The big idea: An alkene contains a carbon–carbon double bond, C=C. A double bond is two shared pairs of electrons — and the π pair sits in an exposed cloud above and below the carbons. That makes the C=C region electron-rich.
An electrophile is an electron-pair acceptor (an electron-poor or positively charged species, e.g. the δ+ H of HBr, or Br2). The electron-rich C=C attacks the electrophile: the π electrons form a new bond, the double bond opens, and a group adds across the two carbons. Because something is added and nothing leaves, this is electrophilic addition.
At HL you must show the mechanism in detail — the curly arrows and the carbocation intermediate — not just the product.
The two ends of the C=C are different, so HBr can add two ways — Markovnikov decides which dominates.
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Key terms: - Alkene — a hydrocarbon with a C=C double bond (general formula CnH2n); it is unsaturated. - π (pi) electrons — the second shared pair of the double bond, held in an exposed cloud; this is the electron-rich part that attacks. - Electrophile — an electron-pair acceptor, attracted to the electron-rich C=C (e.g. δ+ H of H–Br, or Br2). - Carbocation — a carbon atom bearing a positive charge (only 3 bonds, 6 electrons); the reactive intermediate formed in step 1. - Heterolytic fission — a bond breaks so both electrons go to one atom (here, Br takes the H–Br pair, leaving as Br⁻).
A curly arrow shows the movement of a pair of electrons: the tail starts at the electrons that move and the head points to where the pair ends up. Electrophilic addition of HBr to an alkene happens in two steps, via a carbocation.
Step 1 — the C=C attacks the electrophile
- Arrow 1 — from the C=C to the H of HBr. The tail is on the C=C π electrons; the head points to the δ+ hydrogen of the polar H–Br bond, forming a new C–H bond.
- Arrow 2 — from the H–Br bond to the bromine. As the H is taken, the H–Br bonding pair shifts onto the bromine, which leaves as a bromide ion, Br⁻ (heterolytic fission).
- Result — a carbocation. One carbon now has only three bonds and a positive charge — the reactive carbocation intermediate.
Step 1: the C=C π electrons attack the δ+ H of HBr, forming a carbocation and releasing Br⁻. Step 2: the Br⁻ attacks the carbocation.
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Step 2 — the nucleophile attacks the carbocation
- Arrow 3 — from the Br⁻ lone pair to the positive carbon. The bromide ion formed in step 1 (an electron-pair donor) attacks the positively charged carbon, forming the second new bond.
- Result — the addition product. Both parts of HBr are now added across what was the C=C, which is now a C–C single bond. (For Br2, the second arrow comes from a Br⁻ in exactly the same way.)
Why the alkene is so reactive: The C=C π electrons are exposed and electron-rich, so the double bond is a strong electron-pair donor — far more reactive than the locked single bonds of a saturated alkane. With Br_{2}, the approaching alkene even induces a dipole in the non-polar Br–Br bond (the near Br becomes δ+), so Br2 can act as the electrophile too.
Common mechanism mistakes: - Drawing arrow 1 from the H (or Br) to the C=C (backwards) — the C=C π electrons move first, so arrow 1 must start at the double bond. - Forgetting the carbocation — HL needs the intermediate shown with its + charge before the halide attacks. - Omitting the arrow that breaks H–Br — that bonding pair must go onto the bromine to release Br⁻. - Calling it substitution — nothing is replaced; the reactant adds across the double bond.
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When an unsymmetrical reagent like HBr adds to an unsymmetrical alkene (e.g. propene, CH3CH=CH2), the H could add to either end — giving two possible carbocations and so two possible products. The major product is the one formed via the more stable carbocation.
Markovnikov's rule: The major product forms via the more stable carbocation. Carbocation stability increases with the number of alkyl groups attached to the positive carbon, because alkyl groups are weakly electron-donating (they push electron density towards the positive centre and spread out the charge):
tertiary (3°) > secondary (2°) > primary (1°)
A shortcut: the H of HX adds to the carbon that already has more hydrogens, so the positive charge (and then the halogen) ends up on the more substituted carbon.
Primary (1°)
- one alkyl group on the C⁺
- e.g. CH3CH2CH2⁺
- least stable → minor route
Secondary (2°)
- two alkyl groups on the C⁺
- e.g. CH3CH⁺CH3
- more stable than 1°
Tertiary (3°)
- three alkyl groups on the C⁺
- e.g. (CH3)3C⁺
- most stable → favoured route
Worked example — propene + HBr (which carbocation?)
Propene, CH3CH=CH2, reacts with HBr. Compare the two possible carbocations and predict the major product.
Solution
- Route A — H adds to the end CH_{2}. The positive charge lands on the middle carbon, which carries two alkyl groups → a secondary (2°) carbocation, CH3CH⁺CH3.
- Route B — H adds to the middle carbon. The positive charge lands on the end carbon, which carries only one alkyl group → a primary (1°) carbocation, CH3CH2CH2⁺.
- Compare: the secondary carbocation (Route A) is more stable (more alkyl groups donating electron density), so it forms preferentially — this is Markovnikov's rule.
- Product: Br⁻ then attacks that middle carbon, giving 2-bromopropane, CH_{3}CHBrCH_{3}, as the major product.
Final answer
Route A gives a secondary carbocation (more stable) → major product 2-bromopropane, CH3CHBrCH3.
Predicting the Markovnikov product fast: Put the H on the carbon that already has more hydrogens. The other atom (Br, OH, etc.) then goes on the more substituted carbon — via the more stable carbocation. Justify it by carbocation stability (3° > 2° > 1°) to earn the explain mark.
Benzene substitutes, it does not add: Benzene, C6H6, has a delocalised π system — six π electrons spread evenly above and below a flat ring of six carbons. This delocalisation makes benzene unusually stable.
An addition would lock two of those π electrons into a new σ bond and break up the delocalisation, destroying that stability. So benzene resists addition. Instead, an electrophile substitutes for a ring hydrogen: the ring attacks the electrophile, then a proton (H⁺) is lost so the delocalised ring is restored (aromaticity regained). The net change is one H swapped for the electrophile — the stable ring survives.
The classic example is nitration: benzene reacts with concentrated nitric acid and a concentrated sulfuric acid catalyst at about 50 °C to give nitrobenzene, C_{6}H_{5}NO_{2}. The active electrophile is the nitronium ion, NO_{2}⁺, generated by the two acids reacting together.
The inner circle shows the six π electrons delocalised evenly around the ring — this stability is why benzene substitutes rather than adds.
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One ring hydrogen has been replaced by –NO₂; the delocalised ring is restored once H⁺ is lost, so the stable aromatic system survives.
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| Stage | What happens (electron-pair movement) |
|---|---|
| Generate the electrophile | Conc. H2SO4 protonates HNO3; loss of water gives the nitronium ion, NO_{2}⁺ (the electrophile). |
| Step 1 — ring attacks NO2⁺ | Two electrons from the delocalised ring attack NO2⁺, forming a C–N bond. This makes an unstable, positively charged intermediate in which the delocalisation is partly broken. |
| Step 2 — lose H⁺ (restore the ring) | The C–H bond on that carbon breaks; the proton (H⁺) leaves and the electrons return to the ring, restoring the full delocalised π system. |
| Regenerate the catalyst | The H⁺ released combines with HSO4⁻ to reform H_{2}SO_{4} — so sulfuric acid is a true catalyst. |
Why substitution, not addition: Addition would permanently destroy the delocalised π system and forfeit benzene's stability. Substitution only temporarily disrupts it (in the intermediate) and then restores the aromatic ring when H⁺ is lost — so the energetically favourable, stable ring is preserved. That is why benzene's typical reaction is electrophilic substitution: the net change is just one ring H swapped for –NO_{2}, as the two diagrams above show.
IB-style question — why benzene substitutes
Explain why benzene undergoes electrophilic substitution rather than electrophilic addition with the nitronium ion, NO2⁺, and identify the organic product of nitration. [3]
How to score the marks
- Mark 1 — the ring is delocalised/stable. Benzene has a delocalised π system spread over all six carbons, which makes it unusually stable (low in energy).
- Mark 2 — addition would cost that stability. Addition would break up the delocalisation permanently and lose the stabilisation; substitution only disrupts it briefly and then restores the delocalised ring (when H⁺ is lost).
- Mark 3 — the product. The product of nitration is nitrobenzene, C_{6}H_{5}NO_{2} (one ring H replaced by –NO2).
Final answer
Benzene's delocalised ring is stable; addition would destroy that delocalisation, whereas substitution restores it — so benzene substitutes. Nitration gives nitrobenzene, C6H5NO2.