Stereochemistry

A compound can be obtained in enantiopure form if and only if its molecules are chiral (non-superimposable on their mirror image).

For an organic molecule, the key test for chirality is the absence of an improper axis of rotation ( ), which includes the absence of a center of inversion ( ) or a plane of symmetry ( ).

Analysis of options:

A. cis-1,2-Dimethylcyclohexane

• Structure: Both methyl groups are on the same side of the ring (cis) and are attached to C1 and C2.
• Chirality: The molecule has two chiral centers (C1 and C2).
• Symmetry Analysis (Chair Conformer):
• The molecule exists as a pair of rapidly interconverting chair conformers (diaxial and diequatorial).
• The cis-1,2-dimethylcyclohexane molecule lacks a plane of symmetry ( ) and a center of inversion ( ) in both chair conformations.
• The entire molecule (as an average structure, or in its individual conformers) is chiral.
• Result: Since the molecule is chiral, it exists as a pair of enantiomers that are non-superimposable mirror images. Therefore, it can be resolved and obtained in enantiopure form.

B. trans-1,2-Dimethylcyclohexane

• Structure: The methyl groups are on opposite sides (trans) at C1 and C2.
• Chirality: The molecule has two chiral centers (C1 and C2).
• Symmetry Analysis (Chair Conformer): The trans-1,2-dimethylcyclohexane exists as a pair of enantiomers, similar to the cis isomer.
• Result: It can be obtained in enantiopure form. However, Option (B) in the image shows trans-1,3-dimethylcyclohexane (based on the way the methyl groups are drawn on adjacent carbons that are not C1 and C2, but if C1 is the top right, then they are at C1 and C2—this drawing is ambiguous, but let's assume it represents trans-1,2-dimethylcyclohexane for simplicity or cis-1,3-dimethylcyclohexane if based on the connectivity).
• If B is trans-1,2: It is chiral, and can be resolved.
• If B is cis-1,3: It is achiral (a meso compound) due to a plane of symmetry that bisects the C2 and C5 bonds and passes through C1 and C4. It cannot be resolved.

C. cis-1,3-Dimethylcyclohexane (Meso)

• Structure: Methyl groups are on the same side (cis) at C1 and C3.
• Chirality: The molecule has two chiral centers (C1 and C3).
• Symmetry Analysis: This molecule is a meso compound. It possesses an internal plane of symmetry ( ) that passes through C2 and C5 and bisects the C1-C6 and C3-C4 bonds (or passes through C1 and C4, depending on the conformation).
• Result: Due to the plane of symmetry, the molecule is achiral and cannot be obtained in enantiopure form.

D. trans-1,4-Dimethylcyclohexane

• Structure: Methyl groups are on opposite sides (trans) at C1 and C4.
• Chirality: The molecule has no chiral centers.
• Symmetry Analysis: It possesses a center of inversion ( ) and a plane of symmetry ( ) in its most stable chair conformation.
• Result: The molecule is achiral and cannot be obtained in enantiopure form.

Answer: (A)

The question asks for the C-2 epimer of D-glucose. In carbohydrate chemistry, epimers are pairs of monosaccharides that differ only at one chiral center. To solve this, we need to identify the C-2 epimer of D-glucose from the given options. Let's go through each option:

1. D-Glucose: The structure of D-glucose is:
   But C-2 epimer of D-glucose would have a different configuration at the C-2 chiral center.
2. D-Mannose: D-Mannose is the C-2 epimer of D-glucose. In the Fischer projection, D-mannose has the hydroxyl group on the left side at the C-2 position, which is opposite to the configuration in D-glucose. Hence, the correct answer is D-Mannose.
3. D-Fructose: D-fructose is a ketohexose and not an epimer of D-glucose as it differs at the carbonyl group (C=O position).
4. D-Galactose: D-Galactose differs from D-glucose at the C-4 position, not the C-2 position, so it is not the C-2 epimer of D-glucose.

Therefore, the C-2 epimer of D-glucose is D-Mannose.

We have the molecule (1,2-diphenyl-1,2-ethanediol, “hydrobenzoin”). There are two stereogenic carbon atoms (the two centers), so the maximum possible number of stereoisomers is

Label the two stereocenters as and . The four configurational combinations are:

However, because the two end groups are identical phenyl rings, the molecule with opposite configurations at the two centers (the and forms) is identical to its mirror image — it has an internal mirror plane and is therefore meso (achiral). Thus and represent the same single meso isomer, not two different stereoisomers.

The remaining two configurations, and , are nonidentical and are a pair of enantiomers.

So the actual distinct stereoisomers are:

Counting these gives

Number of Compounds with -Configuration

We apply the Cahn-Ingold-Prelog (CIP) rules to each stereocenter.

The compounds with the -configuration are 2, 4, and 5.

The total number of compounds having the -configuration is 4.

To determine which of the given compounds can have achiral stereoisomers, we need to understand the concept of chirality and identify if a compound can exist in a form that does not possess chirality (i.e., a meso compound).

Let's analyze each compound:

1. Option (A): The compound is 2-butyne, a simple alkyne with symmetry along the triple bond axis. However, this molecule does not have any chiral centers, which means it is already achiral in its current form and no separate achiral stereoisomer would be derived.
2. Option (B): This compound is 2,2'-dibromo-1,1'-binaphthyl-2,2'-dicarboxylic acid. This structure is typically helical and can have axial chirality. The presence of stereogenic centers might lead to chiral isomers, making it unlikely to have an achiral stereoisomer.
3. Option (C): The given compound is tartaric acid. Tartaric acid can exist as a meso compound due to the internal plane of symmetry, making it achiral despite having stereogenic centers. Thus, one of its stereoisomers is achiral.
4. Option (D): The compound is 1,3-dimethylcyclopentane. This can achieve an achiral conformation due to the possibility of symmetric positions about the plane of the cyclopentane ring.

Conclusion: The correct options representing compounds that can have achiral stereoisomers are (C) and (D).

The reaction involves an alkyne (PhC≡CH) reacting with HBr. In the presence of one equivalent of HBr, an addition reaction occurs. This can form different structural isomers based on the position of the bromine atom.

Step 1: Addition of HBr to Alkyne

The reaction proceeds via Markovnikov’s rule, where HBr adds across the triple bond. The hydrogen atom preferentially bonds to the carbon with more hydrogens, leading to the formation of a vinyl bromide (PhCHBr=CH₂).

Step 2: Formation of Stereoisomers

The product PhCHBr=CH₂ has a double bond, meaning it can exhibit cis-trans isomerism. The addition of H and Br across the triple bond can lead to the formation of:

• Cis isomer: Both hydrogen atoms on the same side of the double bond.
• Trans isomer: Hydrogen atoms on opposite sides of the double bond.

Therefore, there are two possible stereoisomers.

Conclusion: The total number of stereoisomers for the major product is 2.

To determine the specific rotation of the synthetic sample, we first need to calculate the ratio of the enantiomers and their contributions to specific rotation. Given is a 4:1 mixture of (S)- and (R)-enantiomers. Let's denote:
(R)-enantiomer: 20%
(S)-enantiomer: 80%
The specific rotation of optically pure (R)-enantiomer is given as +40.
For the (S)-enantiomer, the specific rotation will be of equal magnitude but opposite in sign, i.e., -40.
To find the specific rotation [α] of the mixture, use the formula:
[α] = (fraction of R) × [α]₍R₎ + (fraction of S) × [α]₍S₎
Substituting the values:
[α] = (0.2 × 40) + (0.8 × (-40))
[α] = (8) + (-32)
[α] = -24
The computed specific rotation of the sample is -24

• (A) The structure in option (A) has a pair of substituents that make the molecule asymmetric and thus chiral.
• (B) Option (B) shows a molecule with symmetry due to identical substituents, making it achiral.
• (C) Option (C) also shows a symmetrical structure with no asymmetry, making it achiral.
• (D) Option (D) contains a chiral center with different substituents attached, making it chiral.

Thus, the correct answers are (A) and (D)

The given compound is an unsaturated hydrocarbon containing two double bonds. Upon exhaustive hydrogenation, both double bonds will be reduced, converting the compound into a saturated hydrocarbon.

The structure of the saturated hydrocarbon will have two stereogenic centers, which will give rise to 3 possible stereoisomers. These include:

1. One meso form (achiral due to symmetry).
2. Two enantiomers (which are non-superimposable mirror images of each other).

Thus, the number of stereoisomers possible for this product is 3

The enantiomeric excess (ee) can be calculated using the following formula:

Given:

• Observed specific rotation = +66.3
• Specific rotation of the optically pure compound = +75.3

Substitute the values:

Thus, the enantiomeric excess (ee) of the synthetic sample is 88%.