Question

Which from following compounds is NOT optically active?

• A. 3 - Bromohexane
• B. 2 - Bromo - 2 - methylbutane
• C. 2 - Bromopentane
• D. 2 - Bromo - 3 - methylbutane

Hint:

A compound is optically active only if it has a carbon bonded to four different groups.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A compound is optically active when it is chiral, which usually means it contains a carbon bonded to four different groups (a chiral centre).

Step 2: Key Formula or Approach:
Draw each structure and check every carbon that carries Br or a branch for four different substituents.

Step 3: Check 3-bromohexane:
\(\text{CH}_3\text{CH}_2\text{CHBr}\text{CH}_2\text{CH}_2\text{CH}_3\): C3 carries H, Br, ethyl and propyl. All four are different, so it is chiral and optically active.

Step 4: Check 2-bromo-2-methylbutane:
\((\text{CH}_3)_2\text{CBr}\text{CH}_2\text{CH}_3\): C2 carries Br, two identical methyl groups and an ethyl group. Two groups are the same, so there is no chiral centre and the molecule is optically inactive.

Step 5: Check the other two:
2-Bromopentane: C2 carries H, Br, methyl and propyl, all different, so it is chiral.
2-Bromo-3-methylbutane, \(\text{CH}_3\text{CHBr}\text{CH}(\text{CH}_3)_2\): C2 carries H, Br, methyl and isopropyl, all different, so it is chiral.

Final Answer:
Only 2-bromo-2-methylbutane has no chiral carbon, so it is not optically active. This is option (B). \[ \boxed{\text{2-Bromo-2-methylbutane (B)}} \]