Question

Explain why tert-butyl bromide reacts with aqueous sodium hydroxide by S_Nl mechanism while n-butyl chloride reacts by S_N2 mechanism.

Answer 1

Tert-butyl bromide reacts with aqueous sodium hydroxide by an S_N1 mechanism because it is a tertiary alkyl halide, which can form a stable tertiary carbocation intermediate. In the first slow step, the C–Br bond breaks heterolytically, releasing the bromide ion and generating a positively charged tertiary carbocation:

\[\text{(CH}_3)_3\text{C–Br} \rightarrow \text{(CH}_3)_3\text{C}^+ + \text{Br}^-\]

This carbocation is stabilized by the three methyl groups through inductive and hyperconjugation effects. In the second fast step, the hydroxide ion (OH⁻) from the aqueous solution attacks the carbocation, forming tert-butyl alcohol:

\[\text{(CH}_3)_3\text{C}^+ + \text{OH}^- \rightarrow \text{(CH}_3)_3\text{C–OH}\]

Because the rate-determining step involves only the alkyl halide, the reaction rate depends solely on the concentration of tert-butyl bromide, consistent with an S_N1 mechanism. The bulky tertiary carbon also hinders direct nucleophilic attack, favoring carbocation formation instead.

On the other hand, n-butyl chloride is a primary alkyl halide, which cannot form a stable carbocation intermediate. Therefore, it undergoes substitution via an S_N2 mechanism, where the nucleophile attacks the electrophilic carbon atom at the same time as the chloride leaves, in a single concerted step:

\[\text{CH}_3\text{–CH}_2\text{–CH}_2\text{–CH}_2\text{–Cl} + \text{OH}^- \rightarrow \text{CH}_3\text{–CH}_2\text{–CH}_2\text{–CH}_2\text{–OH} + \text{Cl}^-\]

This occurs via a backside attack, which leads to inversion of configuration if the carbon is chiral. The reaction rate depends on both the alkyl halide and hydroxide concentrations, reflecting the bimolecular nature of S_N2.

The primary carbon has less steric hindrance, allowing the direct nucleophilic approach necessary for S_N2.