Nitrogen compounds

All four subtopics here are printed under "A Level subject content" in the 9701 syllabus and every objective carries the tier "A Level". None of it is AS, and it is a separate topic from the AS topic 19 of the same name. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper and the whole 9701 bank on this site comes from it, so no practice here is tagged to this topic.

Syllabus points

34.1 Primary and secondary amines

• Recall the reactions (reagents and conditions) by which primary and secondary amines are produced: reaction of halogenoalkanes with NH₃ in ethanol heated under pressure; reaction of halogenoalkanes with primary amines in ethanol, heated in a sealed tube or under pressure; the reduction of amides with LiAlH₄; the reduction of nitriles with LiAlH₄ or H₂/Ni.
• Describe the condensation reaction of ammonia or an amine with an acyl chloride at room temperature to give an amide.
• Describe and explain the basicity of aqueous solutions of amines.

34.2 Phenylamine and azo compounds

• Describe the preparation of phenylamine via the nitration of benzene to form nitrobenzene, followed by reduction with hot Sn and concentrated HCl, followed by NaOH(aq).
• Describe the reaction of phenylamine with Br₂(aq) at room temperature, and the reaction of phenylamine with HNO₂, or NaNO₂ and dilute acid below 10 °C, to produce the diazonium salt, with further warming of the diazonium salt with H₂O to give phenol.
• Describe and explain the relative basicities of aqueous ammonia, ethylamine and phenylamine.
• Recall the following about azo compounds: describe the coupling of benzenediazonium chloride with phenol in NaOH(aq) to form an azo compound; identify the azo group; state that azo compounds are often used as dyes; state that other azo dyes can be formed via a similar route.

34.3 Amides

• Recall the reactions (reagents and conditions) by which amides are produced: the reaction between ammonia and an acyl chloride at room temperature; the reaction between a primary amine and an acyl chloride at room temperature.
• Describe the reactions of amides: hydrolysis with aqueous alkali or aqueous acid; the reduction of the CO group in amides with LiAlH₄ to form an amine.
• State and explain why amides are much weaker bases than amines.

34.4 Amino acids

• Describe the acid and base properties of amino acids and the formation of zwitterions, to include the isoelectric point.
• Describe the formation of amide (peptide) bonds between amino acids to give di- and tripeptides.
• Interpret and predict the results of electrophoresis on mixtures of amino acids and dipeptides at varying pHs.

Making amines

Four routes, and the conditions distinguish them.

Why ethanol and not water. Water would hydrolyse the halogenoalkane to an alcohol, so an ethanolic solution is used. Why a sealed tube. Ammonia is a gas and would escape on heating.

Why excess ammonia. The primary amine produced is itself a nucleophile, and a better one than ammonia, so it attacks a second molecule of halogenoalkane to give a secondary amine, then a tertiary amine, then a quaternary ammonium salt. Using a large excess of ammonia makes it statistically more likely that the halogenoalkane meets ammonia than an amine, so the yield of the primary amine is maximised. That is the whole reason the condition is specified, and it is a common short question.

The nitrile route is worth noticing for a different reason. A halogenoalkane converted to a nitrile with KCN gains a carbon, and the nitrile then reduces to an amine, so the two steps together are the way to lengthen a carbon chain by one and finish with an amine. Synthesis questions in topic 36 turn on this.

Basicity of amines

An amine is a base because the nitrogen has a lone pair which can accept a proton. In water:

CH₃CH₂NH₂ + H₂O ⇌ CH₃CH₂NH₃⁺ + OH⁻

so an aqueous solution of an amine is alkaline. Amines also react with acids to give salts:

CH₃CH₂NH₂ + HCl → CH₃CH₂NH₃⁺Cl⁻

The order: ethylamine, ammonia, phenylamine

ethylamine > ammonia > phenylamine, from most basic to least.

The argument is a single question asked three times: how available is the lone pair on the nitrogen?

• Ethylamine is the strongest base. The ethyl group is electron donating, so it pushes electron density towards the nitrogen. The lone pair is therefore more available to accept a proton, and the ammonium ion formed is stabilised by the same effect.
• Ammonia is the reference, with nothing attached but hydrogen.
• Phenylamine is by far the weakest. The nitrogen's lone pair is delocalised into the benzene ring, overlapping with the ring's pi system. It is therefore much less available to accept a proton, and phenylamine is only very slightly basic.

That difference is large, not marginal: phenylamine is roughly a million times weaker a base than ethylamine, which is why it does not turn damp litmus paper blue in the way an aliphatic amine does.

Note the symmetry with topic 32. In phenol, delocalisation of the oxygen's lone pair into the ring made the compound more acidic. In phenylamine, delocalisation of the nitrogen's lone pair into the ring makes it less basic. It is the same structural effect producing opposite-looking results, because in one case the lone pair was being given up and in the other it was being offered.

Amines with acyl chlorides

Ammonia or an amine reacts with an acyl chloride at room temperature to give an amide, by the addition-elimination mechanism of topic 33:

CH₃COCl + 2NH₃ → CH₃CONH₂ + NH₄Cl

CH₃COCl + CH₃NH₂ → CH₃CONHCH₃ + HCl

An excess of the amine is used, because the HCl produced protonates it.

Phenylamine

Preparation

Two steps from benzene:

1. Nitration. Concentrated HNO₃ with concentrated H₂SO₄ at 25 to 60 °C gives nitrobenzene.
2. Reduction. Heat with tin and concentrated hydrochloric acid, then add aqueous sodium hydroxide.

The second step needs explaining, because the sodium hydroxide is not decoration. The reduction in acid produces the phenylammonium salt, C₆H₅NH₃⁺, since phenylamine is a base and the solution is strongly acidic. Adding alkali liberates the free phenylamine. Leaving that step out is a common omission.

Reaction with bromine water

Phenylamine with aqueous bromine at room temperature gives an immediate white precipitate of 2,4,6-tribromophenylamine, and the bromine water is decolourised.

This is exactly the phenol result, and for exactly the same reason: the -NH₂ group releases electron density into the ring by delocalising its lone pair, so the ring is strongly activated and 2,4-directing, and three bromines substitute in cold water with no catalyst.

Diazotisation

Phenylamine with nitrous acid, made in situ from NaNO₂ and dilute HCl, at a temperature below 10 °C, gives a benzenediazonium salt:

C₆H₅NH₂ + HNO₂ + HCl → C₆H₅N₂⁺Cl⁻ + 2H₂O

The temperature is the examinable condition: diazonium salts decompose above about 10 °C, so an ice bath is essential.

Warming the diazonium salt with water decomposes it to phenol, with nitrogen evolved, which is the route in topic 32.

Azo coupling

Below 10 °C, the diazonium salt is added to phenol dissolved in aqueous sodium hydroxide, and a brightly coloured azo compound forms, typically yellow or orange.

The azo group is -N=N-, joining two aromatic rings.

Azo compounds are intensely coloured because the delocalised system extends across both rings and the azo linkage, and such an extended system absorbs visible light. They are widely used as dyes, and changing either the diazonium salt or the coupling compound gives a different colour, so the same route produces a whole family of dyes.

Amides

An amide contains the -CONH₂ group, or -CONHR if substituted.

Hydrolysis

Amides are hydrolysed by heating under reflux with either acid or alkali, and the products differ because the acid or base formed reacts with the reagent.

• With aqueous acid: gives the carboxylic acid and the ammonium salt, since the ammonia produced is protonated.

CH₃CONH₂ + H₂O + HCl → CH₃COOH + NH₄Cl

• With aqueous alkali: gives the carboxylate salt and ammonia gas, which can be detected with damp red litmus.

CH₃CONH₂ + NaOH → CH₃COO⁻Na⁺ + NH₃

Predicting which form the products take is what the question tests, so decide first which reagent is in excess and then protonate or deprotonate accordingly.

Reduction

LiAlH₄ in dry ether reduces the C=O group of an amide to CH₂, giving an amine:

CH₃CONH₂ → CH₃CH₂NH₂

Note that the nitrogen is retained and the carbon count is unchanged.

Why amides are much weaker bases than amines

An amide has a nitrogen with a lone pair, so on the face of it it should be basic. It is not, and the reason is the one running through this whole topic.

The nitrogen's lone pair is delocalised into the adjacent C=O group, overlapping with the pi bond of the carbonyl. It is therefore not available to accept a proton. The oxygen is highly electronegative and pulls that electron density further towards itself, which makes the effect stronger than the delocalisation into a benzene ring in phenylamine.

So amides are neutral in water and do not form stable salts with dilute acids. Evidence for the delocalisation is that the C-N bond in an amide is shorter than a normal C-N single bond and there is no free rotation about it, which is what holds a protein backbone rigid.

Amino acids

An amino acid contains both an -NH₂ group and a -COOH group. The 2-amino acids of proteins have both on the same carbon, with the general formula RCH(NH₂)COOH.

Zwitterions

Having an acid and a base in the same molecule, an amino acid transfers its own proton internally: the COOH gives its proton to the NH₂. The result is a zwitterion, with a -COO⁻ and an -NH₃⁺ group in the same molecule and no overall charge.

That explains the physical properties that otherwise look wrong for a small organic molecule: amino acids are crystalline solids with high melting points, and are soluble in water and insoluble in non-polar solvents, because the zwitterion is effectively an ionic species.

The effect of pH

In acid, the added H⁺ protonates the -COO⁻ group, so the ion carries a positive charge overall. In alkali, the added OH⁻ removes a proton from the -NH₃⁺ group, so the ion carries a negative charge overall.

The isoelectric point is the pH at which the amino acid exists predominantly as the zwitterion and has no overall charge. It differs from one amino acid to another, because the R group may itself be acidic or basic: an acidic side-chain lowers the isoelectric point and a basic one raises it.

Peptide bonds

The -COOH of one amino acid and the -NH₂ of another condense, losing water, and the link formed is an amide bond, called a peptide bond in this context:

-CO-NH-

Two amino acids give a dipeptide, three a tripeptide, many a polypeptide.

One point that appears in questions: two different amino acids can form two different dipeptides, depending on which one supplies the COOH and which the NH₂. Glycine and alanine give both Gly-Ala and Ala-Gly.

Peptides are hydrolysed back to their amino acids by heating under reflux with 6 mol dm⁻³ hydrochloric acid, exactly as any amide is.

Electrophoresis

A mixture is spotted onto a support soaked in a buffer solution of chosen pH, and a voltage is applied. Each species then moves according to its charge at that pH:

• Species carrying a positive charge move towards the cathode, the negative electrode.
• Species carrying a negative charge move towards the anode, the positive electrode.
• Species at their isoelectric point carry no overall charge and do not move.

The distance moved depends on the size of the charge and on the size of the molecule, so a doubly charged ion moves further than a singly charged one of similar size.

Worked prediction. A mixture of three amino acids with isoelectric points 3.0, 6.0 and 9.7 is run at pH 6.0.

• The one with isoelectric point 6.0 is at its isoelectric point, so it is a zwitterion with no overall charge and stays on the baseline.
• The one with isoelectric point 3.0 is in a solution more alkaline than its isoelectric point, so it has lost a proton and is negative. It moves to the anode.
• The one with isoelectric point 9.7 is in a solution more acidic than its isoelectric point, so it has gained a proton and is positive. It moves to the cathode.

The rule to carry: buffer pH above the isoelectric point gives a negative species; buffer pH below it gives a positive one. Working from that rather than memorising cases handles any combination the question offers, including dipeptides.

Common mistakes

• Using aqueous rather than ethanolic ammonia with a halogenoalkane, which hydrolyses it to an alcohol.
• Failing to explain the excess of ammonia, or saying it is to speed the reaction up rather than to limit further substitution.
• Leaving out the NaOH step in the preparation of phenylamine, so the product is the phenylammonium salt.
• Putting phenylamine as a stronger base than ammonia. Delocalisation of the lone pair into the ring makes it much weaker.
• Explaining basicity by the presence of a lone pair without saying how available that lone pair is.
• Letting a diazotisation warm above 10 °C.
• Giving the same products for acid and alkaline hydrolysis of an amide. Acid gives the acid and an ammonium salt; alkali gives the carboxylate salt and ammonia.
• Saying amides are basic because they contain nitrogen. The lone pair is delocalised into the carbonyl.
• Drawing a zwitterion with an overall charge, or with COOH and NH₂ still intact.
• Saying the isoelectric point is pH 7 for every amino acid.
• Getting the direction of movement in electrophoresis backwards. Negative species move to the positive electrode.
• Forgetting that two different amino acids give two different dipeptides.