Question

What is the leaving group in the hydrolysis of tert-butyl bromide with water forming tert-butanol?

Answer 1

The `S_N1` reaction of t-butyl bromide and water gives bromide as the leaving group. You can tell simply from the switching of bromide with the hydroxyl group of water, even if you don't know the mechanism.

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In a list of primary, secondary, and tertiary carbocations, the tertiary one is the most stable due to the most hyperconjugation stabilization between the methyl p orbitals and the empty p orbital in the center.

Therefore, it is reasonable to expect the bromide to leave when it coordinates with the water molecule in a first-order substitution (`S_N1`).

1. The bromine "falls off" slowly due to a stable carbocation intermediate, but slowly because nothing else happens at the same time other than water coordinating with the bromine and slightly polarizing it
2. Water becomes a nucleophile and adds onto the carbocation intermediate because the intermediate is now electrophilic enough
3. Another water molecule (not `Br^(-)`) grabs a proton off of the protonated water substituent to finish the reaction since `HBr` is a much stronger acid than `H_3O^(+)` is