Past exam of the chemistry B course of the University of Cambridge / 2024 / Paper 1 / B3 / b / Solution

For both substrates, the alkene pi bond attacks bromine and polarizes/breaks Br–Br. A bromine lone pair completes a three-membered bromonium ion, and bromide subsequently attacks a carbon from the opposite face, cleaving that C–Br bridge bond. The two new C–Br bonds are anti. A free carbocation that rotates before capture is not needed.

In trans-fused $A$, the fixed conformation makes the alternative anti-opening pathways diastereomeric. A chair-like, trans-diaxial opening is favored over a geometrically strained alternative that requires a twist/boat-like arrangement of the six-membered ring. The Fürst-Plattner rule describes this conformational preference. One depicted enantiomer of $A$ therefore gives essentially one dibromide diastereomer, with the two bromines occupying the trans-diaxial arrangement in its favored chair. If a racemic sample of $A$ were used, its other enantiomer would of course give the mirror product as well.

For cis-fused $B$, the accessible mirror-related folds and the equivalent anti pathways produce mirror-related dibromides with equal rates in an achiral environment. Adding the two bromines destroys the symmetry that made the starting configuration achiral. $B$ gives a racemic dibromide pair. The racemate follows from equivalent enantiomer-forming pathways, not from a loss of anti stereospecificity.