Back to Topic 6.4 — Electron-pair sharing reactions
6.4.2Chemistry SL12 flashcards

Electrophilic addition to alkenes

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Card 1 of 126.4.2
6.4.2
Question

What is an electrophile?

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All 12 Flashcards — Electrophilic addition to alkenes

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Card 1definition

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. Examples: Br_{2}, HBr, H⁺.

Card 2concept

Question

Why is the C=C double bond reactive?

Answer

It is **electron-rich** (two shared pairs / exposed π electrons), so it readily donates electrons and attacks electrophiles.

Card 3definition

Question

What is an addition reaction?

Answer

Two molecules join to form **one** product; the C=C opens to a single bond and the reactant adds across it. **Nothing leaves**.

Card 4comparison

Question

Saturated vs unsaturated?

Answer

Saturated = only **single** bonds (alkane); unsaturated = has a **C=C** (alkene) so more atoms can be **added**.

Card 5process

Question

Describe the two curly arrows in electrophilic addition (Br_{2}).

Answer

Arrow 1: the **C=C π electrons → a bromine** (new bond). Arrow 2: the **Br–Br bond → the other bromine**, which leaves as Br⁻.

Card 6example

Question

Product of ethene + Br_{2}?

Answer

**1,2-dibromoethane, CH_{2}BrCH_{2}Br** — one bromine adds to each carbon.

Card 7example

Question

Product of ethene + HBr?

Answer

**Bromoethane, CH_{3}CH_{2}Br** — H and Br add across the C=C.

Card 8example

Question

Product of ethene + steam (H_{2}O)?

Answer

**Ethanol, CH_{3}CH_{2}OH** — water adds across the C=C with an H_{3}PO_{4} catalyst at high T and P.

Card 9example

Question

Product of ethene + H_{2}?

Answer

**Ethane, CH_{3}CH_{3}** (saturated) — hydrogen adds across the C=C with a Ni catalyst.

Card 10concept

Question

What is the test for unsaturation?

Answer

Add **bromine water**: an **alkene decolourises** it (orange → colourless); an **alkane** gives **no change**.

Card 11concept

Question

What is Markovnikov's rule?

Answer

For an unsymmetrical alkene + HX, the **H adds to the carbon that already has more hydrogens**, giving the **major** product.

Card 12example

Question

Major product of propene + HBr?

Answer

**2-bromopropane, CH_{3}CHBrCH_{3}** — H goes to the CH_{2} end, Br to the middle carbon (Markovnikov).

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