2024 H2 Chem P1 Q19 — Alkene Solution

The answer is D (\(\text{C}_{15}\text{H}_{26}\text{Br}_2\), 3).
\(\beta\)-selinene has two C=C bonds (the ring \(=\text{CH}_2\) group and the isopropenyl group), so complete reaction adds two HBr:
\(\text{C}_{15}\text{H}_{24} + 2\text{HBr} \rightarrow \text{C}_{15}\text{H}_{26}\text{Br}_2\)
In each C=C, one carbon is a \(=\text{CH}_2\) carbon and the other is bonded to two further carbon atoms. Br can add to either carbon of each C=C:
• Br on the more substituted carbon (Markovnikov, via the more stable tertiary carbocation) gives a tertiary C-Br, because that carbon ends up bonded to three carbon atoms.
• Br on the \(\text{CH}_2\) carbon gives a primary C-Br (the minor route).
With two C=C bonds there are four combinations (ignoring optical isomers): tertiary/tertiary, tertiary (ring)/primary, primary/tertiary (side chain) and primary/primary. Three of these contain a tertiary C-Br, as shown in the explanation figure; only the primary/primary product is not a tertiary bromoalkane.
• A and B are wrong: \(\text{C}_{15}\text{H}_{24}\text{Br}_2\) would be the product of adding \(\text{Br}_2\), not HBr. Adding two HBr also adds two H atoms, giving \(\text{C}_{15}\text{H}_{26}\text{Br}_2\).
• C is wrong: the formula is right, but 2 counts only the products with one tertiary C-Br (Br on the ring carbon or on the side-chain carbon) and misses the product in which both C-Br bonds are tertiary.