but iron is cheaper and lets HCl act catalytically, so the running cost per tonne is far lower — decisive at industrial scale.

Why is the whole point of making cheap nitrobenzene to then reduce it?

Nitrobenzene is one nitration away from benzene; reducing it gives aniline, the gateway to diazonium salts, dyes and drugs. Nitro is "amine in disguise."

Why can

L i A l H_{4}

handle aliphatic nitro but not aromatic nitro cleanly (in ether/reflux)?

The aromatic ring stabilises intermediate nitrogen species enough for them to couple into azo compounds; a plain aliphatic chain offers no such stabilisation, so reduction proceeds straight to

R - N H_{2}

Why does silver, specifically, steer

N O_{2}^{-}

to bond through nitrogen?

Silver is a soft, polarising cation that ties up the oxygen ends and favours attack through the harder N-end, giving the C–N nitroalkane instead of the O-bonded nitrite ester.

Edge cases

Edge case: a molecule has both a

C = C

double bond and a

- N O_{2}

group, and you must keep the double bond. Which reagent?

Not

H_{2} / P t

— figure s04 shows it hydrogenates the

C = C

too. Use a metal/acid (Sn/HCl, Fe/HCl) or a sulfide, which reduce the nitro group but leave the alkene intact.

Edge case: p-dinitrobenzene treated with excess Sn/HCl. Product?

Sn/HCl is a full

6 [H]

-per-group reducer with no selectivity, so both nitro groups reduce → p-phenylenediamine. (For only one, switch to

(N H_{4})_{2} S

Edge case: ethyl nitro (

C H_{3} C H_{2} N O_{2}

), aliphatic. Which reagents are safe?

Both

H_{2} / N i

and

L i A l H_{4}

work, giving ethylamine — because with an aliphatic substrate the azo-coupling failure of

L i A l H_{4}

does not apply.

Edge case: you accidentally form

R - O - N = O

instead of

R - N O_{2}

. Will reducing it give the same amine?

No — the nitrite ester bonds through oxygen and has different chemistry; you would not obtain the clean

R - N H_{2}

you'd get from the true nitroalkane. Getting the right regiochemistry (via

A g N O_{2}

) matters before reduction.

Edge case: nitrogen's oxidation state in the nitroso intermediate

A r - N O

— is reduction finished there?

No. In

A r - N O

nitrogen is

+ 1

, still well above the

- 3

of the amine, so several more

[H]

are needed; nitroso is only a midpoint. See Oxidation states of nitrogen.

Recall

Recall Quick self-check

Which salt gives the true nitroalkane, and why? ::: $A g N O_{2}$ ; the soft silver cation steers ambident $N O_{2}^{-}$ to bond through N.
Why is Fe/HCl the industrial favourite? ::: Cheap iron and effectively catalytic HCl.
What does $L i A l H_{4}$ give with aromatic nitro in ether/reflux? ::: Azobenzene, not aniline.
What stops $(N H_{4})_{2} S$ at one nitro group? ::: Its mildness — not enough driving power for both.
What controls whether you get the amine or azo products in reduction? ::: The acidity of the medium.

Connections

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