Amines

In a nutshell

An amine is ammonia, NH₃, with one or more of its hydrogen atoms replaced by a carbon group (an alkyl group, or an aryl group such as a benzene ring).

Replacing a hydrogen leaves the important feature of ammonia untouched: a lone pair of electrons on the nitrogen. That lone pair is the whole subtopic. It lets an amine act as a weak base (it accepts a proton) and as a nucleophile (it attacks electron-poor carbon).

Assumed knowledge: Halogenoalkanes (nucleophilic substitution), Aromatic chemistry (delocalisation), Carboxylic acids and derivatives (acyl chlorides and addition-elimination).

Core content

The four classes of amine

How many carbon groups sit on the nitrogen decides the class. This matters because the nucleophilic reactions keep adding more groups to the same nitrogen.

• Primary (1°) amine: one carbon group on N, e.g. ethylamine CH₃CH₂NH₂.
• Secondary (2°) amine: two carbon groups on N, e.g. (CH₃CH₂)₂NH.
• Tertiary (3°) amine: three carbon groups on N, e.g. (CH₃CH₂)₃N.
• Quaternary ammonium ion: four carbon groups on N, e.g. (CH₃CH₂)₄N⁺.

A quaternary ammonium ion has used up all four bonds on nitrogen, so it has no lone pair left. It carries a permanent positive charge, and because it has no lone pair it is not a base and not a nucleophile (a quaternary ammonium salt is not classed as an amine).

Preparing amines

The spec gives three routes. Note which product each one gives and, crucially, how clean that product is.

Route 1: primary aliphatic amine from a halogenoalkane and ammonia.

Ammonia is a nucleophile, so it attacks the carbon of a halogenoalkane in a nucleophilic substitution reaction:

CH₃CH₂Br + 2NH₃ ⟶ CH₃CH₂NH₂ + NH₄Br

• Use excess ammonia (dissolved in ethanol, heated in a sealed tube) to maximise the primary amine.
• The problem: the primary amine formed still has a lone pair, so it attacks more halogenoalkane, giving secondary and tertiary amines and a quaternary ammonium salt as well. The product is a mixture that must be separated, so the yield of the primary amine is poor.

Route 2: primary amine from reduction of a nitrile.

A nitrile is reduced to a primary amine:

CH₃CN + 4[H] ⟶ CH₃CH₂NH₂

• Reagent and conditions: hydrogen gas with a nickel catalyst (catalytic hydrogenation), or LiAlH₄ in dry ether.
• This route gives a clean primary amine (no further substitution can happen), so a much better yield than Route 1. The nitrile itself is made from a halogenoalkane and KCN, which also adds one carbon to the chain.

Route 3: aromatic amine from reduction of a nitro compound.

Nitrobenzene is reduced to phenylamine:

C₆H₅NO₂ + 6[H] ⟶ C₆H₅NH₂ + 2H₂O

• Reagents and conditions: tin (Sn) and concentrated hydrochloric acid, heated (iron and HCl, or H₂ with a nickel catalyst, also work).
• Because the reaction is in acid, you first get the salt C₆H₅NH₃⁺Cl^-; adding NaOH then releases the free phenylamine.
• Aromatic amines like phenylamine are used to make dyes (phenylamine is a feedstock for dyes, it is not itself a dye).

Amines as weak bases

An amine is a Brønsted-Lowry base: a proton (H⁺) acceptor. The lone pair on the nitrogen accepts the proton, forming a new coordinate (dative) bond.

With water, an amine sets up an equilibrium, producing hydroxide ions:

CH₃NH₂(aq) + H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH^-(aq)

Amines are weak bases because this equilibrium lies well to the left: only a small fraction of the amine reacts, so only a low concentration of OH^- is produced.

Amines react with acids to form ammonium salts:

CH₃CH₂NH₂ + HCl ⟶ CH₃CH₂NH₃⁺Cl^-

(ethylammonium chloride). Adding a stronger base such as NaOH to the salt regenerates the free amine. This is why phenylamine, almost insoluble in water, dissolves in acid: it forms the ionic phenylammonium salt, C₆H₅NH₃⁺.

Base strength: ammonia vs primary aliphatic vs primary aromatic

Base strength depends on how available the lone pair on the nitrogen is to accept a proton. The more available the lone pair, the stronger the base.

So the order is phenylamine < ammonia < ethylamine (weakest to strongest). Adding more alkyl groups pushes even more electron density onto N, which is why a secondary amine is a stronger base still.

Still don't get it? · why the lone pair decides base strength

Picture the lone pair on nitrogen as a pair of hands held out, ready to grab a passing proton. A stronger base grabs more eagerly. Anything that makes the hands fuller and freer makes it grab better; anything that keeps the hands busy makes it grab worse.

Now add the two effects one at a time.

An alkyl group (like the CH3CH2 in ethylamine) gently pushes some of its own electron density along the bond towards the nitrogen. The hands get fuller, so they reach out for a proton more readily. That is why ethylamine grabs protons better than ammonia, which has nothing pushing towards it.

A benzene ring does the opposite. Its own ring of delocalised electrons pulls the nitrogen's lone pair partly into itself, so the lone pair is shared out around the ring instead of waiting on the nitrogen. The hands are now busy in the crowd and cannot grab a proton properly. That is why phenylamine is the weakest base of the three.

Exam wording to land on: base strength depends on the availability of the lone pair on the nitrogen. In ethylamine the alkyl group has a positive inductive effect, making the lone pair more available, so it is a stronger base. In phenylamine the lone pair is delocalised into the benzene ring, making it less available, so it is a weaker base.

Amines and ammonia as nucleophiles: reaction with halogenoalkanes

Ammonia and amines are nucleophiles: the lone pair on nitrogen is donated to the electron-deficient carbon of a halogenoalkane, substituting out the halogen. Each product still has a lone pair, so the reaction runs on in a series of steps:

NH₃ ⟶ primary amine RNH₂ ⟶ secondary amine R₂NH ⟶ tertiary amine R₃N ⟶ quaternary ammonium salt R₄N⁺X^-

(each arrow is a further reaction with the halogenoalkane RX). You control where it stops with the proportions:

• Excess ammonia maximises the primary amine.
• Excess halogenoalkane drives it all the way to the quaternary ammonium salt.

The mechanism (nucleophilic substitution). Shown below for ammonia with bromoethane. It is a two-stage mechanism:

1. The lone pair on the nitrogen of ammonia attacks the carbon bonded to Br; the C-Br bond breaks and Br^- leaves. This gives a positively charged salt, CH₃CH₂NH₃⁺.
2. A second ammonia molecule removes a proton (H⁺) from that salt, giving the neutral amine and NH₄⁺.

Quaternary ammonium salts as cationic surfactants

A quaternary ammonium salt, R₄N⁺X^-, with one long hydrocarbon chain, is used as a cationic surfactant.

• A surfactant lowers the surface tension of a liquid.
• The positive nitrogen head is attracted to negatively charged surfaces such as hair, fabric fibres and glass, while the long hydrocarbon tail points away from them.
• This makes quaternary ammonium salts useful as fabric softeners and hair conditioners.

Reaction with acyl chlorides and acid anhydrides (addition-elimination)

Ammonia and primary amines react with acyl chlorides and acid anhydrides to form amides, in a nucleophilic addition-elimination reaction. A primary amine gives an N-substituted amide.

The mechanism you must be able to outline is the one with the acyl chloride. It has two stages:

1. Addition: the lone pair on the nitrogen attacks the carbonyl carbon (which is δ+); the C=O double bond breaks, pushing the electrons onto the oxygen. This gives a tetrahedral intermediate with a positive charge on N and a negative charge on O.
2. Elimination: the C=O double bond reforms, pushing out the chloride ion, and a second amine (or ammonia) molecule removes the H⁺ from the nitrogen, giving the amide.

Worked examples

Model answer 1: a base-strength "explain" question.

Explain why ethylamine is a stronger base than ammonia, but phenylamine is a weaker base than ammonia. (3 marks)

A full-mark answer makes three distinct, linked points, all referring to the lone pair on the nitrogen:

1. Base strength depends on the availability of the lone pair on the nitrogen atom to accept a proton (H⁺).
2. In ethylamine the alkyl group has a positive inductive effect, pushing electron density onto the nitrogen, so its lone pair is more available than ammonia's: stronger base.
3. In phenylamine the nitrogen's lone pair is delocalised into the benzene ring, so it is less available than ammonia's: weaker base.

The marks are lost by writing only "inductive effect" (you must say it is positive), by saying only "delocalised electrons" (you must say the lone pair on N is delocalised into the ring), or by never mentioning the lone pair at all.

Model answer 2: choosing a synthesis route.

Propylamine, CH₃CH₂CH₂NH₂, can be made from bromoethane in two steps. Give the reagents, conditions and equation for each step and name each reaction. Then explain why this route gives a better yield of propylamine than reacting 1-bromopropane with ammonia.

Step 1 (nucleophilic substitution): warm bromoethane with potassium cyanide, KCN, in aqueous ethanol under reflux.

CH₃CH₂Br + KCN ⟶ CH₃CH₂CN + KBr

(propanenitrile; the CN group adds one carbon to the chain.)

Step 2 (reduction): reduce the nitrile with hydrogen and a nickel catalyst (or LiAlH₄ in dry ether).

CH₃CH₂CN + 4[H] ⟶ CH₃CH₂CH₂NH₂

Why it beats the ammonia route: reducing a nitrile can only give the primary amine, so the product is clean and the yield is high. Reacting 1-bromopropane with ammonia also makes propylamine, but the primary amine's lone pair attacks more 1-bromopropane, so you also get secondary and tertiary amines and a quaternary salt. Propylamine is then only one component of a mixture that must be separated, lowering the yield.

The reagent in Step 1 must be an ionic cyanide (KCN), not HCN, and the reducing agent in Step 2 must not be NaBH₄.

Common exam mistakes

• Drawing the ammonia or amine nucleophile with a negative charge in a mechanism. The nucleophile is neutral, and the curly arrow starts from the lone pair on its nitrogen. A minus sign on the amine loses the first mark.
• Not naming the mechanism. With a halogenoalkane it is nucleophilic substitution; with an acyl chloride it is (nucleophilic) addition-elimination. Writing just "substitution", or leaving the name out, loses a mark.
• In step 2 of the substitution mechanism, using Br^- to remove the proton, or drawing the loss-of-H⁺ arrow the wrong way round. The proton is removed by a second ammonia (or amine): the arrow runs from that base's lone pair to the H, and the N-H bonding pair returns to the nitrogen.
• Writing only "inductive effect" for base strength. You must say the alkyl group has a positive inductive effect (it pushes electron density onto N).
• For phenylamine, saying only "it has delocalised electrons". You must say the lone pair on the nitrogen is delocalised into the ring, so it is less available to accept a proton. Do not write "dissociated" for "delocalised".
• Thinking a base is a proton donor. A Brønsted-Lowry base is a proton acceptor.
• Using NaBH₄ to reduce a nitrile or a nitro compound: it is not accepted. Use H₂/Ni (both), LiAlH₄ (nitriles), or Sn/HCl or Fe/HCl (nitro compounds).
• Giving HCN instead of KCN for making the nitrile, or forgetting the aqueous-ethanol solvent.
• Swapping the two "excess" conditions: excess ammonia maximises the primary amine; excess halogenoalkane gives the quaternary ammonium salt. Stopping at the singly substituted cation when the quaternary salt was asked for is a common slip.
• Saying phenylamine (or the aromatic amine) "is a dye". It is used to make dyes; it is not itself a dye.
• Missing the N-substituted amide name of the addition-elimination product, e.g. calling the product of methylamine and ethanoyl chloride "ethanamide" rather than N-methylethanamide.

Key definitions

• Brønsted-Lowry base - a proton (H⁺) acceptor.
• Weak base - a base that reacts only partially with water (the equilibrium lies to the left), producing a low concentration of hydroxide ions.
• Nucleophile - a species with a lone pair of electrons that it can donate to an electron-deficient carbon atom to form a new covalent bond.
• Nucleophilic substitution - a reaction in which a nucleophile replaces an atom or group on a carbon atom.
• (Nucleophilic) addition-elimination - a nucleophile adds to a carbonyl carbon, then a small molecule (here HCl) is eliminated.
• Quaternary ammonium salt - a salt containing an ion in which four carbon groups are bonded to a positively charged nitrogen atom (R₄N⁺).
• Cationic surfactant - a positively charged species that lowers the surface tension of a liquid and is attracted to negatively charged surfaces.

Specification

• I can state that amines are compounds based on ammonia with H atoms replaced by alkyl or aryl groups.
• I can describe the preparation of a primary aliphatic amine from a halogenoalkane and ammonia, and by the reduction of a nitrile.
• I can describe the preparation of an aromatic amine by the reduction of a nitro compound, and state that aromatic amines are used to make dyes.
• I can state that amines are weak bases.
• I can explain the difference in base strength between ammonia, a primary aliphatic amine and a primary aromatic amine, in terms of the availability of the lone pair of electrons on the nitrogen.
• I can state that amines are nucleophiles.
• I can outline the mechanism of the nucleophilic substitution reactions of ammonia and amines with halogenoalkanes to form primary, secondary and tertiary amines and quaternary ammonium salts.
• I can state the use of quaternary ammonium salts as cationic surfactants.
• I can outline the mechanism of the nucleophilic addition-elimination reactions of ammonia and primary amines with acyl chlorides, and give the products with acyl chlorides and acid anhydrides.

Related notes

• Halogenoalkanes
• Aromatic chemistry (3.3.10)
• Carboxylic acids and derivatives (3.3.9)
• Amino acids, proteins and DNA (3.3.13)
• Polymers (3.3.12)
• Organic synthesis (3.3.14)