Question

Question #d32c3

Answer 1

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

Explanation:

`("CH"_3)_3"CCH"_2"Br"`, also known as neopentyl bromide, is unreactive to `"S"_text(N)2` attack on the α-carbon because of steric hindrance by the bulky tert-butyl group.

It is also unreactive to `"S"_"N"1` reactions because the substrate is a primary halide.

Nevertheless, the `"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 `"OH"^"-"` removes a β-hydrogen to form the more stable alkene.

These three steps constitute an `"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 `"HSO"_4^"-"` to the carbocation.

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