Stereochemistry Questions and Answers

Answer: C

The product is (C), meso-3,4-dibromohexane.

Addition of bromine through a bromonium ion is anti: the two bromine atoms add to opposite faces of the double bond. Anti addition to a trans (E) alkene gives the meso product, while anti addition to the cis alkene gives the racemic pair. With identical ethyl groups at both ends, (E)-3-hexene therefore gives (3R,4S)-3,4-dibromohexane, which is meso.

Assigning configurations to the drawings: in C one bromine is wedged and the other hashed on the zigzag chain, which works out as R at one centre and S at the other, the meso form. A (both hashed) and D (both wedged) are each one enantiomer of the chiral (R,R)/(S,S) pair, the product a syn addition would give. B, 2,5-dibromohex-3-ene, would come from allylic bromination, which does not occur here.

Answer: D

The product is (D), with the two deuteriums from the dienophile cis to each other.

The Diels-Alder reaction is a concerted [4+2] cycloaddition that is suprafacial on the dienophile: both new bonds form on the same face of the alkene at the same time, so the relative stereochemistry of the dienophile is kept in the product. The dienophile here is cis-1,2-dideuterioethylene, so its two deuteriums end up cis on adjacent ring carbons, both drawn with wedges in (D). The deuteriums on the diene termini become the two CD₂ groups.

(B) and (C) both show the two deuteriums trans, the product a trans dienophile would give. A 1:1 mixture of cis and trans (A) would require a stepwise reaction through an intermediate that can rotate, which the concerted cycloaddition does not have.

Answer: A

The correct statement is that addition specifically from the A face gives the S isomer of malic acid.

Fumaric acid is planar, and its two faces are enantiotopic: adding water to one face gives one enantiomer of malic acid, and adding it to the other face gives the other. In the drawing, the A face arrow is drawn bold, in front of the page, and the B face arrow hashed, behind it.

Take OH arriving from the A face at the carbon on the left, bonded to the HOOC group. The new stereocentre carries OH (priority 1), COOH (2), CH₂COOH (3) and H (4). With the OH in front and the H in the plane of the page, pointing up and to the left, the sequence 1 \(\to\) 2 \(\to\) 3 viewed from the side opposite the hydrogen runs anticlockwise, so the centre is S. Addition from the B face therefore gives R.

That rules out 'A face gives R' and 'B face gives S'. Addition specifically from one face cannot give a racemic mixture, which needs both faces to react equally. (The enzyme fumarase does this addition stereospecifically and gives L-malic acid, which is S.)

Answer: C

The order is II \(\gt\) III \(\gt\) I.

The t-butyl group locks each ring in the chair with t-Bu equatorial. In II the ester is trans to t-Bu (one wedge, one hash, 1,4), so it is also equatorial. In III the ester is cis to t-Bu (both wedges), so it is forced axial.

Alkaline hydrolysis goes through a crowded tetrahedral intermediate at the carbonyl carbon. An equatorial ester is less hindered than an axial one, which suffers 1,3-diaxial interactions, so II reacts faster than III. Compound I is an amide: the nitrogen lone pair is strongly delocalised into the carbonyl and NH₂⁻ is a very poor leaving group, so amides are hydrolysed far more slowly than esters whatever their conformation.

Any order placing I first, or III ahead of II, contradicts these two effects.

Answer: A

The product is (A), (S)-2-methyl-3-phenylpropan-1-ol.

This is an Evans asymmetric alkylation. The auxiliary drawn, with the isopropyl group on a wedge, is (R)-4-isopropyloxazolidin-2-one. LDA forms the (Z)-enolate, held rigid by lithium chelated between the enolate oxygen and the ring carbonyl oxygen. The isopropyl group then covers one face of the enolate, and benzyl bromide alkylates it from the other face, giving essentially one diastereomer.

Excess LiAlH₄ reduces the exocyclic imide carbonyl and releases the auxiliary, giving the primary alcohol with the new stereocentre intact. With this auxiliary the stereocentre is (S), drawn in (A) with the benzyl bond hashed.

(B) is the enantiomer, the product of the (S)-auxiliary. (C) and (D) are amides from attack on the ring carbonyl, which is not the course of a hydride cleavage.

Answer: B

The answer is (B): P has the new stereocentre with the butenyl chain on a hashed bond, and Q is (S)-cyclopent-2-en-1-ylmethanol with CH₂OH on a hashed bond.

LDA forms the Z-lithium enolate of the N-acyloxazolidinone, chelated between the enolate oxygen and the ring carbonyl. The isopropyl group of the (S)-auxiliary blocks one face, so 4-bromobut-1-ene alkylates the opposite face (Evans alkylation). With the CIP priorities in the order C(O)N, then vinyl, then butenyl, then H, the new centre in P is S, as drawn in (A) and (B).

Grubbs-II then joins the vinyl and butenyl alkenes by ring-closing metathesis, releasing ethylene and forming a cyclopentene, and LiBH₄ cleaves the auxiliary to the primary alcohol. The stereocentre is untouched, and the priorities keep the same order (CH₂OH, then the ring CH=, then the ring CH₂, then H), so Q is also S, the hashed CH₂OH of (B).

(A) has the right P but draws Q as the R alcohol. (C) and (D) show the alkylation on the face shielded by the isopropyl group (P is R); (C) also pairs it with an S alcohol that it cannot give.

Answer: B

The relationships are Homotopic, Enantiotopic and Enantiotopic.

• X is a chiral tartaric acid: in the Fischer projection the two hydrogens lie on opposite sides (and so do the two OH groups). This molecule has a C₂ axis that exchanges the two CH(OH) carbons, so H_a and H_b are interchanged by a rotation: homotopic.
• Y is [2.2]paracyclophane. H_a on the upper ring and H_b on the lower ring sit at the same position relative to the bridges, so the mirror plane between the two rings maps one onto the other, but no rotation does: enantiotopic.
• Z, ClCH=C=CH_aH_b, is an allene whose CH₂ plane is perpendicular to the plane of the CHCl end. That CHCl plane is a mirror plane bisecting the H-C-H angle and exchanging H_a and H_b, and no axis does so: enantiotopic.

Calling X diastereotopic would fit meso-tartaric acid, not the chiral isomer drawn. Calling Y homotopic or Z homotopic overlooks that only a reflection relates the two hydrogens.

Answer: B

The chiral compound is Q.

Q is penta-2,3-diene, an allene with two different groups (H and CH₃) on each terminal carbon. In an allene the two terminal planes are perpendicular, so the molecule has no mirror plane and is axially chiral.

R has three cumulated double bonds; with an odd number the terminal substituents lie in one plane, giving cis-trans isomers but not chirality. P would be axially chiral only if the two groups on C4 differed, but both are isopropyl. S is a spiro[3.3]heptane, which is axially chiral only when each terminal carbon carries two different groups; here one carbon has two identical acetyl groups. So P, R and S are achiral.

Answer: A

M is (A), (R)-1-phenylethyl acetate.

Candida antarctica lipase B carries out a kinetic resolution: it acetylates one enantiomer of racemic 1-phenylethanol much faster than the other. By Kazlauskas' rule for secondary alcohols, the lipase acylates the (R) enantiomer, in which the large (phenyl) and medium (methyl) groups fit its pockets, leaving the (S) alcohol behind. Isopropenyl acetate is the acyl donor; its enol by-product tautomerises to acetone, which makes the transfer irreversible.

In (A) the OAc is on a wedge with methyl to the left and phenyl to the right, which is the (R) configuration; (B) is the (S) acetate. Styrene (C) would need elimination, and phenyl acetate (D) has lost the CH(CH₃) carbon, neither of which a lipase does.

Answer: D

The answer is Re face of enolate on to the Si face of aldehyde.

Bu₂BOTf and Hunig's base give the Z-boron enolate of the Evans imide, which adds to benzaldehyde through a chair (Zimmerman-Traxler) transition state on the face left open by the benzyl group, giving the syn aldol. Reading the product as drawn, both new centres are S: (2S,3S)-3-hydroxy-2-methyl-3-phenylpropanoic acid.

At C3 the priorities are OH \(\gt\) Ph \(\gt\) C2 \(\gt\) H, so the incoming carbon ranks third; attack on the Si face of PhCHO then gives S. At C2 the priorities in the adduct are C(O)N \(\gt\) CH(O)Ph \(\gt\) CH₃ \(\gt\) H, so the new bond ranks second; bond formation on the Re face of the enolate carbon then gives S. Each of the other three options would give at least one centre of the opposite configuration.