2020 H2 Chem P1 Q18 — Introduction to Organic Solution

The answer is D (8).

Androstenolone contains a secondary alcohol, a ketone and one C=C bond. Cold \(\text{KMnO}_4\) (mild oxidation) reacts only with the C=C bond, adding an OH group to each of its two carbon atoms to form a diol, Z. This fits the \(M_\text{r}\) data: \(322 - 288 = 34 = 2 \times 17\), the gain of two OH groups.

Chiral centres in Z (each marked with an asterisk in the explanation figure):
• six that are already present in androstenolone: the carbon bearing the secondary OH group, the two ring-junction carbons that carry the \(\text{CH}_3\) groups, and the three ring-junction CH carbons;
• two new ones: the two carbon atoms of the former C=C bond. One is bonded to OH, H and two different ring carbons; the other (at the ring junction) is bonded to OH and three different carbon groups.
Total = 6 + 2 = 8.

• A is wrong: androstenolone itself already has six chiral centres, and cold \(\text{KMnO}_4\) does not remove any of them, so Z must have more than 5.
• B is wrong: 6 is the number in androstenolone; it leaves out the two new chiral centres formed at the former C=C carbons.
• C is wrong: 7 counts only one of the two new diol carbons, but both are bonded to four different groups.