# Crossed Cannizzaro reaction

What is the major product of the following reaction:

I think methanol will be formed. Although formaldehyde is more reactive to nucleophic addition, the rate determining step of the Cannizzaro reaction is the hydride transfer. So instead of a nucleophilic attack of $\ce{OH-}$ on formaldehyde, nucleophilic attack takes place on the aromatic aldehyde.

Is that correct?

• Yes apparently, molcalc.org/calculation/f795351741aefc756661f83735c228f7#/… , molcalc.org/calculation/1358b63dfd8439726a48e1c402dd4cbd#/… (you can get the charges by clicking on the atoms) – Sujith Sizon Dec 27 '15 at 12:52
• No, the formaldehyde is used as a "sacrificial" reagent. The formaldehyde is oxidized to formic acid and the other aldehyde is reduced to the corresponding alcohol. Since the formaldehyde carbonyl is very reactive, more of it will form the tetrahedral intermediate then the other carbonyl compound and although this step is an equilibrium, it stacks the deck in favor of formaldehyde being oxidized. – ron Dec 27 '15 at 17:17
• @Sujith Sizon bro.... our teacher was wrong. He made us learn the wrong products. – Aditya Dev Dec 27 '15 at 21:50

Background

It seems that you understand the mechanism of the Canizzarro and Crossed Cannizzaro reaction, but let me supply a link to an earlier answer on the Cannizzaro reaction itself for others who might want to review the reaction. Also, here is the mechanism for the Cannizzaro reaction from the same link (but note, as we'll discuss below, the first step is really an equilibrium).

In the Cannizzaro reaction an aldehyde disproportionates into the corresponding alcohol and acid. The maximum yield of the alcohol or acid is therefore only 50%. If you've worked in the lab for 2 weeks synthesizing 50 mg of the aldehyde, and if your next step is the Cannizzaro reaction to prepare the alcohol from it, then a 50% yield is fairly distressing!

Here is where the Crossed Cannizzaro reaction can come in handy. If it's the alcohol that your after, then using formaldehyde as a sacrificial reagent allows you to achieve a much higher yield of the desired alcohol. In the Crossed Cannizzaro the formaldehyde is always oxidized to the acid and the other aldehyde is always reduced to the alcohol - you could get a 100% yield of alcohol!

Your Question - Why is that?

The first step in either the Cannizzaro or Crossed Cannizzaro reaction is an equilibrium between the aldehyde(s) and hydroxide ion to form a tetrahedral intermediate as pictured below. In the case of Crossed Cannizzaro with formaldehyde, the formaldehyde carbonyl is extremely reactive and the equilibrium will lie strongly on the side of the formaldehyde tetrahedral intermediate, and consequently most of the other aldehyde will remain present as the aldehyde.

You are correct to note that the rate determining step is the second step where hydride is transferred from the tetrahedral intermediate to any remaining carbonyl compound. But look at the 2 possible transition states for the Crossed Cannizzaro with formaldehyde (pictured above) - they are virtually the same, just some minor differences. Hence the 2 possible rate determining steps should have similar activation energies. This means that the energetics of the rate determining step will not control the reaction outcome (e.g product distribution), rather the initial equilibrium will be controlling.

If most of the formaldehyde is converted to the tetrahedral intermediate, then most of the other aldehyde remains as the aldehyde; and if the 2 possible rate determining steps from the tetrahedral intermediates have similar rates (they do as explained above), then the most likely reaction will involve the formaldehyde tetrahedral intermediate (major tetrahedral intermediate present at equilibrium) and the other aldehyde (major remaining carbonyl compound present at equilibrium).

Hence, in a Crossed Cannizzaro the formaldehyde is always oxidized to formic acid and the other aldehyde is always reduced to the corresponding alcohol.

• And one more thing. What if the reaction was between p-nitro benzaldehyde and p-methoxy benzaldehyde? – Aditya Dev Dec 27 '15 at 22:15
• Most know that the formaldehyde carbonyl is much more reactive than most other carbonyls. The rds transition states are pictured above so you can decide for your self if you think the energy difference between them is large enough to control product formation. Here is a [link] (en.wikipedia.org/wiki/Cannizzaro_reaction) that notes that the formaldehyde is always oxidized. – ron Dec 27 '15 at 22:15
• If the reactants were p-nitro benzaldehyde and p-methoxy benzaldehyde, then p-nitro benzaldehyde is more reactive. But if nucleophilic attack takes place on p-nitro benzaldehyde , the hydride transfer step would be difficult. – Aditya Dev Dec 27 '15 at 22:18
• It would be a mess, you'd get 2 alcohols and 2 acids. But as you say, the nitro benzaldehyde tetrahedral intermediate would predominate at equilibrium. Since the 2 rds transition states should again be comparable in energy, I'd expect a slight preponderance of the nitrobenzoic acid and methoxy-benzyl alcohol. – ron Dec 27 '15 at 22:31
• @ron: does this apply to benzoin condensation? Does the nucleophilic attack determine the yeild or the hydride transfer? – Aditya Dev Dec 30 '15 at 15:27