Carbocation Stability and Product Formation Revision Notes | A Level AQA Chemistry

1. Why Carbocation Stability Controls What You Get

When an unsymmetrical alkene reacts with a reagent like HBr, the reaction does not produce equal amounts of two possible products. One product dominates — and the reason lies entirely in which carbocation intermediate forms more readily.

This is one of those topics where understanding the why makes the prediction automatic. Once you see how electron density stabilises a positive charge, you will never need to memorise which product forms — you will be able to work it out from first principles.

2. What Is a Carbocation and How Does It Form?

Picture propene reacting with HBr. The pi bond acts as a nucleophile and attacks the partially positive hydrogen of HBr. Both electrons from the pi bond move onto that hydrogen, breaking the H–Br bond and generating Br⁻. The carbon that did not bond to hydrogen is left with only three bonds and a positive charge — that is your carbocation intermediate.

The key question is: which carbocation — secondary or primary — forms more readily? The answer determines which product dominates.

3. Primary, Secondary, and Tertiary Carbocations — The Stability Ladder

Carbocations are classified by how many carbon groups are directly attached to the positively charged carbon. A primary carbocation has one alkyl group attached; secondary has two; tertiary has three.

The stability order arises from the inductive effect of alkyl groups. Each alkyl group donates electron density through the sigma bonds towards the electron-deficient carbon, partially dispersing the positive charge and lowering the energy of the intermediate.

More alkyl groups means more electron donation, which means greater dispersal of the positive charge, which means a lower-energy (more stable) carbocation. A tertiary carbocation has three alkyl groups doing this job simultaneously; a primary has only one.

With the stability ladder established, you now have the tool to predict which product forms in the greatest yield whenever an unsymmetrical alkene undergoes electrophilic addition.

4. Major and Minor Products from Unsymmetrical Alkenes

An unsymmetrical alkene is one where the two carbon atoms of the double bond carry different substituents — so adding a reagent like HBr across the double bond can give two structurally different products depending on which carbon the H attaches to.

The reaction proceeds through whichever carbocation intermediate is more stable. Because the more stable carbocation forms faster (it has a lower activation energy), the product derived from it is produced in greater yield — this is the major product. The product from the less stable carbocation is the minor product.

Worked example: Predicting Major and Minor Products: HBr Addition to Propene

Propene (CH₃CH=CH₂) reacts with HBr. Identify the major and minor products, justifying your answer by reference to carbocation stability.

1. C-1: CH₂= (two H atoms attached) C-2: =CHCH₃ (one H atom, one methyl group attached) — These two carbons are different, so propene is an unsymmetrical alkene. Adding H to either carbon gives a different carbocation.
2. Route A — H adds to C-1: CH₃–CH⁺–CH₃ (secondary carbocation at C-2) Route B — H adds to C-2: CH₃–CH₂–CH₂⁺ (primary carbocation at C-1) — In Route A, the positive charge sits on C-2, which has two alkyl groups (two methyl/ethyl fragments) attached. In Route B, the positive charge sits on C-1, which has only one alkyl group attached.
3. Route A carbocation: secondary (2°) — two alkyl groups donate electron density to C⁺ Route B carbocation: primary (1°) — one alkyl group donates electron density to C⁺ — The secondary carbocation in Route A is more stable because two alkyl groups inductively donate electron density, dispersing the positive charge more effectively than one group can.
4. Major product: Br⁻ attacks the secondary carbocation (Route A) → 2-bromopropane (CH₃CHBrCH₃) Minor product: Br⁻ attacks the primary carbocation (Route B) → 1-bromopropane (CH₃CH₂CH₂Br) — The reaction proceeds predominantly through the lower-energy (secondary) carbocation intermediate, so 2-bromopropane is the major product.
5. Major product: 2-bromopropane Minor product: 1-bromopropane — The major product forms because the secondary carbocation intermediate is more stable than the primary carbocation, due to greater inductive electron donation from two alkyl groups rather than one.

Answer: Major product: 2-bromopropane (CH₃CHBrCH₃), formed via the more stable secondary carbocation intermediate. Minor product: 1-bromopropane (CH₃CH₂CH₂Br), formed via the less stable primary carbocation intermediate.

5. Markovnikov's Rule — The Shortcut That Follows from the Mechanism

Markovnikov's rule states that when HX adds to an unsymmetrical alkene, the hydrogen attaches to the carbon that already has the greater number of hydrogen atoms. This is not an arbitrary rule — it is a direct consequence of carbocation stability.

The carbon with more hydrogens is typically the less substituted carbon. Adding H there places the positive charge on the more substituted carbon, which has more alkyl groups to stabilise it. The rule and the mechanism always agree.

At A-level, examiners expect you to go beyond quoting the rule. You must explain the outcome using carbocation stability and the inductive effect of alkyl groups. Stating Markovnikov's rule alone will not earn full marks on an 'explain' question.

6. Extending the Principle — A Tertiary Carbocation Example

The same logic applies whenever a tertiary carbocation can form. Because tertiary carbocations are the most stable of the three types, reactions that can access a tertiary intermediate will do so overwhelmingly, making the corresponding product the dominant one.

Worked example: Predicting Major Product: HBr Addition to 2-methylbut-2-ene

2-methylbut-2-ene (CH₃C(CH₃)=CHCH₃) reacts with HBr. Identify the major product and justify your answer.

1. C-2: bonded to two methyl groups and the double bond C-3: bonded to one methyl group, one H, and the double bond — The two carbons of the double bond carry different substituents, confirming this is an unsymmetrical alkene.
2. Route A — H adds to C-3: positive charge at C-2, which has two methyl groups and one ethyl fragment → tertiary carbocation Route B — H adds to C-2: positive charge at C-3, which has one methyl group → secondary carbocation — Route A generates a tertiary carbocation because C-2 is bonded to three carbon-containing groups once H has added to C-3.
3. Tertiary carbocation (Route A): three alkyl groups donate electron density inductively to C⁺ Secondary carbocation (Route B): two alkyl groups donate electron density to C⁺ — The tertiary carbocation is more stable. Three alkyl groups disperse the positive charge more effectively than two, lowering the activation energy for Route A.
4. Br⁻ attacks the tertiary carbocation at C-2 → 2-bromo-2-methylbutane — Because the tertiary carbocation forms preferentially, 2-bromo-2-methylbutane is the major product.

Answer: Major product: 2-bromo-2-methylbutane, formed via the more stable tertiary carbocation intermediate at C-2.

7. Exam Technique — Writing Answers That Earn Full Marks

AQA questions on this topic typically carry 3–5 marks and use the command word 'explain'. The mark scheme rewards a clear chain of reasoning: identify the two possible carbocations, state which is more stable and why, then link that stability to the major product.

A common error is to name the major product correctly but then give no mechanistic justification. Examiners award marks for the reasoning, not the product name alone. Always include the type of carbocation (primary, secondary, or tertiary) and the reason for its stability (inductive electron donation from alkyl groups).

Carbocation stability and the formation of major and minor products are directly linked: the more stable the carbocation intermediate, the lower its activation energy, and the greater the proportion of product formed through that route. Mastering this reasoning chain — from alkene structure, through carbocation classification, to product prediction — covers both AQA specification requirements for this topic.