The following is a table from a J. Chem Ed. Paper (find it online in experiment
ID: 539719 • Letter: T
Question
The following is a table from a J. Chem Ed. Paper (find it online in experiment 11 folder) Compare the results of 1-bromobutane, 2-bromobutane and 2-bromo-2-methylbutane with sodium methoxide. (a) What conditions/combinations favor SN2? (b) What conditions/combinations favor E2? (c) Do these results match theory? Compare the results of the reactions of each alkyl halide with sodium methoxide versus sodium t-butoxide. Do these results match the expected theory? If we were to run the kinetic experiments on our reaction with KOH as well as the one in this article with methoxide and 2-methyl-2-bromobutane, which would give a higher rate (faster)? Show your answer.Explanation / Answer
4. Comparing the given alkyl haldes reaction with sodium methoxide,
(a) For an Sn2 reaction, primary alkyl halide is the preferred choice due to less steric crowding in the transition state.
(b) For an E2 reaction, as the steric crowding increases, the elimination reaction proceeds at faster rate so as to reduce the steric bulk of the molecule.
(c) Yes the reactions shown do match the theory.
5. Reaction of alkyl halide with less crowded base/nucleophile sodium methoxide or crowded base sodium tert-butoxide shows, with crowded base sodium t-Butoxide E2 is the preferred mechanism with almost no Sn2 product and least substituted alkene is formed as the major product. The results shown above confirm the theory.
6. Kinetic expeiments generate products which are having lower thermodynamic stability (higher energy), therefore, reaction with KOH would give a faster rate as less crwoded base reacts at a faster rate under kinetic controlled experiments than the more crowded base reaction which require higher energy.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.