Given the following SN1 reaction: $C_4{H}_9\mathrm{Br}$ (tert-butyl bromide) reacts with $H_{2}O$, what are the major organic products formed?

• A. $C_4{H}_9\mathrm{Br}$ (tert-butyl bromide) and $H_{2}O$
• B. $C_4{H}_9\mathrm{OBr}$ (tert-butyl hypobromite)
• C. $C_4{H}_9\mathrm{OH}$ (tert-butyl alcohol) and $H_2\mathrm{Br}$
• D. $C_4{H}_9\mathrm{OH}$ (tert-butyl alcohol) and $H_{3}O+$

Answer: (D)

Verified step by step guidance

1. Identify the type of reaction: Since tert-butyl bromide is a tertiary alkyl halide and the solvent is water (a polar protic solvent), the reaction proceeds via an SN1 mechanism.
2. Understand the SN1 mechanism involves two main steps: first, the departure of the leaving group (Br-) to form a tertiary carbocation intermediate; second, nucleophilic attack by the solvent (H2O) on the carbocation.
3. Write the first step: the heterolytic cleavage of the C-Br bond to form the tert-butyl carbocation and bromide ion: \(\mathrm{(CH_3)_3C-Br \rightarrow (CH_3)_3C^+ + Br^-}\).
4. Write the second step: nucleophilic attack of water on the carbocation to form a protonated alcohol intermediate: \(\mathrm{(CH_3)_3C^+ + H_2O \rightarrow (CH_3)_3C-OH_2^+}\).
5. Finally, deprotonation of the protonated alcohol by a water molecule yields the neutral tert-butyl alcohol and a hydronium ion: \(\mathrm{(CH_3)_3C-OH_2^+ + H_2O \rightarrow (CH_3)_3C-OH + H_3O^+}\).