i. Which one of the following molecules will have the ability to rotate the plane of polarized light? (Diagrams I, II, III, IV showing different organic molecules)
Answer: c) I, II and IV only
Explanation: A molecule is optically active (chiral) if it has a non-superimposable mirror image, usually due to a chiral carbon. Structures I, II, and IV possess chiral centers, while structure III does not.
ii. Which one of the following molecules shows optical isomerism?
Answer: c) CH₃=CHCH(Cl)CH₃ (Note: This formula appears to have a typo. It should likely be CH₂=CHCH(Cl)CH₃, which has a chiral carbon at C3.) The presence of a chiral carbon (C bonded to four different groups) leads to optical isomerism.
iii. How many chiral centers are present in this molecule? (Diagram of a cyclic molecule with a CH₃ group)
Answer: b) 2
Explanation: The molecule shown (likely a substituted cyclohexane derivative) has two carbon atoms bonded to four different groups, making them chiral centers.
iv. Geometric isomerism is shown by
Answer: a) alkenes
Explanation: Geometric (cis-trans) isomerism arises due to restricted rotation around a carbon-carbon double bond (C=C), which is present in alkenes.
v. Which one is correct statement about cis-trans isomerism?
Answer: a) It is shown by alkenes due to restricted rotation of double bond
Explanation: The presence of a pi bond in alkenes restricts rotation around the C=C bond, allowing for cis-trans (geometric) isomerism when different groups are attached to each carbon.
vi. Which compounds need to be designated by EZ system?
Answer: c) highly substituted alkenes
Explanation: The EZ system is used when there are more than two different groups attached to the carbons of the double bond, making cis/trans notation ambiguous or inapplicable.
vii. Which one is the least stable compound?
Answer: a) cis-but-2-ene
Explanation: Cis isomers generally have higher steric hindrance between the substituent groups, making them less stable than their trans counterparts. Among the options, cis-but-2-ene has the most significant steric hindrance relative to its size.
viii. Which molecule is chiral and can rotate the plane of polarized light?
Answer: d) pentan-2-ol
Explanation: Pentan-2-ol has a chiral carbon at C2 (bonded to H, OH, CH₃, and C₃H₇), making it optically active. 2-methylbutan-2-ol has no chiral carbon (C2 is bonded to two methyl groups).
ix. Which one of the following is optically active?
Answer: c) all chiral molecules
Explanation: Chiral molecules are optically active. A racemic mixture is optically inactive (0° rotation) because the rotations cancel out. Meso compounds are optically inactive due to internal symmetry.
x. Which one is incorrect statement about optical isomers?
Answer: d) the double bond carbon in alkenes acts chiral center
Explanation: A double bond carbon is sp² hybridized and planar; it cannot be a chiral center. Chiral centers require sp³ hybridization and four different groups.
i. Which of the following compounds can and cannot show cis-trans isomerism and why?
Answer: Yes, it can show cis-trans isomerism. The double bond carbons have two different groups attached (C1 has H and Cl; C2 has H and CH₃).
Answer: No, it cannot show cis-trans isomerism. The terminal carbon (C1) of the double bond has two identical hydrogen atoms.
Answer: Yes, it can show cis-trans isomerism. The double bond carbons have two different groups attached (C2 has H and CH₃; C3 has H and C₃H₇).
ii. Draw the cis-trans isomers of hex-3-ene.
Answer: (Diagram Description)
cis-hex-3-ene: The two ethyl groups (C₂H₅) are on the same side of the double bond. trans-hex-3-ene: The two ethyl groups (C₂H₅) are on opposite sides of the double bond.
iii. Highlight the importance of chirality in drugs. Why do we need to separate optical isomers in the preparation of drugs?
Answer: Chirality is crucial in drugs because biological receptors are themselves chiral. Different enantiomers of a drug can have drastically different effects. One enantiomer may be therapeutic, while the other could be inactive or even harmful (e.g., Thalidomide). Separating optical isomers ensures that the drug is pure and only contains the therapeutically active enantiomer, improving efficacy and safety.
iv. What is meant by levo-rotatory and dextro-rotatory isomers? Give one example.
Answer:
Dextro-rotatory (d or +): An isomer that rotates the plane of polarized light to the right (clockwise). Levo-rotatory (l or -): An isomer that rotates the plane of polarized light to the left (counterclockwise). Example: Lactic acid has two enantiomers: (+)-lactic acid (dextro) and (-)-lactic acid (levo).
v. Why can't geometric isomerism occur in alkenes where identical atoms or groups are attached to the same carbon atom of the double bond?
Answer: Geometric isomerism requires that each carbon of the double bond be attached to two different groups. If identical groups are attached to the same carbon, then swapping the groups on that carbon does not produce a different molecule. For example, in
propene (CH₂=CHCH₃), C1 has two identical H atoms, so cis-trans isomerism is not possible.
vi. How does the presence of a π (pi) bond in alkenes restrict rotation, and what energy implication does this have for isomerism?
Answer: The π bond in alkenes is formed by the sideways overlap of parallel p-orbitals. This overlap locks the two carbon atoms in a
planar arrangement and prevents rotation around the C=C bond. To rotate, the π bond must be broken, which requires a significant amount of energy (approximately 270 kJ/mol). This high energy barrier allows the cis and trans isomers to exist as separate, isolable compounds.
vii. Why are cis-isomers generally more reactive but less stable than their trans-isomer counterparts?
Answer: Cis-isomers are less stable because the substituent groups on the same side of the double bond experience steric hindrance (repulsion). This makes the cis isomer higher in energy. They are often more reactive because the steric strain makes them more susceptible to reactions that can relieve the strain, such as addition reactions.
viii. How can a molecule be chiral yet remain optically inactive? Give the term used for such a molecule.
Answer: A molecule can be chiral yet optically inactive if it exists as a racemic mixture (racemate). A racemic mixture contains equal amounts of both enantiomers (d and l forms). The optical rotations of the two enantiomers cancel each other out, resulting in a net rotation of zero. The term used is a racemic mixture or racemate.
ix. Why is it not effective to administer only the (+)-thalidomide enantiomer to patients?
Answer: It is not effective to administer only the (+)-thalidomide enantiomer because the human body can interconvert the enantiomers (a process called chiral inversion). The (+)-enantiomer can racemize in the body to form the (-)-enantiomer, which is teratogenic (causes birth defects). Therefore, administering only one enantiomer does not prevent the harmful effects.
x. Why is the use of chiral catalysts preferred in the pharmaceutical industry for synthesizing optical isomers of drugs?
Answer: Chiral catalysts are preferred because they allow for the synthesis of a specific enantiomer (asymmetric synthesis) rather than producing a racemic mixture. This is more efficient and cost-effective because it eliminates the need for separation of enantiomers. It also reduces waste and ensures that the final drug product is pure and contains only the desired enantiomer.
i. Define geometric isomerism. Draw cis-trans isomers of any alkene.
Answer:
Geometric isomerism (cis-trans isomerism) is a type of stereoisomerism where atoms or groups are arranged differently in space due to restricted rotation around a double bond or in a ring structure.
Example: But-2-ene
cis-but-2-ene: The two methyl groups (-CH₃) are on the same side of the double bond. trans-but-2-ene: The two methyl groups (-CH₃) are on opposite sides of the double bond.
(Diagram Description: Draw the C=C double bond. For cis, place both CH₃ groups above the double bond. For trans, place one CH₃ above and one below.)
ii. Why propene cannot show cis-trans isomerism?
Answer: Propene (CH₂=CH-CH₃) cannot show cis-trans isomerism because one of the double-bonded carbon atoms (C1) has two identical hydrogen atoms attached to it. For cis-trans isomerism to occur, each carbon of the double bond must have two different groups attached. Since C1 has two H atoms, swapping them does not create a new isomer.
iii. Explain the way of separating a racemic mixture of thalidomide.
Answer: A racemic mixture of thalidomide can be separated by several methods:
1. Chiral Chromatography: Using a chiral stationary phase (e.g., chiral HPLC) to selectively interact with one enantiomer, causing it to elute at a different time.
2. Diastereomer Formation: Reacting the racemic mixture with a chiral resolving agent to form diastereomers, which have different physical properties (e.g., solubility, melting point) and can be separated by conventional methods (e.g., crystallization). The diastereomers are then converted back to the pure enantiomers.
3. Enzymatic Resolution: Using enzymes that selectively react with one enantiomer.
iv. How a catalyst can play its role in this separation.
Answer: A chiral catalyst can play a role in separation (or synthesis) by selectively interacting with one enantiomer or by facilitating the formation of one enantiomer over the other. In kinetic resolution, a chiral catalyst selectively reacts with one enantiomer of a racemic mixture, converting it to a different product while leaving the other enantiomer unchanged. The two products can then be separated by conventional means (e.g., distillation, extraction). This allows for the isolation of the desired enantiomer.
v. Explain why the trans-isomer of but-2-ene is more stable than the cis-isomer. What role do steric interactions play?
Answer: The trans-isomer of but-2-ene is more stable than the cis-isomer because of steric interactions. In the cis-isomer, the two methyl groups are on the same side of the double bond, causing significant steric hindrance (repulsion) between them. This repulsion raises the energy of the cis-isomer, making it less stable. In the trans-isomer, the methyl groups are on opposite sides, minimizing steric hindrance and resulting in a lower energy, more stable molecule.
vi. Given the compound 1,2-dichloroethene, identify and draw all possible geometrical isomers.
Answer: 1,2-dichloroethene (ClCH=CHCl) has two geometrical isomers:
cis-1,2-dichloroethene: Both chlorine atoms are on the same side of the double bond. trans-1,2-dichloroethene: The chlorine atoms are on opposite sides of the double bond.
(Diagram Description: Draw the C=C double bond. For cis, place both Cl atoms above the double bond. For trans, place one Cl above and one below.)
vii. Indicate which isomer would likely have a higher boiling point and why.
Answer: The cis-isomer would likely have a higher boiling point. Cis-isomers are generally more polar than trans-isomers because the substituent groups are on the same side, creating a net dipole moment. This increased polarity leads to stronger dipole-dipole interactions between molecules, requiring more energy to separate them, resulting in a higher boiling point. Trans-isomers are often more symmetrical, less polar, and have lower boiling points.
viii. What structural features must a molecule have to exhibit optical isomerism? Give an example and explain why it is optically active.
Answer: To exhibit optical isomerism, a molecule must be chiral, meaning it must have a non-superimposable mirror image. The most common structural feature is the presence of a chiral center (asymmetric carbon), which is a carbon atom bonded to four different groups.
Example: Lactic acid (CH₃CH(OH)COOH). The central carbon (C2) is bonded to four different groups: -H, -OH, -CH₃, and -COOH.
Why it is optically active: The presence of a chiral center means the molecule lacks an internal plane of symmetry. It can exist in two enantiomeric forms (d and l) that rotate the plane of polarized light in opposite directions. Therefore, it is optically active.
1. Which of the following compounds can and cannot show cis-trans isomerism and why?
i. 1-chloroprop-1-ene
Answer: Can show cis-trans isomerism. The double bond carbons have different groups attached (C1: H, Cl; C2: H, CH₃).
ii. 3-chloroprop-1-ene
Answer: Cannot show cis-trans isomerism. The terminal carbon (C1) has two identical hydrogen atoms.
iii. hex-2-ene
Answer: Can show cis-trans isomerism. The double bond carbons have different groups attached (C2: H, CH₃; C3: H, C₃H₇).
Separate chiral and achiral molecules in the following.
Answer: Achiral. No chiral carbon.
Answer: Chiral. C2 is bonded to H, F, CF₂H, and CCl₃ (four different groups).
Answer: Chiral. The carbon bearing the OH group is bonded to H, OH, CH₃, and CHCl₂(CH₂)₂ (four different groups).
Answer: Chiral. The carbon bearing OH and Cl is bonded to H, OH, Cl, and CH₂CH₃ (four different groups). e. CH₂ClCH(OH)CH₃
Answer: Chiral. C2 is bonded to H, OH, CH₃, and CH₂Cl (four different groups). f. CF₃CH₂CCl₃
Answer: Achiral. Same as (a), no chiral carbon.
