Which of the following alkyl halides would be most reactive in an $S_{\mathrm{N1}}$ reaction?

• A. Isopropyl bromide (${\left({\mathrm{CH}}_3\right)}_2\mathrm{CHBr}$)
• B. Methyl bromide (${\mathrm{CH}}_3\mathrm{Br}$)
• C. Tert-butyl bromide (${\left({\mathrm{CH}}_3\right)}_3\mathrm{CBr}$)
• D. Ethyl bromide (${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}$)

Answer: (C)

Verified step by step guidance

1. Recall that the rate-determining step in an SN1 reaction is the formation of a carbocation intermediate after the leaving group departs.
2. Identify the type of carbocation formed from each alkyl halide: methyl bromide forms a methyl carbocation, ethyl bromide forms a primary carbocation, isopropyl bromide forms a secondary carbocation, and tert-butyl bromide forms a tertiary carbocation.
3. Understand that carbocation stability increases in the order: methyl < primary < secondary < tertiary, due to alkyl group electron-donating effects and hyperconjugation.
4. Since the SN1 reaction rate depends on carbocation stability, the alkyl halide that forms the most stable carbocation will be the most reactive in SN1.
5. Conclude that tert-butyl bromide, which forms a tertiary carbocation, will be the most reactive alkyl halide in an SN1 reaction.