2019 H2 Chem P1 Q18 — Alkane Solution

The answer is A (1, 2 and 3).

In free-radical substitution of propane, C-H bonds are replaced by C-Cl bonds (and Cl-Cl bonds break); the C-C bonds of propane stay intact. The radicals present are \(\text{Cl}\bullet\) and propyl radicals, \(\text{CH}_3\text{CH}_2\text{CH}_2\bullet\) and \(\text{CH}_3\text{CH}\bullet\text{CH}_3\) (and chlorinated propyl radicals).
• 1 \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}\): possible, a monosubstitution product: \(\text{CH}_3\text{CH}_2\text{CH}_2\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{Cl}\bullet\).
• 2 \(\text{CH}_2\text{ClCH}_2\text{CH}_2\text{Cl}\): possible, a disubstitution product formed when \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}\) is itself attacked by \(\text{Cl}\bullet\) and chlorinated again.
• 3 hexane: possible, from termination between two propyl radicals: \(\text{CH}_3\text{CH}_2\text{CH}_2\bullet + \bullet\text{CH}_2\text{CH}_2\text{CH}_3 \rightarrow \text{CH}_3(\text{CH}_2)_4\text{CH}_3\).
• 4 pentane: not possible. A 5-carbon chain would need a 2-carbon (ethyl) radical to join a propyl radical, and an ethyl radical could only come from breaking a C-C bond, which does not happen.

• B is wrong: it includes 4, pentane, which cannot form, and leaves out 2.
• C is wrong: it leaves out 3; hexane does form in the termination step.
• D is wrong: it leaves out 1, the main monosubstitution product.