Question

Arrange the following compounds in increasing order of their reactivity towards \( S_N2 \) displacement: 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane.

Hint:

For \( S_N2 \) reactions, remember that less steric hindrance at the reactive center increases the reaction rate. Primary halides are typically more reactive than secondary and tertiary halides.

Answer 1

To solve the problem, we need to arrange the compounds 2-Bromo-2-methylbutane, 1-Bromopentane, and 2-Bromopentane in increasing order of their reactivity towards Sₓ2 displacement.

1. Understand Sₓ2 Reaction Mechanism:
Sₓ2 reactions involve a backside attack by the nucleophile, requiring access to the carbon attached to the leaving group (Br⁻). Reactivity decreases with increasing steric hindrance: primary (1°) > secondary (2°) > tertiary (3°).

2. Classify the Compounds:
- 2-Bromo-2-methylbutane: The carbon with Br is bonded to three other carbons (two methyl groups and one ethyl group), making it a tertiary (3°) halide.
- 1-Bromopentane: The carbon with Br is bonded to one other carbon, making it a primary (1°) halide.
- 2-Bromopentane: The carbon with Br is bonded to two other carbons, making it a secondary (2°) halide.

3. Assess Steric Hindrance:
- Tertiary (2-Bromo-2-methylbutane): High steric hindrance due to three alkyl groups, least reactive in Sₓ2.
- Secondary (2-Bromopentane): Moderate steric hindrance, more reactive than tertiary but less than primary.
- Primary (1-Bromopentane): Low steric hindrance, most reactive in Sₓ2.

4. Arrange in Increasing Order:
Reactivity increases as steric hindrance decreases: tertiary < secondary < primary.

Final Answer:
The compounds in increasing order of reactivity towards Sₓ2 displacement are: 2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane.