This mechanism is found for substitutions at the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: of H2O is about 15) rather than the alkyne anion. If we use the second hydroxide ion to deprotonate the intermediate, only one leaving group remains, though it is a poor one, and the decomposition of the dianion must be the rate- determining step. This mechanism is found for substitutions at the carbonyl group with very bad leaving groups, as in the hydrolysis of amides (p. 213 in the textbook). O OH Ph fast O OH OH Ph Ph Ph fast OO rate-determining step Ph O + Ph O Ph Ph Solutions for Chapter 12 – Equilibria, Rates and Mechanisms 3 PROBLEM 3 Draw an energy profile diagram for this reaction. You will of course need to draw the mechanism first. Suggest which step in this mechanism is likely to be the slow step and what kinetics would be observed O HO CN NaCN H2O, HCl Purpose of the problem Practice at drawing energy profile diagrams as one way to present the energetics of mechanisms. Suggested solution The first thing is to draw the mechanism of the reaction. ■ Thismechanism is described in detail on p. 127 of the textbook. O CN NC O H O H H NC OH The first step is bimolecular and forms a new C–C bond. The second step is just a proton transfer between oxygen atoms and is certainly fast. The first step must be the rate- determining step and the intermediate must have a higher energy than the starting material or the product. In this answer we have used the style of energy- profile diagrams used in the book (e.g. p. 252) but there is nothing sacre...
View Full Document

This document was uploaded on 02/10/2014.

Ask a homework question - tutors are online