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9476 / 2027
Organic Chemistry

Lesson 9 of 34

An alkene donates its pi electrons to an electrophile

Write bromine addition as two steps: a slow step that makes a carbocation, then a fast step in which a nucleophile attacks it.

A-Level 9476 (2026-2027)

  • I can explain alkene reactivity towards electrophiles.Syllabus

    Syllabus 9476, 11.3(d). Explain alkene reactivity towards electrophiles.

  • I can describe electrophilic addition of bromine to ethene.Syllabus

    Syllabus 9476, 11.3(l). Describe electrophilic addition of bromine to ethene.

  • I can describe alkene chemistry using ethene and analogous structures.Syllabus

    Syllabus 9476, 11.4(b). Describe alkene chemistry using ethene and analogous structures.

  • I can apply and explain Markovnikov addition.Syllabus

    Syllabus 9476, 11.4(c). Apply and explain Markovnikov addition.

An alkene donates its pi electrons to an electrophile

Write bromine addition as two steps: a slow step that makes a carbocation, then a fast step in which a nucleophile attacks it.

Make a guess. You are not marked.

Propene, CH3CH=CH2, reacts with hydrogen bromide. What is the major product?

  1. 2-bromopropane, via the secondary carbocation
  2. An equal mixture of 1- and 2-bromopropane
  3. 1-bromopropane, via the primary carbocation
Show the answer

2-bromopropane, via the secondary carbocation

H+ adds to the end carbon to make the more stable secondary carbocation, which Br- then attacks.

An alkene reacts with electrophiles because of its π bond. The π electrons sit above and below the C=C bond, further from the carbon nuclei than the σ electrons. They are held less tightly, so the C=C bond is an electron-rich site that is easy to reach. An electrophile is an electron-pair acceptor. It is attracted to this electron-rich region, and the π bond donates a pair of electrons to it.

Step 1 (slow). Br2 is non-polar on its own. As it comes close to ethene, the electron-rich π bond repels the electrons in the Br-Br bond. This gives Br2 an induced dipole: the nearer Br becomes δ+ and the further Br becomes δ-. The π electrons move to form a new C-Br bond with the δ+ bromine. At the same time the Br-Br bond breaks by heterolytic fission: both bonding electrons go to the δ- bromine, which leaves as Br-. The other carbon has lost its share of the π pair, so it now has only three bonds and a positive charge. This ion is a carbocation. The step is slow because it breaks the π bond and the Br-Br bond and creates two ions.

Step 1: the pi bond attacks the delta+ bromine

Ethene, H2C=CH2, sits on the left and Br2 on the right, labelled delta plus on the nearer bromine and delta minus on the further one. One curly arrow starts at the middle of the C=C double bond and ends at the delta plus bromine. A second curly arrow starts at the middle of the Br-Br bond and ends at the delta minus bromine. Below, the products of this slow step are drawn: a carbocation in which the left carbon carries the new Br and the right CH2 carries a plus charge, and a bromide ion with a lone pair.

Mark points: δ+ and δ- on Br2, an arrow from the C=C bond to δ+ Br, an arrow from the Br-Br bond to δ- Br, the + charge on the carbon that does not carry Br, Br- with a lone pair, and the word slow.

Step 2 (fast). The bromide ion donates a lone pair to the positive carbon. This forms the second C-Br bond. Oppositely charged ions attract strongly and no bond has to break, so this step is fast. The product is BrCH2CH2Br, 1,2-dibromoethane.

Step 2: bromide attacks the carbocation

The carbocation BrCH2-CH2+ is on the left, with Br hanging below the left carbon and a plus charge on the right carbon. A bromide ion with a lone pair is on the right. A curly arrow starts at the lone pair on bromide and ends at the positive carbon. The product written underneath is BrCH2CH2Br, 1,2-dibromoethane.

The arrow starts at the lone pair on Br-, not at the minus sign, and ends at C+. Label this step fast.

Addition of HX follows the same two-step pattern. The π bond accepts H+ from the δ+ H of H-X while the H-X bond pair moves onto X. This forms a carbocation and X-. Then X- donates a lone pair to the positive carbon. For propene + HBr, the step that gives the secondary carbocation is favoured over the one that gives the primary carbocation, so 2-bromopropane is the major product.

Apply Markovnikov's rule by comparing intermediates
  1. Draw both choices for step 1

    For CH3CH=CH2 + HBr, H can add to C1 to give CH3-C+H-CH3, or to C2 to give CH3CH2CH2+.

  2. Compare carbocation stability

    Alkyl groups donate electron density and disperse the positive charge. The secondary carbocation has two alkyl groups on C+ and the primary has one, so the secondary carbocation is more stable and forms faster.

  3. Add the nucleophile to the more stable positive centre

    The major product is CH3CHBrCH3. H has added to the carbon that already had more H atoms.

Markovnikov's rule predicts the major product. It does not mean the minor product is impossible: a little 1-bromopropane still forms through the primary carbocation. The same carbocation argument decides orientation when the electrophile is δ+ Br from Br2, as the worked example below shows. Radical addition in the presence of peroxides is outside this mechanism.

Worked exampleBromine water and bromine in sodium chloride solution

Propene is shaken with (a) Br2(aq), and (b) Br2 dissolved in concentrated NaCl(aq). Use the mechanism to predict the organic products and explain where each group ends up.

Try it first. Then show the worked stepsHide the worked steps
  1. Step 1 is the same in both mixtures, because Br2 is the only electrophile present. The π bond attacks the δ+ Br. If Br bonds to C1 (the CH2 end), the positive charge sits on C2: CH3-C+H-CH2Br, a secondary carbocation. If Br bonds to C2, the charge sits on C1: a primary carbocation.
  2. The secondary carbocation has two carbon groups bonded to C+; the primary has only one. Alkyl groups donate electrons and disperse the positive charge, so the secondary carbocation is more stable and forms faster. Br therefore ends up on C1 and the nucleophile on C2.
  3. In step 2, any nucleophile in the mixture can attack C+. In Br2(aq), water is present in far larger amounts than Br-. A water lone pair attacks C2, and the oxygen then loses H+. This gives CH3CH(OH)CH2Br, 1-bromopropan-2-ol. Some Br- also attacks, giving a little CH3CHBrCH2Br, 1,2-dibromopropane.
  4. In concentrated NaCl(aq), Cl- is present at high concentration, so it also competes for C+. This gives CH3CHClCH2Br, 1-bromo-2-chloropropane, alongside the products from water and Br-.
  5. NaCl supplies only Cl-, a nucleophile. There is no δ+ Cl electrophile to start step 1, so no product has Cl on C1 and 1,2-dichloropropane cannot form.
Answer

(a) Mainly 1-bromopropan-2-ol, CH3CH(OH)CH2Br, with a little 1,2-dibromopropane. (b) 1-Bromo-2-chloropropane, CH3CHClCH2Br, as well as those two products. In every product Br is on C1 and the nucleophile is on C2, because both mixtures go through the more stable secondary carbocation.

Check your understanding

Bromine in concentrated NaCl(aq) is shaken with 2-methylpropene, (CH3)2C=CH2. Which compound is not formed?

  1. (CH3)2CBrCH2Br
  2. (CH3)2C(OH)CH2Br
  3. (CH3)2CBrCH2Cl
  4. (CH3)2CClCH2Br
Show the answer

(CH3)2CBrCH2Cl

The only electrophile is delta+ Br from Br2. Br adds to the CH2 end to give the tertiary carbocation (CH3)2C+-CH2Br. A Cl on the CH2 carbon would need a Cl electrophile, and NaCl supplies only Cl-.

Exam-style question

But-2-ene, CH3CH=CHCH3, reacts with bromine dissolved in CCl4. Name and describe the mechanism. Show all partial charges, charges, relevant lone pairs and curly arrows. [3]

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Put these in order

Put the events in the addition of bromine to ethene in order.

Show the answer

1. Br2 approaches the electron-rich pi bond and becomes polarised The delta+ Br end is now an electrophile that the pi electrons can attack.

2. The pi electrons attack the delta+ Br, and the Br-Br bond breaks heterolytically This slow step leaves a positive carbon and a bromide ion.

3. A carbocation intermediate and a Br- ion are formed The carbocation is electron-poor, so a nucleophile is attracted to it.

4. Br- uses a lone pair to bond to the carbocation, giving 1,2-dibromoethane