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.
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.
The first one is the correct answer: <span>The potential energy of the products is greater than that of the reactants and the change in enthalpy is positive.<span> </span></span>