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2021 H2 Chem P1 Q16 — Alkene Solution

Alkene 2021 A-Level H2 Chemistry · Paper 1 · Q16

Question

Which statement about propene explains how it reacts with bromine?
  1. AElectrons in the carbon-carbon \(\pi\) bond are donated to an electrophile.
  2. BElectrons in the carbon-carbon \(\sigma\) bond are donated to an electrophile.
  3. CThe \(sp^2\) hybridised carbon is an electrophile and accepts a pair of electrons.
  4. DThe \(\sigma\) bond between the \(sp^2\) hybridised carbon atoms is weak and readily broken.
Correct answer: A

Video Solution

Explanation

A ✓: The carbon-carbon \(\pi\) bond is a region of high electron density lying above and below the plane of the \(sp^2\) carbons. These \(\pi\) electrons are donated to an electrophile (the \(\delta+\) bromine of a polarised \(\text{Br}_2\) molecule), initiating electrophilic addition.

B ✗: The carbon-carbon \(\sigma\) bond electrons lie between the two nuclei and are held too tightly to be donated to an electrophile.

C ✗: The \(sp^2\) carbon is electron-rich because of the \(\pi\) bond, so it behaves as a nucleophile, not an electrophile.

D ✗: It is the weaker \(\pi\) bond, not the \(\sigma\) bond, that breaks during the addition; the \(\sigma\) framework stays intact.

Worked solution diagram

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