Question

The optically inactive compound from the following is:

• A. 2-Bromopropanal
• B. 3-Bromopropanal
• C. 3-Bromo-2-iodopropanal
• D. 2-Bromo-3-iodopropanal

Hint:

$\textbf{How to identify chiral centers:}$
1. Look for carbons with four different substituents
2. Sp$^3$ hybridized carbons only
3. Terminal CH$_3$ and CH$_2$ groups are never chiral
4. Aldehyde carbons (CHO) are never chiral

Answer 1

The Correct Option is B

A compound is optically inactive if it either: 1. Lacks a chiral center (no carbon with four different substituents), or
2. Exists as a racemic mixture (which isn't relevant for this question)
Analysis of each option: 1. 2-Bromopropanal \(CH_3-C(=O)-CH_2Br\)
- Contains a chiral center at C-2 (attached to -H, -Br, -CHO, -CH$_3$)
- Optically active 2. 3-Bromopropanal \(Br-CH_2-CH_2-CHO\)
- No chiral center (C-1: attached to two H's; C-2: attached to two CH$_2$ groups)
- Optically inactive 3. 3-Bromo-2-iodopropanal \(Br-CH_2-CH(I)-CHO\)
- Chiral center at C-2 (attached to -H, -I, -CHO, -CH$_2$Br)
- Optically active 4. 2-Bromo-3-iodopropanal \(I-CH_2-CH(Br)-CHO\)
- Chiral center at C-2 (attached to -H, -Br, -CHO, -CH$_2$I)
- Optically active Conclusion:
Only 3-Bromopropanal lacks any chiral center and is therefore optically inactive.

Answer 2

Step 1: Understand optical activity.
A compound is optically active if it has one or more chiral centers (carbon atoms with four different groups attached) and lacks a plane of symmetry.
Optically inactive compounds either do not have chiral centers or possess an internal plane of symmetry (meso compounds), making them optically inactive.

Step 2: Analyze 3-Bromopropanal.
3-Bromopropanal has the structure:
\[ \text{Br–CH}_2–CH_2–CHO, \] where the bromine is attached to the terminal carbon, and the aldehyde group is at the other end.
The middle carbon atoms do not have four different groups attached, so there is no chiral center.

Step 3: Conclusion.
Since 3-Bromopropanal lacks a chiral center, it is optically inactive.