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tert-Butylbromide reacts with aq. NaOH by S_N1 mechanism while n-butylbromide reacts by S_N2 mechanism. Why?

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In general, the S_N1 reaction goes forward by the formation of carbocation. The tert-butyl bromide will form 3^{\circ} carbocation by losing the Br^{-} ion, which is actually stable. Therefore, it reacts with aqueous KOH by S_N1 mechanism as:

On the other hand, due to its unstable in nature, n-butyl bromide does not form 1° n-butyl carbocation by ionization. Therefore, it follows the mechanism of S_N2 mechanism. This happens through a transition state where OH^{-} ions form (nucleophilic attack) in the remote side with simultaneous expulsion of Br^{-} ions from the front side.

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