Cambridge International AS & A Level Chemistry 9701 · A Level topic 34

Nitrogen compounds (A Level)

What this chapter covers34.1–34.4

The smell of rotting fish is due to amines; so is the physiological action of many drugs, from local anaesthetics to antihistamines. Proteins, the working molecules of every cell, are long chains of amino acids joined by amide bonds, and the dyes that coloured the first synthetic fabrics were azo compounds made from phenylamine. All of these compounds contain nitrogen with a lone pair of electrons, and how available that lone pair is decides much of their chemistry.

This chapter covers four families. Amines are derivatives of ammonia in which hydrogen atoms are replaced by carbon-containing groups; they are bases and nucleophiles. Phenylamine is an aromatic amine whose lone pair is shared with the benzene ring, giving it distinctive chemistry and making it the starting point for azo dyes. Amides contain the –CONH– group and are neutral. Amino acids carry both an amine and a carboxylic acid group; they exist as zwitterions, join to form peptides, and can be separated by electrophoresis.

What topic 34 asks you to do

34.1 Primary and secondary amines — recall their production from halogenoalkanes with NH₃ or a primary amine (in ethanol, heated under pressure / in a sealed tube), and by reduction of amides (LiAlH₄) and nitriles (LiAlH₄ or H₂/Ni); describe the condensation of ammonia or amines with acyl chlorides; describe and explain the basicity of aqueous amines.

34.2 Phenylamine and azo compounds — the preparation of phenylamine from benzene; its reactions with Br₂(aq) and with nitrous acid below 10 °C, and warming the diazonium salt with water; the relative basicities of ammonia, ethylamine and phenylamine; coupling of benzenediazonium chloride with phenol in NaOH(aq), the azo group, and azo dyes.

34.3 Amides — their formation from acyl chlorides; hydrolysis with aqueous acid or alkali; reduction with LiAlH₄; why amides are much weaker bases than amines.

34.4 Amino acids — acid–base properties, zwitterions and the isoelectric point; formation of peptide bonds in di- and tripeptides; interpreting electrophoresis of amino acids and dipeptides at different pH.

What you are assumed to know already

Nucleophilic substitution of halogenoalkanes, including with ammonia; nitriles and their reduction (topics 15 and 18).
Brønsted–Lowry acids and bases; buffers (topics 7 and 26).
Electrophilic substitution and nitration of benzene; phenol and diazonium salts (topics 30 and 32).
Acyl chlorides and the addition–elimination mechanism (topic 33).

Amines: structure and classification34.1

Replacing one hydrogen atom of ammonia with an alkyl or aryl group gives a primary amine, RNH₂; replacing two gives a secondary amine, R₂NH; replacing three gives a tertiary amine, R₃N. The nitrogen atom keeps one lone pair and, in an amine with three single bonds, is sp³ hybridised with a pyramidal shape. Simple amines are named from the alkyl group: CH₃NH₂ is methylamine, CH₃CH₂NH₂ ethylamine, (CH₃CH₂)₂NH diethylamine, C₆H₅NH₂ phenylamine.

The lone pair gives amines their two characteristic roles: as bases (the lone pair accepts a proton) and as nucleophiles (the lone pair attacks an electron-deficient carbon atom).

Making amines34.1.1

From halogenoalkanes

Ammonia is a nucleophile. When a halogenoalkane is heated with ammonia dissolved in ethanol, the lone pair on nitrogen attacks the δ+ carbon and displaces the halide ion (nucleophilic substitution). The reaction is carried out in a sealed tube or under pressure, because ammonia is a gas and would otherwise escape from the hot mixture. The ammonium salt formed first loses a proton to more ammonia, releasing the amine:

CH₃CH₂Br + NH₃ → CH₃CH₂NH₂ + HBr NH₃ in ethanol, heat under pressure

The product is itself a nucleophile, so it competes with ammonia for the halogenoalkane. Ethylamine reacts with bromoethane to give diethylamine, which reacts again to give triethylamine, and finally a quaternary ammonium salt, (CH₃CH₂)₄N⁺Br⁻. An excess of ammonia makes the primary amine the main product; an excess of halogenoalkane pushes the substitution further. For the same reason, a secondary amine is made by heating a halogenoalkane with a primary amine in ethanol in a sealed tube:

CH₃CH₂Br + CH₃CH₂NH₂ → (CH₃CH₂)₂NH + HBr

AnimationAmines as nucleophiles

An amine attacks a halogenoalkane; the product amine can attack again, giving secondary and tertiary amines and a quaternary ammonium salt.

By reduction of nitriles and amides

Nitriles are reduced to primary amines by lithium tetrahydridoaluminate, LiAlH₄, in dry ether, or by hydrogen with a nickel catalyst. Because the nitrile carbon becomes a CH₂ group, the amine has one more carbon than the halogenoalkane used to make the nitrile — a useful way of lengthening a chain.

CH₃CN + 4[H] → CH₃CH₂NH₂ LiAlH₄ in dry ether, or H₂/Ni

Amides are reduced by LiAlH₄: the C=O group becomes CH₂, and the oxygen leaves as water.

CH₃CONH₂ + 4[H] → CH₃CH₂NH₂ + H₂O LiAlH₄ in dry ether

Figure 34.1 Four routes to amines. Substitution of halogenoalkanes gives mixtures; reduction of nitriles and amides gives a single primary amine.

Balancing reductions with [H]

Count the hydrogen atoms needed. A nitrile gains four H atoms (C≡N → CH₂–NH₂). An amide gains four H atoms and loses its oxygen as water — writing ½O₂ as the product is a common error.

Amines with acyl chlorides34.1.2

As nucleophiles, ammonia and primary and secondary amines react with acyl chlorides at room temperature by addition–elimination (chapter 33), forming amides and hydrogen chloride. The reaction is a condensation: two molecules join and a small molecule (HCl) is lost.

CH₃COCl + CH₃NH₂ → CH₃CONHCH₃ + HCl

Because the HCl reacts with any excess amine, forming an alkylammonium salt, two moles of amine are often used for each mole of acyl chloride.

AnimationAmines with acyl chlorides

The addition–elimination reaction of an amine with an acyl chloride, giving an N-substituted amide.

Amines as bases34.1.3, 34.2.3

An amine dissolves in water to give an alkaline solution. The lone pair on nitrogen accepts a proton from water, forming a coordinate bond, and hydroxide ions are left in the solution:

CH₃CH₂NH₂(aq) + H₂O(l) ⇌ CH₃CH₂NH₃⁺(aq) + OH⁻(aq)

Amines are weak bases: the equilibrium lies well to the left. With strong acids they react completely to form salts, which are ionic, crystalline and soluble in water:

CH₃CH₂NH₂ + HCl → CH₃CH₂NH₃⁺Cl⁻ ethylammonium chloride

AnimationAmines as Brønsted–Lowry bases

The nitrogen lone pair accepts a proton, forming an alkylammonium ion and leaving hydroxide ions in solution.

Explaining relative base strength

The strength of a base depends on how available the nitrogen lone pair is to accept a proton — in other words, on the electron density on the nitrogen atom.

Ethylamine > ammonia. An alkyl group is electron-donating (a positive inductive effect). It pushes electron density onto the nitrogen, so the lone pair is more available to accept a proton. The alkylammonium ion formed is also stabilised by the same effect. Two alkyl groups increase the effect further: diethylamine is a stronger base than ethylamine.
Ammonia > phenylamine. In phenylamine the lone pair on nitrogen occupies a p-type orbital that overlaps with the π system of the ring. The lone pair is partly delocalised into the ring, so the electron density on nitrogen is lower and the lone pair is less available to accept a proton.
Electron-withdrawing groups on the ring weaken the base further. A nitro group on the ring of phenylamine draws even more electron density from the nitrogen: 4-nitrophenylamine is a weaker base than phenylamine.

diethylamine > ethylamine > ammonia > phenylamine > 4-nitrophenylamine

Figure 34.2 The availability of the nitrogen lone pair decides base strength. Alkyl groups increase it; delocalisation into a ring or onto a C=O group decreases it.

Loading the model…

The words that score

Every explanation of base strength should mention the lone pair on nitrogen and its ability to accept a proton (or form a coordinate bond to one). "Attract a proton" or "be attacked by a proton" do not score. Then give, for each compound, the group and its effect on the electron density on N: alkyl groups donate; the ring and the C=O group withdraw by delocalisation; NO₂ withdraws.

Quick check 34.1

Give the reagent and conditions for making ethylamine from bromoethane, and name the mechanism.
answer
Ammonia in ethanol, heated under pressure (sealed tube); nucleophilic substitution.
Why does this reaction also give diethylamine?
answer
Ethylamine has a lone pair and is itself a nucleophile, so it substitutes into more bromoethane.
Write the equation for the reduction of propanenitrile with LiAlH₄, using [H].
answer
CH₃CH₂CN + 4[H] → CH₃CH₂CH₂NH₂
Write the equation for methylamine acting as a base in water.
answer
CH₃NH₂ + H₂O ⇌ CH₃NH₃⁺ + OH⁻
Explain why phenylamine is a weaker base than ammonia.
answer
The N lone pair is delocalised into the benzene ring, lowering the electron density on N, so it is less able to accept a proton.

Past-paper practice · Set 34A · Making amines and comparing their basicity

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 34A.1[7]

Parts (c) and (d) of the same question are 34B.2.

Answer and marking guidance

(a)(i) CH₃CONH₂ + 4[H] → CH₃CH₂NH₂ + H₂O ✔. (ii) mechanism: nucleophilic substitution ✔; conditions: (ammonia in) ethanol, heated under pressure or in a sealed tube ✔. (iii) (CH₃CH₂)₂NH or (CH₃CH₂)₃N ✔. (b) Order: phenylamine < ammonia < ethylamine ✔; explanation, two points 1 mark, three 2, all four 3 ✔✔✔: basicity is the ability to accept a proton, or to donate the N lone pair to a proton; in phenylamine the lone pair (p orbital) on N is delocalised into / overlaps with the ring; in ethylamine the ethyl group is electron-donating (+I); this increases the electron density on N (ethylamine), or decreases it (phenylamine). Examiner insight: in (a)(i) a common error was CH₃CONH₂ + 2[H] → CH₃CH₂NH₂ + ½O₂. In (a)(ii) most named the mechanism but the conditions were less well known. (a)(iii) was found difficult: H₂NCH₂CH₂NH₂ and (CH₃)₂NH were common errors. In (b) most recognised lone-pair delocalisation in phenylamine and the electron-donating ethyl group, but the explanation of their effect on the electron density on N was normally absent.

Question 34A.2[4]

Answer and marking guidance

Order: ethylamine > ammonia > phenylamine ✔; the ethyl group is electron-donating ✔; the p orbital of N in phenylamine overlaps with the π ring system, or the lone pair on N is delocalised into the ring ✔; basicity is linked to the ability of N to accept a proton ✔. Examiner insight: many good answers. Specific vocabulary was needed: basicity depends on the tendency of the lone pair to be protonated, or to receive a proton; "the tendency of a substance to be attacked by a proton" was not accepted. A few answered about acidity instead.

Question 34A.3[3]

Part (b) of the same question is 34C.2.

Answer and marking guidance

Order: cyclohexylamine > ammonia > phenylamine ✔; two of three for 1 mark, all three for 2 ✔✔: basicity linked to the ability of N to accept a proton / donate its lone pair; the cyclohexyl (alkyl) group is electron-donating and increases the electron density on N; the lone pair (p orbital) of N in phenylamine is delocalised into the ring, decreasing the electron density on N. Examiner insight: challenging and discriminating; answers were often not detailed enough. For each substance a clear statement linked to the ability of the N lone pair to accept a proton was required. The most common error was leaving the "lone pair" out of the explanation.

Question 34A.4[4]

Part (b) of the same question is 34B.3.

Answer and marking guidance

M1 order: ethylamine > phenylamine > 4-nitrophenylamine ✔. Two points 1 mark, three 2, four 3 ✔✔✔: basicity linked to the lone pair on N accepting (coordinating to) a proton; the ethyl group is electron-donating (+I), so the lone pair is more available; in phenylamines the lone pair (p orbital) on N is delocalised into the ring, so it is less available; the NO₂ group is electron-withdrawing, so the lone pair of 4-nitrophenylamine is even less available. Examiner insight: challenging; for each substance a clear statement about the ability of the N lone pair to accept a proton was needed. Errors included omitting "lone pair", placing 4-nitrophenylamine above phenylamine, and calling NO₂ electron-donating.

Question 34A.5[7]

Answer and marking guidance

(a) M1 order: U (phenylmethanamine, C₆H₅CH₂NH₂) > T (phenylamine) > S (benzamide) ✔; two of ✔✔: the alkyl (CH₂) group is electron-donating, so the lone pair is more able to accept a proton; in T the lone pair on N overlaps with the delocalised ring, so it is less able to accept a proton; in S the electron-withdrawing oxygen / carbonyl group means the lone pair is not available to accept a proton, or amides are neutral. (b)(i) reaction 1: LiAlH₄ ✔; reaction 2: NH₃ heated under pressure or in a sealed tube ✔. (ii) reaction 1: reduction ✔; reaction 2: nucleophilic substitution ✔.

Examiner's overall observation · Making amines and explaining basicity

Answered well: the mechanism for making amines from halogenoalkanes (nucleophilic substitution); the order of basicity in most comparisons; recognising that the alkyl group is electron-donating and that the N lone pair of phenylamine is delocalised into the ring.

Found difficult: the conditions for making amines from halogenoalkanes — many omitted the pressure or sealed tube; suggesting a secondary or tertiary amine that forms as a by-product; explaining basicity fully. For each compound a clear statement linked to the ability of the N lone pair to accept a proton was needed, and the effect of the group on the electron density on N was normally absent.

Recurring errors: CH₃CONH₂ + 2[H] → CH₃CH₂NH₂ + ½O₂; leaving "lone pair" out of the explanation; "attract a proton" or "attacked by a proton"; placing 4-nitrophenylamine above phenylamine or calling NO₂ electron-donating; answering about acidity instead of basicity.

What successful answers did: began from the lone pair's ability to accept a proton, and then explained each compound's position by the electron-donating or electron-withdrawing effect of its groups on the electron density on nitrogen.

Making phenylamine34.2.1

Phenylamine is made from benzene in two stages.

Nitration. Benzene is warmed with a mixture of concentrated nitric and concentrated sulfuric acids between 25 °C and 60 °C, giving nitrobenzene by electrophilic substitution (chapter 30).
Reduction. Nitrobenzene is heated under reflux with tin and concentrated hydrochloric acid. The nitro group is reduced to an amine group; because the mixture is strongly acidic, the product is present as the phenylammonium ion, C₆H₅NH₃⁺. Sodium hydroxide solution is then added to remove the proton and release phenylamine, which is separated from the mixture.

C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O hot Sn and concentrated HCl, then NaOH(aq)

Figure 34.3 Making phenylamine from benzene, and its reactions with bromine water and with nitrous acid. The diazonium salt is the starting point both for phenol and for azo dyes.

Conditions that are often incomplete

Nitration: concentrated HNO₃ and concentrated H₂SO₄; HNO₂ is a different reagent.
Reduction: tin and concentrated HCl, heated; then NaOH(aq).
Reduction equation: six [H] and two H₂O, not O₂ as a product.
The intermediate is nitrobenzene, a compound — not the nitronium ion or the arenium intermediate.

Phenylamine with bromine water34.2.2(a)

The nitrogen lone pair that is delocalised into the ring makes the ring far more electron-rich than benzene, just as the oxygen lone pair does in phenol. Phenylamine therefore reacts at once with bromine water at room temperature, without a catalyst. The orange colour is removed and a white precipitate of 2,4,6-tribromophenylamine forms: the –NH₂ group directs substitution to the 2-, 4- and 6-positions.

C₆H₅NH₂ + 3Br₂ → C₆H₂Br₃NH₂ + 3HBr

The equation must be balanced with 3HBr, and the bromine atoms go to positions 2, 4 and 6 — not 3 and 5.

Diazotisation and the diazonium ion34.2.2(b)

Phenylamine reacts with nitrous acid, HNO₂, generated in the mixture from sodium nitrite and dilute hydrochloric acid, at a temperature below 10 °C. The product is benzenediazonium chloride, C₆H₅N₂⁺Cl⁻, in which the positive charge is on the nitrogen atom bonded to the ring (–N⁺≡N):

C₆H₅NH₂ + HNO₂ + HCl → C₆H₅N₂⁺Cl⁻ + 2H₂O below 10 °C

The diazonium salt decomposes above about 10 °C, which is why the mixture is kept in ice. When the solution is warmed with water, nitrogen is evolved and phenol forms (chapter 32):

C₆H₅N₂⁺ + H₂O → C₆H₅OH + N₂ + H⁺

Azo compounds and dyes34.2.4

Kept cold, the diazonium ion is a weak electrophile. It attacks the very electron-rich ring of a phenoxide ion, so when benzenediazonium chloride is added to phenol dissolved in sodium hydroxide solution below 10 °C, a yellow-orange precipitate of an azo compound forms immediately. This coupling reaction is an electrophilic substitution, usually at the 4-position of the phenol:

C₆H₅N₂⁺ + C₆H₅OH → C₆H₅–N=N–C₆H₄OH + H⁺ NaOH(aq), below 10 °C

The product contains the azo group, –N=N–, linking two aromatic rings. The azo group joins the delocalised systems of both rings into one extended system, which absorbs visible light: azo compounds are strongly coloured and are widely used as dyes. Other azo dyes are made by the same two-step route, changing the aromatic amine (for example 4-nitrophenylamine) or the coupling partner (for example 1-naphthol or 2-naphthol).

Worked example 34.1 · Designing an azo dye synthesis

Problem	The dye O₂NC₆H₄–N=N–(naphthol) is made from 4-nitrophenylamine in two steps. Give the intermediate and the reagents and conditions.
Reasoning	Work backwards from the azo group: one side came from a diazonium ion, the other from the coupling partner, which carries the –OH group — here a naphthol. The amine side must be the one that was diazotised.
Answer	Step 1: NaNO₂ and dilute HCl (HNO₂), below 10 °C, giving the diazonium salt O₂NC₆H₄N₂⁺Cl⁻. Step 2: add to the naphthol dissolved in NaOH(aq), below 10 °C.
Check	The + charge of the diazonium ion is on the N attached to the ring; the product contains –N=N–, not –N≡N–, and the –OH is kept.

Quick check 34.2

Give the reagents and conditions for converting nitrobenzene into phenylamine.
answer
Heat under reflux with tin and concentrated HCl, then add NaOH(aq).
Write the equation for phenylamine with excess bromine water, and give two observations.
answer
C₆H₅NH₂ + 3Br₂ → C₆H₂Br₃NH₂ + 3HBr; bromine decolourised, white precipitate.
Why must the diazotisation be carried out below 10 °C?
answer
The diazonium salt decomposes above about 10 °C, giving phenol and nitrogen.
Identify the functional group responsible for the colour of azo dyes.
answer
The azo group, –N=N–, linking two aromatic rings.
What conditions are needed for the coupling reaction with phenol?
answer
Phenol dissolved in NaOH(aq) (alkaline), below 10 °C.

Past-paper practice · Set 34B · Phenylamine, diazonium salts and azo dyes

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 34B.1[8]

Answer and marking guidance

(b)(iv) tin and HCl ✔; concentrated, and heat (reflux) ✔. (c)(i) C₆H₅NH₂ + 3Br₂ → C₆H₂Br₃NH₂ + 3HBr ✔. (ii) 2,4,6-tribromophenylamine ✔. (iii) the bromine is decolourised AND a white precipitate forms ✔. (d) Order: phenylamine < ammonia < ethylamine ✔; the lone pair on N of phenylamine is delocalised into the ring, and the alkyl group of ethylamine is electron-donating ✔; a correct statement about the availability of the lone pair to accept a proton ✔. Examiner insight: (b)(iv) discriminated well. In (c)(i) many knew the formula of 2,4,6-tribromophenylamine but did not balance the equation with 3HBr; (c)(ii) and (iii) discriminated well — both the decolourisation and the white precipitate are seen. (d) was difficult for many, but good answers were built on one central theme: the relative ability of lone pairs in different environments to accept H⁺.

Question 34B.2[8]

Parts (a) and (b) of the same question are 34A.1.

Answer and marking guidance

(c) intermediate: nitrobenzene ✔; step 1: concentrated HNO₃ and concentrated H₂SO₄ (25–60 °C) ✔; step 2: (reduction with) Sn and concentrated HCl, heat ✔. (d)(i) W is 2,4,6-tribromophenylamine ✔. (ii) nitrous acid, HNO₂, or NaNO₂ and dilute HCl ✔. (iii) X is phenol ✔. (iv) NaOH / alkali ✔. (v) dyes ✔. Examiner insight: in (c), common errors were omitting "concentrated", using HNO₂ instead of HNO₃, and drawing an intermediate ion instead of the intermediate compound. In (d)(i) the most common error was 3,5-dibromophenylamine. In (d)(iv) many omitted the alkaline conditions and gave only a low temperature, which was not enough.

Question 34B.3[3]

Part (a) of the same question is 34A.4.

Answer and marking guidance

(i) Q is the diazonium salt O₂NC₆H₄N₂⁺Cl⁻ (–N⁺≡N, with the + on the N bonded to the ring) ✔. (ii) step 1: HNO₂, or NaNO₂ and HCl(aq), at or below 10 °C ✔; step 2: 1-naphthol in NaOH / alkaline solution ✔. Examiner insight: many diagrams of Q were well drawn; the most common error was the + charge on the terminal nitrogen. Step 1 was usually correct, but only stronger candidates identified 1-naphthol and alkaline conditions for step 2; many suggested phenol or gave a molecular formula for 1-naphthol.

Examiner's overall observation · Phenylamine and azo compounds

Answered well: the product of phenylamine with bromine water and the name 2,4,6-tribromophenylamine; the reagents for diazotisation; phenol as the product of warming the diazonium salt; dyes as the use of azo compounds.

Found difficult: the conditions for coupling — many gave only a low temperature and omitted the alkaline conditions; identifying a naphthol as the coupling partner of an unfamiliar dye.

Recurring errors: omitting "concentrated" for nitration or reduction; HNO₂ instead of HNO₃ for nitration; drawing an intermediate ion instead of nitrobenzene; not balancing the bromination with 3HBr; 3,5-dibromophenylamine; the + charge on the terminal nitrogen of the diazonium ion; suggesting phenol when a naphthol was needed.

What successful answers did: gave every part of each set of conditions, balanced every equation, and placed substituents at the 2-, 4- and 6-positions of the activated ring.

Making amides34.3.1

An amide contains the group –CONH₂ (a primary amide) or –CONHR (an N-substituted amide). Amides are made at room temperature from acyl chlorides and ammonia or a primary amine, by addition–elimination (chapter 33):

CH₃COCl + NH₃ → CH₃CONH₂ + HCl ethanamide

CH₃COCl + CH₃CH₂NH₂ → CH₃CONHCH₂CH₃ + HCl N-ethylethanamide

If a molecule contains both an amine group and an acyl chloride group, it can react with itself. When the two groups are far enough apart to form a five- or six-membered ring, a cyclic amide forms on warming; otherwise molecules link end to end to form a polyamide (chapter 35).

Reactions of amides34.3.2

Hydrolysis

The amide bond is much less reactive than the C–Cl bond of an acyl chloride, and amides are hydrolysed only on heating under reflux with aqueous acid or aqueous alkali. The C–N bond is broken and water is added across it.

With dilute acid the products are the carboxylic acid and an ammonium salt (the NH₃ released is protonated by the acid):

CH₃CONH₂ + H₂O + HCl → CH₃COOH + NH₄Cl

With aqueous alkali the products are the carboxylate salt and ammonia, which is given off as a gas that turns damp red litmus blue:

CH₃CONH₂ + NaOH → CH₃COONa + NH₃

N-substituted amides give the corresponding amine (or its salt in acid) instead of ammonia. The same reaction breaks the peptide bonds of proteins, and the amide links of polyamides such as nylon.

Reduction

Lithium tetrahydridoaluminate in dry ether reduces the C=O group of an amide to CH₂, giving an amine with the same number of carbon atoms:

CH₃CONH₂ + 4[H] → CH₃CH₂NH₂ + H₂O

A cyclic amide is reduced in the same way to a cyclic amine, the ring staying intact.

Figure 34.4 Hydrolysis and reduction of ethanamide.

Why amides are such weak bases34.3.3

The nitrogen atom of an amide still has a lone pair, but it is next to a carbonyl group. The lone pair occupies a p-type orbital that overlaps with the π bond of C=O and is delocalised onto the C=O group, where the electronegative oxygen draws it away from nitrogen. The electron density on nitrogen is much lower than in an amine, and the lone pair is not available to accept a proton. Amides are therefore neutral in water: ethanamide does not turn red litmus blue.

diethylamine > ethylamine > ethanamide (base strength)

Amides are neutral

Do not protonate an amide group. In an amino acid such as asparagine, whose side chain is –CH₂CONH₂, the side chain stays –CONH₂ even at low pH; only the amine group is protonated. Writing –CONH₃⁺ is a common error.

Worked example 34.2 · A cyclic amide and its reduction

Problem	The compound H₂NCH₂CH₂CH₂CH₂COCl forms S, C₅H₉NO, when warmed. S is then reduced by LiAlH₄ to T. Suggest S and T and name the types of reaction.
Reasoning	The –NH₂ and –COCl groups in one molecule are separated by four CH₂ groups, so the N can attack the carbonyl carbon to close a six-membered ring (N, four CH₂ and C=O), losing HCl. C₅H₁₀NOCl − HCl = C₅H₉NO. LiAlH₄ reduces C=O to CH₂.
Answer	S: a six-membered ring containing –NH–C(=O)– and four CH₂ groups (a cyclic amide). T: the same ring with C=O reduced to CH₂ — a cyclic secondary amine, C₅H₁₁N. Step I: condensation; step II: reduction.
Check	The molecular formula of S matches; the amide in S is not basic, the amine in T is.

Quick check 34.3

Write the equation for the hydrolysis of propanamide by NaOH(aq).
answer
CH₃CH₂CONH₂ + NaOH → CH₃CH₂COONa + NH₃
What are the products when N-methylethanamide is heated with dilute HCl?
answer
Ethanoic acid and methylammonium chloride, CH₃NH₃⁺Cl⁻.
Name the reagent that converts ethanamide into ethylamine.
answer
LiAlH₄ (in dry ether).
Explain why ethanamide is not basic.
answer
The N lone pair is delocalised onto the C=O group, so it is not available to accept a proton.

Past-paper practice · Set 34C · Amides: formation, reactions and basicity

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 34C.1[4]

Parts (b)–(d) of the same question are 34D.1.

Answer and marking guidance

M1 order: diethylamine > ethylamine > ethanamide ✔; M2 basicity linked to the ability of the lone pair on N to accept a proton ✔; M3 the electron-donating ethyl groups increase the electron density on N / make the lone pair more available ✔; M4 in ethanamide the lone pair on N is delocalised into the C=O group ✔. Examiner insight: many gave the correct order; the explanation was challenging for weaker candidates. "Attract a proton" is not equivalent to "accept a proton": accept a proton, or form a coordinate bond to a proton, are the correct terms. The lack of basicity of an amide is due to delocalisation of the N lone pair into the C=O group.

Question 34C.2[6]

Part (a) of the same question is 34A.3.

Answer and marking guidance

(i) (nucleophilic) addition–elimination ✔. (ii) M1/M2 (two points 1 mark, four 2 marks): lone pair on N; arrow from the N lone pair to the C of C=O; dipole on C=O; arrow from the C=O bond to O ✔✔; M3 correct intermediate ✔; M4 arrow from the lone pair on O⁻ to the C–O bond AND arrow from the C–Cl bond to Cl ✔. (iii) M: both H atoms on N replaced by ethanoyl groups, C₆H₁₁N(COCH₃)₂ ✔. Examiner insight: the mechanism name was not well known (electrophilic and nucleophilic substitution were given). Errors in (ii) included omitting the N lone pair, directing the N arrow to the wrong atom, omitting the C=O dipole or the arrow on C=O, wrong intermediates, the first arrow in the intermediate going to C rather than to the C–O bond, and the second starting on C rather than on the C–Cl bond. In (iii) some wrongly substituted CH₃CO– into the cyclohexyl ring.

Question 34C.3[4]

Answer and marking guidance

(i) S is the cyclic amide (lactam) formed when the NH₂ group attacks the acyl chloride in the same molecule: a five-membered ring containing –NH–C(=O)–, with a CH₃ on the carbon next to N ✔; T is the cyclic amine formed when LiAlH₄ reduces the C=O of S to CH₂: 2-methylpyrrolidine ✔. (ii) step I: condensation ✔; step II: reduction ✔. Examiner insight: the correct amide S was rarely seen; many answers involved impossible rearrangements or did not match C₅H₉NO. More candidates drew a T consistent with the S they had drawn. "Condensation" was rarely seen for step I, but many identified step II as reduction.

Examiner's overall observation · Amides

Answered well: the order diethylamine > ethylamine > ethanamide; the structure of the amide formed from an amine and an acyl chloride; reduction as the type of reaction with LiAlH₄.

Found difficult: explaining the lack of basicity of amides — the key point is that the lone pair on nitrogen is delocalised into the C=O group. Naming the addition–elimination mechanism, and drawing it with every lone pair, dipole and curly arrow. Deducing a cyclic amide from a molecular formula, and recognising its formation as a condensation.

Recurring errors: "attract a proton" instead of "accept a proton"; protonating amide groups (–CONH₃⁺); curly arrows in the intermediate going to the carbon atom instead of the C–O bond, or starting at the carbon instead of the C–Cl bond; substituting an acyl group into a cyclohexyl ring; structures that do not match the given molecular formula.

What successful answers did: treated the amide nitrogen as non-basic throughout, and checked each proposed structure against the molecular formula given.

Amino acids as acids and bases: zwitterions34.4.1

The amino acids that make up proteins are 2-amino acids (α-amino acids): an amine group and a carboxylic acid group are attached to the same carbon atom, which also carries a hydrogen atom and a side chain R. The general formula is H₂NCH(R)COOH. In glycine R is H; in alanine R is CH₃. Except for glycine, the central carbon is chiral, so these amino acids are optically active.

Each molecule contains an acidic group (–COOH) and a basic group (–NH₂). In the solid, and in aqueous solution near neutral pH, a proton is transferred from the carboxylic acid group to the amine group of the same molecule, giving a zwitterion — an ion with a positive and a negative charge but no overall charge:

H₂NCH(R)COOH → H₃N⁺CH(R)COO⁻

This explains the physical properties of amino acids. The solids consist of zwitterions held together by strong ionic attractions, so they are crystalline, have high melting points, and are more soluble in water than in non-polar solvents.

AnimationWhat is a zwitterion?

Proton transfer from the carboxyl group to the amine group gives a zwitterion with no net charge.

The effect of pH

Because it has both acidic and basic groups, an amino acid is amphoteric, and it can act as a buffer. The form present depends on the pH:

in acidic solution (low pH), the –COO⁻ group accepts a proton: the amino acid is a cation, H₃N⁺CH(R)COOH;
in alkaline solution (high pH), the –NH₃⁺ group loses a proton: the amino acid is an anion, H₂NCH(R)COO⁻;
at one particular pH, the isoelectric point, the zwitterion predominates and the amino acid has no overall charge.

Isoelectric point

The isoelectric point of an amino acid is the pH at which it exists as a zwitterion, with no overall charge.

Amino acids with a neutral side chain have isoelectric points close to 6 (alanine and valine 6.0 in examination data). A second –COOH group in the side chain lowers the isoelectric point (glutamic acid, about 3); a second –NH₂ group raises it (lysine, 9.8). An amide group in a side chain, as in asparagine (–CH₂CONH₂), is neither acidic nor basic and is never protonated.

Figure 34.5 An amino acid below, at and above its isoelectric point. Adding acid protonates –COO⁻; adding alkali removes H⁺ from –NH₃⁺.

AnimationThe effect of pH on amino acids

Decrease or increase the pH to see the amino acid gain or lose a proton, and the direction it would move in electrophoresis.

Writing buffer equations for an amino acid

To show how an amino acid solution resists pH change, write the neutral form reacting with H⁺ and with OH⁻: H₂NCH(R)COOH + H⁺ → H₃N⁺CH(R)COOH and H₂NCH(R)COOH + OH⁻ → H₂NCH(R)COO⁻ + H₂O. A buffer resists changes in pH when small amounts of acid or alkali are added; it does not keep the pH constant.

Peptide bonds34.4.2

The –COOH group of one amino acid can react with the –NH₂ group of another, forming an amide link and eliminating water. In proteins this link, –CO–NH–, is called a peptide bond, and the reaction is a condensation. Two amino acids give a dipeptide, three a tripeptide, and many a polypeptide.

H₂NCH(CH₃)COOH + H₂NCH₂COOH → H₂NCH(CH₃)CONHCH₂COOH + H₂O

Two different amino acids can combine in two orders. Dipeptides are named from the free –NH₂ end: ala-gly has alanine at the –NH₂ end, gly-ala has glycine there. A dipeptide still has a free –NH₂ at one end and a free –COOH at the other. When asked to draw a dipeptide, show those terminal groups (not continuation bonds) and display the peptide bond in full.

Figure 34.6 Formation of the dipeptide ala-gly. The peptide bond is an amide link formed by condensation.

AnimationThe peptide bond

Two amino acids join by condensation, forming a peptide bond and water; hydrolysis reverses the reaction.

Peptides and proteins are hydrolysed back to amino acids by heating under reflux with aqueous acid or alkali, or by protease enzymes at about body temperature. The amino acids are obtained in the form that suits the conditions: as cations after acid hydrolysis, as anions (carboxylate salts) after alkaline hydrolysis.

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Electrophoresis34.4.3

Electrophoresis separates ions by their movement in an electric field. A strip of paper or gel is soaked in a buffer solution, which fixes the pH. The mixture is applied to the centre of the strip, and a d.c. voltage is applied across it. Each species moves according to its charge at that pH:

a species that is positive (pH below its isoelectric point) moves to the negative electrode (cathode);
a species that is negative (pH above its isoelectric point) moves to the positive electrode (anode);
a species at its isoelectric point is a zwitterion and does not move.

How far it moves depends on the balance of two factors: a larger charge gives a larger force and a greater distance; a larger ion (higher M_r) moves more slowly. Species with the same charge are separated by size — the smaller one travels further. The positions are revealed afterwards with a locating agent, and the amino acids identified by comparison with standards run under the same conditions.

Figure 34.7 Electrophoresis. The buffer pH fixes the charge on each species; charge sets the direction, and charge and size together set the distance moved.

Worked example 34.3 · Predicting an electrophoresis result

Problem	A mixture of alanine (pI 6.0), lysine (pI 9.8) and the dipeptide ala-lys is run at pH 6.0. Predict the positions of the three spots.
Reasoning	At pH 6.0 alanine is at its isoelectric point, so it is a zwitterion. Lysine has an extra –NH₂ in its side chain; at pH 6.0, below its pI, it is positive. The dipeptide also contains lysine's side-chain –NH₃⁺ and has the same +1 charge as lysine, but a larger M_r.
Answer	Alanine stays at the starting line. Lysine and ala-lys both move towards the negative electrode; lysine (M_r 146) moves further than ala-lys (M_r 217).
Check	Direction comes from the sign of the charge; distance from charge and M_r together. Both factors are needed in an explanation.

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Quick check 34.4

Define the isoelectric point of an amino acid.
answer
The pH at which the amino acid exists as a zwitterion, with no overall charge.
Draw the form of glycine present at pH 1.
answer
H₃N⁺CH₂COOH
Give the structures of the two dipeptides formed from glycine and alanine.
answer
H₂NCH₂CONHCH(CH₃)COOH (gly-ala) and H₂NCH(CH₃)CONHCH₂COOH (ala-gly).
At pH 11, alanine carries a 1− charge and glutamic acid a 2− charge. Which moves further towards the anode, and why?
answer
Glutamic acid: its greater charge outweighs its greater M_r.
Why is a buffer used in electrophoresis?
answer
To keep the pH, and so the charge on each species, constant.

Past-paper practice · Set 34D · Amino acids, peptides and electrophoresis

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 34D.1[9]

Part (a) of the same question is 34C.1.

Answer and marking guidance

(b)(i) resists a change in pH ✔ when a small amount of acid or alkali is added ✔. (ii) H₂NCH(CH₃)COOH + H⁺ → H₃N⁺CH(CH₃)COOH ✔; H₂NCH(CH₃)COOH + OH⁻ → H₂NCH(CH₃)COO⁻ + H₂O ✔. (c)(ii) the dipeptide with the peptide bond –CO–NH– displayed between the two amino acids ✔, rest correct with terminal –NH₂ and –COOH ✔. (d) spots: glu furthest towards +, ala-glu between, ala on the start line ✔; ala is a zwitterion (neutral, at its isoelectric point) at pH 6, while ala-glu and glu are negatively charged ✔; glu has the lower M_r (ala-glu the higher), so glu moves further ✔. Examiner insight: (b)(i) was answered well — a buffer resists changes in pH, it does not keep pH constant. In (b)(ii) many used H₂NCH(CH₃)COO⁻ or H₃N⁺CH(CH₃)COOH as reactants. In (c)(ii) tripeptides and polymer sections were common; a dipeptide needs terminal –NH₂ and –COOH. In (d) the commonest error was sending glu and ala-glu to the negative electrode.

Question 34D.2[5]

Answer and marking guidance

(c)(i) the pH at which an amino acid exists as a zwitterion / has no overall charge ✔. (ii) at pH 1.0: H₃N⁺CH(CH₂CONH₂)COOH — the amine group is protonated and the carboxyl group un-ionised; the side-chain amide is not protonated ✔. (e) E = Asn, F = Lys-Asn, G = Lys ✔; Lys and Lys-Asn are positively charged at pH 5.0, while Asn is (nearly) uncharged ✔; Lys-Asn has the higher M_r, so moves less far than Lys ✔. Examiner insight: in (c)(i) good responses stated that the isoelectric point is a pH. Correct diagrams in (c)(ii) were rare: the commonest error was protonating the amide group to –CONH₃⁺; amides are neutral. Good answers to (e) considered both charge and M_r; many considered only M_r.

Question 34D.3[7]

Answer and marking guidance

(a) val-lys: the peptide bond displayed between the COOH of valine and the α-NH₂ of lysine ✔, rest of structure correct ✔. (b) valine on the cross ✔; Val-Lys and Lys both on the negative side ✔; Lys further than Val-Lys ✔; explanation, [1] × 2: Val does not move as it is a zwitterion (neutral) at pH 6, or Lys and Val-Lys move towards the negative electrode as they are positively charged; Lys moves furthest as it has the lower M_r with the same positive charge ✔✔. Examiner insight: (a) was well answered, though some drew a polymer rather than the dipeptide. (b) discriminated: many saw that valine would not move and lysine would move furthest, but some explained only by charge or only by mass, and some suggested that Lys and Val-Lys have different overall charges.

Question 34D.4[4]

Answer and marking guidance

(i) use a buffer (at pH 11) ✔. (iii) alanine: the anode (+); glutamic acid: the anode (+) ✔. (iv) M1 alanine carries a 1− charge and glutamic acid a 2− charge ✔; M2 alanine is lighter (lower M_r) ✔ — the greater charge of glutamic acid outweighs its greater mass. Examiner insight: (i) was generally correct. The number of correct answers to (iii) was disappointing — many forgot the information given that both exist as negative ions at pH 11. For (iv), answers had to explain the difference, referring to the 1− and 2− charges and to the smaller mass of alanine, not merely state that glutamic acid travels further.

Question 34D.5[6]

Answer and marking guidance

(i) zwitterion: the ring N–H becomes NH₂⁺ and –CO₂H becomes –CO₂⁻ ✔; a proton is transferred from the carboxylic acid group to the amine (N) ✔. (ii) M1 glutamic acid towards the + end ✔; M2 proline and alanine towards the − end ✔; M3 at pH 4.0 glutamic acid (pI 3.1) is negatively charged (contains COO⁻), while proline and alanine (pI above 4) are positively charged (contain NH₂⁺ / NH₃⁺) ✔; M4 alanine moves further than proline because of its lower M_r ✔. Examiner insight: drawing the zwitterion was easier than explaining how it forms. In (ii) many did not use the isoelectric points to deduce the charge on each amino acid at pH 4.0, and so could not relate this to the direction of movement.

Question 34D.6[3]

Answer and marking guidance

(i) any one complete set ✔: NaOH(aq) or dilute aqueous alkali, heat under reflux; or dilute aqueous HCl / H₂SO₄, heat under reflux; or a protease enzyme with water at 30–40 °C. (ii) the three amino acids released by breaking both peptide bonds ✔, each shown in the correct ionic form for the conditions chosen — as cations (–NH₃⁺, –COOH) in acid, as anions (–NH₂, –COO⁻) in alkali ✔. Examiner insight: common errors were omitting heat or aqueous conditions. Most hydrolysed the tripeptide correctly but did not show the amino acids in the correct ionised form for their chosen conditions.

Examiner's overall observation · Amino acids, peptides and electrophoresis

Answered well: the definition of a buffer; the structure of zwitterions; the definition of the isoelectric point when it was stated to be a pH; the use of a buffer in electrophoresis.

Found difficult: explaining how a zwitterion forms (the proton transfers from –COOH to the amine group of the same molecule); deducing from isoelectric points the charge on each species at the buffer pH; explaining electrophoresis results — good answers considered both charge and M_r, while many considered only one. Drawing the apparatus with a d.c. power supply, and recalling information given in the question about the ions present, were also weak.

Recurring errors: protonating the amide group of asparagine to –CONH₃⁺; using the cation or anion rather than the neutral amino acid as the reactant in buffer equations; drawing tripeptides or polymer sections when a dipeptide was asked for; forming the peptide bond through a side-chain group; sending negatively charged species to the negative electrode; suggesting different overall charges for species that carry the same charge; omitting heat or aqueous conditions for hydrolysis; showing hydrolysis products in the wrong ionic form for the conditions.

What successful answers did: compared the buffer pH with each isoelectric point to fix the sign of the charge, then used the size of the charge and M_r together to explain the distances.

Misconceptions and how the topic is assessed34.1–34.4

Table 34.1 Misconceptions in topic 34.
misconception	why it is wrong	correct model	examination consequence
Phenylamine is more basic than ethylamine because it has a ring.	The ring withdraws the lone pair by delocalisation.	ethylamine > ammonia > phenylamine.	Order and explanation marks lost.
Bases "attract" protons.	Basicity is about donating a lone pair to form a bond.	The N lone pair accepts a proton.	Explanation not credited.
Amides are basic because they contain NH₂.	The lone pair is delocalised onto C=O.	Amides are neutral.	Wrong orders; –CONH₃⁺ drawn.
Ammonia and a halogenoalkane react at room temperature in water.	NH₃ escapes; water hydrolyses the halogenoalkane.	NH₃ in ethanol, heated under pressure.	Conditions mark lost.
Coupling only needs a low temperature.	The phenoxide ion is the reactive species.	Phenol in NaOH(aq), below 10 °C.	Conditions mark lost.
In electrophoresis the heaviest species always moves least.	Direction and distance depend on charge first.	Compare pH with pI for sign; then charge and M_r.	Explanation marks lost.

Table 34.2 How topic 34 appears in examination questions.
question family	typical demand	what the answer needs
Making amines	reagents, conditions, mechanism; by-products	NH₃ in ethanol under pressure; nucleophilic substitution; LiAlH₄ for nitriles and amides
Relative basicity	order three compounds and explain, 3–4 marks	lone pair accepts a proton; group effect on electron density on N, for each compound
Phenylamine	two-step preparation; bromination; diazotisation	conc. acids; Sn/conc. HCl, heat; 3Br₂ and 3HBr; NaNO₂/HCl <10 °C
Azo dyes	intermediate, coupling partner and conditions	diazonium salt; phenol or naphthol in NaOH(aq), <10 °C; –N=N–
Amides	hydrolysis products; reduction; mechanism of formation	acid → RCOOH + NH₄⁺; alkali → RCOO⁻ + NH₃; LiAlH₄ → amine
Amino acids	zwitterion; form at a given pH; dipeptide; electrophoresis	pH vs pI; displayed peptide bond with terminal groups; charge and M_r

Self-test34.1–34.4

Ten questions on the whole chapter. Each gives its reason once you answer.

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Definitions to learn34.1–34.4

term	definition
primary / secondary / tertiary amine	one / two / three hydrogen atoms of NH₃ replaced by carbon-containing groups
Brønsted–Lowry base	a proton acceptor
diazonium ion	an ion containing –N⁺≡N bonded to an aromatic ring
azo group	–N=N–, linking two aromatic rings in an azo compound
zwitterion	an ion with both a positive and a negative charge and no overall charge
isoelectric point	the pH at which an amino acid exists as a zwitterion with no overall charge
peptide bond	the amide link –CO–NH– between two amino acid residues
electrophoresis	separation of ions by their movement in an electric field on paper or gel soaked in a buffer

Summary

Amines are made from halogenoalkanes with NH₃ (or a primary amine) in ethanol heated under pressure, giving mixtures; and by reducing nitriles (LiAlH₄ or H₂/Ni) or amides (LiAlH₄).
Amines are weak bases: RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻. Base strength: secondary alkylamine > primary alkylamine > NH₃ > phenylamine > amide.
Phenylamine: benzene → nitrobenzene (conc. HNO₃/H₂SO₄) → phenylamine (Sn/conc. HCl, heat, then NaOH). With Br₂(aq): white precipitate of 2,4,6-tribromophenylamine.
Phenylamine + HNO₂ (NaNO₂ + dilute HCl) below 10 °C → benzenediazonium chloride; warm with water → phenol + N₂; with phenol in NaOH(aq) → azo dye (–N=N–).
Amides are made from acyl chlorides and NH₃ or amines; hydrolysed by hot aqueous acid or alkali; reduced by LiAlH₄ to amines; neutral because the N lone pair is delocalised onto C=O.
Amino acids exist as zwitterions; below the pI they are cations, above it anions. Peptide bonds form by condensation.
In electrophoresis cations move to −, anions to +, zwitterions stay; greater charge and lower M_r give greater distance.

Examination checklist

Can I give reagents and conditions for four ways of making amines?
Can I order and explain the basicities of alkylamines, ammonia, phenylamine, substituted phenylamines and amides?
Can I describe the preparation of phenylamine and its reaction with bromine water, with an equation?
Can I describe diazotisation and coupling, with conditions, and identify the azo group?
Can I write equations for the hydrolysis of amides in acid and alkali, and for their reduction?
Can I draw an amino acid at low pH, at its pI and at high pH, and write buffer equations?
Can I draw a dipeptide with a displayed peptide bond?
Can I predict and explain electrophoresis results using pI values and M_r?

Knowledge organiser

idea	key facts	must-remember distinctions and common errors
Amines from R–X	NH₃ in ethanol, heat under pressure; nucleophilic substitution	mixture of 1°, 2°, 3° amines and quaternary salt
Amines by reduction	RCN + 4[H] → RCH₂NH₂; RCONH₂ + 4[H] → RCH₂NH₂ + H₂O	LiAlH₄ (dry ether); H₂/Ni for nitriles
Basicity	lone pair on N accepts H⁺	alkyl +I ↑; ring or C=O delocalisation ↓; NO₂ ↓
Phenylamine	nitration; Sn/conc. HCl, heat; NaOH	C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O
Bromination	Br₂(aq), rt, 2,4,6-tribromophenylamine	3HBr; white ppt
Diazotisation and coupling	NaNO₂/HCl <10 °C; phenol/NaOH <10 °C	–N⁺≡N (+ on N next to ring); product –N=N–
Amides	hydrolysis: acid → RCOOH + NH₄⁺; alkali → RCOO⁻ + NH₃	neutral; never –CONH₃⁺
Amino acids	H₃N⁺CH(R)COO⁻ at pI; cation below, anion above	pI is a pH
Electrophoresis	buffer; d.c. supply; charge sets direction	charge and M_r both explain distance