RBSE Solutions Class 12 Chemistry Chapter 16 Stereo Chemistry

RBSE Class 12 Chemistry Chapter 16 Text Book Questions

RBSE Class 12 Chemistry Chapter 16 Multiple Choice Questions

 

Question 1. Which of the following is not involved in a branch of stereochemistry?
(a) Geometrical isomerism
(b) Conformational isomerism
(c) Functional group isomerism
(d) Optical isomerism
Answer: (c) Functional group isomerism
In simple words: Stereochemistry looks at the 3D shapes of molecules and how they affect properties. Functional group isomerism is about different types of molecules with the same formula but different basic groups, not about their 3D shape.

🎯 Exam Tip: Remember that stereochemistry specifically deals with the spatial arrangement of atoms, while functional group isomerism focuses on different functional groups.

 

Question 2. Which of the following compounds do not show geometrical isomerism?
Answer: (d) None of the options
In simple words: Geometrical isomerism happens when parts of a molecule cannot rotate freely because of a double bond or a ring. If a compound has free rotation, it won't show this type of isomerism. All given options likely show this property.

🎯 Exam Tip: To determine if a compound shows geometrical isomerism, check for restricted rotation around a double bond or in a cyclic structure, and ensure different groups are attached to the same carbon atoms.

 

Question 3. Which of the following statements is not true about optical isomerism?
(a) Axis of symmetry is present in the molecule
(b) Plane of symmetry is present in the molecule
(c) Centre of symmetry is present in the molecule
(d) None of the above
Answer: (d) None of the above
In simple words: Optical isomers are special molecules that can rotate plane-polarized light. This happens because they do not have any symmetry elements like a plane, axis, or center of symmetry, making them non-superimposable on their mirror images.

🎯 Exam Tip: For a molecule to be optically active, it must be chiral, meaning it lacks a plane of symmetry, a center of symmetry, and an alternating axis of symmetry.

 

Question 4. Meso tartaric acid does not show optical activity because
(a) It has two chiral centres
(b) It shows external compensation
(c) It has a plane of symmetry
(d) It has an erythro form
Answer: (c) It has a plane of symmetry
In simple words: Meso tartaric acid has parts that cancel out each other's light-rotating effect because the molecule can be split down the middle into two identical halves. This internal balancing makes it optically inactive, even though it has chiral centers.

🎯 Exam Tip: The key reason for optical inactivity in meso compounds is internal compensation due to a plane of symmetry, even if chiral centers are present.

 

Question 5. Which of the following do not show optical isomerism?
(a) Ethyl Alcohol
(b) 2-Butanol
(c) 2-Chloropropane
(d) Lactic acid
Answer: (a) Ethyl Alcohol
In simple words: Ethyl alcohol doesn't have a carbon atom with four different groups attached, which is needed for optical isomerism. It's a simple molecule that can be perfectly matched with its mirror image.

🎯 Exam Tip: Optical isomerism requires a chiral carbon, which is a carbon atom bonded to four different groups. Ethyl alcohol lacks such a carbon, unlike the other options.

 

Question 6. Which of the following statement is not true for writing Fischer Projection formula?
(a) Two perpendicular lines intersecting each other are drawn
(b) First number carbon is placed on the left side.
(c) Both the groups of horizontal surface are directed above.
(d) Plane of the molecule can be rotated by 180°.
Answer: (b) First number carbon is placed on the left side.
In simple words: When drawing a Fischer projection, the main carbon chain should be drawn vertically. The most oxidized carbon, usually the one with the lowest number in IUPAC naming, goes at the top. The statement saying the first number carbon is placed on the left side is incorrect.

🎯 Exam Tip: In a Fischer projection, the highest priority functional group or the carbon with the lowest number in the main chain is typically placed at the top of the vertical line.

 

Question 7. Which molecule is considered as standard in respect to relative configuration?
(a) Lactic acid
(b) Tartaric acid
(c) Glyceraldehyde
(d) 2-Butanol
Answer: (c) Glyceraldehyde
In simple words: Glyceraldehyde is used as a reference to figure out the 3D arrangements of other similar molecules. It helps chemists compare and name the different forms of molecules.

🎯 Exam Tip: Glyceraldehyde is crucial as a reference for assigning D/L configurations to other chiral compounds, especially sugars, based on the position of its hydroxyl group.

 

Question 8. What is the method used for the resolution of enantiomers?
(a) Biochemical method
(b) Mechanical method
(c) Effective distillation
(d) Column chromatography
Answer: (c) Effective distillation
In simple words: Effective distillation is one way to separate enantiomers, which are mirror-image molecules. This method works by taking advantage of their slightly different boiling points.

🎯 Exam Tip: While distillation can separate compounds with different boiling points, resolution of enantiomers often involves converting them to diastereomers first, which have different physical properties. Mechanical methods (like Pasteur's) or biochemical methods are also used.

 

Question 9. Which of the following statements is not true about conformational isomerism?
(a) These can be represented by Newman and Sawhorse projections.
(b) The number of conformational isomers is infinite.
(c) The eclipsed form is most stable.
(d) Conformational isomerism is also found in the cyclic compound.
Answer: (c) The eclipsed form is most stable.
In simple words: The eclipsed form is actually the least stable arrangement of atoms because the atoms are very close together, causing strain. The staggered form, where atoms are spread out more, is the most stable.

🎯 Exam Tip: Always remember that steric hindrance and torsional strain make eclipsed conformations higher in energy and less stable than staggered conformations.

 

Question 10. Which of the following statements is not true about diastereomers?
(a) They do not show optical activity.
(b) These isomers show the difference in physical properties.
(c) They show internal compensation.
(d) They scatter plane of polarised light.
Answer: (d) They scatter plane of polarised light.
In simple words: Diastereomers are not mirror images of each other. They usually have different physical properties and can be optically active, meaning they can rotate plane-polarized light. The statement that they scatter plane of polarized light is generally true if they are chiral, so it's not the "not true" statement. Option (a) is incorrect, as diastereomers can show optical activity.

🎯 Exam Tip: Diastereomers often possess optical activity, meaning they rotate plane-polarized light, and unlike enantiomers, they have different physical and chemical properties due to their non-mirror-image relationship.

RBSE Class 12 Chemistry Chapter 16 Very Short Answer Type Questions

 

Question 1. Define isomerism.
Answer: Isomerism is when two or more compounds have the exact same chemical formula but different arrangements of atoms. These different arrangements give them different properties. For example, glucose and fructose both have the formula \( \text{C}_{6}\text{H}_{12}\text{O}_{6} \) but are different molecules.
In simple words: Isomerism is about molecules that share the same chemical recipe but look different in their atom arrangement.

🎯 Exam Tip: Clearly state that isomers have the same molecular formula but different structural or spatial arrangements of atoms, leading to different properties.

 

Question 2. What do you understand by stereochemistry?
Answer: Stereochemistry is a part of chemistry that studies molecules in three dimensions. It looks at how the spatial arrangement of atoms affects the molecule's physical and chemical properties. This field helps us understand why molecules with the same formula can behave differently.
In simple words: Stereochemistry is the study of how molecules are shaped in 3D space and how that shape changes what they do.

🎯 Exam Tip: Emphasize that stereochemistry focuses on the three-dimensional structure of molecules and its impact on their characteristics.

 

Question 3. What is an optical activity?
Answer: Optical activity is a special property of some compounds that can rotate the plane of polarized light. Molecules that do this are called optically active compounds. This rotation happens because these molecules are chiral, meaning they cannot be perfectly matched with their mirror image.
In simple words: Optical activity is when a compound makes special light (polarized light) turn as it passes through.

🎯 Exam Tip: Define optical activity as the ability of a compound to rotate the plane of plane-polarized light, and mention that such compounds are called optically active.

 

Question 4. What are the essential conditions for optical isomerism?
Answer: For a compound to show optical isomerism, it must be asymmetric, meaning it cannot be perfectly superimposed on its mirror image. This usually means it must have at least one chiral center, which is a carbon atom bonded to four different groups.
In simple words: A molecule needs to be "lopsided" (asymmetric) and usually have a carbon with four different things attached to it to show optical isomerism.

🎯 Exam Tip: The primary condition for optical isomerism is chirality, which means the molecule is asymmetric and non-superimposable on its mirror image, often due to chiral centers.

 

Question 5. What do you understand by symmetry?
Answer: In chemistry, symmetry refers to certain elements within a molecule that help decide if it is chiral or not. These elements include planes of symmetry, centers of symmetry, or axes of symmetry. If a molecule has a plane of symmetry, it means it can be divided into two identical halves, making it superimposable on its mirror image and thus achiral.
In simple words: Symmetry is like a rule to see if a molecule can be folded in half perfectly or if it has matching parts. If it does, it's not chiral.

🎯 Exam Tip: Connect symmetry elements (plane, center, axis) directly to determining chirality: the presence of any of these elements usually makes a molecule achiral.

 

Question 6. Meso form is optically inactive, Why?
Answer: A meso form is optically inactive because its molecules have an internal plane of symmetry. This plane divides the molecule into two halves that rotate polarized light in equal but opposite directions, causing the overall optical rotation to cancel out internally. This property is known as internal compensation.
In simple words: Meso forms are optically inactive because they have a hidden mirror inside, canceling out any light-turning effect from within the molecule itself.

🎯 Exam Tip: The key explanation for meso compound optical inactivity is "internal compensation" due to a plane of symmetry, even if chiral centers are present.

 

Question 7. What is racemisation?
Answer: Racemization is the process where an optically active compound loses its ability to rotate plane-polarized light because it changes into a racemic mixture. A racemic mixture contains equal amounts of two enantiomers (mirror-image isomers) that rotate light in opposite directions, thus canceling out each other's optical activity. This mixing results in zero net rotation.
In simple words: Racemization is when a light-turning chemical turns into a balanced mix of its mirror forms, so it stops turning the light.

🎯 Exam Tip: Define racemization as the conversion of an optically active compound into an optically inactive racemic mixture, emphasizing the equal amounts of enantiomers.

 

Question 1. Define optical activity. Explain with the example which type of molecules show optical activity.
Answer: Optical activity is the property of some substances to rotate the plane of plane-polarized light. Compounds that exhibit this property are called optically active. This phenomenon occurs in molecules that are chiral, meaning they are non-superimposable on their mirror images and lack elements of symmetry like a plane or center of symmetry. An example is 2-bromobutane:
\( \text{CH}_{3}\text{-CH}_{2}\text{-}\overset{\ast}{\text{CH}}\text{-CH}_{3} \)
\( \text{Br} \)
This molecule contains one chiral center (marked with an asterisk) and is therefore optically active. Lactic acid \( \text{(CH}_{3}\text{-}\overset{\ast}{\text{CH}}\text{-COOH}\text{)} \) is another example of a compound with a chiral center that shows optical activity.
In simple words: Optical activity means a molecule can twist special light. This happens if the molecule's shape is "lopsided" (chiral), like 2-bromobutane, which has a carbon atom with four different groups attached.

🎯 Exam Tip: Clearly define optical activity and provide a simple example of a chiral molecule, pointing out the chiral center as the reason for its optical activity.

 

Question 2. 1-Butanol does not show optical isomerism whereas 2-Butanol shows, Why?
Answer: 1-Butanol does not show optical isomerism because it lacks a chiral carbon atom. A chiral carbon atom must be bonded to four different groups. In 1-Butanol \( \text{(CH}_{3}\text{-CH}_{2}\text{-CH}_{2}\text{-CH}_{2}\text{-OH}\text{)} \), all carbon atoms either have two identical hydrogen atoms or are part of a symmetrical chain. However, 2-Butanol \( \text{(CH}_{3}\text{-CH}_{2}\text{-}\overset{\ast}{\text{CH}}\text{(OH)}\text{-CH}_{3}\text{)} \) has a chiral carbon at the second position (marked with an asterisk). This carbon is bonded to a methyl group \( \text{(-CH}_{3}\text{)} \), an ethyl group \( \text{(-CH}_{2}\text{CH}_{3}\text{)} \), a hydroxyl group \( \text{(-OH)} \), and a hydrogen atom \( \text{(-H)} \). Since these four groups are different, 2-butanol is a chiral molecule and thus exhibits optical isomerism.
In simple words: 1-Butanol doesn't have a special carbon with four different arms, so it's not "lopsided." But 2-Butanol does, making it a "lopsided" molecule that can turn light.

🎯 Exam Tip: When comparing molecules for optical isomerism, always identify the presence or absence of a chiral carbon (bonded to four unique substituents).

 

Question 3. What are the necessary conditions for optical activity? Give an example of optically active molecules.
Answer: Necessary conditions for optical activity are:

• The compound must contain an asymmetrical carbon atom (chiral center).
• The molecule must be chiral, meaning it is non-superimposable on its mirror image. This means it must not possess any elements of symmetry like a plane of symmetry, a center of symmetry, or an alternating axis of symmetry.
• The molecule must possess a chiral axis (for some specific types of molecules).
• The molecule must possess a chiral plane (for some specific types of molecules).

An example of an optically active molecule is Lactic Acid \( \text{(CH}_{3}\text{-}\overset{\ast}{\text{CH}}\text{(OH)}\text{-COOH}\text{)} \), which has one chiral carbon atom.
In simple words: For a molecule to be optically active, it generally needs a "lopsided" carbon (a chiral center) and its mirror image shouldn't fit exactly on top of it. Lactic acid is a good example.

🎯 Exam Tip: Focus on the most common condition: the presence of a chiral carbon and the absence of any symmetry elements that would make the molecule superimposable on its mirror image.

 

Question 4. Interpret two optical isomers of Lactic acid?
Answer: Lactic acid has one asymmetric carbon atom, which means it has two optical isomers, also known as enantiomers. These are D-lactic acid and L-lactic acid. D-lactic acid rotates the plane of plane-polarized light to the right (clockwise), which is called dextrorotatory (+). L-lactic acid rotates the plane of plane-polarized light to the left (anticlockwise), which is called levorotatory (-). The structures are:
\[ \begin{array}{cc} \text{COOH} & \text{COOH} \\ | & | \\ \text{C} & \text{C} \\ \text{OH H CH}_{3} & \text{CH}_{3} \text{ H OH} \\ d\text{-Lactic acid} & l\text{-Lactic acid} \end{array} \] These two forms are mirror images of each other and cannot be superimposed, demonstrating optical isomerism.
In simple words: Lactic acid has two mirror-image forms. One (D-lactic acid) turns light right, and the other (L-lactic acid) turns light left.

🎯 Exam Tip: When describing optical isomers like those of lactic acid, always mention their ability to rotate plane-polarized light in opposite directions (dextro and levo forms).

 

Question 5. Describe racemisation giving suitable examples.
Answer: Racemisation is the process where an optically active compound converts into an optically inactive racemic mixture. This mixture contains equal amounts of both dextrorotatory and laevorotatory isomers. Because these isomers rotate plane-polarized light in opposite directions by the same amount, their optical activities cancel each other out, making the overall mixture optically inactive. This is a common method for preparing specific enantiomers.
Racemisation can occur through various methods:
1. **By simple heating (Thermal Racemisation):** Some optically active compounds, like tartaric acid or lactic acid, can undergo racemisation when heated. The energy from heating allows the chiral molecules to interconvert between their enantiomeric forms, leading to a racemic mixture.
2. **Auto racemisation:** This is when racemisation happens spontaneously at room temperature without any external heating. For instance, dimethyl succinate can undergo auto racemisation over time. In these cases, the conditions allow for the gradual conversion and establishment of a racemic equilibrium.
In simple words: Racemisation is when a compound that can turn light becomes a balanced mix of its left- and right-turning forms, so it stops turning light. This can happen from heating or even by itself over time.

🎯 Exam Tip: Define racemisation and provide at least two distinct methods (e.g., thermal, auto) with examples to illustrate the process clearly.

 

Question 6. Explain erythro and threo isomers giving suitable examples.
Answer: Erythro and threo isomers are types of diastereomers that differ in the relative configurations of two adjacent chiral centers in an acyclic molecule.
* **Erythro isomer:** When two similar groups (or identical substituents) on adjacent chiral carbons are on the *same side* of the vertical line in a Fischer projection, the isomer is called erythro.
* **Threo isomer:** When two similar groups (or identical substituents) on adjacent chiral carbons are on *opposite sides* of the vertical line in a Fischer projection, the isomer is called threo.
This naming convention comes from the sugars erythrose and threose.
For example, consider 2,3-dibromobutane:
Erythro forms (similar groups on the same side):
\[ \begin{array}{cc} \text{CHO} & \text{CHO} \\ | & | \\ \text{H--OH} & \text{OH--H} \\ | & | \\ \text{H--OH} & \text{OH--H} \\ | & | \\ \text{CH}_2\text{OH} & \text{CH}_2\text{OH} \\ \text{(-)-Erythro} & \text{(+)-Erythro} \end{array} \]
Threo forms (similar groups on opposite sides):
\[ \begin{array}{cc} \text{CHO} & \text{CHO} \\ | & | \\ \text{HO--H} & \text{H--OH} \\ | & | \\ \text{H--OH} & \text{HO--H} \\ | & | \\ \text{CH}_2\text{OH} & \text{CH}_2\text{OH} \\ \text{(-)-Threo} & \text{(+)-Threo} \end{array} \] Erythro and threo isomers are important for differentiating between diastereomers with multiple chiral centers.
In simple words: Erythro and threo isomers describe how two neighboring "lopsided" parts of a molecule are arranged. If matching groups are on the same side, it's erythro; if they're on opposite sides, it's threo.

🎯 Exam Tip: Clearly distinguish erythro and threo based on the relative positions of similar groups on adjacent chiral carbons in Fischer projections (same side for erythro, opposite for threo).

 

Question 8. What are the differences between conformation and configuration?
Answer: The key differences between conformation and configuration are:

The main distinction is the energy barrier for interconversion; conformations can interconvert easily, while configurations require significant energy.
In simple words: Conformation is about different shapes a molecule can take by simply twisting its bonds, which is easy to do. Configuration is about the fixed arrangement of atoms that needs bonds to break and reform to change.

🎯 Exam Tip: The critical distinction is the interconversion barrier: conformations interconvert easily by bond rotation, while configurations require bond breaking/making.

 

Question 9. What do you understand by geometrical isomerism? Write essential conditions for geometrical isomerism.
Answer: **Geometrical Isomerism:** This type of isomerism occurs when a molecule has restricted rotation around a bond, often a double bond or within a cyclic structure. Because of this restricted rotation, atoms or groups attached to the carbons involved in the restricted bond can be arranged in different fixed positions relative to each other, leading to distinct isomers called cis-trans isomers.
**Conditions for Geometrical Isomerism:** For a compound to show geometrical isomerism, it must satisfy two main conditions:
1. **Restricted Rotation:** There must be restricted rotation around a bond in the molecule. This typically occurs in compounds with carbon-carbon double bonds (alkenes) or in cyclic compounds.
2. **Different Groups on Each Atom:** Each carbon atom involved in the restricted rotation must be bonded to two *different* atoms or groups. For example, in an alkene, if the two groups on one carbon are identical, geometrical isomerism will not be observed.
This isomerism is crucial in determining the properties and reactions of many organic molecules.
In simple words: Geometrical isomerism happens when a molecule can't twist freely around a bond, usually a double bond or in a ring. For this to happen, each end of that bond must have two different parts attached.

🎯 Exam Tip: Emphasize the two critical conditions: restricted rotation (double bond or ring) AND different groups attached to each carbon of the restricted bond.

 

Question 10. Explain the stereochemistry of oximes.
Answer: Oximes are compounds formed when aldehydes or ketones react with hydroxylamines. They have a carbon-nitrogen double bond (\( \text{C=N} \)). Due to the nature of this double bond, there is restricted rotation between the carbon and nitrogen atoms. This restricted rotation gives rise to geometrical isomerism in oximes, leading to 'syn' and 'anti' isomers.
For an oxime \( \text{R}_{1}\text{R}_{2}\text{C=N-OH} \):
* **Syn isomer:** If the hydroxyl group \( \text{(-OH)} \) on the nitrogen atom is on the *same side* as the higher priority group (or hydrogen) attached to the carbon atom of the \( \text{C=N} \) double bond, it's called the syn isomer.
* **Anti isomer:** If the hydroxyl group \( \text{(-OH)} \) on the nitrogen atom is on the *opposite side* as the higher priority group (or hydrogen) attached to the carbon atom, it's called the anti isomer.
Example: Acetaldoxime \( \text{(CH}_{3}\text{CH = NOH)} \)
\[ \begin{array}{cc} \text{CH}_{3}\text{ H} & \text{CH}_{3}\text{ H} \\ \parallel & \parallel \\ \text{C=N} & \text{N=C} \\ \text{OH} & \text{(HO)} \\ \text{Syn} & \text{Anti} \end{array} \]
Benzaldoxime \( \text{(C}_{6}\text{H}_{5}\text{CH = NOH)} \)
\[ \begin{array}{cc} \text{C}_{6}\text{H}_{5}\text{ H} & \text{C}_{6}\text{H}_{5}\text{ H} \\ \parallel & \parallel \\ \text{C=N} & \text{N=C} \\ \text{OH} & \text{(HO)} \\ \text{Syn} & \text{Anti} \end{array} \] The sp² hybridization of both carbon and nitrogen in the \( \text{C=N} \) bond is responsible for the planarity and restricted rotation that enables these isomers.
In simple words: Oximes have a carbon-nitrogen double bond that doesn't allow free twisting. This creates two forms, 'syn' and 'anti', depending on whether the \( \text{OH} \) group is on the same side or opposite side as a specific group on the carbon.

🎯 Exam Tip: For oximes, the crucial point is the restricted rotation around the \( \text{C=N} \) double bond, which leads to 'syn' and 'anti' isomers based on the relative positions of the \( \text{-OH} \) group and the carbon substituents.

 

Question 11. Draw the Newman projection formula for different conformations of ethane.
Answer: **Conformations:** Conformations are different spatial arrangements of atoms in a molecule that can be easily changed into one another by rotating around single bonds. These different arrangements are called conformers and are generally not separable because they interconvert rapidly at room temperature.
**Newman Projection Formula for Ethane:** Ethane \( \text{(CH}_{3}\text{-CH}_{3}\text{)} \) has two main conformations: staggered and eclipsed. These can be visualized using Newman projections, which look down a carbon-carbon bond.
* **Staggered Conformation:** In this conformation, the hydrogen atoms on the front carbon are as far as possible from the hydrogen atoms on the back carbon. This arrangement minimizes repulsion, making it the most stable form.
* **Eclipsed Conformation:** In this conformation, the hydrogen atoms on the front carbon directly overlap or "eclipse" the hydrogen atoms on the back carbon. This maximizes repulsion, making it the least stable form.
The interconversion between these forms happens through a 60° rotation around the carbon-carbon single bond.
Newman projections of ethane:
\[ \begin{array}{cc} \text{H} & \text{H} \\ | & | \\ \text{H--C--H} & \text{H--C--H} \\ | & | \\ \text{H} & \text{H} \\ \text{Staggered conformation} & \text{Eclipsed conformation} \\ \text{of ethane} & \text{of ethane} \end{array} \] (This would be represented visually as a circle with bonds, for example:
Staggered: Front carbon with three H's spread out, back carbon with three H's spread out in between the front ones.
Eclipsed: Front carbon with three H's, back carbon with three H's directly behind them. Given the instruction, a textual representation is provided above, reflecting a simplified chemical drawing.)
The staggered conformation is more stable than the eclipsed conformation due to lower torsional strain.
In simple words: Ethane can twist its atoms into different shapes. The "staggered" shape, where atoms are spread out, is more stable. The "eclipsed" shape, where atoms line up behind each other, is less stable. Newman projections are like looking straight down the carbon-carbon bond to see these shapes.

🎯 Exam Tip: When drawing Newman projections, clearly label the front and back carbons and accurately depict the spatial relationships of the substituents for both staggered and eclipsed forms, noting their relative stabilities.

 

Question 12. Describe the salt formation method for the separation of enantiomers.
Answer: The salt formation method is a common chemical technique used to separate a racemic mixture (a 50:50 mixture of two enantiomers) into its individual enantiomers. This process is called resolution.
**Chemical method (Salt Formation Method):**
1. **Formation of Diastereomeric Salts:** A racemic mixture of an enantiomeric organic acid (e.g., \( \text{HA} \), which includes both \( \text{d-HA} \) and \( \text{l-HA} \)) is reacted with an optically active base (e.g., \( \text{l-B} \)). The reaction forms two diastereomeric salts: \( \text{d-HA} \cdot \text{l-B} \) and \( \text{l-HA} \cdot \text{l-B} \).
\[ \text{d-HA} + \text{l-B} \rightarrow \text{d-HA} \cdot \text{l-B} \] \[ \text{l-HA} + \text{l-B} \rightarrow \text{l-HA} \cdot \text{l-B} \]
2. **Separation of Diastereomeric Salts:** Unlike enantiomers, diastereomeric salts have different physical properties (such as melting points, boiling points, and crucially, solubilities in a given solvent). Because of these different properties, they can be separated using simple techniques like fractional crystallization. In this process, one diastereomeric salt will crystallize out first, leaving the other in solution.
3. **Regeneration of Enantiomers:** After separation, each diastereomeric salt is treated with an acid (if a base was used) or a base (if an acid was used) to regenerate the pure enantiomeric acid and the original optically active resolving agent (base).
\[ \text{d-HA} \cdot \text{l-B} + \text{Acid} \rightarrow \text{d-HA (pure)} + \text{l-B} \] \[ \text{l-HA} \cdot \text{l-B} + \text{Acid} \rightarrow \text{l-HA (pure)} + \text{l-B} \] This method is effective because it temporarily converts difficult-to-separate enantiomers into easier-to-separate diastereomers.
In simple words: To separate mirror-image molecules (enantiomers), we turn them into "not-quite-mirror-image" molecules (diastereomers) by reacting them with a special, already pure molecule. These "not-quite-mirror-image" molecules are easier to separate, for example, by crystallization. After separating, we can get back the original pure mirror-image molecules.

🎯 Exam Tip: Explain that the salt formation method works by converting enantiomers into diastereomers, which have different physical properties (especially solubility), allowing for their separation, followed by regeneration of the pure enantiomers.

RBSE Class 12 Chemistry Chapter 16 Long Answer Type Questions

 

Question 1. Discuss optical isomerism in tartaric acid.
Answer: Tartaric acid \( \text{(HOOC-CH(OH)-CH(OH)-COOH)} \) is an excellent example to illustrate optical isomerism because it has two asymmetric carbon atoms. According to the \( 2^n \) formula (where \( n \) is the number of chiral centers), there should be \( 2^2 = 4 \) stereoisomers. However, due to its symmetry, tartaric acid exists in only three distinct stereoisomeric forms:
1. **D-(+)-Tartaric acid:** This isomer is dextrorotatory, meaning it rotates plane-polarized light to the right (clockwise).
2. **L-(-)-Tartaric acid:** This isomer is levorotatory, meaning it rotates plane-polarized light to the left (anticlockwise).
3. **Meso-Tartaric acid:** This form is optically inactive. Despite having two chiral centers, it possesses an internal plane of symmetry. This symmetry causes the optical rotation of one half of the molecule to cancel out the rotation of the other half, leading to internal compensation.
The D-(+)-tartaric acid and L-(-)-tartaric acid are enantiomers (mirror images of each other). The meso-tartaric acid is a diastereomer to both D and L forms.
These forms can be represented by Fischer projections:
\[ \begin{array}{ccc} \text{COOH} & \text{COOH} & \text{COOH} \\ | & | & | \\ \text{H--OH} & \text{HO--H} & \text{H--OH} \\ | & | & | \\ \text{HO--H} & \text{H--OH} & \text{H--OH} \\ | & | & | \\ \text{COOH} & \text{COOH} & \text{COOH} \\ \text{(I) D-(+)-Tartaric Acid} & \text{(II) L-(-)-Tartaric Acid} & \text{(III) Meso-Tartaric Acid} \end{array} \] Here, (I) and (II) are enantiomers, while (III) is the meso compound. A racemic mixture of D- and L-tartaric acid is also possible, which is optically inactive due to external compensation. Tartaric acid is a classical example used to understand the relationship between molecular structure and optical activity.
In simple words: Tartaric acid is a special molecule with two "lopsided" carbons. It has two mirror-image forms that twist light in opposite ways, plus a third form called "meso" that doesn't twist light at all because it has a hidden mirror inside itself.

🎯 Exam Tip: When discussing tartaric acid, highlight the two chiral centers, the three possible stereoisomers (D, L, and meso), and explicitly explain why the meso form is optically inactive (internal compensation due to a plane of symmetry).

 

Question 2. Explain diastereoisomerism with an example. What are the characteristics of diastereoisomers?
Answer: **Diastereoisomerism:** Diastereoisomers are stereoisomers that are not mirror images of each other. They occur in compounds that have two or more chiral centers but are not enantiomers. This means that while they have the same molecular formula and sequence of bonded atoms, their atoms are arranged differently in three-dimensional space, and they cannot be superimposed.
**Characteristics of Diastereoisomers:**
1. **Non-Mirror Images:** Diastereoisomers are not mirror images of each other.
2. **Multiple Chiral Centers:** Diastereoisomerism typically occurs in compounds with at least two chiral carbon centers.
3. **Different Physical Properties:** Unlike enantiomers, which have identical physical properties (except for optical rotation), diastereoisomers have different physical properties such as melting points, boiling points, densities, solubilities, and specific rotations.
4. **Separation:** Because of their different physical properties, diastereoisomers can be separated by common physical methods like fractional distillation, fractional crystallization, and column chromatography.
5. **Different Chemical Properties:** They also exhibit different chemical properties, although often similar, making their reactions distinct.
**Example: Cinnamic acid dibromide**
Cinnamic acid dibromide \( \text{(C}_{6}\text{H}_{5}\text{-CH(Br)-CH(Br)-COOH)} \) has two chiral centers. It can exist in four stereoisomeric forms. If we consider the (R,R) and (S,S) forms as enantiomers, then the (R,S) form (meso compound) would be a diastereomer to both of them.
Let's consider the four forms of 2,3-dihydroxybutanoic acid (from another example often used for diastereomers):
\[ \begin{array}{cc} \text{CH}_{3} & \text{CH}_{3} \\ | & | \\ \text{H--OH} & \text{HO--H} \\ | & | \\ \text{H--OH} & \text{HO--H} \\ | & | \\ \text{COOH} & \text{COOH} \\ \text{(I)} & \text{(II)} \end{array} \] (I) and (II) are enantiomers.
\[ \begin{array}{cc} \text{CH}_{3} & \text{CH}_{3} \\ | & | \\ \text{H--OH} & \text{HO--H} \\ | & | \\ \text{HO--H} & \text{H--OH} \\ | & | \\ \text{COOH} & \text{COOH} \\ \text{(III)} & \text{(IV)} \end{array} \] (III) and (IV) are also enantiomers.
However, (I) and (III) are diastereomers, as are (I) and (IV), (II) and (III), and (II) and (IV). They are stereoisomers but not mirror images of each other. Diastereomers are very useful in resolution processes.
In simple words: Diastereoisomers are special molecules that have the same formula but different 3D shapes, and they are NOT mirror images of each other. They have different melting points, boiling points, and other properties, so they can be easily separated.

🎯 Exam Tip: The key points for diastereoisomers are that they are stereoisomers but not mirror images, possess different physical properties, and can be separated by conventional methods.

 

Question 3. What is stereoisomerism? Discuss stereoisomerism in tartaric acid and write the structural formula of all its isomers. Explain when tartaric acid is synthesised then it is optically inactive.
Answer: **Stereoisomerism:** Stereoisomerism is a type of isomerism where compounds have the same molecular formula and the same sequence of bonded atoms but differ in the three-dimensional orientation of their atoms in space. These differences in spatial arrangement can lead to distinct physical and chemical properties.
**Stereoisomerism in Tartaric Acid:**
Tartaric acid \( \text{(HOOC-CH(OH)-CH(OH)-COOH)} \) has two asymmetric (chiral) carbon atoms. Based on the \( 2^n \) rule (where \( n \) is the number of chiral centers), we might expect \( 2^2 = 4 \) stereoisomers. However, due to the presence of an internal plane of symmetry in one of its forms, tartaric acid exists in only three stereoisomeric forms:
1. **D-(+)-Tartaric acid:** Optically active, rotates plane-polarized light to the right.
2. **L-(-)-Tartaric acid:** Optically active, rotates plane-polarized light to the left.
3. **Meso-Tartaric acid:** Optically inactive due to internal compensation.
**Structural Formulas of Tartaric Acid Isomers (Fischer Projections):**
\[ \begin{array}{ccc} \text{COOH} & \text{COOH} & \text{COOH} \\ | & | & | \\ \text{H--OH} & \text{HO--H} & \text{H--OH} \\ | & | & | \\ \text{HO--H} & \text{H--OH} & \text{H--OH} \\ | & | & | \\ \text{COOH} & \text{COOH} & \text{COOH} \\ \text{(I) D-(+)-Tartaric acid} & \text{(II) L-(-)-Tartaric acid} & \text{(III) Meso-Tartaric acid} \end{array} \] (I) and (II) are enantiomers (mirror images). (III) is a meso compound, which is achiral.
**Why Tartaric Acid is Optically Inactive when Synthesized:**
When tartaric acid is synthesized in a laboratory from achiral starting materials, the process typically produces equal amounts of D-(+)-tartaric acid and L-(-)-tartaric acid. This 50:50 mixture of enantiomers is known as a **racemic mixture**. A racemic mixture is optically inactive because the dextrorotation caused by one enantiomer is precisely canceled out by the levorotation caused by the other enantiomer. This phenomenon is called **external compensation**. Since both enantiomers are produced in equal proportions, the net optical rotation observed for the synthesized product is zero, making it optically inactive.
In simple words: Stereoisomerism means molecules have the same parts but different 3D arrangements. Tartaric acid has three main forms: two that are mirror images (D and L, which twist light) and one "meso" form that doesn't twist light because of internal balance. When you make tartaric acid in a lab, you get an equal mix of the D and L forms, which cancel each other out, making the whole batch optically inactive.

🎯 Exam Tip: For tartaric acid, clearly state the definition of stereoisomerism, draw the three isomers, and explain why a synthetic (racemic) mixture is optically inactive (external compensation of D and L forms).

 

Question 4. What do you understand by optical isomerism? Discuss optical isomerism in compounds having two asymmetric carbon atoms.
Answer: **Optical Isomerism:** Optical isomerism is a type of stereoisomerism where compounds have the same molecular formula and sequence of bonded atoms but differ in their ability to rotate the plane of plane-polarized light. Such compounds are called optical isomers or enantiomers, and they are non-superimposable mirror images of each other.
**Optical Isomerism in Compounds with Two Asymmetric Carbon Atoms:**
When a compound has two asymmetric carbon atoms, the number of possible stereoisomers is \( 2^n \), where \( n \) is the number of chiral centers. So, for two chiral centers, there can be up to \( 2^2 = 4 \) stereoisomers.
These four isomers can be categorized as:
1. **Pair of Enantiomers:** Two isomers that are non-superimposable mirror images of each other. They will have equal but opposite specific rotations.
2. **Diastereomers:** The other isomers will be diastereomers to the first pair. Diastereomers are stereoisomers that are not mirror images of each other. They have different physical and chemical properties.
**Case 1: Both asymmetric carbons are identical** (e.g., Tartaric Acid)
In this case, despite having two chiral centers, there are only three unique stereoisomers:
* D-(+)-form
* L-(-)-form
* Meso form (optically inactive due to internal plane of symmetry).
The D and L forms are enantiomers. The meso form is a diastereomer to both D and L forms.
**Case 2: Both asymmetric carbons are non-identical** (e.g., 2,3-dihydroxybutanoic acid)
In this case, all four possible stereoisomers exist and are distinct:
* (R,R) form
* (S,S) form (enantiomer of R,R)
* (R,S) form
* (S,R) form (enantiomer of R,S)
Here, (R,R) and (S,S) are enantiomers. (R,S) and (S,R) are also enantiomers. However, (R,R) is a diastereomer to (R,S) and (S,R).
The existence of multiple chiral centers increases the complexity of optical isomerism and leads to a wider range of stereoisomers.
**Example Structures (for 2,3-dihydroxybutanoic acid):**
\[ \begin{array}{cc} \text{COOH} & \text{COOH} \\ | & | \\ \text{H--Cl} & \text{Cl--H} \\ | & | \\ \text{Cl--H} & \text{H--Cl} \\ | & | \\ \text{CH}_{3} & \text{CH}_{3} \\ \text{(I) (2R,3R)} & \text{(II) (2S,3S)} \end{array} \] (I) and (II) are enantiomers.
\[ \begin{array}{cc} \text{COOH} & \text{COOH} \\ | & | \\ \text{H--Cl} & \text{Cl--H} \\ | & | \\ \text{H--Cl} & \text{Cl--H} \\ | & | \\ \text{CH}_{3} & \text{CH}_{3} \\ \text{(III) (2R,3S)} & \text{(IV) (2S,3R)} \end{array} \] (III) and (IV) are also enantiomers.
Structures (I) and (III) are diastereomers, as are (I) and (IV), (II) and (III), and (II) and (IV).
In simple words: Optical isomerism is when molecules twist light because they are mirror images that can't be stacked perfectly. If a molecule has two "lopsided" (asymmetric) carbons, it can have up to four different forms. Sometimes, like with tartaric acid, one form has an internal mirror and doesn't twist light. Other times, all four forms are unique.

🎯 Exam Tip: Define optical isomerism and then clearly differentiate between compounds with identical vs. non-identical chiral centers when discussing two asymmetric carbons, including the number and types of isomers for each case.

 

Question 5. Explain stereoisomerism in tartaric acid. How many optical isomers are possible for tartaric acid? What are the differences between meso tartaric acid and racemic mixture?
Answer: **Stereoisomerism in Tartaric Acid:**
Tartaric acid \( \text{(HOOC-CH(OH)-CH(OH)-COOH)} \) features two asymmetric carbon atoms. This allows for stereoisomerism, meaning different spatial arrangements of its atoms.
**Number of Optical Isomers for Tartaric Acid:**
Although it has two chiral centers, the actual number of optically active isomers is less than the theoretical maximum of \( 2^2 = 4 \). Tartaric acid has three stereoisomers:
1. **D-(+)-Tartaric acid:** This form rotates plane-polarized light clockwise.
2. **L-(-)-Tartaric acid:** This form rotates plane-polarized light anti-clockwise.
3. **Meso-Tartaric acid:** This form is optically inactive. It possesses an internal plane of symmetry, which cancels out the optical activity from its two chiral centers through internal compensation.
The D and L forms are enantiomers. The meso form is a diastereomer to both the D and L forms.
**Structural Formulas:**
\[ \begin{array}{ccc} \text{COOH} & \text{COOH} & \text{COOH} \\ | & | & | \\ \text{H--OH} & \text{HO--H} & \text{H--OH} \\ | & | & | \\ \text{HO--H} & \text{H--OH} & \text{H--OH} \\ | & | & | \\ \text{COOH} & \text{COOH} & \text{COOH} \\ \text{(I) D-(+)-Tartaric acid} & \text{(II) L-(-)-Tartaric acid} & \text{(III) Meso-Tartaric acid} \end{array} \] **Differences between Meso Tartaric Acid and Racemic Mixture:**
While both meso tartaric acid and a racemic mixture of tartaric acid are optically inactive, the reason for their inactivity is different.

These distinctions are crucial for understanding the behavior of molecules with multiple chiral centers.
In simple words: Tartaric acid has three main forms: two mirror-image forms that twist light, and one "meso" form that doesn't twist light because it's balanced inside itself. A "racemic mixture" also doesn't twist light, but that's because it's an equal mix of the two mirror-image forms, which cancel each other out. They are different in how they become optically inactive.

🎯 Exam Tip: When comparing meso compounds and racemic mixtures, highlight that both are optically inactive, but meso compounds achieve this through internal compensation (molecular symmetry), while racemic mixtures do so via external compensation (equal mixture of enantiomers).

 

Question 6. The physical properties of geometrical isomers are different whereas physical properties of optical isomers are the same. Explain with reason.
Answer: The physical properties of geometrical isomers are generally different because their distinct spatial arrangements lead to differences in molecular polarity and symmetry. For example, cis-1,2-dichloroethene is polar, while trans-1,2-dichloroethene is non-polar due to its symmetrical structure. These differences affect the intermolecular forces and how molecules pack in a solid state, leading to varied boiling and melting points. These distinct arrangements also influence their reactivity in some chemical reactions. In contrast, enantiomers, which are a type of optical isomer, have identical physical properties such as melting point, boiling point, and solubility in achiral solvents. Their only difference is the direction in which they rotate plane-polarized light. However, other forms related to optical activity, such as meso compounds and racemic mixtures, can have different physical properties. For instance, a meso compound is a single molecule that is optically inactive due to internal symmetry, while a racemic mixture is an equal blend of two enantiomers that is also optically inactive due to external compensation.

In simple words: Geometrical isomers have different physical properties because their shapes affect how they interact and pack. Optical isomers that are mirror images (enantiomers) have almost identical physical properties, only differing in how they bend light. However, other special cases like meso compounds and racemic mixtures, while related to optical activity, can have different physical properties. Understanding these differences helps scientists separate and identify various organic compounds.

🎯 Exam Tip: When discussing isomers, always distinguish between enantiomers, diastereomers, meso compounds, and racemic mixtures, as their physical property relationships vary. Be sure to provide specific examples like cis/trans for geometrical isomers.