The shape of the lesson
By the end of the lesson, learners can
- The mechanism: Step through the mechanism in the simulator.
- Which carbocation wins?: With an unsymmetrical alkene the electrophile can add to either carbon.
- Arrow 1: from C=C to Hδ+ or Brδ+.
- Arrow 2: from H–Br or Br–Br bond to Br.
- Arrow 3: from Br− lone pair to C+.
- Stability: tertiary > secondary > primary.
One picture every idea returns to
In electrophilic addition the π electrons of the C=C attack an electrophile. With Br2, the approaching π cloud induces a dipole in Br–Br; the π pair forms a bond to the Brδ+ atom while Br–Br breaks heterolytically, giving a carbocation and Br−, which then attacks the positive carbon. With HBr, Hδ+ is the electrophile. Alkyl groups push electron density towards a positive carbon (the positive inductive effect), so carbocation stability is tertiary > secondary > primary. With an unsymmetrical alkene the major product comes from the more stable carbocation: propene + HBr gives mainly 2-bromopropane.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Electrons go looking for a partner.. The π electrons of an alkene attract electron-poor species. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: The mechanism | Step through the mechanism in the simulator. Project tab 4 of the lab (Hydrocarbons lab, electrophilic addition) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Which carbocation wins? | With an unsymmetrical alkene the electrophile can add to either carbon. | Mix-Pair-Share |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 14 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Which carbocation wins?, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The first curly arrow must start from the C=C double bond (the π electrons) and point to the electrophile, not start at the bromine. | Draw the carbocation with a full + charge on carbon, and the bromide ion with a lone pair and a − charge. The major product comes from the more substituted carbocation. |
Electrophilic addition: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
Describe, with curly arrows, the mechanism of the reaction between ethene and bromine.
Explain why a non-polar bromine molecule can act as an electrophile.
Propene reacts with HBr. Name the major and minor products and explain the difference.
Rank primary, secondary and tertiary carbocations in order of increasing stability.
Explain why bromine water with ethene gives some 2-bromoethanol as well as 1,2-dibromoethane.
Predict the major product of but-1-ene with HBr and give a reason.
Answer key and marking notes
- The π electrons induce a dipole in Br2 (Brδ+–Brδ−).
- Arrow from the C=C to Brδ+; arrow from the Br–Br bond to Brδ−.
- Carbocation CH2BrCH2+ and Br− form.
- Arrow from a lone pair on Br− to C+, giving CH2BrCH2Br.
- As Br2 approaches the electron-rich π bond, its electrons are repelled.
- This induces a dipole, so the nearer Br becomes δ+ and accepts the electron pair.
- Major: 2-bromopropane; minor: 1-bromopropane.
- The major product forms via the secondary carbocation CH3CH+CH3.
- Secondary carbocations are more stable than primary because two alkyl groups donate electron density (positive inductive effect).
primary < secondary < tertiary.
- Water molecules are present in large excess and have lone pairs.
- They compete with Br− to attack the carbocation; after loss of H+ this gives BrCH2CH2OH.
- 2-bromobutane.
- H adds to C1, forming the more stable secondary carbocation on C2.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The diagram shows the first step of the reaction between ethene and bromine. Where does curly arrow 1 start and finish?
- Ait starts at the C=C (the pi bond) and finishes at the δ+ bromine atom
- Bit starts at the δ+ bromine atom and finishes at the C=C
- Cit starts at the δ− bromine atom and finishes at the δ+ bromine atom
- Dit starts at a C-H bond and finishes at the δ+ bromine atom
Bromine, Br2, is a non-polar molecule. Why can it act as an electrophile towards ethene?
- Athe electron-rich pi bond repels the electrons in Br-Br, inducing a dipole so the nearer Br becomes δ+
- Bbromine is permanently polar because one Br atom is more electronegative than the other
- CUV light splits Br2 into radicals, which then add to the C=C
- Dthe C=C is electron-deficient, so it attracts the lone pairs on bromine
Ethene reacts with hydrogen bromide, HBr, by electrophilic addition. What is the intermediate?
- Athe carbocation CH3CH2+
- Bthe radical CH3CH2•
- Cthe carbanion CH3CH2−
- Dthe cation Br+
Propene, CH3CH=CH2, reacts with hydrogen bromide. What is the major product?
- A2-bromopropane, CH3CHBrCH3
- B1-bromopropane, CH3CH2CH2Br
- C1,2-dibromopropane, CH3CHBrCH2Br
- D2-bromopropene, CH3CBr=CH2
Which order of carbocation stability is correct?
- Atertiary > secondary > primary
- Bprimary > secondary > tertiary
- Csecondary > tertiary > primary
- Dall three are equally stable, since each carries a single positive charge
Why is a tertiary carbocation more stable than a primary carbocation?
- Athree alkyl groups each have a positive inductive effect, pushing electron density towards C+ and spreading the charge
- Balkyl groups withdraw electrons from C+, so the positive charge is removed
- Cthe tertiary carbocation has more hydrogen atoms directly bonded to C+ to donate electrons
- Dthe positive carbon in a tertiary carbocation has a lone pair of electrons
Ethene is bubbled through bromine water. As well as 1,2-dibromoethane, a large amount of 2-bromoethanol, BrCH2CH2OH, forms. Why?
- Awater molecules, present in large excess, act as nucleophiles and attack the carbocation intermediate in competition with Br−
- B1,2-dibromoethane is formed first and then rapidly hydrolysed by water
- Cbromine radicals and hydroxyl radicals add to the C=C in a chain reaction
- Dwater attacks the δ+ bromine atom first, before the C=C reacts
2-methylpropene, (CH3)2C=CH2, reacts with hydrogen bromide. What is the major product and why?
- A2-bromo-2-methylpropane, because it forms via the tertiary carbocation (CH3)3C+
- B1-bromo-2-methylpropane, because it forms via the primary carbocation (CH3)2CHCH2+
- C1-bromo-2-methylpropane, because the bulky Br adds to the less crowded end carbon
- D2-bromobutane, because the chain rearranges into a straight chain
Quiz key
The pi bond is the region of high electron density, so the pair of pi electrons moves from the C=C towards the electron-deficient δ+ Br, forming a new C-Br bond. At the same time the Br-Br bonding pair moves onto the δ− Br (arrow 2), which leaves as Br−. Drawing the arrow from Br to the C=C is the classic error: it shows electrons moving from the electrophile to the electron-rich alkene, the wrong direction. The arrow from the Br-Br bond goes to the δ− atom, not from it. C-H bonds are not involved in addition.
As Br2 approaches the C=C, the pi electrons repel the Br-Br bonding pair towards the far Br atom. This induced dipole makes the nearer Br δ+, so it can accept a pair of electrons from the pi bond. The two Br atoms are identical, so there is no permanent dipole. Addition of bromine needs no UV light and happens in the dark, so it is not a radical mechanism. The C=C is electron-rich, not electron-deficient; that is why it attracts electrophiles rather than nucleophiles.
HBr is polar (H δ+, Br δ−). The pi electrons form a bond to the H, the H-Br pair moves to Br giving Br−, and the other carbon is left with only three bonds and a positive charge: CH3CH2+. Br− then attacks it to give bromoethane. The bonds break heterolytically, so no radicals form. A carbanion would need the carbon to gain electrons, but it has lost its share of the pi pair. Free Br+ ions are not formed; the electrophile is the δ+ H of the HBr molecule.
If H+ adds to the end CH2, the intermediate is the secondary carbocation CH3CH+CH3, stabilised by two alkyl groups. Br− then bonds to the middle carbon, so 2-bromopropane is the major product (Markovnikov addition). 1-bromopropane forms via the less stable primary carbocation, CH3CH2CH2+, so it is only a minor product; choosing it is the classic error. 1,2-dibromopropane is the product with Br2, not HBr. Addition removes the C=C, so the product cannot still be an alkene.
Each alkyl group has a positive inductive effect, releasing electron density towards the C+ and spreading the positive charge. A tertiary carbocation has three alkyl groups, a secondary two and a primary one, so stability is tertiary > secondary > primary. Saying the primary carbocation is most stable is the classic error; it has the least electron donation and the most concentrated charge. Secondary is not more stable than tertiary, which has one more donating group. The charge is the same, but how well it is spread out differs, so the stabilities are not equal.
Alkyl groups release electron density through the sigma bonds (positive inductive effect). With three alkyl groups on C+, the positive charge is reduced and spread over more atoms, lowering the energy of the ion. Alkyl groups donate electrons, not withdraw them; withdrawing would make the cation less stable. A tertiary carbocation has no H atoms on C+; it is the primary carbocation that has two. The positive carbon has only six outer electrons in three bonds and no lone pair; having a lone pair would make it a carbanion.
In step 1 the pi electrons attack the δ+ Br, giving the carbocation BrCH2CH2+. In step 2 any nucleophile can attack it. In bromine water, H2O molecules far outnumber Br− ions, so water often bonds to the carbocation, then loses H+, giving 2-bromoethanol. Hydrolysis of 1,2-dibromoethane is far too slow at room temperature to explain this. Addition of bromine water is ionic and happens in the dark; it is not a radical chain reaction. Water does not react with Br2 before the alkene does; the oxygen ends up on carbon because it attacks the carbocation in the second step.
Adding H+ to the CH2 end gives (CH3)3C+, a tertiary carbocation stabilised by three alkyl groups; Br− then bonds to that carbon, giving (CH3)3CBr. Adding H+ to the other carbon would give the primary carbocation (CH3)2CHCH2+, which is much less stable, so 1-bromo-2-methylpropane is only a minor product. The position of Br is decided by carbocation stability, not by crowding. The carbon skeleton is unchanged in this addition, so a straight-chain product such as 2-bromobutane does not form.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.