Electrophilic addition and substitution (HL)
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Flip to reveal answersWhy is an alkene's C=C reactive in electrophilic addition?
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All 12 Flashcards — Electrophilic addition and substitution (HL)
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Question
Why is an alkene's C=C reactive in electrophilic addition?
Answer
It is **electron-rich** — the exposed **π electrons** make it an electron-pair **donor** that attacks an electrophile.
Question
What is an electrophile?
Answer
An **electron-pair acceptor** — an electron-poor (often positive) species attracted to an electron-rich centre such as a C=C (e.g. δ+ H of HBr, NO_{2}⁺).
Question
What is a carbocation?
Answer
A carbon atom bearing a **positive charge** (only 3 bonds / 6 electrons) — the reactive **intermediate** in electrophilic addition.
Question
Describe step 1 of HBr adding to an alkene.
Answer
The **C=C π electrons → the δ+ H** of HBr (arrow 1); the **H–Br bond → Br** (arrow 2), which leaves as **Br⁻**. A **carbocation** forms.
Question
Describe step 2 of HBr adding to an alkene.
Answer
The **bromide ion (Br⁻) attacks the positive carbon** of the carbocation, forming the final C–Br bond — the reagent has now **added across** the C=C.
Question
What is Markovnikov's rule (HL explanation)?
Answer
The major product forms via the **more stable carbocation**; equivalently, the **H of HX adds to the C with more H's**.
Question
Order of carbocation stability?
Answer
**tertiary (3°) > secondary (2°) > primary (1°)** — more alkyl groups donate electron density and spread the positive charge.
Question
Major product of propene + HBr, and why?
Answer
**2-bromopropane, CH_{3}CHBrCH_{3}** — via the more stable **secondary** carbocation (Markovnikov).
Question
Major product of 2-methylpropene + HBr?
Answer
**2-bromo-2-methylpropane, (CH_{3})_{3}CBr** — via the most stable **tertiary** carbocation.
Question
Why does benzene substitute rather than add?
Answer
Its **delocalised** ring is very **stable**; addition would destroy the delocalisation, while **substitution restores** the aromatic ring.
Question
Electrophile, catalyst and product in benzene nitration?
Answer
Electrophile = **nitronium ion, NO_{2}⁺**; catalyst = conc. **H_{2}SO_{4}**; product = **nitrobenzene, C_{6}H_{5}NO_{2}** (plus H⁺ lost).
Question
Addition vs substitution — the key difference?
Answer
**Addition**: reagent adds across C=C, nothing leaves (alkenes). **Substitution**: one group replaces a ring H, ring restored (benzene).
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Full study notes for Electrophilic addition and substitution (HL)
Topic 6.4 hub
Electron-pair sharing reactions
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