Question

Which isomer of \( C_4H_9Br \) is most reactive towards \( S_N1 \) reaction?

Hint:

For \( S_N1 \), stability of carbocation is key. For \( S_N2 \), lack of steric hindrance is key.

Answer 1

Concept:

• The \( S_N1 \) (Substitution Nucleophilic Unimolecular) reaction proceeds via the formation of a carbocation intermediate.

• The rate of an \( S_N1 \) reaction depends directly on the stability of the carbocation formed in the rate-determining step.

• Carbocation stability follows the order: \( 3^\circ \gt 2^\circ \gt 1^\circ \gt \text{Methyl} \).
Step 1: List the isomers of \( C_4H_9Br \)
The four isomers are:
1. n-butyl bromide (1-bromobutane): Primary (\( 1^\circ \))
2. isobutyl bromide (1-bromo-2-methylpropane): Primary (\( 1^\circ \))
3. sec-butyl bromide (2-bromobutane): Secondary (\( 2^\circ \))
4. tert-butyl bromide (2-bromo-2-methylpropane): Tertiary (\( 3^\circ \))

Step 2: Compare carbocation stability
Tert-butyl bromide forms a tertiary (\( 3^\circ \)) carbocation: \( (CH_3)_3C^+ \).
This carbocation is highly stabilized by nine hyperconjugative structures and the +I effect of three methyl groups.

Step 3: Conclusion
Since the tertiary carbocation is the most stable among all isomers, tert-butyl bromide reacts the fastest in an \( S_N1 \) mechanism. The final answer is tert-butyl bromide (2-bromo-2-methylpropane).