Final answer:
Tertiary butyl bromide (R-Br) undergoes SN1 hydrolysis most rapidly due to the formation of a stable tertiary carbocation and because bromide is a good leaving group.
Step-by-step explanation:
The rate of hydrolysis in an SN1 mechanism is influenced by the stability of the carbocation that forms during the reaction. Among the given options of alkyl halides (I, Br, Cl, or F), the one that would undergo SN1 hydrolysis most rapidly is tertiary butyl bromide ((CH3)3C-Br).
This is due to the tertiary carbocation that is formed being more stable due to increased alkyl substitution which stabilizes the positive charge through hyperconjugation and inductive effects. Bromine (Br) is a good leaving group because its bond with carbon is polar and it leaves easily, making the carbocation formation the rate-determining step in the SN1 reaction. Furthermore, bromide anions are a good leaving group and better than chloride (Cl) and fluoride (F) in this context.