Part 3: Alkene bonding and addition mechanisms

Use π-bond electron density to predict addition, show where electron pairs move, and explain why an unsymmetrical alkene can give a major product.

The double bond contains one σ bond and one π bond

Acyclic alkenes with one C=C have formula CₙH₂ₙ. Cycloalkenes also contain a C=C and are unsaturated, but a single ring plus one double bond gives CₙH₂ₙ₋₂. Never apply the open-chain formula to every molecule containing an alkene.

A σ bond has electron density along the line joining the nuclei, formed by head-on orbital overlap. In ethene, each carbon makes three σ bonds in a trigonal planar arrangement, with angles near 120°. The remaining p orbital on each carbon overlaps sideways to form one π bond, with electron density above and below the plane. Those two regions are parts of one π bond, not two extra bonds.

Rotating one end of C=C would destroy the sideways overlap, explaining restricted rotation and E/Z isomerism. The π electrons are more exposed than the σ electrons and can form a new bond to an electrophile. During addition the π bond is replaced by two new σ bonds while the carbon–carbon σ bond remains.

Reagents and conditions determine the product

An electrophile is an electron-pair acceptor. It may be positively charged, such as H⁺, or be the partially positive end of a polar or induced-polar molecule. Not every addition uses an ionic electrophilic mechanism: hydrogenation occurs on a metal surface. Learn the overall transformation and the required mechanism separately.

Nickel-catalysed hydrogenation reduces the number of C=C bonds in unsaturated vegetable oils, increasing saturation and generally increasing melting temperature; this is used to modify oils for margarine manufacture. For industrial ethene hydration, steam and phosphoric acid give ethanol; elevated temperature gives a practical rate and pressure supports the equilibrium yield. Exact operating values are not a substitute for naming steam and the acid catalyst.

Edexcel Topic 6 explicitly treats acidified manganate(VII) addition of two OH groups as diol formation. Use the supplied conditions: strong, prolonged or hot oxidation can lead to further oxidation/cleavage, which is not the simple diol transformation. The 2023 AS paper’s Q3(d)(ii) expected the diol from but-1-ene.

Track both electrons when HBr adds

HBr is polar, Hδ+–Brδ−. A full-headed curly arrow from the C=C π bond to H shows the pair used to form C–H. Simultaneously, a second arrow from the H–Br bond to Br shows heterolytic fission: Br takes both electrons and becomes Br⁻. One alkene carbon gains H; the other becomes the positively charged carbocation.

Then draw a curly arrow from a lone pair on Br⁻ to the positive carbon. A C–Br bond forms and the product is neutral. Arrows move electron pairs, not atoms or positive charge. They must start at a bond or lone pair and end at the atom or bond being formed. Put the positive charge on the correct intermediate carbon and show the bromide charge and lone pair.

Why unsymmetrical alkenes give different amounts

In propene, protonation can give a secondary carbocation, CH₃C⁺HCH₃, or a primary carbocation, CH₃CH₂CH₂⁺. Alkyl groups donate electron density towards an electron-deficient carbon and stabilise it; the usual order is tertiary > secondary > primary. The pathway through the more stable carbocation is favoured, so 2-bromopropane is the major product and 1-bromopropane the minor product.

Worked example: for 2-methylbut-1-ene, CH₂=C(CH₃)CH₂CH₃, placing H on the terminal CH₂ leaves a tertiary carbocation at C2. Bromide attack gives 2-bromo-2-methylbutane as the major product. The alternative gives a primary carbocation and 1-bromo-2-methylbutane. Draw the intermediate before naming the product; counting groups on the wrong carbon reverses the explanation.

Do not assign a universal 90:10 ratio or assume two secondary carbocations give exactly 50:50. The simple stability classification predicts a useful direction, not a measured composition. Compare structures and use numerical product data only when supplied.

Apply the electron-pair model to new reagents

Br₂ is non-polar in isolation. Near the electron-rich double bond its electrons are repelled, inducing Brδ+–Brδ−. In the Edexcel AS simplified mechanism, draw the π-bond arrow to the nearer Brδ+, the Br–Br bond arrow to the farther Br, a bromo-substituted carbocation and then a bromide lone-pair arrow to C⁺. The product has one Br on each former double-bond carbon.

This carbocation drawing is a simplified assessment model. Real halogen addition is commonly described using a bridged halonium ion; that extension is not needed to reproduce the AS model. It would be misleading to claim every alkene addition proceeds through an identical free carbocation.

For an unfamiliar binary molecule, use any supplied polarity to identify the electrophilic end, then track both halves into the product. For Iδ+–Clδ−, for example, the alkene π pair can form C–I while I–Cl breaks towards Cl, followed by chloride attack. Keep atom counts, charges and the given polarity consistent rather than copying the letters H and Br from memory.

Describe what is observed before making an inference

Shake a small sample with bromine water at room temperature, avoiding UV conditions. Loss of its orange/brown colour supports the presence of C=C in an appropriate simple organic unknown. Say colourless, not clear: a coloured solution can still be transparent. An alkane does not rapidly decolourise bromine water under these conditions.

Bromine in a non-aqueous medium gives a vicinal dibromoalkane. In bromine water, water can participate and bromohydrin products are possible; the qualitative test is the loss of bromine colour, so do not claim its only possible aqueous product is the dibromoalkane. Other reducing or reactive groups can also remove bromine colour: combine this evidence with the candidate structures and spectroscopy.

Worked amount and atom economy

Suppose 4.20 g propene reacts with excess bromine, and 16.2 g pure 1,2-dibromopropane is isolated. Using M(propene) = 42.0 and M(product) = 202.0 g mol⁻¹, n(propene) = 4.20/42.0 = 0.100 mol. The 1:1 equation gives a theoretical product mass of 0.100 × 202.0 = 20.2 g. Percentage yield = 16.2/20.2 × 100 = 80.2%.

The balanced addition has one product and 100% atom economy: all reactant atoms appear in it. The experimental yield is lower because not all material is converted, retained and isolated. High atom economy and high yield measure different aspects of a route.

Quick checks

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

Sources