In an SN1 reaction, tert-butyl bromide treated in water yields which product?

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Multiple Choice

In an SN1 reaction, tert-butyl bromide treated in water yields which product?

In this case the reaction proceeds by an SN1 mechanism, where a stable tertiary carbocation forms first and then is attacked by a nucleophile. The tert-butyl bromide easily loses the bromide ion in water to give a tert-butyl carbocation. Water, present in excess, acts as the nucleophile and attacks that carbocation to form a tert-butyl oxonium ion. A quick deprotonation by water then yields tert-butanol, while the leaving bromide stays in solution (often forming HBr with available protons). So the organic product is tert-butanol. That’s why the other options aren’t the product here: you don’t introduce chloride or formate from the solvent, and there’s no reason for the molecule to revert to the starting bromide. The key idea is capturing the carbocation with water to give the alcohol.

In this case the reaction proceeds by an SN1 mechanism, where a stable tertiary carbocation forms first and then is attacked by a nucleophile. The tert-butyl bromide easily loses the bromide ion in water to give a tert-butyl carbocation. Water, present in excess, acts as the nucleophile and attacks that carbocation to form a tert-butyl oxonium ion. A quick deprotonation by water then yields tert-butanol, while the leaving bromide stays in solution (often forming HBr with available protons). So the organic product is tert-butanol.

That’s why the other options aren’t the product here: you don’t introduce chloride or formate from the solvent, and there’s no reason for the molecule to revert to the starting bromide. The key idea is capturing the carbocation with water to give the alcohol.

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