Stereochemistry Questions and Answers

Answer: E

The achiral molecule is 2-Bromo-2-methoxypropane.

Its C-2 carries Br, \(OCH_{3}\) and two identical methyl groups, \((CH_{3})_{2}C(Br)OCH_{3}\), so it is not a stereocentre and the molecule is superimposable on its mirror image.

Each of the others has a carbon with four different groups: 2-chlorobutane (H, Cl, \(CH_{3}\), \(C_{2}H_{5}\)), glyceraldehyde (2,3-dihydroxypropanal), 2-bromopropionic acid and butan-2-ol.

Answer: C

The eclipsed and staggered forms of n-butane are rotamers (conformers).

They differ only by rotation about the central C-C single bond. Since that rotation is easy at room temperature, the forms interconvert rapidly and are different conformations of the same molecule, not separable isomers.

Chain, position and metamers are structural isomers, which differ in how atoms are bonded. Geometrical isomers differ in the arrangement round a double bond or ring, where rotation is restricted.

Answer: A

The compound has 4 stereoisomers.

CH₃CH=CH-CH(OH)-CH=CHCH₃ has two stereogenic double bonds (E or Z) and the carbinol carbon. Because the molecule is symmetric, the two halves are identical when both double bonds have the same geometry:

• (E,E): the carbinol carbon carries two identical E-propenyl groups, so it is not a stereocentre: 1 isomer.
• (Z,Z): likewise 1 isomer.
• (E,Z): the two propenyl groups now differ, so the carbinol carbon is a stereocentre: 2 enantiomers.

That gives 1 + 1 + 2 = 4. The formula 2³ = 8 overcounts because it treats the molecule as unsymmetrical, and 6 or 7 come from partial corrections for the symmetry.

Answer: C

Both chiral centres are R: 1R, 4R.

The compound is a trimethylbicyclo[2.2.1]heptan-2-one in which C1 carries H and is bonded to the carbonyl carbon, and C4 carries a methyl group. Because the bridgeheads are tied together by the bicyclic frame, the two centres are fixed relative to each other.

At C1 the priorities are C=O carbon (O,O,C) > C7, the gem-dimethyl bridge (C,C,C) > ring CH₂ (C,H,H) > H. At C4 they are C7 (C,C,C) > the CH₂ next to the carbonyl (it leads to O) > the other ring CH₂ > CH₃. Reading the perspective drawing with H at C1 and CH₃ at C4 pointing away, the sequence 1 → 2 → 3 runs clockwise at each centre, so both are R.

In the bridged ring the two centres cannot be inverted independently, so a mixed pair such as 1S,4R or 1R,4S does not exist for this constitution, and 1S,4S is the mirror image of the compound drawn.

Answer: B

The least stable conformation is (B).

Viewed along the C2-C3 bond, C2 carries two methyl groups and an H, and C3 carries one methyl (C4) and two H. Options C and D are staggered and so are more stable than either eclipsed form. Comparing the eclipsed forms with approximate torsional costs of CH₃/CH₃ about 11, CH₃/H about 6 and H/H about 4 kJ mol^-1:

\[ \begin{aligned} \text{(A)}&: 3(\text{CH}_3/\text{H}) \approx 18 \\ \text{(B)}&: 11 + 6 + 4 \approx 21\ \text{kJ mol}^{-1} \end{aligned} \]

(B) has a methyl eclipsing a methyl, the most costly interaction, so it has the highest energy. (A) spreads three methyls against hydrogens and is lower.

Answer: B

L-Ribose is structure B.

L-Ribose is the enantiomer of D-ribose, so every stereocentre must be inverted. In the zigzag drawing of D-ribose the OH groups at C4, C3 and C2 are hashed, wedged and hashed. Inverting all three gives wedged, hashed, wedged, with the same carbon skeleton and the CHO and CH₂OH ends unchanged, which is exactly structure B.

Structure A inverts only C4, structure C inverts C4 and C3, and structure D inverts C2. Inverting some but not all centres gives a diastereomer (another pentose), not the enantiomer.

Answer: B

The configurations are 2E, 4Z.

Number the chain from the carboxyl carbon (C-1). At the C-2=C-3 double bond, the carboxyl group on C-2 and the rest of the chain on C-3 lie on opposite sides of the bond as drawn, so it is E.

At C-4=C-5, compare the substituents by CIP priority. On C-4: the cyclobutyl carbon (C, C, H) versus C-3 of the chain (C, C, H with a duplicated carbon); at the next sphere the chain wins because its branch leads to C-2 bearing (C, C, H), while the ring branch leads only to CH₂ groups. On C-5: phenyl (C, C, C) beats cyclobutyl (C, C, H). The higher-priority groups, the C-3 chain and the phenyl, are on the same side of the double bond, so it is Z.

Answer: B

The correct Newman projection is (B).

Convert the Fischer projection by looking down the C2-C3 bond with C-2 in front. In the Fischer eclipsed form, CHO on C-2 and CH₂OH on C-3 both point back; viewing from above, the front carbon carries CHO at the top, HO lower left and H lower right, and the back carbon carries CH₂OH (eclipsed with CHO), H and OH.

Rotating the back carbon by 180° to put CH₂OH at the bottom, as in all four options, moves its H to the upper right and its OH to the upper left. Projection (B) has exactly this arrangement: front HO lower left and H lower right; back HO upper left and H upper right.

(A) swaps the back H and OH, which corresponds to the erythro sugar, and (C) and (D) place H and OH on the front carbon differently, changing the configuration at C-2.

Answer: A

The configuration is 1R, 4R.

At C-1 the substituents are C-7 (C=O), C-2 (C=O), C-6 (CH₂) and H. C-7 and C-2 tie at the first sphere (O, O, C), but at the next sphere C-7 leads to C-4, a CH with two carbon neighbours, while C-2 leads to C-3, a CH₂. So the priorities are C-7 \(\gt\) C-2 \(\gt\) C-6 \(\gt\) H.

In the drawing C-1 is the front bridgehead, its H points towards the viewer, C-7 is up, C-2 lower left and C-6 lower right. Going 7 → 2 → 6 is anticlockwise as seen, and since the lowest-priority H points towards the viewer the descriptor is reversed: R. The molecule has a C₂ axis exchanging C-1 with C-4, so C-4 is also R. Options with mixed descriptors are impossible for this symmetric trione, and 1S, 4S is its enantiomer.

Answer: D

In the lowest energy conformation 0 alkyl substituents are axial.

Number the ring with the wedged methyl at C-1, the hashed methyl at C-2 and the hashed isopropyl at C-4. C-1 and C-2 carry groups on opposite faces: a trans-1,2 relationship, which can be diequatorial. C-1 and C-4 are also on opposite faces: trans-1,4, again diequatorial. C-2 and C-4 are on the same face: cis-1,3, which is also diequatorial.

All three pairwise relationships are compatible with equatorial positions, so one chair has every alkyl group equatorial, and that chair is far more stable than its ring-flipped partner with all three axial. (This substitution pattern is that of the menthane skeleton with all groups equatorial.)