7.1 Geometric Isomerism

In organic chemistry, molecules with the same molecular formula but different arrangements of atoms are called isomers. Stereoisomerism is a type of isomerism where molecules have the same molecular formula and the same connectivity of atoms (same structural formula) but differ in the three-dimensional arrangement of atoms in space. Geometric isomerism is a specific type of stereoisomerism that arises due to restricted rotation around a bond, most commonly a carbon-carbon double bond in alkenes.

Definition

Geometric isomerism (also called cis-trans isomerism) is a form of stereoisomerism describing the orientation of functional groups within a molecule. It occurs when there is restricted rotation about a bond (such as a double bond) and each of the two atoms involved in the bond is attached to two different groups.

Simple Explanation: Geometric isomers have the same atoms connected in the same order, but they are locked into different spatial arrangements because they cannot rotate freely around a double bond.

Core Concept

The core concept is that the carbon-carbon double bond in an alkene consists of one sigma (σ) bond and one pi (π) bond. The π bond is formed by the side-to-side overlap of unhybridized p orbitals. This π bond locks the two carbon atoms in place, preventing free rotation around the double bond at room temperature. Because rotation is restricted, the groups attached to each carbon atom are fixed in their relative positions, leading to different isomers (cis and trans).

Detailed Explanation

The Nature of the Double Bond

Some organic compounds, like alkenes, have the same connectivity of atoms in their molecules, same structural formulae but different arrangement of atoms or groups of atoms around a double bond. Such molecules are called geometric isomers, and this phenomenon is called geometric isomerism.

• One sigma (σ) bond: Formed by head-on overlap between one of the three sp² hybrid orbitals on each carbon atom of the double bond.
• One pi (π) bond: Formed by parallel overlap between unhybridized "p" orbitals on each carbon atom of the double bond.

Restricted Rotation

Since the pi (π) bond between carbon atoms in alkenes restricts rotation about the double bond at room temperature and does not allow free rotation of substituents around. To rotate the carbon atoms in a double bond of alkenes relative to each other, the pi (π) bond must be broken. Breaking the pi (π) bond in ethene requires approximately 264 kJ/mol of energy. This restricted rotation of carbon atoms of the double bond is responsible for geometric isomerism in alkenes.

Comparison with Alkanes

Alkanes, on the other hand, cannot show this type of isomerism because a single bond can easily be rotated at room temperature. For example, rotating a carbon-carbon single bond in ethane requires almost 12 kJ/mol of energy. In the light of this discussion, we can say that one major difference between alkanes and alkenes is the degree of freedom of rotation about carbon-carbon single and double bonds.

Conditions for Geometric Isomerism

Remember that alkene molecules with same atoms or groups of atoms on the same carbon atom of the double bond cannot show geometric isomerism. This is because we get identical molecules even after breaking the pi (π) bond and rotating the molecule across the carbon-carbon double bond.

For a molecule to show geometric isomerism, each carbon atom of the double bond must be attached to two different groups.

Cis and Trans Isomers

When the two identical groups are on the same side of the double bond, the isomer is called cis. When they are on opposite sides, the isomer is called trans.

Intermediate Level

Geometric isomerism in alkenes is due to the restricted rotation around the carbon-carbon double bond. The double bond consists of a sigma bond and a pi bond. The pi bond is formed by the overlap of p orbitals and is relatively weak compared to the sigma bond, but it is strong enough to prevent free rotation at room temperature. To rotate around the double bond, the pi bond must be broken, which requires a significant amount of energy (about 264 kJ/mol). This restriction means that the substituents on the double bond are fixed in their positions. For a molecule to exhibit geometric isomerism, each carbon of the double bond must be attached to two different groups. If the two identical groups are on the same side, it is the cis isomer; if they are on opposite sides, it is the trans isomer. Alkanes, with their single bonds, have free rotation (about 12 kJ/mol) and cannot show geometric isomerism.

Important Points

• - Geometric Isomerism: Also called cis-trans isomerism.
• - Cause: Restricted rotation around a carbon-carbon double bond.
• - Double Bond: Consists of one σ bond and one π bond.
• - Rotation Barrier: ∼264 kJ/mol for ethene (high), ∼12 kJ/mol for ethane (low).
• - Requirement: Each carbon of the double bond must be attached to two different groups.
• - Cis Isomer: Identical groups on the same side.
• - Trans Isomer: Identical groups on opposite sides.
• - Alkanes cannot show geometric isomerism due to free rotation around single bonds.

Remember

• - Geometric isomerism is shown by alkenes due to restricted rotation of the double bond.
• - Double bond = one σ + one π bond.
• - π bond restricts rotation.
• - Each carbon of double bond must have two different groups for geometric isomerism.
• - Cis = same side; Trans = opposite sides.
• - Alkanes cannot show geometric isomerism (free rotation).
• - Rotation energy barrier: ∼264 kJ/mol for alkenes, ∼12 kJ/mol for alkanes.

Common Mistakes

• - Thinking that all alkenes show geometric isomerism. Only those with two different groups on each double-bond carbon do.
• - Confusing cis and trans. Cis = same side, Trans = opposite sides.
• - Believing that alkanes can show geometric isomerism. They cannot, because single bonds rotate freely.
• - Forgetting that the double bond must have two different groups on each carbon. For example, 1-butene (where one carbon of the double bond has two hydrogen atoms) cannot show geometric isomerism.
• - Assuming that geometric isomers have the same properties. They have different physical properties (e.g., boiling points, melting points) and different chemical reactivity.

Quick Revision

Geometric isomerism (cis-trans isomerism) is a type of stereoisomerism that arises from the restricted rotation around a carbon-carbon double bond. The double bond consists of a sigma bond and a pi bond. The pi bond prevents free rotation at room temperature, locking the substituents in their positions. For a molecule to show geometric isomerism, each carbon atom of the double bond must be attached to two different groups. If the two identical groups are on the same side, it is the cis isomer; if they are on opposite sides, it is the trans isomer. Alkanes, with single bonds, have free rotation and cannot show this type of isomerism. The energy barrier to rotation is about 264 kJ/mol for alkenes, compared to about 12 kJ/mol for alkanes.

60-Second Revision

Geometric isomerism (cis-trans isomerism) occurs in alkenes because rotation around the carbon-carbon double bond is restricted by the pi bond (energy barrier ~264 kJ/mol). The two isomers (cis and trans) have different spatial arrangements of groups. For a molecule to show this isomerism, each carbon of the double bond must be attached to two different groups. Alkanes cannot show geometric isomerism because their single bonds rotate freely (energy barrier ~12 kJ/mol).

Frequently Asked Questions (FAQs)

Q1. Define geometric isomerism. Answer: Geometric isomerism, also known as cis-trans isomerism, is a type of stereoisomerism in which molecules have the same molecular formula and connectivity but different spatial arrangements of atoms or groups due to restricted rotation around a double bond. The two isomers are called cis (identical groups on the same side) and trans (identical groups on opposite sides).

Q2. Why is geometric isomerism shown by alkenes but not by alkanes? Answer: Geometric isomerism is shown by alkenes because the carbon-carbon double bond consists of a sigma and a pi bond. The pi bond restricts rotation around the double bond (energy barrier ~264 kJ/mol), locking the substituents in place. Alkanes, which have only single bonds (sigma bonds), can rotate freely (energy barrier ~12 kJ/mol), so their substituents are not fixed and they cannot show geometric isomerism.

• A double bond that restricts rotation.
• Two different groups attached to each of the two carbon atoms of the double bond. If one carbon has two identical groups, the molecule cannot show geometric isomerism.

Q4. Why can't propene show cis-trans isomerism? Answer: Propene (CH₃-CH=CH₂) cannot show cis-trans isomerism because one of the carbon atoms of the double bond (the terminal CH₂) is attached to two identical hydrogen atoms. For geometric isomerism, each carbon of the double bond must be attached to two different groups.

Q5. What is the difference between cis and trans isomers? Answer: In a cis isomer, the two identical (or similar) groups are on the same side of the double bond. In a trans isomer, the two identical (or similar) groups are on opposite sides of the double bond. The two isomers have different physical and chemical properties.

Q6. How does the presence of a π bond in alkenes restrict rotation, and what energy implication does this have for isomerism? Answer: The π bond is formed by the side-to-side overlap of unhybridized p orbitals on the two carbon atoms. This overlap creates a region of electron density that locks the carbon atoms in place. To rotate around the double bond, this π bond must be broken, which requires about 264 kJ/mol of energy. This high energy barrier prevents free rotation at room temperature, allowing the cis and trans isomers to exist as stable, separate molecules.

Q7. Why are cis-isomers generally more reactive but less stable than their trans-isomers? Answer: Cis-isomers have a higher energy (are less stable) than trans-isomers due to steric hindrance (repulsion between bulky groups on the same side of the double bond). This higher energy makes them more reactive, as they are closer to the transition state energy for reactions. Trans-isomers have less steric hindrance and are generally more stable.

Q8. What is the rotation energy barrier for a typical carbon-carbon single bond? Answer: The rotation energy barrier for a typical carbon-carbon single bond (e.g., in ethane) is about 12 kJ/mol. This is low enough that free rotation occurs at room temperature.

Multiple Choice Questions (MCQs)

• Q1. Geometric isomerism is shown by:
• A. Alkanes
• B. Alkenes
• C. Alkynes
• D. Benzene
• Correct Answer: B
• Explanation: Alkenes show geometric isomerism because the double bond restricts rotation, allowing for cis and trans isomers.

• Q2. What is the approximate energy required to break the pi bond in ethene?
• A. 12 kJ/mol
• B. 120 kJ/mol
• C. 264 kJ/mol
• D. 500 kJ/mol
• Correct Answer: C
• Explanation: Breaking the pi bond in ethene requires approximately 264 kJ/mol of energy.

Q3. Which of the following compounds can show cis-trans isomerism? A. Propene B. 1-butene C. 2-butene D. 1-pentene Correct Answer: C Explanation: 2-butene has two different groups (CH₃ and H) on each carbon of the double bond, so it can show cis-trans isomerism. Propene and 1-butene have two identical groups (H) on one of the double-bond carbons.

• Q4. What is the main reason alkanes do not show geometric isomerism?
• A. They have a double bond.
• B. They have free rotation around single bonds.
• C. They have a pi bond.
• D. They are highly reactive.
• Correct Answer: B
• Explanation: Alkanes have single bonds that allow free rotation, so they cannot have fixed spatial arrangements of groups.

• Q5. In the cis isomer, the identical groups are:
• A. On opposite sides of the double bond.
• B. On the same side of the double bond.
• C. Always on the same carbon atom.
• D. Not present.
• Correct Answer: B
• Explanation: In a cis isomer, identical groups are on the same side of the double bond.

• Q6. Which of the following is the correct structure for trans-2-butene?
• A. The two methyl groups on the same side of the double bond.
• B. The two methyl groups on opposite sides of the double bond.
• C. Both methyl groups on the same carbon atom.
• D. No methyl groups.
• Correct Answer: B
• Explanation: In trans-2-butene, the two methyl groups are on opposite sides of the double bond.

• Q7. The rotation around a carbon-carbon single bond in ethane requires about:
• A. 12 kJ/mol
• B. 120 kJ/mol
• C. 264 kJ/mol
• D. 500 kJ/mol
• Correct Answer: A
• Explanation: The energy required for rotation around a single bond in ethane is about 12 kJ/mol.

• Q8. What type of bond is formed by the side-to-side overlap of p orbitals?
• A. Sigma bond
• B. Pi bond
• C. Ionic bond
• D. Hydrogen bond
• Correct Answer: B
• Explanation: A pi bond is formed by the side-to-side overlap of unhybridized p orbitals.

• Q9. Which of the following cannot show geometric isomerism?
• A. 2-pentene
• B. 2-hexene
• C. 1-hexene
• D. 3-hexene
• Correct Answer: C
• Explanation: 1-hexene has two hydrogen atoms on the terminal carbon of the double bond, so it cannot show geometric isomerism.

• Q10. What happens to the energy of the molecule when it is in a cis configuration with bulky groups?
• A. The energy is lower than the trans isomer.
• B. The energy is higher than the trans isomer due to steric hindrance.
• C. The energy is the same as the trans isomer.
• D. The molecule is more stable.
• Correct Answer: B
• Explanation: The cis isomer has higher energy (is less stable) due to steric hindrance between bulky groups on the same side.

Multiple Choice Questions (MCQs)

• Q1. Geometric isomerism is shown by:
• A. Alkanes
• B. Alkenes
• C. Alkynes
• D. Benzene
• Correct Answer: B
• Explanation: Alkenes show geometric isomerism due to restricted rotation around the double bond.

• Q2. What is the approximate energy required to break the pi bond in ethene?
• A. 12 kJ/mol
• B. 120 kJ/mol
• C. 264 kJ/mol
• D. 500 kJ/mol
• Correct Answer: C
• Explanation: The pi bond in ethene requires approximately 264 kJ/mol to break.

• Q3. Which of the following compounds can show cis-trans isomerism?
• A. Propene
• B. 1-butene
• C. 2-butene
• D. 1-pentene
• Correct Answer: C
• Explanation: 2-butene can show cis-trans isomerism because each carbon of the double bond has two different groups.

• Q4. What is the main reason alkanes do not show geometric isomerism?
• A. They have a double bond.
• B. They have free rotation around single bonds.
• C. They have a pi bond.
• D. They are highly reactive.
• Correct Answer: B
• Explanation: Alkanes have single bonds that allow free rotation, so they cannot have fixed spatial arrangements.

• Q5. In the cis isomer, the identical groups are:
• A. On opposite sides of the double bond.
• B. On the same side of the double bond.
• C. Always on the same carbon atom.
• D. Not present.
• Correct Answer: B
• Explanation: In a cis isomer, identical groups are on the same side.

• Q6. Which of the following is the correct structure for trans-2-butene?
• A. The two methyl groups on the same side.
• B. The two methyl groups on opposite sides.
• C. Both methyl groups on the same carbon.
• D. No methyl groups.
• Correct Answer: B
• Explanation: In trans-2-butene, the two methyl groups are on opposite sides.

• Q7. The rotation around a carbon-carbon single bond in ethane requires about:
• A. 12 kJ/mol
• B. 120 kJ/mol
• C. 264 kJ/mol
• D. 500 kJ/mol
• Correct Answer: A
• Explanation: The energy required for rotation around a single bond in ethane is about 12 kJ/mol.

• Q8. What type of bond is formed by the side-to-side overlap of p orbitals?
• A. Sigma bond
• B. Pi bond
• C. Ionic bond
• D. Hydrogen bond
• Correct Answer: B
• Explanation: A pi bond is formed by the side-to-side overlap of p orbitals.

• Q9. Which of the following cannot show geometric isomerism?
• A. 2-pentene
• B. 2-hexene
• C. 1-hexene
• D. 3-hexene
• Correct Answer: C
• Explanation: 1-hexene has two hydrogen atoms on the terminal carbon, so it cannot show geometric isomerism.

• Q10. What happens to the energy of the molecule when it is in a cis configuration with bulky groups?
• A. The energy is lower than the trans isomer.
• B. The energy is higher than the trans isomer due to steric hindrance.
• C. The energy is the same as the trans isomer.
• D. The molecule is more stable.
• Correct Answer: B
• Explanation: The cis isomer has higher energy due to steric hindrance.

Multiple Choice Questions (MCQs)

• Q1. Which of the following molecules can show geometric isomerism?
• A. Ethene
• B. Propene
• C. But-2-ene
• D. But-1-ene
• Correct Answer: C
• Explanation: But-2-ene has two different groups on each double-bond carbon, so it can show geometric isomerism.

• Q2. The energy barrier to rotation around a carbon-carbon double bond is approximately:
• A. 12 kJ/mol
• B. 264 kJ/mol
• C. 500 kJ/mol
• D. 1000 kJ/mol
• Correct Answer: B
• Explanation: The energy barrier is approximately 264 kJ/mol.

• Q3. In a trans isomer, the identical groups are:
• A. On the same side of the double bond.
• B. On opposite sides of the double bond.
• C. On the same carbon atom.
• D. Not present.
• Correct Answer: B
• Explanation: In a trans isomer, identical groups are on opposite sides.

• Q4. Which of the following bonds is formed by head-on overlap of orbitals?
• A. Pi bond
• B. Sigma bond
• C. Ionic bond
• D. Hydrogen bond
• Correct Answer: B
• Explanation: A sigma bond is formed by head-on overlap of orbitals.

• Q5. What is the hybridization of the carbon atoms in a double bond?
• A. sp
• B. sp²
• C. sp³
• D. sp³d
• Correct Answer: B
• Explanation: The carbon atoms in a double bond are sp² hybridized.

• Q6. Which of the following statements is true about geometric isomerism?
• A. It is shown by all alkenes.
• B. It is shown by alkenes with two different groups on each double-bond carbon.
• C. It is shown by alkanes.
• D. It is not a type of stereoisomerism.
• Correct Answer: B
• Explanation: Geometric isomerism is shown by alkenes with two different groups on each double-bond carbon.

• Q7. The energy barrier to rotation around a carbon-carbon single bond is approximately:
• A. 12 kJ/mol
• B. 264 kJ/mol
• C. 500 kJ/mol
• D. 1000 kJ/mol
• Correct Answer: A
• Explanation: The energy barrier for a single bond is approximately 12 kJ/mol.

• Q8. Which of the following compounds cannot show cis-trans isomerism?
• A. 2-pentene
• B. 3-hexene
• C. 1-butene
• D. 2-butene
• Correct Answer: C
• Explanation: 1-butene has two hydrogen atoms on the terminal carbon, so it cannot show geometric isomerism.

• Q9. What is the relationship between the stability of cis and trans isomers?
• A. Cis isomers are generally more stable than trans isomers.
• B. Trans isomers are generally more stable than cis isomers due to less steric hindrance.
• C. Both isomers have the same stability.
• D. Stability depends only on the molecular weight.
• Correct Answer: B
• Explanation: Trans isomers are generally more stable due to less steric hindrance between bulky groups.

• Q10. Which of the following is a correct statement about geometric isomerism?
• A. It is a type of structural isomerism.
• B. It arises due to restricted rotation around a double bond.
• C. It is shown by alkanes.
• D. It involves the breaking of sigma bonds.
• Correct Answer: B
• Explanation: Geometric isomerism arises due to restricted rotation around a double bond.

Multiple Choice Questions (MCQs)

• Q1. Which of the following molecules can show geometric isomerism?
• A. Ethene
• B. Propene
• C. But-2-ene
• D. But-1-ene
• Correct Answer: C
• Explanation: But-2-ene can show geometric isomerism.

• Q2. The energy barrier to rotation around a carbon-carbon double bond is approximately:
• A. 12 kJ/mol
• B. 264 kJ/mol
• C. 500 kJ/mol
• D. 1000 kJ/mol
• Correct Answer: B
• Explanation: The energy barrier is approximately 264 kJ/mol.

• Q3. In a trans isomer, the identical groups are:
• A. On the same side of the double bond.
• B. On opposite sides of the double bond.
• C. On the same carbon atom.
• D. Not present.
• Correct Answer: B
• Explanation: In a trans isomer, identical groups are on opposite sides.

• Q4. Which of the following bonds is formed by head-on overlap of orbitals?
• A. Pi bond
• B. Sigma bond
• C. Ionic bond
• D. Hydrogen bond
• Correct Answer: B
• Explanation: A sigma bond is formed by head-on overlap.

• Q5. What is the hybridization of the carbon atoms in a double bond?
• A. sp
• B. sp²
• C. sp³
• D. sp³d
• Correct Answer: B
• Explanation: The carbon atoms in a double bond are sp² hybridized.

• Q6. Which of the following statements is true about geometric isomerism?
• A. It is shown by all alkenes.
• B. It is shown by alkenes with two different groups on each double-bond carbon.
• C. It is shown by alkanes.
• D. It is not a type of stereoisomerism.
• Correct Answer: B
• Explanation: Geometric isomerism is shown by alkenes with two different groups on each double-bond carbon.

• Q7. The energy barrier to rotation around a carbon-carbon single bond is approximately:
• A. 12 kJ/mol
• B. 264 kJ/mol
• C. 500 kJ/mol
• D. 1000 kJ/mol
• Correct Answer: A
• Explanation: The energy barrier for a single bond is approximately 12 kJ/mol.

• Q8. Which of the following compounds cannot show cis-trans isomerism?
• A. 2-pentene
• B. 3-hexene
• C. 1-butene
• D. 2-butene
• Correct Answer: C
• Explanation: 1-butene cannot show geometric isomerism.

• Q9. What is the relationship between the stability of cis and trans isomers?
• A. Cis isomers are generally more stable than trans isomers.
• B. Trans isomers are generally more stable than cis isomers due to less steric hindrance.
• C. Both isomers have the same stability.
• D. Stability depends only on the molecular weight.
• Correct Answer: B
• Explanation: Trans isomers are generally more stable.

• Q10. Which of the following is a correct statement about geometric isomerism?
• A. It is a type of structural isomerism.
• B. It arises due to restricted rotation around a double bond.
• C. It is shown by alkanes.
• D. It involves the breaking of sigma bonds.
• Correct Answer: B
• Explanation: Geometric isomerism arises due to restricted rotation around a double bond.

Multiple Choice Questions (MCQs)

• Q1. Geometric isomerism is shown by alkenes due to:
• A. Free rotation around the double bond.
• B. Restricted rotation around the double bond.
• C. The presence of a single bond.
• D. The absence of a pi bond.
• Correct Answer: B
• Explanation: Geometric isomerism is due to restricted rotation around the double bond.

• Q2. Which of the following is a correct statement about cis-trans isomerism?
• A. It is shown by alkenes due to restricted rotation of the double bond.
• B. It is shown by alkenes due to free rotation of the sigma bond.
• C. It is not shown by alkenes because they cannot have different atoms/groups.
• D. Alkanes cannot show geometric isomerism due to free rotation of sigma bonds.
• Correct Answer: A
• Explanation: Cis-trans isomerism is shown by alkenes due to restricted rotation of the double bond.

• Q3. Which compounds need to be designated by the EZ system?
• A. Low substituted alkanes.
• B. High substituted alkenes.
• C. Highly substituted alkanes.
• D. Low substituted alkenes.
• Correct Answer: B
• Explanation: The EZ system is used for alkenes with different groups on each carbon, which are often highly substituted.

• Q4. Which one is the least stable compound?
• A. cis-but-2-ene
• B. trans-but-2-ene
• C. trans-hex-3-ene
• D. trans-pent-2-ene
• Correct Answer: A
• Explanation: Cis-but-2-ene is less stable due to steric hindrance between the two methyl groups on the same side.

• Q5. Which molecule is chiral and can rotate the plane of polarized light?
• A. cyclohexane
• B. benzene
• C. 2-methylbutan-2-ol
• D. pentan-2-ol
• Correct Answer: D
• Explanation: Pentan-2-ol has a chiral carbon (C2) with four different groups attached, so it is chiral and can rotate the plane of polarized light.

• Q6. Which one of the following is optically active?
• A. racemic mixture
• B. meso-compounds
• C. all chiral molecules
• D. all asymmetric molecules
• Correct Answer: C
• Explanation: All chiral molecules are optically active (they rotate the plane of polarized light).

• Q7. Which one is an incorrect statement about optical isomers?
• A. They cause different rotation of plane polarized light.
• B. They have no functional groups.
• C. Alkanes cannot show optical isomerism.
• D. The double bond carbon in alkenes acts as a chiral center.
• Correct Answer: D
• Explanation: The double bond carbon in alkenes is sp² hybridized and cannot be a chiral center (which requires sp³ hybridization with four different groups).

Q8. Which of the following compounds can and cannot show cis-trans isomerism and why? (i) 1-chloroprop-1-ene, (ii) 3-chloroprop-1-ene, (iii) hex-2-ene. A. (i) Yes, (ii) Yes, (iii) Yes. B. (i) Yes, (ii) No, (iii) Yes. C. (i) No, (ii) No, (iii) Yes. D. (i) No, (ii) Yes, (iii) No. Correct Answer: B Explanation: (i) 1-chloroprop-1-ene has two different groups on each double-bond carbon (Cl and H on C1; CH₃ and H on C2), so it can show cis-trans isomerism. (ii) 3-chloroprop-1-ene has two hydrogens on the terminal carbon, so it cannot. (iii) Hex-2-ene has two different groups on each double-bond carbon, so it can.

• Q9. Which of the following is a correct statement about the stability of geometric isomers?
• A. Cis isomers are always more stable than trans isomers.
• B. Trans isomers are generally more stable than cis isomers due to less steric hindrance.
• C. Both isomers have exactly the same stability.
• D. Stability depends only on the number of carbon atoms.
• Correct Answer: B
• Explanation: Trans isomers are generally more stable than cis isomers due to less steric hindrance between bulky groups.

• Q10. Which of the following molecules can show geometric isomerism?
• A. CH₂=CH-CH₃
• B. CH₃-CH=CH-CH₃
• C. CH₃-CH₂-CH=CH₂
• D. CH₂=CH-CH₂-CH₃
• Correct Answer: B
• Explanation: CH₃-CH=CH-CH₃ (2-butene) has two different groups on each double-bond carbon, so it can show geometric isomerism.

Multiple Choice Questions (MCQs)

• Q1. Geometric isomerism is shown by alkenes due to:
• A. Free rotation around the double bond.
• B. Restricted rotation around the double bond.
• C. The presence of a single bond.
• D. The absence of a pi bond.
• Correct Answer: B
• Explanation: Geometric isomerism is due to restricted rotation around the double bond.

• Q2. Which of the following is a correct statement about cis-trans isomerism?
• A. It is shown by alkenes due to restricted rotation of the double bond.
• B. It is shown by alkenes due to free rotation of the sigma bond.
• C. It is not shown by alkenes because they cannot have different atoms/groups.
• D. Alkanes cannot show geometric isomerism due to free rotation of sigma bonds.
• Correct Answer: A
• Explanation: Cis-trans isomerism is shown by alkenes due to restricted rotation of the double bond.

• Q3. Which compounds need to be designated by the EZ system?
• A. Low substituted alkanes.
• B. High substituted alkenes.
• C. Highly substituted alkanes.
• D. Low substituted alkenes.
• Correct Answer: B
• Explanation: The EZ system is used for alkenes with different groups on each carbon.

• Q4. Which one is the least stable compound?
• A. cis-but-2-ene
• B. trans-but-2-ene
• C. trans-hex-3-ene
• D. trans-pent-2-ene
• Correct Answer: A
• Explanation: Cis-but-2-ene is less stable due to steric hindrance.

• Q5. Which molecule is chiral and can rotate the plane of polarized light?
• A. cyclohexane
• B. benzene
• C. 2-methylbutan-2-ol
• D. pentan-2-ol
• Correct Answer: D
• Explanation: Pentan-2-ol has a chiral carbon.

• Q6. Which one of the following is optically active?
• A. racemic mixture
• B. meso-compounds
• C. all chiral molecules
• D. all asymmetric molecules
• Correct Answer: C
• Explanation: All chiral molecules are optically active.

• Q7. Which one is an incorrect statement about optical isomers?
• A. They cause different rotation of plane polarized light.
• B. They have no functional groups.
• C. Alkanes cannot show optical isomerism.
• D. The double bond carbon in alkenes acts as a chiral center.
• Correct Answer: D
• Explanation: The double bond carbon in alkenes is sp² hybridized and cannot be a chiral center.

• Q8. Which of the following compounds can and cannot show cis-trans isomerism and why? (i) 1-chloroprop-1-ene, (ii) 3-chloroprop-1-ene, (iii) hex-2-ene.
• A. (i) Yes, (ii) Yes, (iii) Yes.
• B. (i) Yes, (ii) No, (iii) Yes.
• C. (i) No, (ii) No, (iii) Yes.
• D. (i) No, (ii) Yes, (iii) No.
• Correct Answer: B
• Explanation: (i) 1-chloroprop-1-ene can show cis-trans isomerism; (ii) 3-chloroprop-1-ene cannot; (iii) hex-2-ene can.

• Q9. Which of the following is a correct statement about the stability of geometric isomers?
• A. Cis isomers are always more stable than trans isomers.
• B. Trans isomers are generally more stable than cis isomers due to less steric hindrance.
• C. Both isomers have exactly the same stability.
• D. Stability depends only on the number of carbon atoms.
• Correct Answer: B
• Explanation: Trans isomers are generally more stable.

• Q10. Which of the following molecules can show geometric isomerism?
• A. CH₂=CH-CH₃
• B. CH₃-CH=CH-CH₃
• C. CH₃-CH₂-CH=CH₂
• D. CH₂=CH-CH₂-CH₃
• Correct Answer: B
• Explanation: CH₃-CH=CH-CH₃ (2-butene) can show geometric isomerism.