Amines and Nitriles

AS · 10 min

At AS, two families of organic nitrogen compounds matter: amines, which are derivatives of ammonia, and nitriles, which contain the $\ce{C#N}$ group. Both are made from halogenoalkanes by nucleophilic substitution, and nitriles are also made from aldehydes and ketones by nucleophilic addition. Their real importance is in synthesis: nitriles are the AS route for adding a carbon atom to a chain, and their hydrolysis turns that extra carbon into a carboxylic acid group. This note covers the structure and naming of both, how they are made, and the hydrolysis of nitriles, with the reagents and conditions for every step.

Amines

Structure and classification

An amine is formed when one or more hydrogen atoms of ammonia, $\ce{NH3}$ , are replaced by carbon groups. Amines are classified by the number of carbon groups on the nitrogen atom (not, as with alcohols, on the carbon):

class	general formula	example	name
primary	$\ce{RNH2}$	$\ce{CH3CH2NH2}$	ethylamine (ethanamine)
secondary	$\ce{R2NH}$	$\ce{(CH3CH2)2NH}$	diethylamine
tertiary	$\ce{R3N}$	$\ce{(CH3)3N}$	trimethylamine

The nitrogen has three bonding pairs and a lone pair, so the arrangement around nitrogen is pyramidal (bond angles about $107^\circ$ ). The lone pair makes amines nucleophiles (and bases), just like ammonia.

At AS you need to name primary amines only: alkyl group + "amine" (methylamine, ethylamine, propylamine), or the IUPAC form alkane + "amine" (methanamine, ethanamine, propan-1-amine). Both are accepted.

Small amines have a fishy smell. Primary and secondary amines can hydrogen bond (through N–H), so their boiling points are higher than those of alkanes of similar $M_r$ but lower than alcohols, because N–H bonds are less polar than O–H bonds.

Making primary amines

Key result

Halogenoalkane + ammonia: excess ammonia in ethanol, heated under pressure in a sealed tube. Nucleophilic substitution.

\ce{CH3CH2Br + 2NH3 -> CH3CH2NH2 + NH4Br}

The reaction happens in two stages. First, ammonia's lone pair attacks the $\delta+$ carbon and bromide leaves, giving an ethylammonium ion; then a second ammonia molecule removes $\ce{H+}$ from it:

\ce{CH3CH2Br + NH3 -> CH3CH2NH3+ + Br-}

\ce{CH3CH2NH3+ + NH3 -> CH3CH2NH2 + NH4+}

The full mechanism is in Nucleophilic substitution and elimination.

Ethanol is the solvent because halogenoalkanes do not dissolve in water, and because water (or hydroxide) would compete as a nucleophile to give an alcohol.
The tube is sealed and heated under pressure because ammonia is a gas: in an open flask it would boil out of the solution.

Further substitution: secondary and tertiary amines

The primary amine still has a lone pair on its nitrogen, and it is actually a better nucleophile than ammonia (the alkyl group pushes electron density onto the nitrogen). So it competes with ammonia for the remaining halogenoalkane:

\ce{CH3CH2NH2 + CH3CH2Br -> (CH3CH2)2NH + HBr}

\ce{(CH3CH2)2NH + CH3CH2Br -> (CH3CH2)3N + HBr}

(The HBr formed reacts with ammonia or amine to give a salt.) A tertiary amine can react once more, giving a quaternary ammonium salt, $\ce{(CH3CH2)4N+ Br-}$ .

Key result

To obtain mainly the primary amine, use a large excess of ammonia. Each halogenoalkane molecule is then far more likely to collide with ammonia than with an amine. To obtain more of the secondary and tertiary amines, use an excess of halogenoalkane.

Even with excess ammonia, some secondary amine forms, so the product must be separated by fractional distillation. This makes the reaction a poor method for a pure primary amine.

Tip

Extension (A Level): primary amines are also made by reducing nitriles with $\ce{LiAlH4}$ or with hydrogen and a nickel catalyst ( $\ce{RCN + 4[H] -> RCH2NH2}$ ). This gives only the primary amine. Amines also act as bases: $\ce{CH3CH2NH2 + HCl -> CH3CH2NH3+Cl-}$ . Both of these are A Level topics.

Nitriles

Structure and naming

A nitrile contains the $\ce{-C#N}$ group. The carbon of the nitrile group is sp hybridised, forming one $\sigma$ bond to carbon and a triple bond (one $\sigma$ , two $\pi$ ) to nitrogen, so the $\ce{C-C#N}$ unit is linear ( $180^\circ$ ). Nitrogen carries a lone pair.

The nitrile carbon is counted in the name, as carbon 1:

formula	name
$\ce{CH3CN}$	ethanenitrile
$\ce{CH3CH2CN}$	propanenitrile
$\ce{CH3CH2CH2CN}$	butanenitrile
$\ce{(CH3)2CHCN}$	2-methylpropanenitrile
$\ce{CH3CH(OH)CN}$	2-hydroxypropanenitrile

Making nitriles

Key result

starting material	reagents and conditions	type of reaction	product
halogenoalkane	KCN in ethanol, heat under reflux	nucleophilic substitution	nitrile
aldehyde or ketone	HCN, with KCN as catalyst, heat	nucleophilic addition	hydroxynitrile

\ce{CH3CH2Br + KCN -> CH3CH2CN + KBr}

\ce{CH3CHO + HCN -> CH3CH(OH)CN}

In both, the cyanide ion attacks through its carbon atom, forming a new C–C bond. The product has one more carbon than the starting material: bromoethane (2 C) gives propanenitrile (3 C); ethanal (2 C) gives 2-hydroxypropanenitrile (3 C). Details of the hydroxynitrile reaction and its mechanism are in Aldehydes and ketones.

Potassium cyanide and hydrogen cyanide are extremely toxic; these reactions are carried out on a small scale in a fume cupboard.

Hydrolysis of nitriles to carboxylic acids

Heating a nitrile with dilute acid or dilute alkali converts $\ce{-C#N}$ into $\ce{-COOH}$ . The carbon count does not change.

Key result

	acid hydrolysis	alkaline hydrolysis
reagents and conditions	dilute HCl (or $\ce{H2SO4}$ ), heat under reflux	dilute NaOH(aq), heat under reflux, then acidify with dilute HCl
organic product before acidifying	carboxylic acid	sodium carboxylate
nitrogen ends up as	$\ce{NH4+}$ (ammonium salt)	$\ce{NH3}$ gas

\ce{CH3CH2CN + 2H2O + HCl -> CH3CH2COOH + NH4Cl}

\ce{CH3CH2CN + NaOH + H2O -> CH3CH2COONa + NH3}

\ce{CH3CH2COONa + HCl -> CH3CH2COOH + NaCl}

In alkaline hydrolysis, ammonia gas is given off, which turns damp red litmus paper blue: this is a useful observation.

Nitriles in synthesis

Putting the reactions together gives two important chain-lengthening routes:

Key result

Halogenoalkane → carboxylic acid with one more carbon

\ce{RBr ->[KCN, ethanol][reflux] RCN ->[dilute HCl][reflux] RCOOH}

Aldehyde or ketone → 2-hydroxycarboxylic acid

\ce{RCHO ->[HCN, KCN][heat] RCH(OH)CN ->[dilute HCl][reflux] RCH(OH)COOH}

Worked examples

Routine: naming and classifying

Name each compound and classify each amine as primary, secondary or tertiary: (a) $\ce{CH3CH2CH2NH2}$ , (b) $\ce{CH3NHCH3}$ , (c) $\ce{CH3CH2CH2CH2CN}$ , (d) $\ce{(CH3)2C(OH)CN}$ .

Solution

(a) Propylamine (propan-1-amine); primary (one carbon group on N).

(b) Dimethylamine; secondary (two carbon groups on N).

(d) 2-hydroxy-2-methylpropanenitrile (from propanone and HCN).

Routine: making a primary amine

Give the reagents and conditions to make propylamine from 1-bromopropane, write the overall equation, and explain why the conditions are chosen.

Solution

Excess ammonia dissolved in ethanol, heated in a sealed tube (under pressure).

\ce{CH3CH2CH2Br + 2NH3 -> CH3CH2CH2NH2 + NH4Br}

Ethanol dissolves the halogenoalkane and avoids hydrolysis by water. The sealed tube keeps the volatile ammonia in solution while heating. A large excess of ammonia makes reaction with ammonia much more likely than reaction with propylamine, so little secondary amine forms.

Standard: a mixture of amines

Bromomethane is heated with ammonia in a sealed tube, with the bromomethane in excess. Give the structures and names of the amines that can form, and explain why they form.

Solution

$\ce{CH3NH2}$ methylamine (primary); $\ce{(CH3)2NH}$ dimethylamine (secondary); $\ce{(CH3)3N}$ trimethylamine (tertiary). (Also the quaternary salt $\ce{(CH3)4N+Br-}$ .)

Each amine has a lone pair on nitrogen and is a nucleophile, so it attacks further bromomethane molecules, replacing another H on nitrogen with a methyl group. With bromomethane in excess, these further substitutions are favoured.

Standard: from bromoethane to propanoic acid

(a) Give the reagents and conditions for converting bromoethane into propanenitrile, and propanenitrile into propanoic acid by alkaline hydrolysis. (b) Write equations for each step. (c) Give one observation during the alkaline hydrolysis.

Solution

(a) Step 1: KCN in ethanol, heat under reflux. Step 2: NaOH(aq), heat under reflux, then acidify with dilute HCl.

(b) $\ce{CH3CH2Br + KCN -> CH3CH2CN + KBr}$

$\ce{CH3CH2CN + NaOH + H2O -> CH3CH2COONa + NH3}$

$\ce{CH3CH2COONa + HCl -> CH3CH2COOH + NaCl}$

Exam-hard: the isomers of C3H9N

There are four amines with the molecular formula $\ce{C3H9N}$ . Give their structures, name the two primary amines, and classify all four.

Solution

Three carbons and one nitrogen; all bonds single.

$\ce{CH3CH2CH2NH2}$ : propylamine (propan-1-amine), primary.
$\ce{CH3CH(NH2)CH3}$ : 1-methylethylamine (propan-2-amine), primary (one carbon group on N, even though it is attached to a secondary carbon).
$\ce{CH3CH2NHCH3}$ : ethylmethylamine, secondary.
$\ce{(CH3)3N}$ : trimethylamine, tertiary.

Note the trap: propan-2-amine is a primary amine, because amines are classified by the number of carbons on the nitrogen.

Exam-hard: identifying a nitrile

Compound N contains $65.5\%$ C, $9.1\%$ H and $25.5\%$ N by mass, and $M_r = 55.0$ . N is made from a halogenoalkane in one step. (a) Find the molecular formula and identify N. (b) Name a halogenoalkane from which N can be made, and give the reagent and conditions. (c) Name the product of acid hydrolysis of N and write the equation. ( $A_r$ : H 1.0, C 12.0, N 14.0)

Solution

(a) C: $65.5 / 12.0 = 5.46$ ; H: $9.1 / 1.0 = 9.1$ ; N: $25.5 / 14.0 = 1.82$ . Divide by 1.82: $3.00 : 5.0 : 1.00$ . Empirical formula $\ce{C3H5N}$ , $M_r = 55.0$ , so the molecular formula is $\ce{C3H5N}$ . A nitrile made from a halogenoalkane: $\ce{CH3CH2CN}$ , propanenitrile.

(b) Bromoethane (or chloroethane or iodoethane), with KCN in ethanol, heated under reflux.

Watch out

Classifying amines like alcohols. Amines are classified by the number of carbon groups on the nitrogen. $\ce{CH3CH(NH2)CH3}$ is a primary amine.
Ammonia conditions. "Heat under reflux" with ammonia is wrong: the ammonia escapes. Use a sealed tube / under pressure, in ethanol.
Excess of the wrong reagent. Excess ammonia favours the primary amine; excess halogenoalkane favours multiple substitution.
Counting nitrile carbons. The $\ce{C#N}$ carbon is counted: $\ce{CH3CN}$ is ethanenitrile. A halogenoalkane with $n$ carbons gives a nitrile (and acid) with $n + 1$ carbons.
Hydrolysis products. Nitrile hydrolysis gives a carboxylic acid (or its salt) and ammonium ions or ammonia. It does not give an amine.
Cyanide in water. KCN must be in ethanol; in water, hydroxide ions form and give the alcohol instead.

Exam tip

The most common questions: reagents and conditions for halogenoalkane → amine and halogenoalkane → nitrile, and "explain why excess ammonia is used".
In synthesis questions, a product with one more carbon than the starting material almost always means a cyanide step (KCN with a halogenoalkane, or HCN with a carbonyl compound), followed by hydrolysis if the target is an acid.
Hydrolysis conditions must say "dilute" and "heat under reflux", and for alkaline hydrolysis "then acidify".
Equations for alkaline hydrolysis give the sodium salt; include the acidification step if the question asks for the carboxylic acid.

Summary

Amines are derivatives of ammonia: primary $\ce{RNH2}$ , secondary $\ce{R2NH}$ , tertiary $\ce{R3N}$ (classified by carbons on N). N has a lone pair: nucleophile.
Primary amines: halogenoalkane + excess $\ce{NH3}$ in ethanol, heated under pressure (sealed tube). Excess ammonia limits further substitution to secondary and tertiary amines.
Nitriles $\ce{RC#N}$ : linear, sp carbon; the nitrile carbon is counted in the name.
Nitriles from halogenoalkanes: KCN in ethanol, reflux (nucleophilic substitution). Hydroxynitriles from aldehydes and ketones: HCN with KCN catalyst, heat (nucleophilic addition). Both add one carbon.
Nitrile hydrolysis: dilute acid, reflux → RCOOH + $\ce{NH4+}$ ; dilute alkali, reflux → $\ce{RCOO-}$ + $\ce{NH3}$ , then acidify.

Practice

Question

Name: (a) $\ce{CH3CH2CH2NH2}$ , (b) $\ce{CH3CH2CN}$ , (c) $\ce{CH3CH(OH)CN}$ , (d) $\ce{(CH3)2CHCN}$ .
State the hybridisation of the nitrile carbon and the C–C≡N bond angle in ethanenitrile.
Give the reagent and conditions, and write an equation, for the conversion of 1-chlorobutane into pentanenitrile.
Give the conditions for making ethylamine from bromoethane. Explain why the reaction is carried out in a sealed tube and why ammonia is used in excess.
Write an equation for the reaction of ethylamine with bromoethane to form a secondary amine, and name the secondary amine.
Write an equation for the hydrolysis of butanenitrile by dilute hydrochloric acid.
Butanenitrile is heated under reflux with aqueous sodium hydroxide. Name the organic product, give an observation, and state what must be done to obtain butanoic acid.
Give the reagents and conditions for converting propanal into 2-hydroxybutanoic acid in two steps, with an equation for each step.
Classify each amine: (a) $\ce{(CH3)2CHNH2}$ , (b) $\ce{(CH3CH2)2NH}$ , (c) $\ce{(CH3)2NCH2CH3}$ .
$4.36\ \text{g}$ of bromoethane is heated with ammonia in ethanol in a sealed tube. (a) Calculate the minimum amount (in mol) of ammonia needed according to the equation, and the volume this would occupy as a gas at room temperature and pressure. (b) The yield of ethylamine is $70.0\%$ . Calculate the mass of ethylamine obtained. (c) Explain why, in practice, a much larger amount of ammonia than your answer to (a) is used. ( $A_r$ : H 1.0, C 12.0, N 14.0, Br 79.9; molar gas volume $24.0\ \text{dm}^3\ \text{mol}^{-1}$ )

Answers

(a) Propylamine (propan-1-amine). (b) Propanenitrile. (c) 2-hydroxypropanenitrile. (d) 2-methylpropanenitrile.
sp; $180^\circ$ (linear).
KCN in ethanol, heat under reflux. $\ce{CH3CH2CH2CH2Cl + KCN -> CH3CH2CH2CH2CN + KCl}$ .
Excess ammonia in ethanol, heated in a sealed tube (under pressure). Sealed tube: ammonia is a gas and would escape from an open, heated vessel. Excess ammonia: the ethylamine formed is also a nucleophile and can react with more bromoethane to give secondary and tertiary amines; excess ammonia makes it much more likely that bromoethane reacts with ammonia instead.
$\ce{CH3CH2NH2 + CH3CH2Br -> (CH3CH2)2NH + HBr}$ ; diethylamine.
$\ce{CH3CH2CH2CN + 2H2O + HCl -> CH3CH2CH2COOH + NH4Cl}$ .
Sodium butanoate, $\ce{CH3CH2CH2COONa}$ . Ammonia gas is released (turns damp red litmus blue). Acidify with dilute hydrochloric acid to convert the salt into butanoic acid.
Step 1: HCN with KCN catalyst, heat: $\ce{CH3CH2CHO + HCN -> CH3CH2CH(OH)CN}$ . Step 2: dilute HCl, heat under reflux: $\ce{CH3CH2CH(OH)CN + 2H2O + HCl -> CH3CH2CH(OH)COOH + NH4Cl}$ .
(a) Primary (one carbon group on N). (b) Secondary. (c) Tertiary.
(a) $n(\ce{C2H5Br}) = 4.36 / 108.9 = 0.0400\ \text{mol}$ . Equation: $\ce{C2H5Br + 2NH3 -> C2H5NH2 + NH4Br}$ , so $n(\ce{NH3}) = 0.0800\ \text{mol}$ ; volume $= 0.0800 \times 24.0 = 1.92\ \text{dm}^3$ . (b) $n(\ce{C2H5NH2}) = 0.0400 \times 0.700 = 0.0280\ \text{mol}$ ; $M_r = 45.0$ ; mass $= 1.26\ \text{g}$ . (c) With only the stoichiometric amount, the ethylamine formed would compete with ammonia for the bromoethane, producing diethylamine, triethylamine and quaternary salts; a large excess of ammonia keeps further substitution to a minimum and maximises the yield of the primary amine.

Amines

Structure and classification

Making primary amines

Further substitution: secondary and tertiary amines

Nitriles

Structure and naming

Making nitriles

Hydrolysis of nitriles to carboxylic acids

Nitriles in synthesis

Worked examples

Practice

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