Stereochemistry Questions and Answers

Answer: B

The new stereocentre has the R configuration (option B).

This is an Evans alkylation. n-BuLi at -75°C makes the Z-enolate of the N-acyl oxazolidinone, chelated through lithium between the enolate oxygen and the ring carbonyl. The bulky tert-butyl group on C-4 shields one face, so allyl chloride attacks from the other face and one diastereomer is formed in large excess. The drawn auxiliary has the (S) configuration at C-4, and with this (S) auxiliary the allylation gives the (R) centre at C-2 of the new 2-ethylpent-4-enoyl group (the same sense as the textbook propionyl case, where the (S)-4-benzyl auxiliary with allyl iodide gives the (R)-2-methylpent-4-enoyl product).

S would need attack on the shielded face. Equal amounts of both diastereomers would mean no facial control, which is the very reason for the auxiliary. The product does have a new chiral centre, since C-2 carries C(=O)N, allyl, ethyl and H.

Answer: A

In the stable conformer the iPr, CH₃ and OH groups are all equatorial.

The arrangement described is that of menthol: OH on C-1, isopropyl on C-2 trans to it, and methyl on C-5 cis to it. On a cyclohexane chair, adjacent (1,2) substituents that are trans can both be equatorial, and 1,3-related substituents that are cis can both be equatorial. So one chair places all three groups equatorial.

The ring flip would put all three axial, with severe 1,3-diaxial interactions, so the all-equatorial chair is strongly preferred. The options putting the OH or the methyl axial describe a less stable arrangement that the relative configuration does not force.

Answer: A

The descriptors are S for the allene axis and E for the alkene (option A).

Allene: Br and Ph sit on the left end, CO₂Et and the vinyl group on the right. On the left carbon Br outranks Ph, and on the right carbon CO₂Et (O,O,O) outranks the vinyl carbon (C,C,H). Looking along the axis from the left end, the left substituents lie up (Br) and down (Ph) and the right substituents lie towards (CO₂Et, bold) and away from (vinyl, hashed) the page. The path Br → Ph → CO₂Et runs anticlockwise, so the axial chirality is S.

Alkene: the carbon at the end bears C₆D₅ and C₆H₅, and C₆D₅ ranks higher because deuterium has a greater atomic mass than hydrogen when all else is equal. On the other alkene carbon, the allene carbon outranks H. The allene carbon and C₆D₅ lie on opposite sides of the double bond, so the alkene is E. The pair S, E is option A; R means the opposite sense and Z the same side.

Answer: A

The descriptors are 2R, 4S; H_A pro-s; H_B pro-r.

C-2: priorities OH > COOH > C-3 > H. With H on a horizontal bond, OH (right) to COOH (top) to C-3 (bottom) runs anticlockwise, which reads S; because the lowest priority group points towards the viewer, the true descriptor is reversed: R.

C-4: priorities OH > COOH (bottom) > C-3 (top) > H. OH to COOH to C-3 runs clockwise, read as R and reversed to S. Both OH groups on the same side make this the meso (2R,4S) diacid.

H_A: replace it by D. At C-3 the R branch ranks above the S branch, so C-2 > C-4 > D > H_B. C-2 (top) to C-4 (bottom) to D (left) runs clockwise, read as r and reversed (H_B is horizontal) to s. So H_A is pro-s and H_B pro-r, which only the first option gives with 2R, 4S.