# Why does a hydride shift take place in the cannizzaro reaction?

Hydride is a poor leaving group so a hydride shift should not take place and instead the attacking hydroxyl should leave the carbonyl group. I don’t understand why this doesn’t happen?

• Well if the hydroxyl group just falls off then you have no reaction, right? – orthocresol Jan 14 '17 at 11:20
• Right but what about the poor leaving hydride group – Sahil Gupta Jan 14 '17 at 12:08
• A discrete hydride ion may not form. An intermolecular hydride transfer takes place instead. Just compare it with normal hydride reagent reduction (like LiAlH4). – Marko Jan 14 '17 at 13:17

You are correct that a hydride ion is a terrible leaving group and that it can under no circumstances be eliminated in e.g. an $\mathrm{S_N2}$ reaction. However, there are quite a few reactions where hydride does leave one atom and subsequently attaches to another; Wagner-Meerwein rearrangements jump to mind as one of the most prominent examples.

The feature of a Cannizzaro reaction is that it is not actually a leaving hydride. Rather, the reactivity can be explained in a cyclic transition state as shown in scheme 1 below.

Scheme 1: Transition state and products of a Cannizzaro reaction.

Instead of seeing the reaction as a displaced hydride that subsequently attacks another atom, we can see it as a type of pseudo-electrocyclic reaction. Breaking and forming the bonds will happen almost simultaneously. At no point in the mechanism do we actually have a ‘free’ hydride anywhere.

Note, by the way, that the Cannizzaro reaction is not the only one that features a hydride shift in a cyclic transition state; you may, for example, also be interested in the Evans-Tishchenko asymmetric reduction.

• I guess, going by their formal definitions, it would not be an electrocyclic reaction (goldbook.iupac.org/E01948.html), but rather a group transfer reaction. Not that it matters. – orthocresol Jan 14 '17 at 20:30
• @orthocresol Hence pseudo ;) – Jan Jan 14 '17 at 20:33