Understanding Nucleophilic Substitution Revision Notes | A Level AQA Chemistry

1. Why Halogenoalkanes React: Bond Polarity

Picture a carbon atom bonded to a chlorine atom. Carbon and chlorine are both non-metals, so you might expect a straightforward covalent bond — electrons shared equally. But chlorine is significantly more electronegative than carbon, and that difference changes everything.

Electronegativity is a measure of how strongly an atom attracts the shared bonding electrons towards itself. Fluorine sits at the top of the electronegativity scale, and all halogens (F, Cl, Br, I) are considerably more electronegative than carbon. The result is a polar covalent bond: the halogen pulls electron density towards itself, acquiring a partial negative charge (δ−), while the carbon is left electron-deficient with a partial positive charge (δ+).

This δ+ carbon is the key to everything that follows. Electron-rich species — nucleophiles — are attracted to it. That attraction drives the entire class of reactions you need to know for AQA: nucleophilic substitution.

The polarity of the C–X bond also explains why halogenoalkanes are far more reactive than alkanes. Alkanes have non-polar C–H and C–C bonds, so there is no δ+ carbon to attract a nucleophile. Swap one hydrogen for a halogen and you immediately create a reactive site.

2. What Is a Nucleophile? The Electron-Donor Principle

Now that you know the carbon in a halogenoalkane is δ+, the next question is: what attacks it? The answer is a nucleophile — any species with a lone pair of electrons that it can donate to an electron-deficient centre.

Think of it like a magnet. The δ+ carbon is the positive pole; the nucleophile, with its lone pair, is the negative pole. The lone pair is donated to form a new covalent bond with the carbon, and simultaneously the C–X bond breaks. The halogen leaves as a halide ion (X⁻), taking both electrons from the original bond with it.

AQA requires you to know three specific nucleophiles: the hydroxide ion (OH⁻), the cyanide ion (CN⁻), and ammonia (NH₃). Each one attacks the δ+ carbon in the same fundamental way, but the conditions and products differ — and that is where the exam marks live.

Notice that curly arrows in mechanisms always start from an electron-rich site — a lone pair or a bond — and point towards an electron-poor site. This is non-negotiable in AQA mark schemes. An arrow starting from the wrong place loses the mark immediately.

3. Reaction with Hydroxide Ions (OH⁻): Making Alcohols

The first nucleophile to master is the hydroxide ion, OH⁻. When a halogenoalkane is heated with aqueous sodium hydroxide (NaOH(aq)), the OH⁻ ion attacks the δ+ carbon and replaces the halogen. The product is an alcohol.

The overall equation for bromoethane reacting with aqueous sodium hydroxide is:

The conditions matter for AQA: you need aqueous NaOH (or KOH) and heat. The aqueous environment ensures OH⁻ is available as the nucleophile. If you use ethanolic NaOH instead, you get elimination rather than substitution — a completely different reaction producing an alkene. Keep that distinction sharp.

4. Reaction with Cyanide Ions (CN⁻): Extending the Carbon Chain

The cyanide ion, CN⁻, is a particularly useful nucleophile because it introduces a new carbon atom into the product. When a halogenoalkane reacts with potassium cyanide (KCN) dissolved in ethanol, the CN⁻ ion attacks the δ+ carbon and the halide departs. The product is a nitrile.

For bromoethane, the reaction produces propanenitrile — a three-carbon nitrile from a two-carbon starting material. This is why organic chemists value this reaction: it is a reliable way to increase the carbon chain length by one carbon atom.

The conditions here are ethanolic KCN and heat under reflux. Ethanol helps keep the organic halogenoalkane and KCN in the same phase, so CN^- can reach the halogenoalkane effectively, and it avoids the competing hydrolysis conditions used with aqueous hydroxide. The carbon end of CN^- is the attacking atom, so the new carbon-carbon bond forms and the chain length increases by one.

The nitrile group (–C≡N) can subsequently be hydrolysed to a carboxylic acid or reduced to an amine, making this reaction a gateway to a wide range of further organic synthesis steps — though those transformations are covered elsewhere in the specification.

5. Reaction with Ammonia (NH₃): Making Amines

Ammonia (NH₃) is a neutral molecule, yet it still acts as a nucleophile. The nitrogen atom carries a lone pair of electrons, and that lone pair can be donated to the δ+ carbon of a halogenoalkane. The product is an amine.

The reaction requires excess concentrated ammonia dissolved in ethanol, heated in a sealed tube. The sealed tube is necessary because ammonia is a gas at room temperature and would otherwise escape before the reaction is complete. Bromoethane reacting with excess ammonia produces ethylamine (CH₃CH₂NH₂), a primary amine.

Here is a subtlety worth noting. After the nitrogen of NH₃ attacks the carbon and Br⁻ departs, the nitrogen in the product initially carries a positive charge (it has four bonds). A second molecule of ammonia then acts as a base, removing a proton from this intermediate to give the neutral amine product and ammonium bromide (NH₄Br). This is why two moles of NH₃ appear in the equation.

Excess ammonia is used to minimise further substitution. The primary amine product (CH₃CH₂NH₂) is itself a nucleophile — its nitrogen also has a lone pair — and can react with another molecule of bromoethane to give a secondary amine, then a tertiary amine, and eventually a quaternary ammonium salt. Using a large excess of ammonia reduces the probability of these further substitutions occurring.

6. Comparing the Three Reactions: Conditions and Products at a Glance

AQA examiners frequently ask you to compare these reactions, so having a clear mental map of the differences is essential. The table below consolidates the key information — use it as a retrieval tool, not a substitute for understanding the mechanisms.

Notice the pattern: all three reactions use heat, all three involve the same δ+ carbon being attacked, and all three release the halide ion as the leaving group. What differs is the nucleophile's identity, the solvent, and the functional group introduced into the product.

7. Exam Technique: Writing Mechanisms and Answering 'Explain' Questions

Mechanism questions are among the most mark-rich in AQA organic chemistry, and they are also among the most penalised for careless errors. The following guidance consolidates the most common pitfalls and the language that secures marks.

Worked example: Drawing a Nucleophilic Substitution Mechanism for AQA

Draw the mechanism for the reaction of 1-bromopropane with aqueous sodium hydroxide. Show all curly arrows, relevant dipoles, and the identity of the leaving group.

1. Draw CH₃CH₂CH₂–Br with δ+ on the carbon bonded to Br and δ− on Br. — Bromine is more electronegative than carbon, so the C–Br bond is polar. Showing δ+ and δ− is required for full marks on mechanism questions — examiners look for this explicitly.
2. Draw OH⁻ with two dots (lone pair) on the oxygen, positioned approaching the δ+ carbon. — The lone pair must be shown on the nucleophile. Without it, the curly arrow has no starting point and the mark is lost.
3. Arrow from the lone pair on O in OH⁻ → to the δ+ carbon of 1-bromopropane. — This arrow represents the donation of the lone pair to form the new O–C bond. The tail of the arrow must start from the lone pair, not from the oxygen atom itself.
4. Arrow from the C–Br bond → to the Br atom. — This arrow represents the heterolytic breaking of the C–Br bond. Both electrons move to bromine, which departs as Br⁻. The tail starts from the C–Br bond, not from the carbon alone.
5. Products: CH₃CH₂CH₂OH (propan-1-ol) and Br⁻. — Both products must be shown. Br⁻ is the leaving group. The alcohol is the organic product. State symbols are not required in mechanism diagrams but are required in full equations.

Answer: The mechanism shows OH⁻ attacking the δ+ carbon of 1-bromopropane via two curly arrows (lone pair → C, and C–Br bond → Br), producing propan-1-ol and Br⁻.

In exam questions, the condition often tells you the product: aqueous hydroxide gives substitution to an alcohol, ethanolic cyanide gives a nitrile with one extra carbon, and excess ethanolic ammonia gives a primary amine. The solvent is part of the chemistry, not just background detail.