The reaction of tert-butyl bromide with aq. NaOH follows an SN1 mechanism because of the stability of the carbocation that is formed. In an SN1 reaction, the rate determining step is the formation of the carbocation. Tert-butyl bromide forms a tertiary carbocation, which is highly stable due to hyperconjugation and inductive effects. Therefore, it follows an SN1 mechanism.

On the other hand, n-butyl bromide reacts via an SN2 mechanism. In an SN2 reaction, the rate determining step is a single concerted step, where the nucleophile attacks the substrate and the leaving group leaves simultaneously. n-Butyl bromide forms a primary carbocation, which is less stable. Therefore, it is more likely to undergo an SN2 reaction, where carbocation stability is not a factor.

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The reaction of tert-butyl bromide with aq. NaOH follows an SN1 mechanism because of the stability of the carbocation that is formed. In an SN1 reaction, the rate determining step is the formation of the carbocation. Tert-butyl bromide forms a tertiary carbocation, which is highly stable due to hyperconjugation and inductive effects. Therefore, it follows an SN1 mechanism.

On the other hand, n-butyl bromide reacts via an SN2 mechanism. In an SN2 reaction, the rate determining step is a single concerted step, where the nucleophile attacks the substrate and the leaving group leaves simultaneously. n-Butyl bromide forms a primary carbocation, which is less stable. Therefore, it is more likely to undergo an SN2 reaction, where carbocation stability is not a factor.

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The reaction of tert-butyl bromide with aq. NaOH follows an SN1 mechanism because of the stability of the carbocation that is formed. In an SN1 reaction, the rate determining step is the formation of the carbocation. Tert-butyl bromide forms a tertiary carbocation, which is highly stable due to hyperconjugation and inductive effects. Therefore, it follows an SN1 mechanism.

On the other hand, n-butyl bromide reacts via an SN2 mechanism. In an SN2 reaction, the rate determining step is a single concerted step, where the nucleophile attacks the substrate and the leaving group leaves simultaneously. n-Butyl bromide forms a primary carbocation, which is less stable. Therefore, it is more likely to undergo an SN2 reaction, where carbocation stability is not a factor.

tert-Butyl bromide reacts with aq.NaOH by S mechanism while n-butyl bromide reacts by Smechanism. Why?

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