Mechanisms and Reaction Rates Revision Notes | A Level AQA Chemistry

1. Why Halogenoalkanes React: The Electrophilic Carbon

Picture a halogenoalkane such as bromoethane, $CH_3CH_2Br$. The bromine atom is far more electronegative than carbon, so it pulls electron density towards itself, leaving the carbon with a partial positive charge: $C^{\delta+}-Br^{\delta-}$. That electron-poor carbon is a magnet for anything carrying a lone pair of electrons.

Any species that donates an electron pair to an electrophilic centre is called a nucleophile. Common nucleophiles you will meet in AQA questions include hydroxide ions ($OH^-$), cyanide ions ($CN^-$), and ammonia ($NH_3$). When a nucleophile attacks the $C^{\delta+}$ of a halogenoalkane and the halide ion departs, the overall process is nucleophilic substitution.

There are two distinct pathways by which nucleophilic substitution can occur, and which one operates depends on the structure of the halogenoalkane. These are the S_N2 (bimolecular) and S_N1 (unimolecular) mechanisms. Understanding both — and knowing when each applies — is a core AQA requirement.

2. The S_N2 Mechanism: One Step, Back-Side Attack

The S_N2 mechanism is a single, concerted step — meaning bond breaking and bond forming happen simultaneously. The nucleophile approaches the electrophilic carbon from the side directly opposite the leaving group (the back of the molecule). This geometry is called back-side attack.

As the nucleophile's lone pair forms a new bond with carbon, the C-X bond stretches and breaks, and the halide ion leaves. There is no intermediate — the transition state is the highest-energy point on the reaction pathway. Because both the nucleophile and the halogenoalkane are involved in this single step, the rate depends on the concentration of both species.

S_N2 is strongly favoured by primary halogenoalkanes (one alkyl group on the carbon bearing the halogen). The carbon is relatively unhindered, so the nucleophile can approach the back face without obstruction. Secondary halogenoalkanes can also undergo S_N2, but more slowly due to greater steric crowding.

3. The S_N1 Mechanism: Two Steps via a Carbocation

When the carbon bearing the halogen is surrounded by three alkyl groups (a tertiary halogenoalkane), the back face is too crowded for a nucleophile to attack directly. Instead, the reaction takes a different route: the C-X bond breaks first, generating a carbocation intermediate, and the nucleophile attacks in a separate second step.

Because the rate-determining step (the slow step) involves only the halogenoalkane breaking apart — not the nucleophile — the rate depends only on the concentration of the halogenoalkane. This is why the mechanism is called unimolecular.

The tertiary carbocation is stabilised by the inductive electron-donating effect of the three surrounding alkyl groups. This stabilisation lowers the activation energy for Step 1, making S_N1 viable for tertiary substrates. Primary carbocations are far too unstable for this route.

4. Carbon-Halogen Bond Enthalpy and Reaction Rate

Now that you understand the two mechanisms, there is a second factor that controls how fast these reactions proceed: the strength of the carbon-halogen bond itself. Bond enthalpy is the energy required to break a bond — a higher value means a stronger, harder-to-break bond.

Across the halogens, bond enthalpy decreases as you move down Group 7. The C-F bond is exceptionally strong because fluorine is very small, giving very effective orbital overlap with carbon and a short strong bond. As atomic radius increases down the group, orbital overlap becomes less effective and the C-X bond becomes weaker.

The weaker the C-X bond, the lower the activation energy required for the bond to break — whether that is the heterolytic cleavage in S_N1 or the simultaneous bond-breaking in S_N2. A lower activation energy means a faster rate at a given temperature. Iodoalkanes therefore react fastest, and fluoroalkanes react slowest.

Here is where many students make a critical error. Fluorine is the most electronegative element, so you might expect the $C^{\delta+}-F^{\delta-}$ bond to be the most polarised and therefore the most reactive towards nucleophiles. The logic sounds reasonable — but it is wrong.

5. Testing Reactivity: The Silver Nitrate Experiment

The relative rates of hydrolysis of different halogenoalkanes can be demonstrated experimentally. Each halogenoalkane is warmed with aqueous ethanol and silver nitrate solution. As the C-X bond breaks and the halide ion is released into solution, it immediately precipitates with silver ions to form a silver halide precipitate.

The time taken for the precipitate to appear is a direct measure of the reaction rate. Iodoalkanes produce a yellow precipitate of silver iodide almost immediately. Bromoalkanes give a cream precipitate of silver bromide more slowly. Chloroalkanes produce a white precipitate of silver chloride most slowly of all. Fluoroalkanes produce no precipitate under these conditions — the C-F bond is too strong to break at this temperature.

6. Putting It Together: A Worked Mechanism Question

Exam questions on mechanisms require you to draw curly arrows correctly, identify the type of mechanism, and justify your choice. The worked example below models the level of detail AQA expects for a 4-mark mechanism question.

Worked example: Drawing the S_N2 Mechanism for the Reaction of 1-Bromopropane with Hydroxide Ions

Draw the mechanism for the reaction of 1-bromopropane with aqueous sodium hydroxide. State the type of mechanism and explain why this mechanism operates rather than S_N1. [4 marks]

1. Substrate: $CH_3CH_2CH_2Br$ (1-bromopropane — a primary halogenoalkane) Nucleophile: $OH^-$ (hydroxide ion, carries lone pairs on oxygen) — 1-Bromopropane is primary: only one alkyl group ($CH_3CH_2CH_2-$) is attached to the carbon bearing the bromine. This is the key structural feature that determines the mechanism.
2. $CH_3CH_2CH_2^{\delta+}Br^{\delta-}$ Bromine is more electronegative than carbon, creating a partial positive charge on the carbon and a partial negative charge on bromine. — The $\delta+$ on carbon is what attracts the nucleophile. Always mark this polarity in your mechanism diagram — it shows the examiner you understand why the nucleophile attacks that specific carbon.
3. Arrow 1: from lone pair on $O$ of $OH^-$ → to the $C^{\delta+}$ (back-side, 180° from the C-Br bond) Arrow 2: from the $C-Br$ bond → to $Br$ — Both arrows are drawn simultaneously — this is the concerted nature of S_N2. Arrow 1 shows the new O-C bond forming; Arrow 2 shows the C-Br bond breaking with both electrons going to bromine. The back-side approach is essential to show.
4. Transition state: $HO^{\delta-}\cdots C \cdots Br^{\delta-}$ with partial bonds shown as dashed lines and a trigonal bipyramidal arrangement around carbon. — The transition state is not an intermediate — it cannot be isolated. Showing it demonstrates understanding of the concerted mechanism and can secure an additional mark in extended questions.
5. $CH_3CH_2CH_2OH + Br^-$ Mechanism: S_N2 (bimolecular nucleophilic substitution) — S_N2 operates here because 1-bromopropane is a primary halogenoalkane. The carbon bearing the bromine has only one alkyl group, so the back face is relatively unhindered and the nucleophile can approach directly. A primary carbocation would be too unstable for S_N1.

Answer: S_N2 mechanism. Products: propan-1-ol ($CH_3CH_2CH_2OH$) and bromide ion ($Br^-$). S_N2 operates because the primary substrate is unhindered, allowing back-side attack; a primary carbocation intermediate (required for S_N1) would be too unstable to form.

7. Exam Technique and Mark-Winning Strategies

Mechanism questions are among the most reliably marked in AQA organic chemistry — the mark scheme is precise, and every curly arrow either scores or does not. The good news is that the same errors appear repeatedly, which means you can train yourself to avoid them.

When a question asks you to explain why one halogenoalkane reacts faster than another, the mark scheme expects a full reasoning chain: bond enthalpy value (or comparison) → activation energy → rate. Stopping at 'the C-I bond is weaker' without connecting it to activation energy and rate will typically cost you one mark.

For mechanism diagrams, AQA mark schemes award marks for: (1) correct curly arrow from nucleophile lone pair to carbon, (2) correct curly arrow from C-X bond to halogen, (3) correct products with charges, and (4) for S_N1, the carbocation intermediate drawn as a separate species. Each of these is an independent mark — you can gain partial credit even if your overall diagram has a flaw.

Across this lesson, two specification requirements have been woven together: the mechanisms of nucleophilic substitution (S_N1 and S_N2, with curly arrow logic and the role of substrate structure) and the influence of C-X bond enthalpy on reaction rate. These two ideas are not separate — they combine in every exam question that asks you to compare the reactivity of different halogenoalkanes, and mastering both gives you the tools to answer any variant the examiner can construct.