Question #d32c3

1 Answer
Ernest Z.
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.

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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 1

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

Step 2

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

Step 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 4

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

Step 5

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

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