# Question d32c3

Jan 13, 2017

The final product is 1,1-dimethylpropyl hydrogen sulfate.

#### Explanation:

("CH"_3)_3"CCH"_2"Br"#, also known as neopentyl bromide, is unreactive to ${\text{S}}_{\textrm{N}} 2$ attack on the α-carbon because of steric hindrance by the bulky tert-butyl group.

It is also unreactive to $\text{S"_"N} 1$ reactions because the substrate is a primary halide.

Nevertheless, the $\text{S"_"N} 1$ ionization will occur slowly.

Here are the steps for your reactions.

Step 1. Slow ionization of the halide.

This generates an unstable primary carbocation.

Step 2. The cation rapidly undergoes a methyl shift to form the more stable tertiary carbocation.

Step 3. The $\text{OH"^"-}$ removes a β-hydrogen to form the more stable alkene.

These three steps constitute an $\text{E1}$ elimination to form 2-methylbut-2-ene.

The concentrated sulfuric acid then undergoes electrophilic addition to the double bond.

Step 4. Protonation of the double bond.

**Step 5. Addition of $\text{HSO"_4^"-}$ to the carbocation.

The final product is 1,1-dimethylpropyl hydrogen sulfate.