Part 1: Double bonds, E/Z and hydrogen bromide addition

Use the electron-rich double bond to build an addition mechanism, then explain major products through the intermediate that forms.

What makes an alkene reactive?

Alkenes contain C=C and are unsaturated hydrocarbons. CₙH₂ₙ applies to an acyclic hydrocarbon containing one double bond; extra rings or extra double bonds change the formula. Name the longest appropriate chain containing C=C and use the lower of the two double-bond carbon numbers, as in hex-2-ene.

A C=C bond contains a σ bond along the internuclear axis and a π bond formed by sideways overlap of p orbitals. The π electron density lies above and below the plane around the double bond. Each alkene carbon has approximately trigonal planar geometry with bond angles near 120°.

An electrophile accepts an electron pair. The accessible electron density in the π bond supplies that pair in electrophilic addition. Do not describe the whole alkene as a negative ion. Addition replaces the π bond while retaining the C–C σ bond and making new bonds to the added groups.

Check E/Z before assigning a label

E/Z requires restricted rotation about C=C and two different substituents on each of its carbons. Compare substituents at each end separately. Higher atomic number of the directly bonded atom gives higher CIP priority; if those atoms tie, compare the next attached atoms in descending atomic-number order.

Priority groups on the same side give Z; opposite sides give E. Two identical groups on one double-bond carbon prevent E/Z. Having matching groups somewhere across the molecule is not sufficient to assign the label.

For CH₃CH₂C(CH₃)=CHCH₂CH₃, ethyl outranks methyl at the left alkene carbon because C,H,H outranks H,H,H. At the right, ethyl outranks H. The two ethyl groups therefore determine E/Z in this example.

Build the HBr mechanism

HBr is polar, Hδ+–Brδ−. Draw an arrow from C=C to H and a second arrow from the H–Br bond to Br. This forms a C–H bond, a carbocation and Br⁻. Then draw an arrow from a lone pair on Br⁻ to the positively charged carbon. The final bromoalkane is neutral.

For propene, adding H to its terminal carbon gives CH₃CH⁺CH₃, a secondary carbocation. Br⁻ attack gives 2-bromopropane. The alternative gives a primary carbocation and leads to 1-bromopropane.

The major pathway uses the more stable intermediate

For the simple alkyl carbocations used here, the familiar stability order is tertiary > secondary > primary. Count carbon groups attached directly to the positively charged carbon. Explain the major product by identifying the more stable intermediate through which it forms; do not call the final bromoalkane a carbocation.

AQA’s historical explanation describes greater stabilisation by electron-releasing alkyl groups. Its guidance of 5 February 2025 acknowledges research challenging the inductive explanation and continues to credit the historical approach during the life of these specifications, while also crediting answers informed by that research. Use the current assessment convention with this qualification, rather than treating the inductive description as an unquestionable universal fact.

Do not predict exact percentages without data. Two secondary carbocations need not be equally stable, so their classification alone cannot prove a 50:50 product ratio.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Sources