Question

$\text{2-chlorobutane} + \text{Cl}_2 \rightarrow \text{C}_4\text{H}_8\text{Cl}_2 \, (\text{isomers})$
$\text{Total number of optically active isomers shown by} \, \text{C}_4\text{H}_8\text{Cl}_2, \, \text{obtained in the above reaction is} \, \_\_\_\_\_\_.$

Answer 1

Correct Answer: 6

When 2-chlorobutane reacts with chlorine (Cl₂) under light (Cl₂/hv), multiple isomers of C₄H₈Cl₂ are formed due to the possible substitution of hydrogen atoms by chlorine at different positions.

To determine the number of optically active isomers, we follow this approach:

The reaction yields several structural isomers of C₄H₈Cl₂ depending on the position where chlorine substitutes. Some of these isomers exhibit chirality due to the presence of chiral centers (carbon atoms bonded to four different groups).

Among the possible isomers, six have chiral centers, making them optically active. These isomers can exist as enantiomers (mirror-image pairs), further contributing to their optical activity.

The total number of optically active isomers formed in this reaction is 6.

Answer 2

The reaction is the chlorination of 2-chlorobutane: \( \text{2-chlorobutane} + \text{Cl}_2 \rightarrow \text{C}_4\text{H}_8\text{Cl}_2 \text{ (isomers)} \). We need to find the total number of optically active isomers among the dichloro products.

Concept Used:

Optical activity arises in molecules that are chiral, meaning they lack an internal plane of symmetry and are not superimposable on their mirror image. A carbon atom is a chiral center if it is attached to four different substituents. In free radical chlorination, chlorine can substitute at different carbon atoms, leading to structural isomers. For each structural isomer, we must identify all chiral centers and check for molecular symmetry (meso forms) to count optically active stereoisomers.

Step-by-Step Solution:

Step 1: Identify the structure of 2-chlorobutane and possible substitution sites.

2-chlorobutane: \( \text{CH}_3-\text{CH}(\text{Cl})-\text{CH}_2-\text{CH}_3 \).
Carbon numbering: C1 (\( \text{CH}_3 \)), C2 (\( \text{CHCl} \)), C3 (\( \text{CH}_2 \)), C4 (\( \text{CH}_3 \)).
Distinct H types for substitution:

• C1 (3H, primary)
• C2 (1H, secondary)
• C3 (2H, primary) – these are equivalent
• C4 (3H, primary) – equivalent to C1

Thus, we have three distinct monochlorination products: substitution at C1, C2, or C3.

 

Step 2: Draw and analyze each possible dichloro structural isomer.

Case 1: 1,2-Dichlorobutane (Cl at C1 and C2)
Structure: \( \text{CH}_2\text{Cl}-\text{CH}(\text{Cl})-\text{CH}_2-\text{CH}_3 \)
Chiral centers: C2 is chiral (four different groups: H, Cl, CH₂Cl, CH₂CH₃).
No plane of symmetry. Optically active: 2 enantiomers.

Case 2: 2,2-Dichlorobutane (Cl at C2 twice)
Structure: \( \text{CH}_3-\text{C}(\text{Cl})_2-\text{CH}_2-\text{CH}_3 \)
C2 has two Cl atoms → not chiral. Molecule has a plane of symmetry. Optically inactive.

Case 3: 2,3-Dichlorobutane (Cl at C2 and C3)
Structure: \( \text{CH}_3-\text{CH}(\text{Cl})-\text{CH}(\text{Cl})-\text{CH}_3 \)
Two chiral centers (C2 and C3).
Stereoisomers:

• (R,R) and (S,S) → meso compound (optically inactive, internal compensation).
• (R,S) and (S,R) → a pair of enantiomers, optically active.

Optically active: 2 enantiomers.

 

Case 4: 1,3-Dichlorobutane (Cl at C1 and C3)
Structure: \( \text{CH}_2\text{Cl}-\text{CH}_2-\text{CH}(\text{Cl})-\text{CH}_3 \)
Chiral centers: C3 is chiral (H, Cl, CH₃, CH₂CH₂Cl). C1 is CH₂Cl, not chiral.
No plane of symmetry. Optically active: 2 enantiomers.

Case 5: 1,1-Dichlorobutane (Cl at C1 twice) – Not possible here since starting material is 2-chlorobutane, not 1-chlorobutane.

Case 6: 3,3-Dichlorobutane (Cl at C3 twice) – Not possible from 2-chlorobutane.

Case 7: 1,4-Dichlorobutane – Not possible from 2-chlorobutane.

Also possible: 1,1-dichloro derivative from C1 – Not possible from 2-chlorobutane.

Wait — we must also consider substitution at C4 (equivalent to C1) giving 1,2 and 1,3 type, but they are already counted as 1,2 and 1,3 above.

Actually, from 2-chlorobutane + Cl₂, the possible structural isomers of C₄H₈Cl₂ are: 1,2- ; 2,2- ; 2,3- ; 1,3- ; and also 1,1- is not possible; but 3,3- is not possible; 1,4- is not possible.

But careful: 1,3-dichlorobutane from 2-chlorobutane? Yes — chlorinate at C1 and C3.

Also: 1,1-dichloro? No, because original Cl is at C2, so to get 1,1-dichloro, need two Cl at C1, but starting material has Cl at C2, so 1,1-dichloro would require replacing all H at C1 with Cl, but starting material is 2-chloro, so 1,1-dichloro-2-chloro? That would be C₄H₆Cl₃, not C₄H₈Cl₂.

So structural isomers: 1,2- ; 1,3- ; 2,2- ; 2,3-.

Let’s check optical activity:

• 1,2-dichlorobutane: 1 chiral center → 2 optically active enantiomers.
• 1,3-dichlorobutane: 1 chiral center (C3) → 2 optically active enantiomers.
• 2,2-dichlorobutane: achiral → 0 optically active.
• 2,3-dichlorobutane: 2 chiral centers → 3 stereoisomers (1 meso + 2 enantiomers) → 2 optically active.

Step 3: Count total optically active isomers.

1,2-dichloro: 2 enantiomers
1,3-dichloro: 2 enantiomers
2,3-dichloro: 2 enantiomers
Total = 2 + 2 + 2 = 6 optically active isomers.

Therefore, the total number of optically active isomers of \( \text{C}_4\text{H}_8\text{Cl}_2 \) obtained is 6.