Explain how the following changes would affect the rate of the reaction of 2-bromo-3-methylbutane with methanol: Part A The alkyl halide is changed to 2-chloro-3-methylbutane. The alkyl halide is changed to 2-chloro-3-methylbutane. The reaction will be slower because the leaving group will be poorer. The reaction will be slower because the leaving group will be better. The reaction will be faster because the leaving group will be better. The rate of the reaction does not change.

Question

Finley Vance

Chemistry

18 July, 02:54

Explain how the following changes would affect the rate of the reaction of 2-bromo-3-methylbutane with methanol: Part A The alkyl halide is changed to 2-chloro-3-methylbutane. The alkyl halide is changed to 2-chloro-3-methylbutane. The reaction will be slower because the leaving group will be poorer. The reaction will be slower because the leaving group will be better. The reaction will be faster because the leaving group will be better. The rate of the reaction does not change.

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Giovanny Quinn · Answer 1

The reaction will be slower because the leaving group will be poorer.

Explanation:

One Important thing to put at the back of our mind is that weak bases are good leaving groups. Another thing to take note is that in the halogen series (which is our main subject in this question) as you go down the group the greater or the more heavy the halide is and the heavier halides are more stable that is Bromine will be more stable than chlorine.

Now, we are now told that the bromo halide is been replaced by the chloro halide which means that the rate of Reaction will surely decrease because the leaving group. The cl^ - is a stronger base.