Draw the mechanism of the slow step that occurs in both first-order substitution and first-order elimination reactions for (R)-3-bromo-2,3-dimethylpentane in methanol with heat applied. Provide curved arrows in Box 1 to depict the flow of electrons and draw the intermediate in Box 2.

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Explanation:

We are given the (R)-3-bromo-2,3-dimethylpentane and asking to draw the curved arrow which is the showing the mechanism for first-order substitution and first-order elimination reactions. We know the formation of carbocation is the rate determining step in the first-order substitution and first-order elimination reactions.

So in the (R)-3-bromo-2,3-dimethylpentane there is –Br gets removed and formed the tertiary carbocation which is more stable, so the curved arrows in Box 1 to depict the flow of electrons and intermediate in Box 2.

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i just learnt this recently but i need to look through my notes again