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Notes/Chemistry/Paper 4/Nitrogen Compounds
CAIEA2 Level9701§34.1–34.4

Nitrogen Compounds

Amines, amides, phenylamine, azo dyes and amino acids — every mark routes through one question: where is the nitrogen lone pair?

330 min read 10 sub-topics
222
question parts
2021–2025 · 37 papers
11 marks
per paper
≈ 11% of the paper
2.0/3
avg difficulty
moderate
#3
most examined
of 15 topics by marks

Nitrogen compounds are Paper 4's third-heaviest topic: over 2021–2025 they filled 222 examined parts worth 409 marks across 37 papers, and the questions recycle one idea so reliably that it is worth stating before any chemistry — every basicity, nucleophilicity and electrophoresis mark in this topic is decided by where the nitrogen lone pair is. Free on the nitrogen (alkylamines): strong base, keen nucleophile. Delocalised into a carbonyl (amides): weakest base on the ladder. Delocalised into a ring (phenylamine, and the diazonium salts built from it): barely a base at all, but the ring becomes so reactive that bromine water slams three bromines on at once. The route through: §01 the four doors into the amine family (two substitutions, two reductions), §02 amides built cold from acyl chlorides, §03 the basicity ladder — the most repeated question in the topic — §04 amide hydrolysis and reduction, §05 phenylamine from benzene and its bromine-water fingerprint, §06 diazonium salts and the phenol they hand you, §07 azo coupling and dyes, §08 zwitterions and the isoelectric point, §09 peptide bonds drawn with the C=O displayed and hydrolysed back, and §10 electrophoresis plates read from two numbers: charge and MrM_r.

By the end of this page you can

  • Write all four amine-making routes with reagents AND conditions — halogenoalkane + NH₃ in ethanol heated under pressure (excess NH₃), halogenoalkane + primary amine in ethanol in a sealed tube, amide + LiAlH₄ (dry ether), nitrile + LiAlH₄ or H₂/Ni — and pick the route that hits a target carbon count (nitrile adds one carbon; amide reduction keeps the count)

  • Write the two amide-forming equations — acyl chloride + NH₃ and acyl chloride + primary amine, both at room temperature — with TWO nucleophile equivalents and the ammonium-salt co-product, and name the N-substituted amide correctly

  • Rank basicity amide < phenylamine < ammonia < alkylamine and give each species its own lone-pair clause: delocalised into C=O / into the ring π-system / plain baseline / alkyl +I push — with the 'lone pair accepts H⁺' link stated first

  • Hydrolyse amides with aqueous acid (carboxylic acid + ammonium salt) and aqueous alkali (carboxylate + amine/ammonia) and reduce the CO group with LiAlH₄ to the amine at the same carbon count

  • Prepare phenylamine from benzene in three stages (conc. HNO₃/conc. H₂SO₄; hot Sn/concentrated HCl; NaOH(aq)) and write the 6[H] reduction equation — then state the bromine-water result: white precipitate of 2,4,6-tribromophenylamine with phenylamine, no change with ethylamine

  • Diazotise phenylamine below 10 °C with HNO₂ (or NaNO₂ + dilute HCl) and hydrolyse the diazonium salt by warming with water to phenol — quoting both steps' reagents and temperatures

  • Couple benzenediazonium chloride with phenol in NaOH(aq) to an azo dye, identify the –N=N– azo group, and extend the route to other dye structures

  • Draw any amino acid's three ionised forms against pH (cation when pH < pI, zwitterion at the isoelectric point, anion when pH > pI) and predict the dominant species at any stated pH, including for diacid and diamino acids

  • Draw di- and tripeptides with the peptide bond FULLY displayed (C=O drawn out), respecting the N-terminal-first naming order, and name the amino acids released on complete hydrolysis

  • Predict electrophoresis spot positions from charge alone (pH vs pI decides cation/zwitterion/anion; charge decides direction, MrM_r decides distance) on printed plates with either electrode orientation

01

Making amines: four routes

Syllabus requirement · §34.1.1

“

recall reactions (reagents and conditions) producing primary and secondary amines: (a) halogenoalkanes + NH₃ in ethanol heated under pressure (b) halogenoalkanes + primary amines in ethanol, heated in a sealed tube / under pressure (c) reduction of amides with LiAlH₄ (d) reduction of nitriles with LiAlH₄ or H₂/Ni

”

Four doors into the amine family

Amines are this topic's gateway group: the amides of §02, the basicity ladder of §03, the diazonium chemistry of §§05–06 and the amino acids of §§08–10 are all amines wearing extra groups. The syllabus lists exactly FOUR ways of making them, and the exam examines them as a set — a typical question prints a two-step scheme and asks you to name the intermediate or quote the reagent. The four doors split into two families, and the families answer different questions: the two substitution doors swap a halogen for a nitrogen group (carbon count unchanged), while the two reduction doors build a −CH2NH2-\text{CH}_2\text{NH}_2 onto an existing nitrogen (and differ in whether a carbon is added on the way).

The two substitution routes — swap X for N

Both routes are the AS Halogenoalkanes note's nucleophilic substitution, now with a nitrogen nucleophile and examinable conditions.

Route (a): halogenoalkane + ammonia → primary amine. Reagents and conditions, in full: NH3\text{NH}_3 in ethanol, heated under pressure (a sealed tube — ammonia is a gas, and pressure lets the mixture be heated above its boiling point). The substitution itself:

CH3CH2CH2Br+2NH3→CH3CH2CH2NH2+NH4Br\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + 2\text{NH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 + \text{NH}_4\text{Br}

Two things the mark scheme always checks:

  • Excess NH3\text{NH}_3. The product amine is ITSELF a nucleophile — if ammonia runs short, the product attacks more halogenoalkane and you get a soup of 2∘2^\circ and 3∘3^\circ amines. A large excess of ammonia makes NH3\text{NH}_3 the nucleophile that wins every collision.
  • The second ammonia. HBr\text{HBr} is expelled in the substitution; a second NH3\text{NH}_3 mops it up as NH4Br\text{NH}_4\text{Br}. Equations written 1:1 with HBr\text{HBr} lose the co-product mark.

Route (b): halogenoalkane + primary amine → secondary amine. Same mechanism, but the nitrogen nucleophile arrives already carrying an R group — so the product's nitrogen carries TWO: conditions are the primary amine in ethanol, heated in a sealed tube / under pressure:

CH3CH2Br+2CH3NH2→CH3CH2NHCH3+CH3NH3Br\text{CH}_3\text{CH}_2\text{Br} + 2\text{CH}_3\text{NH}_2 \rightarrow \text{CH}_3\text{CH}_2\text{NHCH}_3 + \text{CH}_3\text{NH}_3\text{Br}

The attacking nitrogen keeps its methyl: the product is the secondary amine CH3CH2NHCH3\text{CH}_3\text{CH}_2\text{NHCH}_3, and the second amine molecule takes the HBr\text{HBr} away as the methylammonium salt — the same mopping-up bookkeeping as route (a).

The two reduction routes — build −CH₂NH₂ onto an existing N

Both reductions add four hydrogens across a multiple bond next to the nitrogen — equations write them as the symbol [H][\text{H}] — and both use the strongest hydride on the course.

Route (c): amide reduction. LiAlH4\text{LiAlH}_4 in dry ether reduces the CO\text{CO} group of an amide to CH2\text{CH}_2 — the nitrogen and its substituents ride through untouched:

RCONH2+4[H]→RCH2NH2+H2O\text{RCONH}_2 + 4[\text{H}] \rightarrow \text{RCH}_2\text{NH}_2 + \text{H}_2\text{O}

Count the carbons: the carbonyl carbon STAYS, so an amide reduction never changes the chain length. A substituted amide reduces the same way, keeping its N-substituent: CH3CONHCH3→CH3CH2NHCH3\text{CH}_3\text{CONHCH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{NHCH}_3 — the reaction §04 returns to.

Route (d): nitrile reduction. LiAlH4\text{LiAlH}_4 in dry ether, or H2\text{H}_2 with a Ni catalyst and heat:

RCN+4[H]→RCH2NH2\text{RCN} + 4[\text{H}] \rightarrow \text{RCH}_2\text{NH}_2

Here the nitrile carbon becomes the −CH2−-\text{CH}_2- — the one door that ADDS a carbon, unlike route (c). Pair it with the AS substitution RX+KCN\text{RX} + \text{KCN} in ethanol →RCN\rightarrow \text{RCN} and you have the standard two-step that takes a halogenoalkane to an amine with ONE EXTRA CARBON:

CH3CH2Br→KCN in ethanolCH3CH2CN→LiAlH4 or H2/NiCH3CH2CH2NH2\text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{KCN in ethanol}} \text{CH}_3\text{CH}_2\text{CN} \xrightarrow{\text{LiAlH}_4 \text{ or } \text{H}_2/\text{Ni}} \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2

Note NaBH4\text{NaBH}_4 is too weak for BOTH reduction routes: nitriles and amides need LiAlH4\text{LiAlH}_4 or H2/Ni\text{H}_2/\text{Ni}.

the four-route map into the amine family1 · SUBSTITUTIONRX + 2NH₃ (EXCESS)ethanol · heat under pressure→ RNH₂ + NH₄XCARBON COUNT UNCHANGED2 · SUBSTITUTIONRX + 2R′NH₂ethanol · sealed tube / pressure→ RNHR′ + R′NH₃XGIVES THE SECONDARY AMINER–NH₂the primary-amine targeta 2° amine RNHR′ exitsROUTE 2 instead3 · REDUCTIONRCONH₂ + LiAlH₄ (dry ether)→ RCH₂NH₂SAME CARBON COUNT4 · REDUCTIONRCN + LiAlH₄ or H₂/Ni→ RCH₂NH₂+1 CARBONdetour in first: RX + KCN (ethanol) → RCNchoose the door by CARBON COUNT — and by which N substituents the target carries

Four-route map into the amine family. Centre: the target hub R–NH₂ (primary amine), with a note that a 2° amine RNHR′ exits route 2 instead. Four numbered cards feed the hub. Route 1, substitution: RX + 2NH₃ (excess), ethanol, heat under pressure → RNH₂ + NH₄X, tagged 'carbon count unchanged'. Route 2, substitution: RX + 2R′NH₂, ethanol, sealed tube → RNHR′ + R′NH₃X, tagged 'gives the SECONDARY amine'. Route 3, reduction: RCONH₂ + LiAlH₄ (dry ether) → RCH₂NH₂, tagged 'same carbon count'. Route 4, reduction: RCN + LiAlH₄ or H₂/Ni → RCH₂NH₂, tagged '+1 carbon', with the KCN detour shown beneath it: RX + KCN in ethanol → RCN. A footer bar reads: choose the door by carbon count and by which N substituents the target carries.

route

reagents and conditions

product

carbon count

(a) RX + NH₃

NH₃ in ethanol, heated under pressure (excess NH₃)

1° amine RNH₂

unchanged

(b) RX + 1° amine

R′NH₂ in ethanol, sealed tube / pressure

2° amine RNHR′

unchanged

(c) amide reduction

LiAlH₄ in dry ether

1° (or substituted) amine

unchanged (CO → CH₂)

(d) nitrile reduction

LiAlH₄ in dry ether, OR H₂/Ni + heat

1° amine RCH₂NH₂

+1 (via KCN detour)

The four doors at a glance. The carbon-count column is the fastest route-picker in a synthesis question.

Choosing the route in a synthesis question
  1. 1

    Compare carbon counts: starting material vs target amine. One extra carbon in the target forces the nitrile detour (RX→RCN\text{RX} \rightarrow \text{RCN} with KCN\text{KCN} in ethanol, then reduce); same count rules it out.

    The count is printed in the formulae — a 10-second comparison decides the entire route before any chemistry is quoted.

  2. 2

    Check the target's nitrogen: a 1∘1^\circ amine RNH2\text{RNH}_2 can come from routes (a), (c) or (d); a 2∘2^\circ amine RNHR′\text{RNHR}' needs route (b) — or the reduction of a substituted amide (§02).

    Route (b) is the ONLY substitution door to a secondary amine — questions exploit it as the odd one out.

  3. 3

    Quote conditions in full: ethanol + heat under pressure for the substitutions; dry ether for LiAlH4\text{LiAlH}_4; heat for the H2/Ni\text{H}_2/\text{Ni} option.

    Conditions marks are separate credit lines — 'NH₃' alone scores the reagent mark and drops the conditions mark every time.

  4. 4

    If an equation is asked for, write it with the mopping-up second equivalent (2NH3→RNH2+NH4X2\text{NH}_3 \rightarrow \text{RNH}_2 + \text{NH}_4\text{X}) or with [H][\text{H}] for the reductions.

    The 2-equivalent form is the credited skeleton in this family; the [H] form is how the reductions are printed in mark schemes.

Worked demo — two plans for one amine

Devise a synthesis of propylamine, CH3CH2CH2NH2\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2:

(a) from 1-bromopropane, CH3CH2CH2Br\text{CH}_3\text{CH}_2\text{CH}_2\text{Br};

(b) from bromoethane, CH3CH2Br\text{CH}_3\text{CH}_2\text{Br}.

In each case give the reagents and conditions of every step, and write the equation for the final step of route (a).

Show full working
  1. 1

    (a) Count carbons: C3→C3\text{C}_3 \rightarrow \text{C}_3, no change — the direct substitution door works: NH3\text{NH}_3 in ethanol, heated under pressure, with the ammonia in excess.

    Same carbon count rules out the nitrile immediately — no KCN anywhere in this route.

  2. 2

    (a) Equation: CH3CH2CH2Br+2NH3→CH3CH2CH2NH2+NH4Br\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + 2\text{NH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 + \text{NH}_4\text{Br}.

    Two NH₃ in, amine plus NH₄Br out — the second ammonia's mopping-up job earns its own credit line.

  3. 3

    (b) Count carbons: C2→C3\text{C}_2 \rightarrow \text{C}_3 — ONE extra carbon is needed, so the nitrile detour is forced: step 1, KCN\text{KCN} in ethanol (substitution): CH3CH2Br→CH3CH2CN\text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2\text{CN}.

    The CN carbon JOINS the chain — this is where the extra carbon enters. Forgetting the KCN step and reducing bromoethane directly is impossible: there is no N to reduce.

  4. 4

    (b) Step 2, reduce the nitrile: LiAlH4\text{LiAlH}_4 in dry ether (or H2/Ni\text{H}_2/\text{Ni}, heat): CH3CH2CN+4[H]→CH3CH2CH2NH2\text{CH}_3\text{CH}_2\text{CN} + 4[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2.

    4[H] across the C≡N gives the −CH₂NH₂ — the nitrile carbon becomes the new CH₂, completing the chain extension.

Answer

(a) excess NH₃ in ethanol, heat under pressure → CH₃CH₂CH₂NH₂ + NH₄Br · (b) KCN in ethanol, then LiAlH₄/dry ether (or H₂/Ni) → CH₃CH₂CH₂NH₂

Carbon count first, nitrogen substituents second, conditions third — the same order the marks are paid in.

Your turn

Two route-planning drills and one equation-writing set — all invented, all on the four-door map.

  1. 12 marks

    Devise a synthesis of butan-1-amine, CH3CH2CH2CH2NH2\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2, starting from 1-bromopropane, CH3CH2CH2Br\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}. Give the reagents and conditions of each step.

    Stuck? Show hint

    Compare carbon counts first — the answer is forced before any chemistry is quoted.

    Show solution
    1. 1

      Count: C3→C4\text{C}_3 \rightarrow \text{C}_4 — one extra carbon, so the nitrile detour is mandatory: CH3CH2CH2Br+KCN\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{KCN} in ethanol →CH3CH2CH2CN\rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CN}.

      KCN's carbon becomes part of the chain — the substitution that ADDS a carbon rather than swapping a group.

    2. 2

      Reduce the nitrile: LiAlH4\text{LiAlH}_4 in dry ether (or H2/Ni\text{H}_2/\text{Ni}, heat): CH3CH2CH2CN+4[H]→CH3CH2CH2CH2NH2\text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + 4[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2.

      The C≡N takes 4[H] to become −CH₂NH₂; the direct route (a) would have kept three carbons — the wrong product.

    Answer

    KCN in ethanol → CH₃CH₂CH₂CN; then LiAlH₄ in dry ether (or H₂/Ni, heat) → CH₃CH₂CH₂CH₂NH₂

  2. 22 marks

    (a) Write the equation for the reaction of bromoethane with an excess of ethylamine, CH3CH2NH2\text{CH}_3\text{CH}_2\text{NH}_2, in ethanol in a sealed tube.

    (b) Name the organic product and state its amine class.

    Stuck? Show hint

    The attacking nitrogen arrives carrying an ethyl group — and keeps it.

    Show solution
    1. 1

      (a) CH3CH2Br+2CH3CH2NH2→CH3CH2NHC2H5+CH3CH2NH3Br\text{CH}_3\text{CH}_2\text{Br} + 2\text{CH}_3\text{CH}_2\text{NH}_2 \rightarrow \text{CH}_3\text{CH}_2\text{NHC}_2\text{H}_5 + \text{CH}_3\text{CH}_2\text{NH}_3\text{Br} — the second amine molecule mops up the HBr\text{HBr}.

      Same skeleton as the ammonia route: two nucleophile equivalents in, product plus ammonium salt out.

    2. 2

      (b) The product is diethylamine — a secondary amine (its nitrogen carries two ethyl groups and one H).

      Route (b) is the syllabus's only substitution door to a 2° amine: the nucleophile's own R group is what makes the product secondary.

    Answer

    (a) C₂H₅Br + 2C₂H₅NH₂ → (C₂H₅)₂NH + C₂H₅NH₃Br · (b) diethylamine, a secondary amine

  3. 32 marks

    Complete each reduction with reagents and conditions, and state what happens to the carbon count in each:

    (a) CH3CH2CONH2→\text{CH}_3\text{CH}_2\text{CONH}_2 \rightarrow an amine;

    (b) CH3CH2CN→\text{CH}_3\text{CH}_2\text{CN} \rightarrow an amine.

    Stuck? Show hint

    One keeps the count, one adds it — and NaBH₄ is too weak for both.

    Show solution
    1. 1

      (a) LiAlH4\text{LiAlH}_4 in dry ether: CH3CH2CONH2+4[H]→CH3CH2CH2NH2+H2O\text{CH}_3\text{CH}_2\text{CONH}_2 + 4[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 + \text{H}_2\text{O} — propanamide to propylamine, carbon count UNCHANGED (the carbonyl C becomes the −CH2−-\text{CH}_2-).

      Amide reduction swaps C=O for CH₂ — the carbon stays aboard, so C₃ in gives C₃ out.

    2. 2

      (b) LiAlH4\text{LiAlH}_4 in dry ether OR H2/Ni\text{H}_2/\text{Ni} with heat: CH3CH2CN+4[H]→CH3CH2CH2NH2\text{CH}_3\text{CH}_2\text{CN} + 4[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 — propanenitrile to propylamine, carbon count +1 relative to the halogenoalkane it came from (the CN carbon joins the chain).

      Both routes land on the same amine here — the difference is invisible in the product and lives entirely in the count bookkeeping.

    Answer

    (a) LiAlH₄/dry ether → CH₃CH₂CH₂NH₂ + H₂O (count unchanged) · (b) LiAlH₄ or H₂/Ni → CH₃CH₂CH₂NH₂ (nitrile route adds a C)

Common mistakes

  • Writing 'NH₃ (aq)' for route (a)

    NH₃ in ETHANOL, heated under pressure.

    The syllabus names the solvent and the pressure — conditions marks are paid per clause, and aqueous ammonia is not the credited card.

  • Writing RX+NH3→RNH2+HBr\text{RX} + \text{NH}_3 \rightarrow \text{RNH}_2 + \text{HBr} with one ammonia

    RX+2NH3→RNH2+NH4X\text{RX} + 2\text{NH}_3 \rightarrow \text{RNH}_2 + \text{NH}_4\text{X} (excess NH₃).

    The second ammonia is both the neutralising agent and the reason the product stays a 1° amine — the equation and the 'why excess' clause are the same mark family.

  • Reducing a nitrile or amide with NaBH₄

    LiAlH₄ in dry ether (or H₂/Ni for the nitrile).

    NaBH₄ reduces aldehydes and ketones only — nitriles and amides need the stronger hydride, and the wrong reagent forfeits the whole step.

  • Using the nitrile detour when the carbon count is unchanged

    Count first: nitrile adds one carbon; direct substitution keeps it.

    The KCN step is the examiner's favourite trap in reverse — inserting it where the count says 'same' gives the next homologue, not the target.

Practise the four amine routes — reagents, conditions and route choiceReal past-paper questions · Amines: preparation and basicity

The rest of this note

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Can you do all of these?

  • I can write all four amine routes with their conditions (ethanol + pressure for the substitutions; dry ether for LiAlH₄) and choose between them by carbon count

  • I can explain why EXCESS NH₃ is used in route (a) and why a second amine/NH₃ molecule appears in every equation (mopping up HX)

  • I can write both acyl-chloride-to-amide equations cold, with NH₄Cl / amine salt as the paid co-product, and name N-substituted amides

  • I can rank amide < phenylamine < ammonia < alkylamine and give all four lone-pair clauses, link clause first

  • I can hydrolyse an amide with aq acid AND aq alkali, switching products correctly (acid + ammonium salt vs carboxylate + amine), and reduce one with LiAlH₄

  • I can take benzene to phenylamine in three stages and write the 6[H] equation, remembering the NaOH step exists because the reduction runs in acid

  • I can state the bromine-water observations for phenylamine (white ppt) and ethylamine (nothing) and name 2,4,6-tribromophenylamine

  • I can diazotise below 10 °C and warm the salt with water to phenol — reagents, temperatures and the reason for the ice

  • I can run the azo coupling in NaOH(aq) and spot the –N=N– group in any printed dye structure

  • I can draw an amino acid's cation, zwitterion and anion against pH, define the isoelectric point, and predict the dominant form at any pH

  • I can draw a dipeptide with a fully displayed peptide bond in the correct N-terminal-first order, and read an electrophoresis plate from charge and MrM_r

Now do the questions
222 real Paper 4 parts from 2021–2025, sorted by difficulty, with mark schemes