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AS 19

Nitrogen compounds

AS Level · Organic chemistry · Papers 1 and 2 · extended in A2 34

At AS Level this is a short topic about three reactions you have already met from the other side — making amines, making nitriles and hydroxynitriles — and one new one: hydrolysing a nitrile to a carboxylic acid. Its real value is in synthesis, because the nitrile route is how you add a carbon atom to a chain. Amines as bases, amides and amino acids come at A Level.

🎯What you need to be able to do

  • Recall how primary amines are made from halogenoalkanes and ammonia.
  • Recall how nitriles are made from halogenoalkanes, and hydroxynitriles from aldehydes and ketones.
  • Describe the hydrolysis of nitriles, by dilute acid or by dilute alkali followed by acidification, to carboxylic acids.

📚The chemistry

19.1 Primary amines

A primary amine has an NH2 group attached to a carbon: R–NH2. They are named with -amine after the alkyl group: CH3CH2NH2 is ethylamine. (The syllabus will not ask you to classify amines as primary, secondary or tertiary at AS.)

They are made by nucleophilic substitution of a halogenoalkane by ammonia in ethanol, heated under pressure in a sealed tube (topic 15):

\[ \mathrm{CH_3CH_2Br + 2NH_3 \rightarrow CH_3CH_2NH_2 + NH_4Br} \]

Two moles of ammonia are used: the first substitutes for the bromine, forming CH3CH2NH3+Br−; the second removes an H+ from that salt to release the free amine, forming NH4Br. The ammonia is in excess for a second reason: the amine product still has a lone pair on its nitrogen, so it is itself a nucleophile and could attack more halogenoalkane. A large excess of ammonia makes it much more likely that the halogenoalkane meets NH3 instead.

19.2 Nitriles and hydroxynitriles

Making them

  • Nitriles — a halogenoalkane with KCN in ethanol, heated under reflux (nucleophilic substitution). The carbon chain gets one carbon longer. \[ \mathrm{CH_3CH_2Br + KCN \rightarrow CH_3CH_2CN + KBr} \] The product, CH3CH2CN, is propanenitrile: three carbons, because the carbon of C≡N counts.
  • Hydroxynitriles — an aldehyde or ketone with HCN, with KCN as catalyst, and heat (nucleophilic addition, topic 17). Again one carbon is added, and an OH group appears on the carbon that was the carbonyl carbon. \[ \mathrm{CH_3CHO + HCN \rightarrow CH_3CH(OH)CN} \]

Hydrolysis of nitriles

Heating a nitrile under reflux with water, catalysed by acid or alkali, converts the C≡N into a carboxyl group. The C of the nitrile becomes the C of COOH, and the nitrogen ends up as ammonium ions or ammonia.

  • With dilute acid (e.g. HCl), heat under reflux — gives the carboxylic acid directly: \[ \mathrm{CH_3CH_2CN + 2H_2O + HCl \rightarrow CH_3CH_2COOH + NH_4Cl} \]
  • With dilute alkali (e.g. NaOH), heat under reflux — gives the carboxylate salt and ammonia gas: \[ \mathrm{CH_3CH_2CN + NaOH + H_2O \rightarrow CH_3CH_2COONa + NH_3} \] then acidify with dilute HCl to obtain the free acid: \[ \mathrm{CH_3CH_2COONa + HCl \rightarrow CH_3CH_2COOH + NaCl} \]

Hydrolysing a hydroxynitrile the same way gives a 2-hydroxy carboxylic acid. Ethanal → 2-hydroxypropanenitrile → 2-hydroxypropanoic acid (lactic acid), the acid that builds up in muscles during anaerobic exercise.

Why this matters: making chains longer

Most organic reactions change a functional group but leave the carbon skeleton alone. Cyanide is the one AS reagent that forms a new C–C bond, so a nitrile is the usual answer when a synthesis question needs the product to have one more carbon than the starting material. The standard sequence:

R–X → (KCN, ethanol, reflux) → R–CN
R–CN → (dilute HCl, reflux) → R–COOH

✏️Worked example

Propanoic acid is to be made from bromoethane in two steps. (a) Give the reagents and conditions, and name the intermediate. (b) Write an equation for each step, using dilute hydrochloric acid in the second. (c) 10.9 g of bromoethane is used and the overall yield is 60.0%. Calculate the mass of propanoic acid obtained. (d) Suggest why the hydrolysis is carried out under reflux. [Mr: C2H5Br 108.9, C2H5COOH 74.0]

(a) Bromoethane has two carbons and propanoic acid three, so a carbon must be added: the nitrile route. Step 1: KCN in ethanol, heat under reflux → propanenitrile. Step 2: dilute HCl, heat under reflux → propanoic acid.

(b)

\[ \mathrm{CH_3CH_2Br + KCN \rightarrow CH_3CH_2CN + KBr} \] \[ \mathrm{CH_3CH_2CN + 2H_2O + HCl \rightarrow CH_3CH_2COOH + NH_4Cl} \]

(c) Each step is 1 : 1, so moles of acid = moles of bromoethane × yield.

\[ n(\mathrm{C_2H_5Br}) = \frac{10.9}{108.9} = 0.1001\ \mathrm{mol} \] \[ m(\text{acid}) = 0.1001 \times 0.600 \times 74.0 = 4.44\ \mathrm{g} \]

(d) Hydrolysis is slow at room temperature, so the mixture must be heated for a long time. Under reflux the vapours condense and run back into the flask, so the volatile nitrile and product are not lost while it is heated.

Check it. Count carbons at every stage: C2 → C3 (the CN adds one) → C3 (hydrolysis changes the group, not the chain). And a 60% yield of 0.100 mol is 0.0600 mol; at 74.0 g mol−1 that is a little under 4.5 g.
Using aqueous KCN, or naming the nitrile by its alkyl group. KCN must be in ethanol: in water, the hydroxide ions present would compete and give ethanol. And CH3CH2CN is propanenitrile, not “ethanenitrile” or “ethyl cyanide” — the nitrile carbon is part of the stem.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Give the reagents and conditions for converting 1-chloropropane into propylamine, and explain why the reagent is used in excess.
Excess ammonia in ethanol, heated under pressure (in a sealed tube). CH3CH2CH2Cl + 2NH3 → CH3CH2CH2NH2 + NH4Cl. The amine product has a lone pair on nitrogen, so it is also a nucleophile and can react with more 1-chloropropane to give further substituted products. A large excess of ammonia makes it far more likely that each chloropropane molecule reacts with ammonia rather than with the amine.
2. Name the nitrile formed when 1-bromobutane reacts with KCN, and the acid formed when it is hydrolysed.
CH3CH2CH2CH2Br + CN− → CH3CH2CH2CH2CN, pentanenitrile (five carbons). Hydrolysis gives CH3CH2CH2CH2COOH, pentanoic acid.
3. Write equations for the hydrolysis of ethanenitrile by (a) dilute hydrochloric acid; (b) aqueous sodium hydroxide, and state what must be done after (b) to obtain ethanoic acid.
(a) CH3CN + 2H2O + HCl → CH3COOH + NH4Cl. (b) CH3CN + NaOH + H2O → CH3COONa + NH3. The product of (b) is sodium ethanoate; to obtain ethanoic acid, acidify with a dilute strong acid: CH3COONa + HCl → CH3COOH + NaCl. Ammonia given off in (b) turns damp red litmus blue.
4. Outline a two-step synthesis of 2-hydroxypropanoic acid from ethanal.
Step 1: HCN with a KCN catalyst, heat (nucleophilic addition): CH3CHO + HCN → CH3CH(OH)CN, 2-hydroxypropanenitrile. Step 2: dilute acid, heat under reflux (hydrolysis): CH3CH(OH)CN + 2H2O + H+ → CH3CH(OH)COOH + NH4+. The product is lactic acid; the chain has grown from two carbons to three.
5. A student wants to make ethylamine, CH3CH2NH2, and propanenitrile, CH3CH2CN, from the same starting material. Name it, and explain what the two reactions have in common.
Both can be made from bromoethane (or another halogenoethane). Ethylamine: excess NH3 in ethanol, heated under pressure. Propanenitrile: KCN in ethanol, heated under reflux. Both are nucleophilic substitutions: a nucleophile (NH3 or CN−) uses its lone pair to attack the δ+ carbon of C–Br, and the bromide leaves. Only the cyanide reaction lengthens the chain.
6. Why does the nitrile route appear so often in synthesis questions?
Because it is the only way in the AS syllabus to form a new carbon–carbon bond, increasing the length of the chain by one. Cyanide adds to a halogenoalkane (giving a nitrile) or to an aldehyde or ketone (giving a hydroxynitrile), and the nitrile can then be hydrolysed to a carboxylic acid (and at A Level reduced to an amine). So whenever a target molecule has one more carbon than the starting material, the route almost certainly passes through a nitrile.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Chemguide (Jim Clark) — the nitriles pages, including their hydrolysis in acid and alkali