Electrophilic Addition Mechanisms Revision Notes | A Level AQA Chemistry

1. Why Alkenes React the Way They Do

Alkenes contain a carbon-carbon double bond made up of a sigma bond and a pi bond. The pi bond sits above and below the plane of the molecule, and its electrons are exposed and relatively loosely held. That electron-rich region is what makes alkenes so reactive towards electrophiles.

An electrophile is an electron-pair acceptor — a species that is attracted to regions of high electron density. When an electrophile approaches the pi bond of an alkene, the pi electrons are donated to the electrophile. This breaks the pi bond and forms a new bond, leaving one carbon with a positive charge. That positively charged carbon species is called a carbocation.

The overall reaction type is called electrophilic addition: the electrophile adds across the double bond, and the pi bond is replaced by two new sigma bonds. Nothing is eliminated — both atoms of the reagent end up in the product. This distinguishes addition from substitution or elimination.

2. Mechanism for Addition of HBr to Alkenes

The addition of hydrogen bromide (HBr) to an alkene is the clearest example to learn first, because the two steps of the mechanism are distinct and straightforward to follow. Take ethene reacting with HBr to form bromoethane.

Notice that curly arrows always start from an electron-rich site — either a bond or a lone pair — and point towards an electron-poor site. A full curly arrow represents the movement of two electrons. Getting the arrow tails and heads in the right place is where most marks are won or lost in mechanism questions.

Worked example: Drawing the HBr Addition Mechanism Step by Step

Describe and draw the mechanism for the reaction of HBr with ethene, CH2=CH2, to form bromoethane.

1. Nucleophile: pi bond of CH2=CH2 (electron-rich) Electrophile: H of H-Br (partially positive, delta+) — The H-Br bond is polar because Br is more electronegative than H. The H carries a partial positive charge, making it the electrophile.
2. Arrow 1: from the C=C pi bond -> to the H of H-Br Arrow 2: from the H-Br bond -> to the Br atom — Arrow 1 shows the pi electrons attacking H. Arrow 2 shows the H-Br bond breaking heterolytically, with both electrons going to Br, forming Br-.
3. CH3-CH2+ (primary carbocation) and Br- are the products of Step 1 — One carbon now has only three bonds and a positive charge. This is the carbocation intermediate. It is high in energy and short-lived.
4. Arrow: from a lone pair on Br- -> to the C+ carbon — Br- is a nucleophile. It donates a lone pair to the electron-deficient carbon, forming a new C-Br sigma bond.
5. CH3CH2Br (bromoethane) — The pi bond has been replaced by a C-H bond and a C-Br bond. Both atoms of HBr have added across the double bond — this is addition, not substitution.

Answer: Bromoethane, CH3CH2Br, formed via a two-step electrophilic addition mechanism involving a carbocation intermediate.

3. Unsymmetrical Alkenes and Markovnikov's Rule

With ethene, both carbons of the double bond are identical, so there is only one possible product. With an unsymmetrical alkene such as propene (CH3-CH=CH2), the electrophile can add to either carbon of the double bond, giving two possible carbocation intermediates — and therefore two possible products. This is where carbocation stability becomes critical.

Alkyl groups (such as CH3-) donate electron density towards a positive charge through a process called inductive effect. The more alkyl groups attached to the positively charged carbon, the more the positive charge is stabilised. A tertiary carbocation (three alkyl groups) is more stable than a secondary (two alkyl groups), which is more stable than a primary (one alkyl group).

When HBr adds to propene, the H+ can add to C1 (giving a secondary carbocation at C2) or to C2 (giving a primary carbocation at C1). The secondary carbocation is more stable, so the reaction proceeds predominantly through that pathway. The Br- then attacks C2, giving 2-bromopropane as the major product. This outcome is summarised by Markovnikov's rule: the H adds to the carbon that already has more hydrogens, because that produces the more stable (more substituted) carbocation.

4. Mechanism for Addition of H2SO4 to Alkenes

Concentrated sulphuric acid (H2SO4) also undergoes electrophilic addition with alkenes. The mechanism follows the same two-step pattern as HBr addition. In Step 1, the pi bond donates electrons to H+ (from H2SO4), forming a carbocation and releasing HSO4-. In Step 2, the HSO4- ion acts as the nucleophile and attacks the carbocation, forming an alkyl hydrogensulphate.

For ethene reacting with concentrated H2SO4, the product is ethyl hydrogensulphate (CH3CH2OSO3H). This intermediate is important industrially: when warmed with water, it hydrolyses to give ethanol. This is one route for the industrial production of ethanol from ethene.

The same Markovnikov selectivity applies to H2SO4 addition across unsymmetrical alkenes: H+ adds to the carbon with more hydrogens, generating the more stable (more substituted) carbocation on the other carbon. HSO4- then attacks that carbocation carbon.

5. Mechanism for Addition of Br2 to Alkenes

Bromine (Br2) is a non-polar molecule, so it might seem odd that it acts as an electrophile. The key is what happens when Br2 approaches the electron-rich pi bond of an alkene. The pi electrons repel the electron cloud of the nearer bromine atom, inducing a temporary dipole: the closer Br becomes delta-positive and the further Br becomes delta-negative.

This induced dipole makes the nearer Br atom electrophilic. The pi bond then attacks it. In Step 1, the pi electrons attack the near (delta+) Br, forming a new C-Br bond; simultaneously, the Br-Br bond breaks heterolytically, with both electrons going to the far Br, which leaves as Br-. A carbocation forms on the other carbon. In Step 2, Br- attacks the carbocation, forming a dibromoalkane.

The product of Br2 addition to ethene is 1,2-dibromoethane (CH2BrCH2Br). Both bromine atoms have added across the double bond, one to each carbon. This is the same two-step pattern as HBr addition, but here the electrophile is a temporarily polarised Br2 molecule rather than a permanently polar H-Br bond.

6. The Bromine Water Test for Unsaturation

One of the most important applications of Br2 addition is as a qualitative test for the presence of a C=C double bond. Bromine water is an aqueous solution of bromine and has a characteristic orange-brown colour. When it is added to a compound containing a C=C double bond, the bromine undergoes electrophilic addition across the pi bond. The bromine is consumed in the reaction, and the solution turns colourless.

The product when bromine water reacts with an alkene is not a simple dibromoalkane. Because water is present as the solvent and in large excess, water molecules and Br- ions compete to attack the carbocation in Step 2. Water attacks as H2O, forming an oxonium ion intermediate (C-OH2+), which then loses H+ to give a bromohydrin (where -OH has been incorporated from water). In practice, the bromohydrin is the predominant product in aqueous conditions. For the purposes of the test, however, the key observation is the decolourisation — the product identity is not required at this stage.

A saturated compound such as an alkane does not react with bromine water at room temperature in the absence of UV light. The orange-brown colour remains. This contrast makes the bromine water test a reliable and rapid way to distinguish between saturated and unsaturated organic compounds.

7. Exam Technique and Common Mistakes

Mechanism questions are among the most mark-rich and most frequently dropped sections in AQA organic chemistry. The marks are very specific: examiners check the position of every curly arrow tail and head. Getting the concept right but drawing the arrow backwards will cost you the mark.

Extended 'explain' questions on mechanisms often ask you to justify why a particular product is the major product. A full-mark answer links the structure of the carbocation intermediate to its stability, then connects that stability to the product distribution. Mentioning 'inductive effect' and 'electron donation from alkyl groups' will strengthen your answer at A-level.

Across all three reactions covered — HBr, H2SO4, and Br2 — the underlying logic is identical: the electron-rich pi bond attacks an electrophile, a carbocation intermediate forms, and a nucleophile completes the addition. Mastering this two-step pattern, and being able to apply it to any alkene and any of these three reagents, is the core skill this topic demands.