Stereochemistry Questions and Answers

Q: Predict stereochemical outcome of the following reaction :

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.

Q: In a stable conformer of the following compound :

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.

Q: Assign the chiral descriptors for the following molecule :

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.

Q: Absolute configuration of the chiral centres (C-2, C-4) and prochirality descriptors (HA and HB) in the following compound is :(Fischer projection, top to bottom:)COOHHC-2OHHAC-3HBHC-4OHCOOH

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

41.

Predict stereochemical outcome of the following reaction :

An N-butanoyl oxazolidin-2-one with a hashed tert-butyl group on C4, treated with nBuLi at -75 C and then allyl chloride (ClCH2CH=CH2)

A.The new stereogenic center formed will have 'S' configuration
B.The new stereogenic center formed will have 'R' configuration
C.Both the diastereomers formed are in equal amount
D.The product formed will have no new chiral center

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

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.

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42.

In a stable conformer of the following compound :

A cyclohexane ring with OH, an isopropyl group on the carbon next to it (trans to the OH) and a methyl group 1,3 to the OH on the other side (cis to the OH), drawn with hashed and wedge bonds

A.Equatorial - iPr, \(-\text{CH}_{3}\), -OH groups
B.Equatorial - iPr and \(-\text{CH}_{3}\) groups; Axial -OH group
C.Equatorial - iPr group; axial \(-\text{CH}_{3}\) and -OH groups
D.Equatorial -iPr and - OH groups; axial - \(\text{CH}_{3}\) group

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

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.

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43.

Assign the chiral descriptors for the following molecule :

A vinylallene: an allene whose one end carries Br and Ph, and whose other end carries a bold CO2Et and a hashed bond to a CH=C(C6D5)(C6H5) group

A.S, E
B.R, E
C.S, Z
D.R, Z

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

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.

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44.

Absolute configuration of the chiral centres (C-2, C-4) and prochirality descriptors (H_A and H_B) in the following compound is :

(Fischer projection, top to bottom:)

	COOH
H	C-2	OH
H_A	C-3	H_B
H	C-4	OH
	COOH

A.2R, 4S; \(\text{H}_{A}\) : Pro-s; \(\text{H}_{B}\) : Pro-r
B.2R, 4R; \(\text{H}_{A}\) : Pro-s; \(\text{H}_{B}\) : Pro-r
C.2S, 4S; \(\text{H}_{A}\) : Pro-r; \(\text{H}_{B}\) : Pro-s
D.2S, 4R; \(\text{H}_{A}\) : Pro-s; \(\text{H}_{B}\) : Pro-r

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

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.

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