ChemistryAdvanced22 min read

Alkyl Halides & Amines

Substitution, elimination, and the basicity of nitrogen

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Practise the full question bank:

01

Alkyl halides (haloalkanes)

An alkyl halide is an alkane with a halogen replacing a hydrogen (R–X, where X = Cl, Br, I). The carbon–halogen bond is polar — the halogen pulls electrons, leaving the carbon slightly positive (δ+) — and that electron-poor carbon is the reactive site, open to attack by nucleophiles.

02

Their two reactions: substitution and elimination

Nucleophilic substitution: a nucleophile (an electron-pair donor) attacks the δ+ carbon and displaces the halogen. With hydroxide you get an alcohol; with ammonia an amine; with cyanide a nitrile (extending the carbon chain). This is a central route for making other functional groups.

Elimination: with a hot alcoholic solution of a strong base, the halide loses HX to form a double bond — an alkene. Substitution and elimination compete, and conditions (reagent, temperature) decide which wins: aqueous base favours substitution, hot alcoholic base favours elimination.

Same reagent, different conditions

A haloalkane with hydroxide: in warm AQUEOUS solution it substitutes to give an alcohol; in HOT ALCOHOLIC solution it eliminates to give an alkene. The conditions, not just the reagent, choose the product.

03

Amines

Amines are organic derivatives of ammonia, with one or more of ammonia's hydrogens replaced by carbon chains: primary (one, R–NH₂), secondary (two, R₂NH) or tertiary (three, R₃N).

Like ammonia, the nitrogen carries a lone pair, so amines are bases — they accept a proton (H⁺) to form a positively charged ion, and react with acids to give salts. That lone pair also makes them nucleophiles, which is how the ammonia-plus-haloalkane substitution above builds them.

04

A worked example

Worked example

Bromoethane is treated with (a) warm aqueous sodium hydroxide, then (b) hot ethanolic sodium hydroxide. What forms each time?

  1. (a) Warm aqueous NaOH → nucleophilic substitution. OH⁻ replaces Br.Aqueous conditions favour substitution.
  2. Product: ethanol (CH₃CH₂OH) + NaBr.Hydroxide substitutes onto the carbon, giving an alcohol.
  3. (b) Hot ethanolic NaOH → elimination. HBr is removed.Hot alcoholic conditions favour elimination.
  4. Product: ethene (CH₂=CH₂) + NaBr + water.Loss of HX forms a carbon–carbon double bond.

(a) ethanol (substitution); (b) ethene (elimination) — the conditions decide which reaction dominates.

05

Nucleophilic substitution in detail

The δ+ carbon of a haloalkane is attacked by whatever nucleophile is supplied, and the halogen leaves — so one reagent opens the door to many products. Hydroxide gives an alcohol; cyanide gives a nitrile, adding a carbon and lengthening the chain; ammonia gives an amine.

Reactivity depends on the carbon–halogen bond strength, not its polarity: the weaker C–I bond breaks most easily, so iodoalkanes react fastest and chloroalkanes slowest (C–I < C–Br < C–Cl in reactivity order, iodo fastest).

06

Amines: basicity and preparation

Amines are organic relatives of ammonia, and like ammonia the lone pair on nitrogen makes them both bases (accepting H⁺ to form a salt) and nucleophiles. They are classified as primary, secondary or tertiary by how many carbon groups sit on the nitrogen.

They are made by nucleophilic substitution of a haloalkane with excess ammonia, or by reducing a nitrile. Reacting an amine with an acid gives an ammonium-type salt — the basis of many drug formulations, which are made as soluble salts.

Lone pair does double duty

Nitrogen's lone pair is why an amine is both a base (it grabs a proton) and a nucleophile (it attacks δ+ carbon). Spot the lone pair and you can predict the chemistry.

07

Two mechanisms: SN1 and SN2

Nucleophilic substitution can happen by two different routes. In SN2 the nucleophile attacks the δ+ carbon as the halogen leaves, in a single concerted step — so its rate depends on both the haloalkane and the nucleophile. This is favoured by primary haloalkanes, where the small carbon is easy to approach.

In SN1 the halogen leaves first, forming a carbocation, which the nucleophile then attacks in a second step — so the rate depends only on the haloalkane. This is favoured by tertiary haloalkanes, because the tertiary carbocation is the most stable. Secondary haloalkanes sit in between and can go by either route.

Which route?

Primary haloalkane → SN2 (one step, backside attack). Tertiary → SN1 (via a stable carbocation). The clue is carbocation stability: only tertiary carbocations are stable enough to form readily.

Practice questions

7 questions · 27 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

4 · 9 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Explain why the carbon–halogen bond in a haloalkane is the site of attack by nucleophiles.
Model answer

The bond is polar; the halogen is more electronegative, leaving the carbon slightly positive (δ+), which attracts electron-pair-donating nucleophiles.

Examiner tip. δ+ carbon = the target for nucleophilic attack.

SQ2[2 marks]
State the product and reaction type when bromoethane is warmed with aqueous sodium hydroxide.
Model answer

Ethanol is formed by nucleophilic substitution (OH⁻ replaces Br⁻).

Examiner tip. Aqueous conditions favour substitution; hot alcoholic conditions favour elimination.

SQ3[3 marks]
Explain why amines act as bases.
Model answer

The nitrogen atom has a lone pair of electrons which can accept a proton (H⁺), forming a positively charged ion; so amines are proton acceptors (bases).

Examiner tip. The lone pair on nitrogen makes amines both basic and nucleophilic.

SQ4[2 marks]
Explain why amines act as bases, and give the product when ethylamine reacts with hydrochloric acid.
Model answer

The nitrogen lone pair can accept a proton (H⁺), so amines are bases. Ethylamine + HCl gives the salt ethylammonium chloride (C₂H₅NH₃⁺Cl⁻).

Examiner tip. Base = proton acceptor; an amine + acid gives an ammonium-type salt.

Long questions

3 · 18 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
Bromoethane can react in two different ways depending on conditions.
  1. State the reagent and conditions for converting bromoethane to ethene, and name the reaction type. [3]
  2. State the reagent for converting bromoethane to ethylamine, and name the reaction type. [3]
Mark scheme
  1. Hot ethanolic (alcoholic) potassium hydroxide[1]
  2. under reflux / heat[1]
  3. This is elimination (loss of HBr forms the C=C)[1]
  4. Excess ammonia[1]
  5. (in a sealed tube / ethanolic solution)[1]
  6. This is nucleophilic substitution[1]

→ ethene: hot alcoholic KOH, elimination; → ethylamine: excess ammonia, nucleophilic substitution.

Examiner tip. Same haloalkane, different reagents/conditions give elimination vs substitution products.

LQ2[6 marks]
Bromoethane can be converted into several products by nucleophilic substitution.
  1. State the reagent and product for converting bromoethane to ethanol. [2]
  2. State the reagent and product for converting bromoethane to a nitrile, and why this is useful. [2]
  3. Explain why iodoethane reacts faster than bromoethane. [2]
Mark scheme
  1. Warm aqueous sodium hydroxide (or water)[1]
  2. gives ethanol[1]
  3. Potassium cyanide (in ethanol) gives propanenitrile[1]
  4. useful because it adds a carbon, lengthening the chain[1]
  5. The C–I bond is weaker than C–Br[1]
  6. so it breaks more easily, making iodoethane react faster[1]

OH⁻ → ethanol; CN⁻ → propanenitrile (adds a carbon); iodoethane is faster because C–I is weaker than C–Br.

Examiner tip. Reactivity follows bond strength: C–I breaks easiest.

LQ3[6 marks]
Nucleophilic substitution of haloalkanes can follow an SN1 or an SN2 mechanism.
  1. Describe the SN2 mechanism and the type of haloalkane it favours. [3]
  2. Describe the SN1 mechanism and explain why tertiary haloalkanes favour it. [3]
Mark scheme
  1. In SN2 the nucleophile attacks the δ+ carbon as the halogen leaves[1]
  2. in a single concerted step[1]
  3. favoured by primary haloalkanes (the carbon is easy to approach)[1]
  4. In SN1 the halogen leaves first to form a carbocation[1]
  5. then the nucleophile attacks the carbocation[1]
  6. tertiary haloalkanes form a stable tertiary carbocation, so this route is favoured[1]

SN2: one-step, nucleophile attacks as halide leaves, favoured by primary. SN1: two-step via a carbocation, favoured by tertiary (stable carbocation).

Examiner tip. Carbocation stability decides the route: tertiary → SN1, primary → SN2.