2017 H2 Chem P1 Q18 — Alkene Solution

The answer is D (\(\text{C}_{15}\text{H}_{26}\text{Br}_2\); 3 tertiary bromoalkanes).
\(\beta\)-selinene is \(\text{C}_{15}\text{H}_{24}\) (15 C, two rings and two C=C, so \(2 \times 15 + 2 - 2 \times 4 = 24\) H). Both C=C bonds (the ring \(=\text{CH}_2\) and the side-chain \(\text{C(CH}_3){=}\text{CH}_2\)) undergo electrophilic addition with HBr. Each addition adds one H and one Br, so reacting completely with 2 HBr gives \(\text{C}_{15}\text{H}_{26}\text{Br}_2\).
In each alkene one carbon is a \(=\text{CH}_2\) and the other is bonded to two other carbons. If Br adds to the \(=\text{CH}_2\) carbon it gives a primary \(\text{CH}_2\text{Br}\); if Br adds to the other carbon it gives a tertiary C-Br. With two alkenes this gives \(2 \times 2 = 4\) products (see the figure): primary/primary, primary/tertiary, tertiary/primary and tertiary/tertiary. Three of the four contain a tertiary bromoalkane group.
• A is wrong: \(\text{C}_{15}\text{H}_{24}\text{Br}_2\) would mean Br is added without H, but addition of HBr adds H as well as Br across each C=C; and only 2 tertiary products is too few.
• B is wrong: the count of 3 is right, but the formula \(\text{C}_{15}\text{H}_{24}\text{Br}_2\) leaves out the two H atoms added from the two HBr molecules.
• C is wrong: the formula is right, but only 2 tertiary products misses one. Of the 4 products, only the one with Br on both \(\text{CH}_2\) carbons has no tertiary C-Br.