Arrange the following compounds in increasing order of their reactivity towards \( S_N2 \) displacement: 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane.
To address the problem, we must order the compounds 2-Bromo-2-methylbutane, 1-Bromopentane, and 2-Bromopentane according to their increasing susceptibility to Sₓ2 displacement.
1. Sₓ2 Mechanism Fundamentals:
Sₓ2 reactions are characterized by a nucleophile's backside attack on the carbon atom bearing the leaving group (Br⁻). The rate of these reactions is inversely proportional to steric hindrance around this carbon. The general trend for reactivity is: primary (1°) > secondary (2°) > tertiary (3°).
2. Compound Classification:
- 2-Bromo-2-methylbutane: The carbon atom attached to bromine is bonded to three other carbon atoms (two methyl groups and one ethyl group), classifying it as a tertiary (3°) halide.
- 1-Bromopentane: The carbon atom attached to bromine is bonded to only one other carbon atom, classifying it as a primary (1°) halide.
- 2-Bromopentane: The carbon atom attached to bromine is bonded to two other carbon atoms, classifying it as a secondary (2°) halide.
3. Steric Hindrance Evaluation:
- Tertiary (2-Bromo-2-methylbutane): Exhibits significant steric hindrance due to the presence of three alkyl groups, rendering it the least reactive in Sₓ2 reactions.
- Secondary (2-Bromopentane): Presents moderate steric hindrance, making it more reactive than tertiary halides but less reactive than primary ones.
- Primary (1-Bromopentane): Possesses minimal steric hindrance, leading to the highest reactivity in Sₓ2 reactions.
4. Ordering by Reactivity:
Reactivity escalates as steric hindrance diminishes: tertiary < secondary < primary.
Final Arrangement:
The compounds, ordered by increasing reactivity towards Sₓ2 displacement, are: 2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane.
Match Column A with Column B: 