Hydrocarbons: Question 6

Worked solution

Step 1: Which hydrogens are available

Propane, $\text{CH}_3\text{CH}_2\text{CH}_3$, has two chemically distinct types of hydrogen atom: six equivalent primary hydrogens (on the two terminal carbons) and two equivalent secondary hydrogens (on the central carbon).

Step 2: The propagation step and where it can occur

In the first propagation step of free-radical substitution, a chlorine radical abstracts a hydrogen atom from propane. This abstraction can happen at either type of carbon, generating either a primary free radical, $\text{CH}_3\text{CH}_2\text{CH}_2^\bullet$, or a secondary free radical, $\text{CH}_3\overset{\bullet}{\text{C}}\text{HCH}_3$. Each radical then reacts with $\text{Cl}_2$ in the second propagation step to give the corresponding chloropropane and regenerate a chlorine radical.

Step 3: Comparing the stability of the two possible radicals

The secondary free radical, $\text{CH}_3\overset{\bullet}{\text{C}}\text{HCH}_3$, is more stable than the primary free radical, $\text{CH}_3\text{CH}_2\text{CH}_2^\bullet$, because it is attached to two alkyl (methyl) groups rather than one. Each alkyl group donates electron density towards the radical carbon (an inductive/hyperconjugative stabilising effect), and this stabilisation is greater with two adjacent alkyl groups than with only one.

Step 4: The outcome

Because both types of hydrogen atom are present and both types of radical can form, chlorination of propane produces a mixture of 1-chloropropane (from the primary radical) and 2-chloropropane (from the secondary radical), rather than a single pure product. The greater stability of the secondary radical means the secondary C-H bonds react somewhat more readily per hydrogen atom than the primary C-H bonds, but since neither pathway is excluded, both isomers are always obtained together.

Why the other options are wrong

• A and C both wrongly claim that only one isomer forms; in reality a chlorine radical can abstract a hydrogen atom from any C-H bond in propane, so both types of radical, and hence both chloropropane isomers, are produced.
• D correctly identifies that a mixture forms, but wrongly claims radical stability plays no role; the differing stabilities of the primary and secondary radicals genuinely affect how readily each type of C-H bond reacts, even though the number of each type of hydrogen atom is also a factor.

Final answer

B. A mixture of 1-chloropropane and 2-chloropropane forms, since chlorination can occur at either carbon, with the more stable secondary radical intermediate reflecting the extra stabilisation from two adjacent alkyl groups.