Stereochemistry Questions and Answers

Answer: B

The correct statement is It has \(S_{4}\)-symmetry, hence achiral.

The ion is a spiro compound: the two pyrrolidine rings share the nitrogen and lie in perpendicular planes. Each ring carries a trans pair of methyl groups, and the two rings carry their pairs in mirror-image configurations. An \(S_{4}\) axis runs through the nitrogen, bisecting both rings: rotating by 90° carries one ring onto the position of the other, and reflecting through the plane perpendicular to the axis then restores the configuration, so the ion is reproduced.

A molecule with any improper axis \(S_{n}\) is superimposable on its mirror image, so the ion is achiral, even though it has neither a mirror plane nor a centre of symmetry. That is why this ion is the standard example of an achiral molecule that has no \(\sigma\) and no i.

It does contain a \(C_{2}\) axis (since \(S_{4}^{2} = C_{2}\)), but 'only \(C_{2}\), hence chiral' misses the \(S_{4}\). 'No symmetry' is wrong for the same reason, and there is no centre of symmetry.

Answer: B

The major product is (B), the hydrindane carrying an exocyclic isopropylidene group.

Acid protonates the OH and water leaves, giving a secondary carbocation next to the quaternary CMe₂ carbon. In the rigid trans-decalin the OH is equatorial, and the group antiperiplanar to an equatorial leaving group is a ring C-C bond, not the axial methyl. So the ring bond from the CMe₂ carbon to the ring-fusion carbon migrates to the cationic centre (a Wagner-Meerwein shift).

That migration contracts the six-membered ring to a five-membered one and leaves the positive charge on the CMe₂ carbon, now outside the ring: a stable tertiary cation. Loss of a proton from the ring carbon gives the tetrasubstituted exocyclic alkene, the isopropylidene hydrindane (B). The trans ring fusion is untouched.

(A) and (D) keep the six-membered ring and would come from methyl migration, which needs the methyl antiperiplanar to the leaving group (an axial OH). (C), with the isopropyl group on an endocyclic alkene, is the less substituted alkene from the same cation, not the major product.

Answer: D

A and B are constitutional isomers.

They have the same molecular formula but different connectivity. In A the sp³ carbon carrying the 4-nitrophenyl group is bonded to the ring nitrogen, and the C=C bond joins the benzo-ring carbon to the enone carbon. In B the sp³ carbon carrying the aryl group is bonded to the benzo-ring carbon instead, and the C=C bond now joins the carbon next to nitrogen to the enone carbon.

Because atoms are joined differently, they cannot be enantiomers or diastereomers, which share connectivity and differ only in spatial arrangement, and they cannot be conformers, which interconvert by bond rotation. B is drawn flipped and with a hashed bond to make it look like a mirror image of A, which is the trap.

Answer: C

The most stable conformer is (C).

In (Z)-pent-2-ene the C2 methyl lies on the same side as C4. Rotation about the C3-C4 bond decides which C4 substituent eclipses the double bond and points towards that cis methyl. If it is the C5 methyl, as in (A), the two methyl groups clash severely (allylic 1,3-strain). In (C) a small hydrogen on C4 takes that position and the C5 methyl is turned away, so (C) is the lowest-energy conformer.

(B) and (D) place C4 trans to the C2 methyl across the double bond, so they are drawings of (E)-pent-2-ene, not conformers of the Z isomer at all.

Answer: D

The achiral molecules are I and IV.

Molecule I is meso-tartaric acid. Its two stereocentres carry the same substituents and are mirror images of each other, so in the zigzag drawing the OH groups sit one on a wedge and one on a hash, on opposite sides of the chain. An internal mirror (centre of inversion) maps one half onto the other, so the molecule is achiral. Molecule IV, 4-tert-butylcyclohexanol, has a mirror plane through C1 and C4, so it is achiral too, whichever way the two groups are placed.

Molecule II has both OH groups on wedges, which gives a chiral (R,R) or (S,S) tartaric acid with no internal mirror. Molecule III is a trans-disubstituted epoxide with two different groups on its two stereocentres, and it has no mirror plane. So II and III are chiral, which rules out the options that contain them.

Answer: A

The enantiopure form can be obtained for (A).

A compound can be isolated as a single enantiomer only if it is chiral. Of the dimethylcyclohexanes, trans-1,2- (and trans-1,3-) are chiral; the cis isomers of 1,2- and 1,3- are meso, and both isomers of the 1,4-dimethyl compound have a plane of symmetry. Drawing (A) shows the two methyl groups on adjacent carbons on opposite faces of the ring, the trans-1,2 compound, which has no mirror plane.

(B) is the cis-1,2 isomer, a meso compound with a mirror plane. (C) and (D) are the 1,4-isomers, cis and trans, both achiral.

Answer: A

The molecule is cis-1,2-dimethylcyclohexane.

In a chair, a cis-1,2 pair must be one axial and one equatorial. Ring flipping turns the axial methyl equatorial and the equatorial one axial, and the new chair is the non-superimposable mirror image of the old one. So each chair is chiral, but the two chairs are enantiomers that interconvert rapidly at room temperature, which is why the compound cannot be resolved and behaves as an achiral (meso) compound.

The other options do not do this. In cis-1,3-dimethylcyclohexane both chairs (diaxial and diequatorial) contain a mirror plane, so neither is chiral. In cis-1,4-dimethylcyclohexane the flip gives an identical, achiral chair. In trans-1,2-dimethylcyclohexane the flip converts the diequatorial chair into the diaxial chair of the same enantiomer, a diastereomeric conformer, not its mirror image.

Answer: D

The molecules X and Y are identical.

Both are the same methyl-substituted cyclohexene fused to a five-membered cyclic anhydride, with the same relative arrangement of the two ring-fusion hydrogens (one wedged, one hashed). In X the methyl-bearing carbon of the double bond is two bonds away from the upper ring-fusion carbon, and in Y it is two bonds away from the lower one. Turn X over by 180° about the horizontal axis in the plane of the page: the upper and lower ring-fusion carbons exchange places, a wedge becomes a hash and a hash a wedge, and the methyl moves to the bottom. That is exactly the drawing of Y.

They are not regioisomers, which would need a different position of the methyl group in the ring, and not diastereomers or enantiomers, because the two drawings do not differ at any stereocentre once X is turned over.

Answer: C

The correct statement is (C): P and Q are diastereomers, and R and S are diastereomers.

The starting material has two arms on one ring carbon: \(CH_{2}CO_{2}Et\) and \(CH_{2}CO_{2}H\). LiBH₄ reduces the ester arm to \(CH_{2}CH_{2}OH\) and leaves the acid; BH₃ reduces the acid arm to \(CH_{2}CH_{2}OH\) and leaves the ester. On the acidic workup, each hydroxy acid or hydroxy ester closes to the same kind of six-membered lactone, a spiro compound in which a \(CH_{2}C(=O)O\) arm and a \(CH_{2}CH_{2}O\) arm are joined at the ring carbon. This is the reading that makes the question well formed: without ring closure, P (a hydroxy acid) and Q (a hydroxy ester) would not even be isomers.

In P the carbonyl arm is the one that was drawn hashed; in Q it is the one that was drawn bold. So at the spiro carbon P and Q differ in configuration while the methyl-bearing carbon is unchanged: they are diastereomers. In the second starting material the methyl group is opposite the spiro carbon, so the carbonyl arm of R is trans (hashed) and that of S is cis (bold) to the methyl group: again diastereomers. Options (A), (B) and (D) each call at least one of the pairs identical.

Answer: E

The optically active compound is 2-Chlorobutane.

A molecule is optically active when it is chiral, and the usual cause is a carbon atom carrying four different groups. In 2-chlorobutane, CH₃CH(Cl)CH₂CH₃, carbon 2 bears H, Cl, CH₃ and C₂H₅: four different groups, so the molecule exists as a pair of non-superimposable mirror images.

None of the others has such a carbon. 1-Chlorobutane has Cl on a CH₂. Neopentyl chloride, (CH₃)₃CCH₂Cl, and isobutyl chloride, (CH₃)₂CHCH₂Cl, have carbons with two identical groups. tert-Butyl chloride, (CH₃)₃CCl, has three identical methyl groups on the carbon bearing chlorine.