Stereochemistry Questions and Answers

Q: Most stable conformation of 5-hydroxy-1, 3-dioxane is :

The most stable conformation is (A), the chair with the OH group axial.In 5-hydroxy-1,3-dioxane the axial OH points toward the two ring oxygens on the same face. It can form an intramolecular hydrogen bond with them, and there are no 1,3-diaxial C-H interactions to oppose it, because the axial positions at C-1 and C-3 are ring oxygens with lone pairs rather than hydrogens. This outweighs the usual preference for equatorial substituents, especially in non-polar solvents.Chair (B) has the OH equatorial, so it loses the hydrogen bond. The boat-like forms (C) and (D) suffer torsional and flagpole strain and are higher in energy than either chair.

Q: In the Fischer projection, the absolute configuration of a chiral centre will not change from R to S when,

The configuration does not change when the priority numbers 1 and 2 are interchanged and then 3 and 4 are interchanged.In a Fischer projection, swapping any two groups on the stereocentre inverts its configuration (R becomes S). Two such swaps invert it twice, which restores the original configuration.Options A, B and D each describe a single interchange, so each of them changes R to S.

Q: Relationship between compound I and II is :

Compounds I and II are diastereomers.Both are trimethylcyclohexanes with the same 1,2,4 substitution pattern, so they are not constitutional isomers. Reading the chairs, in I all three methyl groups lie on the same face of the ring. In II the methyl at the position corresponding to one of the adjacent pair points to the opposite face from the other two.Ring flipping changes axial and equatorial positions but never which face a group is on, so II cannot be a conformer of I. A different relative configuration that is not a mirror image is diastereomerism, so they are diastereomers rather than enantiomers.

Q: The correct statement for the following structures is that they are :

The two structures are structural isomers.Read each Newman projection carbon by carbon. In the left one, the front carbon carries two CH3 groups and one H, and the rear carbon carries three H atoms. That is (CH3)2CH-CH3, 2-methylpropane (isobutane). In the right one, each carbon carries two H atoms and one CH3 group, which is CH3CH2-CH2CH3, butane.Both are C4H10, but their atoms are connected differently, which is what makes them structural (chain) isomers.Conformational isomers would be the same molecule turned about a single bond, and enantiomers must be chiral mirror images; neither butane nor isobutane is chiral. 'Not isomers' is wrong because the molecular formula is the same.

Q: In the most stable conformation the molecule having the circled Me group equatorial is :

The circled methyl is equatorial in molecule (C).(C) is a 1,2,3-trimethylcyclohexane: circled Me hashed at C1, Me wedged at C2, Me wedged at C3. C1 and C2 are trans, so both can be equatorial; C2 and C3 are cis, so one of them must be axial. The chair with C1 and C2 equatorial and C3 axial has one axial methyl; its ring flip has two. The preferred chair therefore holds the circled methyl equatorial.In (A) the tert-butyl group locks the ring with itself equatorial; the circled methyl is cis to it (both wedged) across a 1,4 relationship, so it must be axial. (B) is a trans-decalin, which cannot flip; the ring-fusion H atoms are axial, and the wedged methyl on the carbon next to the hashed fusion H points the axial way. In (D) the circled methyl and the phenyl are cis on adjacent carbons, so one is axial; the much larger phenyl takes the equatorial position and the methyl is axial.

Q: Which of the following molecule has enantiotopic protons ?

Butane has enantiotopic protons.Two protons are enantiotopic when replacing first one and then the other by a test group gives a pair of enantiomers. On C2 of butane the carbon carries H, H, CH3 and CH2CH3. Replacing either H by, say, D gives 2-deuteriobutane, which is chiral, and the two replacements give the two enantiomers. So the C2 (and C3) methylene protons are enantiotopic.In propane the central CH2 carries two identical CH3 groups; replacing either H gives the same achiral molecule, and the protons are homotopic. All the protons of ethane are homotopic, and so are those of cyclopropane, where a C2 axis interchanges the two H atoms of each CH2.

Q: Which statement is correct for the following four molecules ?

The correct statement is I, II, III and IV are chiral.All four molecules are sulfoxides. A sulfoxide sulfur carries an oxygen, two different carbon groups (here the aromatic ring carbon and the vinylic C2) and a lone pair, in a pyramidal arrangement that does not invert at room temperature. So each sulfur is a stereogenic centre, and none of the molecules has a mirror plane through it.That is why even I and III are chiral, although C4 in each carries two identical groups (two Cl in I, two CH3 in III) and is not a stereocentre. II and IV have a stereogenic C4 as well as the sulfur. The options listing only some of the molecules overlook the sulfoxide stereocentre.

Q: What is the relationship between Ha and Hb in the product P formed in the following reaction ?

Ha and Hb are enantiotopic in P.Methoxide removes HCl from the side chain (E2, taking the hydrogen on the isopropyl carbon), turning CH(Cl)-CH(CH3)2 into CH=C(CH3)2, while the ring CH2 carrying Ha and Hb is untouched. The chlorine-bearing stereocentre disappears, and P, a cyclopentadiene with a conjugated 2-methylpropenyl group, is achiral and essentially planar.The ring CH2 lies in that plane with Ha above it and Hb below. The molecular plane is a mirror plane that exchanges them, but no rotation axis does. Protons exchanged only by reflection are enantiotopic: replacing one or the other gives a pair of enantiomers.They would be diastereotopic if a stereocentre remained in P, as in the starting chloride. They are not homotopic because no rotation interchanges them, and vicinal describes atoms on adjacent carbons, which these geminal protons are not.

Q: Which of the following alcohols is NOT suitable substrate for sharpless asymmetric epoxidation reaction ?

The alcohol that is not a suitable substrate is option C.Sharpless asymmetric epoxidation works on allylic alcohols: the hydroxyl group binds titanium, which delivers the oxygen to the neighbouring double bond. In C the double bond is the ring-fusion double bond and the hydroxyl-bearing carbon is separated from it by one more carbon, so the OH is homoallylic and cannot direct the epoxidation. Options A and B are cinnamyl alcohols (Z and E), and D is a cyclic allylic alcohol with the double bond next to the carbinol carbon, so all three are allylic alcohols.The Z isomer in A reacts less well than the E isomer in B, but it is still an allylic alcohol and so a substrate.

Q: Assign the chiral descriptors for the following molecule :

The descriptors are 2S, 3R, E, option C.C-2: the two ester carbons both rank above C-3 (O,O,O against N,C,H). They are tied until the isotope is reached, and the carbon carrying 18O outranks the one with 16O, so the order is C(18O ester) > C-1 > C-3 > H. With H towards the viewer the turn is reversed, giving S.C-3: N is first. C-2 and C-4 are both (C,C,H), so go one sphere further: C-2 holds two ester carbons (O,O,O) while C-4 holds C-5 (C,C,C) and a duplicate carbon, so C-2 outranks C-4. The order is N > C-2 > C-4 > H, and with H pointing away N to C-2 to C-4 runs clockwise, so 3R.C-4=C-5: on C-5 phenyl (C,C,C) outranks vinyl (C,C,H), and on C-4 C-3 outranks H. They sit on opposite sides, so E.

31.

Most stable conformation of 5-hydroxy-1, 3-dioxane is :

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The most stable conformation is (A), the chair with the OH group axial.

In 5-hydroxy-1,3-dioxane the axial OH points toward the two ring oxygens on the same face. It can form an intramolecular hydrogen bond with them, and there are no 1,3-diaxial C-H interactions to oppose it, because the axial positions at C-1 and C-3 are ring oxygens with lone pairs rather than hydrogens. This outweighs the usual preference for equatorial substituents, especially in non-polar solvents.

Chair (B) has the OH equatorial, so it loses the hydrogen bond. The boat-like forms (C) and (D) suffer torsional and flagpole strain and are higher in energy than either chair.

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32.

In the Fischer projection, the absolute configuration of a chiral centre will not change from R to S when,

A.the priority numbers 1 and 2 are interchanged
B.the priority numbers 2 and 3 are interchanged
C.the priority numbers 1 and 2 are interchanged and then 3 and 4 are interchanged
D.the priority numbers 3 and 4 are interchanged

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The configuration does not change when the priority numbers 1 and 2 are interchanged and then 3 and 4 are interchanged.

In a Fischer projection, swapping any two groups on the stereocentre inverts its configuration (R becomes S). Two such swaps invert it twice, which restores the original configuration.

Options A, B and D each describe a single interchange, so each of them changes R to S.

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33.

Relationship between compound I and II is :

Two chair cyclohexanes I and II, each carrying three methyl groups in a 1,2,4 pattern with different axial and equatorial placements

A.Diastereomers
B.Enantiomers
C.Conformational isomers
D.Constitutional isomers

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

Compounds I and II are diastereomers.

Both are trimethylcyclohexanes with the same 1,2,4 substitution pattern, so they are not constitutional isomers. Reading the chairs, in I all three methyl groups lie on the same face of the ring. In II the methyl at the position corresponding to one of the adjacent pair points to the opposite face from the other two.

Ring flipping changes axial and equatorial positions but never which face a group is on, so II cannot be a conformer of I. A different relative configuration that is not a mirror image is diastereomerism, so they are diastereomers rather than enantiomers.

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34.

The correct statement for the following structures is that they are :

Two Newman projections. Left: front carbon bears CH3, CH3 and H, rear carbon bears three H. Right: front carbon bears H, H and CH3, rear carbon bears H, H and CH3

A.Not isomers
B.Conformational isomers
C.Enantiomers
D.Structural isomers

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The two structures are structural isomers.

Read each Newman projection carbon by carbon. In the left one, the front carbon carries two CH₃ groups and one H, and the rear carbon carries three H atoms. That is (CH₃)₂CH-CH₃, 2-methylpropane (isobutane). In the right one, each carbon carries two H atoms and one CH₃ group, which is CH₃CH₂-CH₂CH₃, butane.

Both are C₄H₁₀, but their atoms are connected differently, which is what makes them structural (chain) isomers.

Conformational isomers would be the same molecule turned about a single bond, and enantiomers must be chiral mirror images; neither butane nor isobutane is chiral. 'Not isomers' is wrong because the molecular formula is the same.

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35.

In the most stable conformation the molecule having the circled Me group equatorial is :

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The circled methyl is equatorial in molecule (C).

(C) is a 1,2,3-trimethylcyclohexane: circled Me hashed at C1, Me wedged at C2, Me wedged at C3. C1 and C2 are trans, so both can be equatorial; C2 and C3 are cis, so one of them must be axial. The chair with C1 and C2 equatorial and C3 axial has one axial methyl; its ring flip has two. The preferred chair therefore holds the circled methyl equatorial.

In (A) the tert-butyl group locks the ring with itself equatorial; the circled methyl is cis to it (both wedged) across a 1,4 relationship, so it must be axial. (B) is a trans-decalin, which cannot flip; the ring-fusion H atoms are axial, and the wedged methyl on the carbon next to the hashed fusion H points the axial way. In (D) the circled methyl and the phenyl are cis on adjacent carbons, so one is axial; the much larger phenyl takes the equatorial position and the methyl is axial.

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36.

Which of the following molecule has enantiotopic protons ?

A.Butane
B.Cyclopropane
C.Ethane
D.Propane

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

Butane has enantiotopic protons.

Two protons are enantiotopic when replacing first one and then the other by a test group gives a pair of enantiomers. On C2 of butane the carbon carries H, H, CH₃ and CH₂CH₃. Replacing either H by, say, D gives 2-deuteriobutane, which is chiral, and the two replacements give the two enantiomers. So the C2 (and C3) methylene protons are enantiotopic.

In propane the central CH₂ carries two identical CH₃ groups; replacing either H gives the same achiral molecule, and the protons are homotopic. All the protons of ethane are homotopic, and so are those of cyclopropane, where a C₂ axis interchanges the two H atoms of each CH₂.

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37.

Which statement is correct for the following four molecules ?

Four 4H-thiochromene S-oxides: I 4,4-dichloro-2,3-dimethyl; II and IV both 2-methyl-3-chloro with chlorine and methyl on C4, drawn with opposite wedges at C4; III 2,3-dichloro-4,4-dimethyl

A.I and II are chiral
B.II and IV are chiral
C.I, II, III and IV are chiral
D.I and IV are chiral

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The correct statement is I, II, III and IV are chiral.

All four molecules are sulfoxides. A sulfoxide sulfur carries an oxygen, two different carbon groups (here the aromatic ring carbon and the vinylic C2) and a lone pair, in a pyramidal arrangement that does not invert at room temperature. So each sulfur is a stereogenic centre, and none of the molecules has a mirror plane through it.

That is why even I and III are chiral, although C4 in each carries two identical groups (two Cl in I, two CH₃ in III) and is not a stereocentre. II and IV have a stereogenic C4 as well as the sulfur. The options listing only some of the molecules overlook the sulfoxide stereocentre.

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38.

What is the relationship between H_a and H_b in the product P formed in the following reaction ?

A 1-substituted cyclopentadiene whose ring CH2 carries Ha and Hb; the substituent is CH(Cl)-CH(CH3)2; treated with NaOMe, MeOH to give P

A.Diastereotopic
B.Enantiotopic
C.Homotopic
D.Vicinal

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

H_a and H_b are enantiotopic in P.

Methoxide removes HCl from the side chain (E2, taking the hydrogen on the isopropyl carbon), turning CH(Cl)-CH(CH₃)₂ into CH=C(CH₃)₂, while the ring CH₂ carrying H_a and H_b is untouched. The chlorine-bearing stereocentre disappears, and P, a cyclopentadiene with a conjugated 2-methylpropenyl group, is achiral and essentially planar.

The ring CH₂ lies in that plane with H_a above it and H_b below. The molecular plane is a mirror plane that exchanges them, but no rotation axis does. Protons exchanged only by reflection are enantiotopic: replacing one or the other gives a pair of enantiomers.

They would be diastereotopic if a stereocentre remained in P, as in the starting chloride. They are not homotopic because no rotation interchanges them, and vicinal describes atoms on adjacent carbons, which these geminal protons are not.

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39.

Which of the following alcohols is NOT suitable substrate for sharpless asymmetric epoxidation reaction ?

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The alcohol that is not a suitable substrate is option C.

Sharpless asymmetric epoxidation works on allylic alcohols: the hydroxyl group binds titanium, which delivers the oxygen to the neighbouring double bond. In C the double bond is the ring-fusion double bond and the hydroxyl-bearing carbon is separated from it by one more carbon, so the OH is homoallylic and cannot direct the epoxidation. Options A and B are cinnamyl alcohols (Z and E), and D is a cyclic allylic alcohol with the double bond next to the carbinol carbon, so all three are allylic alcohols.

The Z isomer in A reacts less well than the E isomer in B, but it is still an allylic alcohol and so a substrate.

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40.

Assign the chiral descriptors for the following molecule :

Dimethyl ester with carbons numbered 1 to 5: an amino group on C3, a hashed ester bond on C2 whose carbonyl oxygen is labelled 18O, and a C4=C5 double bond carrying a vinyl and a phenyl group

A.2S, 3S, E
B.2R, 3S, Z
C.2S, 3R, E
D.2S, 3R, Z

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The descriptors are 2S, 3R, E, option C.

C-2: the two ester carbons both rank above C-3 (O,O,O against N,C,H). They are tied until the isotope is reached, and the carbon carrying ¹⁸O outranks the one with ¹⁶O, so the order is C(¹⁸O ester) > C-1 > C-3 > H. With H towards the viewer the turn is reversed, giving S.

C-3: N is first. C-2 and C-4 are both (C,C,H), so go one sphere further: C-2 holds two ester carbons (O,O,O) while C-4 holds C-5 (C,C,C) and a duplicate carbon, so C-2 outranks C-4. The order is N > C-2 > C-4 > H, and with H pointing away N to C-2 to C-4 runs clockwise, so 3R.

C-4=C-5: on C-5 phenyl (C,C,C) outranks vinyl (C,C,H), and on C-4 C-3 outranks H. They sit on opposite sides, so E.

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