The stem carries a drawn structure; the four options are text and are reproduced in full above.

The question combines two related assignments.

Absolute configuration at C-2 and C-4 follows the ordinary CIP procedure: rank the four groups at each centre, orient the lowest away, and read the sense of 1→2→3. At a carbon bearing OH, COOH, CH₂ and H, the ranking is OH > COOH > CH₂ > H, since oxygen beats carbon directly, and the carboxyl carbon (O,O,O by duplication) beats a methylene carbon.

Prochirality at C-3 is the more interesting half. The two hydrogens H_A and H_B sit on the same carbon and look equivalent, but they are diastereotopic here because the molecule already contains stereocentres. The test is a thought experiment: replace one hydrogen with a group of higher priority (conventionally deuterium) and assign the resulting centre. If replacing it gives R, that hydrogen is pro-R; if S, it is pro-S. Then do the same for the other hydrogen.

Because the two hydrogens must give opposite answers, one is always pro-R and the other pro-S — which is why every option pairs them that way and the real work is deciding which is which.

This distinction is not merely formal. Enzymes routinely discriminate between the pro-R and pro-S hydrogens of a CH₂ group, removing one and leaving the other untouched, and it was exactly this kind of stereospecific labelling that established the mechanisms of reactions in the citric acid cycle — the classic case being aconitase, which distinguishes the two apparently identical CH₂COOH arms of citrate.

Per the official final answer key the answer is option (A), 2R, 4S with H_A pro-s and H_B pro-r.