The rate law for the reaction of HO- with tert-butyl bromide to form an elimination product in
Question:
rate = 7.1 × 10-5[tert-butyl bromide][HO-] + 1.5 × 10-5[tert-butyl bromide]
What percentage of the reaction takes place by the E2 pathway when these conditions exist?
a. [HO-] = 5.0 M
b. [HO-] = 0.0025 M
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a E2E2 E1 71 10 5 5071 10 5 50 15 10 5 355 10 5 355 50 15 105 ...View the full answer
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