Electrophilic addition reactions of alkenes

Electrophilic addition is the main reaction of alkenes. An electrophile is attracted to the electron-rich C=C double bond, and two atoms or groups add across it, so the double bond becomes a single bond. Alkenes react this way with hydrogen bromide, bromine and sulfuric acid. The mechanism has two steps and goes through a carbocation, which decides the major product when the alkene is unsymmetrical.

Why alkenes undergo electrophilic addition

A C=C double bond is made of two pairs of shared electrons, so it is a region of high electron density. An electrophile is an electron pair acceptor: a positive ion, or the slightly positive (δ+) end of a polar molecule. It is attracted to the double bond and accepts a pair of electrons from it.

Addition means two molecules join to make one product, with nothing else made. Alkenes are unsaturated, so they have room to add atoms; alkanes are saturated and can't do this.

Electrophilic addition mechanism: ethene and HBr

Hydrogen bromide is polar: bromine is more electronegative than hydrogen, so the hydrogen end of the bond is slightly positive (δ+) and the bromine end slightly negative (δ−). The δ+ hydrogen is the electrophile.

Step 1: the electrophile attacks

• Draw a curly arrow from the C=C double bond to the δ+ H atom of HBr. A pair of electrons from the double bond forms a new C-H bond.
• Draw a second curly arrow from the H-Br bond to the Br atom. The bond breaks heterolytically and both electrons go to bromine, which leaves as a bromide ion, Br⁻.
• The other carbon has lost its share of the double bond, so it is now positively charged: a carbocation, CH₃CH₂⁺.

Step 2: the bromide ion attacks

• Draw a curly arrow from a lone pair on Br⁻ to the positive carbon. The pair of electrons forms the C-Br bond.
• The product is bromoethane, CH₃CH₂Br.

Electrophilic addition of bromine

Bromine, Br₂, isn't polar. But as a bromine molecule gets close to the electron-rich double bond, the electrons in the Br-Br bond are repelled away from the alkene. The near bromine atom becomes δ+ and the far one δ−: the double bond has induced a dipole. The mechanism then runs exactly like HBr:

1. A curly arrow from the C=C to the δ+ Br; another from the Br-Br bond to the δ− Br, which leaves as Br⁻.
2. A carbocation forms, with a bromine atom on the neighbouring carbon.
3. Br⁻ attacks the carbocation, giving a dibromoalkane.

The product from ethene is 1,2-dibromoethane. This reaction is the test for an alkene: shake the compound with bromine water, and if it contains a C=C double bond the orange bromine water turns colourless.

Try the bromine water test on a set of compounds in this question from the Organic Chemistry course.

Electrophilic addition of sulfuric acid

Cold, concentrated sulfuric acid adds across the double bond in the same way. The δ+ H of H₂SO₄ is the electrophile, and the hydrogensulfate ion, HSO₄⁻, attacks the carbocation:

The product, ethyl hydrogensulfate, reacts with water when it is warmed, giving ethanol and turning back into sulfuric acid. Overall, water has been added across the double bond, and the sulfuric acid acts as a catalyst.

Industrially, alcohols are made by the hydration of alkenes with steam in the presence of an acid catalyst, through the same kind of carbocation mechanism. AQA asks you to outline that mechanism too (3.3.5.1).

Unsymmetrical alkenes: major and minor products

Propene, CH₃CH=CH₂, is unsymmetrical: the two carbons of the double bond have different groups. When HBr adds, the hydrogen can go on either carbon, so there are two possible carbocations and two possible products:

Why: carbocation stability

A carbocation is classified by how many carbon (alkyl) groups are attached to the positive carbon: one for primary, two for secondary, three for tertiary. Alkyl groups push electron density towards the positive carbon, spreading out its charge. The more alkyl groups, the more stable the carbocation:

The more stable carbocation forms more easily, so more of the product goes through it. That is why 2-bromopropane is the major product. This pattern is known as Markovnikov's rule, which has its own section below.

Example 1: a tertiary carbocation

Predict the major product when HBr reacts with 2-methylbut-2-ene, (CH₃)₂C=CHCH₃.

If H adds to the CH end, the positive charge is left on the carbon joined to two methyl groups and an ethyl group. Three alkyl groups make it a tertiary carbocation, (CH₃)₂C⁺CH₂CH₃. If H adds to the other carbon, the cation is only secondary. The tertiary one is more stable, so the major product is 2-bromo-2-methylbutane, (CH₃)₂CBrCH₂CH₃.

Example 2: a symmetrical alkene

But-2-ene, CH₃CH=CHCH₃, is symmetrical. Whichever carbon the hydrogen joins, the carbocation is the same secondary ion, so there is only one product: 2-bromobutane. Major and minor products only arise with unsymmetrical alkenes.

Check your understanding.

Markovnikov's rule

Markovnikov's rule says that when a compound H-X, such as HBr, adds to an unsymmetrical alkene, the hydrogen goes to the carbon of the double bond that already has more hydrogen atoms, and X goes to the carbon with fewer. It is named after the Russian chemist Vladimir Markovnikov.

The rule tells you what happens; carbocation stability tells you why. In step 1 the hydrogen can bond to either carbon of the double bond, and the positive charge is left on the other one. If the hydrogen joins the carbon with more hydrogens, the positive charge sits on the carbon with more alkyl groups. Alkyl groups push electron density towards it and spread out the charge, so that carbocation is more stable. The more stable carbocation forms more easily, so most of the product comes from it.

Worked example: 2-methylpent-1-ene and HBr

Predict the major and minor products when HBr adds to 2-methylpent-1-ene, CH₂=C(CH₃)CH₂CH₂CH₃.

1. Find the two carbons of the double bond. Carbon 1 (CH₂) has two hydrogens; carbon 2 has none, because it carries a methyl group and a propyl group.
2. If H adds to carbon 1, the positive charge is on carbon 2, which is joined to three alkyl groups: a tertiary carbocation, (CH₃)₂C⁺CH₂CH₂CH₃.
3. If H adds to carbon 2, the positive charge is on carbon 1, which is joined to only one alkyl group: a primary carbocation, ⁺CH₂CH(CH₃)CH₂CH₂CH₃.
4. The tertiary carbocation is far more stable, so Br⁻ mostly attacks there.

The major product is 2-bromo-2-methylpentane, (CH₃)₂CBrCH₂CH₂CH₃, and the minor product is 1-bromo-2-methylpentane, BrCH₂CH(CH₃)CH₂CH₂CH₃. Both are C₆H₁₃Br. That matches the rule: the hydrogen went to carbon 1, the carbon that already had more hydrogens. The same reasoning says that steam with an acid catalyst turns this alkene mainly into 2-methylpentan-2-ol.

With peroxides present, HBr can add the other way round, by a free-radical mechanism. This is called anti-Markovnikov addition, and it isn't in AQA's A-level specification.

Electrophilic addition reactions you need

Common mistakes

• Starting the first arrow at the H. It starts at the C=C double bond and points to the H.
• Forgetting the second arrow in step 1, from the H-Br bond to Br.
• Missing charges. Show δ+ and δ−, the + on the carbocation, and the lone pair and − charge on Br⁻.
• Calling it nucleophilic addition. The first attack is by an electrophile on the alkene; nucleophilic addition is the reaction of carbonyl compounds.
• Putting the + on the wrong carbon. It goes on the carbon that did not gain the hydrogen.
• Explaining the major product with the rule alone. Refer to primary, secondary and tertiary carbocation stability.

Practice questions

1. Define an electrophile.
2. Name the product when ethene reacts with bromine.
3. Name the major and minor products when propene reacts with HBr.
4. Explain why bromine, a non-polar molecule, can act as an electrophile.
5. Why does but-2-ene give only one product with HBr?

Answers: 1) An electron pair acceptor. 2) 1,2-dibromoethane. 3) Major: 2-bromopropane; minor: 1-bromopropane. 4) The high electron density of the C=C bond repels the electrons in the Br-Br bond, inducing a dipole, so the nearer bromine atom becomes δ+. 5) It is symmetrical, so both ways of adding H give the same secondary carbocation.

Practise with a short timed paper

Alkenes and their mechanisms are in the first year of A-level Chemistry. The daily A-level Chemistry: Year 1 paper is a short timed paper of 8 questions, with a new one every day. For the other mechanisms in the course, see nucleophilic substitution and nucleophilic addition.

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