Question #d32c3

1 Answer
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 "S"_text(N)2 attack on the α-carbon because of steric hindrance by the bulky tert-butyl group.

www.kshitij-school.comwww.kshitij-school.com

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 1Step 1

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

Step 2Step 2

Step 3. The "OH"^"-" removes a β-hydrogen to form the more stable alkene.

Step 3Step 3

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 4Step 4

**Step 5. Addition of "HSO"_4^"-" to the carbocation.

Step 5Step 5

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