Question

1) CH₃COCH₃ + OH^– → CH₃COCH₂^– + H₂O

rate₁ = k₁[CH₃COCH₃][OH^–]

2) CH₃COCH₂^– + I₂ → CH₃COCH₂I + I^–

rate₂ = k₂[CH₃COCH₂^–][I₂]

1. The experimentally observed rate law of the reaction is written below (obs = observed). Which of the two steps (1 or 2) is the rate-determining step?

rate_obs = k_obs[CH₃COCH₃][OH^–]

2.The initial concentration of I₂ is doubled. What effect does this have on the overall rate of the reaction? Explain your answer.

3.Write the overall (or net) chemical reaction for the iodination of acetone based on the two-step mechanism.

Answer 1

Answer:

While writing the overall reaction, we need to mention all the reagents used. So in this iodination reaction of acetone, acetone is the substrate used and iodine is the reagent. Then for deprotonation, we need the OH- base, so we need to mention this also. Among the two steps given in the question, the second one is the final step and gives the final product. And we mention the products on the right side of the chemical equation.

So, why didn't we write the products obtained in the first step? We did not write it because in that step the base abstracts the proton from acetone to form the carbanion intermediate. We don't write intermediates in overall chemical equation. Intermediates are transient species due to high reactivity. Here, as soon as the carbanion intermediate forms, they attack iodine molecule and at the end of the reaction there is no intermediate left in the reaction mixture.