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

Step 1: Understanding \( S_N2 \) Mechanism. The \( S_N2 \) mechanism involves a single step where the nucleophile attacks the electrophilic carbon, displacing the leaving group, all happening simultaneously. This mechanism is highly sensitive to steric hindrance; thus, less substituted carbons are more reactive due to easier accessibility for the nucleophile.
2-Bromo-2-methylbutane: This is a tertiary bromide and thus highly hindered, making \( S_N2 \) reactions extremely slow.
2-Bromopentane: As a secondary bromide, it is less hindered than the tertiary but more so than the primary, placing it in the middle for reactivity.
1-Bromopentane: This is a primary bromide with the least steric hindrance, facilitating the fastest \( S_N2 \) reaction.
Therefore, the order of increasing reactivity towards \( S_N2 \) displacement is: \[ \text{2-Bromo-2-methylbutane} < \text{2-Bromopentane} < \text{1-Bromopentane}. \]