2021 H2 Chem P1 Q15 — Alkane Solution

A ✓: The propagation steps are \((\text{CH}_3)_3\text{CH} + \text{X}\bullet \rightarrow (\text{CH}_3)_3\text{C}\bullet + \text{HX}\) and \((\text{CH}_3)_3\text{C}\bullet + \text{X}_2 \rightarrow (\text{CH}_3)_3\text{CX} + \text{X}\bullet\), so the halogenoalkane \(\text{C}_4\text{H}_9\text{X}\) is formed in a propagation step.

B ✗: A termination step only removes radicals by combining them, it does not produce \((\text{CH}_3)_3\text{C}\bullet\).

C ✗: \(\Delta H = \text{BE(C-H)} + \text{BE(X-X)} - \text{BE(C-X)} - \text{BE(H-X)}\). Although the Br-Br bond is weaker than the Cl-Cl bond, the C-Br and H-Br bonds are also weaker than the C-Cl and H-Cl bonds, so the reaction with bromine is less exothermic.

D ✗: The initiation step produces halogen radicals by homolytic cleavage of the X-X bond, not a reactive halide ion.