1. Overview
A nitrile contains the -C≡N group, and a hydroxynitrile contains both an -OH group and a -C≡N group. This section covers the two cyanide reactions that make them, from a halogenoalkane and from an aldehyde or ketone, and the hydrolysis of nitriles to carboxylic acids. Both cyanide reactions make a new C-C bond, so they are the AS way to make a carbon chain one atom longer, and they appear often in synthesis questions (topic 21.1).
Key Definitions
- Nitrile: an organic compound containing the -C≡N group, with the carbon atom of the group bonded to the rest of the molecule; general formula RCN.
- Hydroxynitrile: an organic compound containing both a hydroxy group, -OH, and a nitrile group, -C≡N; those made from aldehydes and ketones have both groups on the same carbon atom.
- Nucleophile: an electron-pair donor; a species that uses a lone pair to form a new covalent bond, usually to a δ+ carbon atom. CN⁻ is the nucleophile in both cyanide reactions.
- Nucleophilic substitution: a reaction in which a nucleophile replaces an atom or group in a molecule, for example a cyanide ion replacing the halogen atom of a halogenoalkane.
- Nucleophilic addition: a reaction in which a nucleophile attacks the δ+ carbon atom of a C=O group and the π bond breaks, so that the reagent adds across the C=O bond and a single product forms.
- Hydrolysis: a reaction in which a bond in a molecule is broken by reaction with water; nitriles are hydrolysed by heating under reflux with a dilute acid or a dilute alkali.
- Heating under reflux: heating a mixture in a flask fitted with a vertical condenser, so that vapour condenses and runs back into the flask; the mixture can be heated for a long time without volatile substances escaping.
- Acidification: adding a dilute strong acid, such as hydrochloric acid, to a solution of a carboxylate salt so that the carboxylate ions take up H⁺ ions and become the carboxylic acid.
Content
The nitrile group and how to name it
A nitrile has the general formula RCN, in full R-C≡N. The carbon atom of the group is bonded to the rest of the molecule, and the nitrogen atom sits at the end with a lone pair. The C-C≡N atoms lie in a straight line.
To name a nitrile, count the carbon of the C≡N group as carbon 1 of the chain, then add "nitrile" to the alkane name, keeping its final "e". So CH₃CH₂CN, made from bromoethane (two carbon atoms), has three carbon atoms and is propanenitrile.
| Structural formula | Name |
|---|---|
| CH₃CN | ethanenitrile |
| CH₃CH₂CN | propanenitrile |
| CH₃CH₂CH₂CN | butanenitrile |
| CH₃CH(OH)CN | 2-hydroxypropanenitrile |
| (CH₃)₂C(OH)CN | 2-hydroxy-2-methylpropanenitrile |
Organic equations on this page are written without state symbols: several of the reactions take place in ethanol, where (aq) would be wrong.
Making nitriles from halogenoalkanes
Reagent and conditions: potassium cyanide, KCN, dissolved in ethanol; heat under reflux.
CH₃CH₂Br + KCN → CH₃CH₂CN + KBr
The potassium ion is a spectator, so the ionic equation is:
CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻
This is nucleophilic substitution. The lone pair on the carbon atom of CN⁻ forms a bond to the δ+ carbon atom of the C-Br bond, and the bromine atom leaves as a bromide ion. The SN1 and SN2 mechanisms are on topic 15.1.
- The chain grows by one carbon. A new C-C bond forms, so bromoethane (2 C) gives propanenitrile (3 C).
- Ethanol is the solvent. It dissolves the halogenoalkane. In water, some of the halogenoalkane would be hydrolysed to an alcohol instead (topic 15.1).
- No acid. The reaction needs free CN⁻ ions. Acid would turn them into hydrogen cyanide, HCN, which is very poisonous and volatile.
Making hydroxynitriles from aldehydes and ketones
Reagents and conditions: hydrogen cyanide, HCN, with a little KCN as a catalyst; heat. The usual examples are ethanal and propanone:
CH₃CHO + HCN → CH₃CH(OH)CN
CH₃COCH₃ + HCN → (CH₃)₂C(OH)CN
The products are 2-hydroxypropanenitrile and 2-hydroxy-2-methylpropanenitrile.
This is nucleophilic addition. The C=O bond is polar, with a δ+ carbon atom. Its π bond breaks: the CN group bonds to the carbon atom, and an H atom bonds to the oxygen atom, which becomes an -OH group. Nothing is lost, so there is a single product.
Why KCN is needed. The nucleophile is the cyanide ion, not the HCN molecule. HCN is a weak acid and only slightly ionised, so on its own it reacts very slowly. KCN supplies CN⁻ ions, and these are regenerated as the reaction goes on, so KCN acts as a catalyst. The mechanism, with curly arrows, is on topic 17.1.
Chiral centres. The carbon atom that was in the C=O group ends up bonded to -OH, -CN and the two groups it already carried. It is a chiral centre (topic 13.4) only when all four groups are different.
- An aldehyde RCHO, where R is an alkyl group, always gives a product with a chiral centre: that carbon atom carries H, R, OH and CN. Methanal (R = H) gives HOCH₂CN, which has no chiral centre.
- A ketone gives a chiral centre only if its two alkyl groups are different. Propanone has two CH₃ groups, so 2-hydroxy-2-methylpropanenitrile has no chiral centre.
HCN is very poisonous and volatile, so this reaction is done in a fume cupboard.
The two cyanide reactions side by side.
| Halogenoalkane | Aldehyde or ketone | |
|---|---|---|
| Reagent | KCN | HCN, KCN catalyst |
| Conditions | in ethanol, heat | heat |
| Reaction type | nucleophilic substitution | nucleophilic addition |
| Other product | halide ion, e.g. Br⁻ | none |
| Product | nitrile | hydroxynitrile |
Hydrolysing nitriles to carboxylic acids
In hydrolysis, water breaks up the C≡N group. Its carbon atom becomes the carbon of a -COOH group, so the chain length does not change, and the nitrogen ends up as ammonia or ammonium ions. Water alone reacts far too slowly, so the nitrile is heated under reflux with a dilute acid or a dilute alkali.
With dilute acid. Heat under reflux with dilute hydrochloric acid (or dilute sulfuric acid). The carboxylic acid forms directly, and the nitrogen ends up as ammonium ions because the acid is in excess:
CH₃CH₂CN + 2H₂O + HCl → CH₃CH₂COOH + NH₄Cl
CH₃CH₂CN + 2H₂O + H⁺ → CH₃CH₂COOH + NH₄⁺
With dilute alkali, then acid. Heat under reflux with dilute sodium hydroxide solution. Ammonia gas is given off (it turns damp red litmus paper blue). The acid that forms is neutralised by the alkali, so the product is its salt:
CH₃CH₂CN + NaOH + H₂O → CH₃CH₂COONa + NH₃
To get the free acid, acidify with a dilute strong acid such as hydrochloric acid:
CH₃CH₂COO⁻ + H⁺ → CH₃CH₂COOH
The -OH group of a hydroxynitrile is unchanged. Only the C≡N group is hydrolysed, so the product is a carboxylic acid with an -OH group on carbon 2. From the ethanal product:
CH₃CH(OH)CN + 2H₂O + HCl → CH₃CH(OH)COOH + NH₄Cl
The product is 2-hydroxypropanoic acid.
Using the cyanide reactions in synthesis
Each cyanide reaction adds one carbon atom, and hydrolysis turns that carbon into -COOH:
- halogenoalkane (n carbons) → nitrile (n + 1) → carboxylic acid (n + 1)
- aldehyde or ketone (n carbons) → hydroxynitrile (n + 1) → carboxylic acid with -OH on carbon 2 (n + 1)
To work backwards from a target acid, count its carbon atoms first: one more than the starting compound means a cyanide step. Then look at carbon 2, next to -COOH. An -OH group there points to the carbonyl route; no -OH group points to the halogenoalkane route. The aldehyde or ketone can come from oxidising a primary or secondary alcohol (topic 17.1).
Worked example 1: a two-step route and its yield
Question: Hexanoic acid, CH₃(CH₂)₄COOH, is made from 1-iodopentane, CH₃(CH₂)₄I, in two steps. (a) Give the reagent and conditions for each step, name the intermediate, and write an equation for each step. (b) 8.91 g of 1-iodopentane gives 3.05 g of hexanoic acid. Calculate the overall percentage yield. (: H 1.0, C 12.0, O 16.0, I 126.9)
Step 1: make the nitrile. The chain grows from 5 to 6 carbon atoms with no -OH group: the halogenoalkane route. Heat 1-iodopentane under reflux with KCN in ethanol (nucleophilic substitution). The intermediate is hexanenitrile.
CH₃(CH₂)₄I + KCN → CH₃(CH₂)₄CN + KI
Step 2: hydrolyse it. Heat hexanenitrile under reflux with dilute hydrochloric acid.
CH₃(CH₂)₄CN + 2H₂O + HCl → CH₃(CH₂)₄COOH + NH₄Cl
Step 3: amount of 1-iodopentane. (C₅H₁₁I) = 5(12.0) + 11(1.0) + 126.9 = 197.9 g mol⁻¹.
Step 4: theoretical mass of acid. The ratio is 1 : 1 : 1 (1-iodopentane : nitrile : acid), so at most 0.04502 mol of acid can form. (C₆H₁₂O₂) = 6(12.0) + 12(1.0) + 2(16.0) = 116.0 g mol⁻¹.
Step 5: percentage yield.
Answer: (a) KCN in ethanol, heat under reflux, giving hexanenitrile; then dilute HCl, heat under reflux (equations in Steps 1 and 2). (b) The overall yield is 58.4% (3 s.f.).
Worked example 2: from an aldehyde to a hydroxy acid
Question: Butanal, CH₃CH₂CH₂CHO, is converted into J, C₅H₉NO. J is then hydrolysed to K, C₅H₁₀O₃. (a) Give the reagents and conditions for making J, and the type of reaction. (b) Give the structural formula and name of J. (c) Write an equation for the hydrolysis of J by dilute hydrochloric acid. (d) Does J have a chiral centre?
Step 1: reagents and type. HCN with KCN as a catalyst, heat. J contains all the atoms of butanal plus HCN (C₄H₈O + HCN = C₅H₉NO), so this is nucleophilic addition to the C=O group.
Step 2: structure of J. CN bonds to the carbonyl carbon and H to the oxygen: CH₃CH₂CH₂CH(OH)CN. Check: 5 C, 9 H, 1 N and 1 O, matching C₅H₉NO. Counting the CN carbon as carbon 1, the -OH group is on carbon 2 of a five-carbon chain, so J is 2-hydroxypentanenitrile.
Step 3: hydrolysis equation. Only the C≡N group changes, and the -OH group stays.
CH₃CH₂CH₂CH(OH)CN + 2H₂O + HCl → CH₃CH₂CH₂CH(OH)COOH + NH₄Cl
K is 2-hydroxypentanoic acid, C₅H₁₀O₃.
Step 4: chirality. Carbon 2 of J is bonded to H, OH, CN and CH₂CH₂CH₃: four different groups, so it is a chiral centre.
Answer: (a) HCN with KCN catalyst, heat; nucleophilic addition. (b) CH₃CH₂CH₂CH(OH)CN, 2-hydroxypentanenitrile. (c) The equation in Step 3. (d) Yes: carbon 2.
Worked example 3: working backwards
Question: Suggest a three-step route from pentan-3-ol, CH₃CH₂CH(OH)CH₂CH₃, to the acid (CH₃CH₂)₂C(OH)COOH. Give the reagents and conditions for each step and the formula of each intermediate.
Step 1: count carbons. Pentan-3-ol has 5 carbon atoms and the target has 6, so one step must be a cyanide reaction.
Step 2: look at carbon 2. It carries -OH, so the acid came from hydrolysing the hydroxynitrile (CH₃CH₂)₂C(OH)CN. That carbon also carries two ethyl groups, so the hydroxynitrile came from HCN adding to the ketone pentan-3-one, CH₃CH₂COCH₂CH₃, which forms when the secondary alcohol pentan-3-ol is oxidised (topic 17.1).
Answer:
- Pentan-3-ol to pentan-3-one, CH₃CH₂COCH₂CH₃: acidified K₂Cr₂O₇, heat and distil.
- Pentan-3-one to (CH₃CH₂)₂C(OH)CN: HCN with KCN catalyst, heat.
- (CH₃CH₂)₂C(OH)CN to (CH₃CH₂)₂C(OH)COOH: dilute HCl, heat under reflux.
Key Equations
Nitrile from a halogenoalkane (R = alkyl group; X = Cl, Br or I; KCN in ethanol, heat):
RX + KCN → RCN + KX
Hydroxynitrile from an aldehyde or ketone (R = alkyl group; R′ = H for an aldehyde or an alkyl group for a ketone; HCN with KCN catalyst, heat):
RCOR′ + HCN → RR′C(OH)CN
Acid hydrolysis (dilute HCl, heat under reflux; the nitrogen ends up as NH₄⁺):
RCN + 2H₂O + HCl → RCOOH + NH₄Cl
Alkaline hydrolysis, then acidification (NaOH(aq), heat under reflux; then dilute HCl):
RCN + NaOH + H₂O → RCOONa + NH₃
RCOONa + HCl → RCOOH + NaCl
Common Mistakes to Avoid
- Mistake: mixing up the two cyanide reagents, adding acid to the KCN, or leaving out the ethanol. Fix: bromoethane to propanenitrile needs KCN in ethanol, heat, with no acid. Propanone to 2-hydroxy-2-methylpropanenitrile needs HCN with KCN as catalyst, heat.
- Mistake: naming from the starting chain, or dropping the "e". Fix: count the CN carbon. CH₃CH₂CN is propanenitrile, not ethanenitrile, "propannitrile" or "ethyl cyanide".
- Mistake: an unbalanced hydrolysis equation, with water missing, or with NH₃ or HCN as a product in acid. Fix: use two H₂O per C≡N group, and give ammonium ions in acid: CH₃CH₂CN + 2H₂O + HCl → CH₃CH₂COOH + NH₄Cl.
- Mistake: stopping at the salt after alkaline hydrolysis. Fix: NaOH(aq) gives sodium propanoate, CH₃CH₂COONa, and ammonia. You get propanoic acid only after acidifying with dilute HCl.
- Mistake: leaving out "dilute" or "aqueous", or using concentrated sulfuric acid, for the hydrolysis. Fix: write, for example, "dilute HCl(aq), heat under reflux".
- Mistake: losing the -OH group when a hydroxynitrile is hydrolysed. Fix: only the C≡N group changes. CH₃CH₂CH₂CH(OH)CN gives 2-hydroxypentanoic acid, CH₃CH₂CH₂CH(OH)COOH, not pentanoic acid.
Exam Tips
- State reagents and conditions in full every time: "KCN in ethanol, heat"; "HCN with KCN catalyst, heat"; "dilute HCl, heat under reflux"; or "NaOH(aq), heat under reflux, then acidify with dilute HCl". For a hydrolysis, also give the nitrogen product (NH₄⁺ in acid, NH₃ gas in alkali).
- Name the type of reaction in full: nucleophilic substitution (halogenoalkane), nucleophilic addition (aldehyde or ketone) and hydrolysis (nitrile). The full name is safer than "substitution" or "addition" alone.
- Construct an equation: use structural formulae unless the question asks for molecular formulae, then count C, H, N and O on both sides.
- Draw a product: show -C≡N with the carbon, not the nitrogen, bonded to the chain (CH₃CH₂CN, never CH₃CH₂NC), and leave every other group unchanged. Mark a chiral centre only on a carbon atom with four different groups.
- Suggest a route or deduce a starting compound: count carbon atoms first (a cyanide step adds one), then check carbon 2 of the target acid for -OH.