Part 2: Addition reactions and mechanisms

Choose reagents and conditions, draw electron movement and explain major products through carbocation stability.

Replace the pi bond with new sigma bonds

Bromine as a test for unsaturation

An alkene decolourises orange/brown bromine water at room temperature as bromine is consumed. Shake a small sample with bromine water under the specified safety controls. No UV is needed, unlike alkane radical substitution.

The standard AS addition model uses Br₂ across C=C to show a vicinal dibromo product. In water, solvent can also react to form bromohydrin products; the colour test itself supports reaction of an unsaturated group, not proof of one pure isolated product. Other reactive substances can also consume bromine, so interpret the test in context.

Electrophilic addition of HBr

An electrophile accepts an electron pair. HBr has Hδ+–Brδ−. Draw an arrow from the C=C π bond to H, and another from the H–Br bond to Br. A C–H bond forms and the other carbon becomes a carbocation. Then draw an arrow from a lone pair on Br⁻ to C⁺, forming the C–Br bond.

For propene, H adding to the terminal CH₂ gives CH₃–C⁺H–CH₃, a secondary carbocation. H adding to the central carbon gives a primary terminal carbocation. The more stable secondary intermediate leads predominantly to 2-bromopropane. In the simple comparison, tertiary > secondary > primary carbocation stability; “the product is more stable” does not explain this pathway.

The common major-product rule is often called Markownikoff’s rule, but the mechanistic explanation is the more stable carbocation. Do not assert an exact ratio or equal proportions from the labels alone.

Induce a dipole in Br₂

The electron-rich double bond polarises approaching Br₂: the nearer Br becomes δ+. Show a π-bond arrow to Brδ+ and a Br–Br bond arrow to the departing Br, followed by attack of Br⁻ on the positively charged intermediate. OCR accepts an appropriate carbocation model or a bromonium-ion intermediate for this AS mechanism. Do not mix the two drawings halfway through.

H032/01 June 2025 Q24(c) checked arrow directions, dipoles, the intermediate, the product and the mechanism name. The associated explanation concerned carbocation stability rather than merely stating a product-selection rule.

Draw an addition product without changing the carbon skeleton

Begin by copying the full carbon skeleton. Change C=C to C–C, then attach one part of the reagent to each former double-bond carbon. For CH₃CH=CHCH₃ plus Br₂ in the simple addition model, put one Br on each middle carbon: CH₃CHBrCHBrCH₃. It is 2,3-dibromobutane, not a product with both Br atoms on one carbon.

For hydrogenation, attach one H to each double-bond carbon. For hydration, attach H and OH. An unsymmetrical alkene can have alternative orientations for addition of HX or water; decide which is major using the relevant reaction model instead of losing the alternative silently.

After constructing a product, recount hydrogens. Each carbon still needs a total bond order of four. Addition uses the π bond while preserving the σ connection between the original double-bond carbons.

Worked major pathway: 2-methylpropene and HCl

For CH₂=C(CH₃)₂, adding H to the terminal CH₂ gives a central C⁺ attached to three methyl groups: a tertiary carbocation. Adding H to the central carbon instead places the positive charge on a terminal carbon attached to only one carbon group: a primary carbocation.

The tertiary intermediate is more stable in this comparison, so the major route leads to (CH₃)₃CCl, 2-chloro-2-methylpropane. Chloride donates a lone pair to the positively charged carbon. The explanation refers to the intermediate’s stability, not simply the number of H atoms on the final product.

For the arrows: π bond → Hδ+; H–Cl bond → Clδ−; then a lone pair on Cl⁻ → C⁺. HCl’s dipole is permanent, whereas the Br₂ dipole in the halogen-addition model is induced on approach. Do not place a permanent δ+/δ− pair on the symmetrical C=C merely to imitate HCl.

A colour test and a preparative reaction answer different questions

Bromine-water decolourisation provides evidence that bromine has reacted. It is particularly useful for distinguishing an alkene from an alkane under controlled room-temperature conditions without UV. It does not itself determine the exact carbon skeleton, the alkene position or a pure product yield.

For preparation, specify the reacting substance, reagent, conditions and intended product. Steam with an acid catalyst hydrates an alkene; aqueous hydroxide hydrolyses a haloalkane. Both can lead to an alcohol but by different starting groups and mechanisms.

A question that asks for a mechanism needs electron movements and intermediates as well as the correct overall product. A question that asks only for reagents and conditions needs those stated clearly; an unlabelled reaction arrow alone is incomplete.

Quick checks

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

Sources