Which compound reacts faster in the Cannizzaro Reaction?

Question

> Which reacts faster in the Cannizzaro Reaction? >a) $\\ce{OHC-C6H4-NO2}$ b) $\\ce{OHC-C6H4-OCH3}$ Obviously, a better hydride releasing group will react faster. Therefore my answer was **b**, as $\\ce{-OCH3}$ shows -I (inductive) effect as well as +R (resonance) effect, as it's in the *para* position. This increases the electron concentration in the $\\ce{C}$ atom of $\\ce{CHO}$ and due to the excess electrons, it is a better donor of an electron to the $\\ce{H}$ attached to it. Thus, leaving a $\\ce{H-}$. But the answer given is **a**. The answer states that $\\ce{C}$ can only give electron to $\\ce{H}$ when it accepts an electron from the $\\ce{O}$ attached to it. For that the $\\ce{C}$ should be electorn deficient. Therefore an electron withdrawing group (in this case $\\ce{NO2}$, which shows -R effect at *ortho* and *para* positions) better releases hydride. But I don't think that happens. That's because if $\\ce{C}$ is electron deficient, then it wouldn't give an electron to $\\ce{H}$ (less electronegative) in the first place, even if it takes an electron from oxygen (which is also tough). Can anyone give a proper explanation to the whole problem?

Answer 1

You said "I don't think that happens"... You need to review the textbook mechanism for the Cannizzaro reaction. Step by step, the initial attack of OH- on the carbonyl would surely be faster in the compound with the electron withdrawing NO2 on it (compound a) as it makes the carbonyl group more electrophilic.

But that's not the RDS, the RDS is the collapse of the tetrahedral intermediate with the expulsion of the hydride to another (nearby) carbonyl group. So what causes that step? The "driving force" is the O- electrons swinging back in to attack the carbon and re-form the carbonyl. If the carbon is more electron deficient, the O electrons would want to come down and attack it more readily.

Let's put it this way - when you look at the actual step of the O electrons swinging down and pushing out the H, you can think of it from a couple perspectives. Which one is easier to push off the hydride? It seems if the situation was electron rich, that might be easier. BUT, we really aren't just pushing off a hydride as a leaving group, hydride is a TERRIBLE leaving group!

You need to look at the transition state of that step, the collapsing tetrahedral intermediate in one molecule has to be right next to another molecule with a carbonyl group to receive the hydride. So we are actually looking at two things at once - how likely is the O to bring its electrons down to the carbon and how likely is the hydride going to shift over to the other carbonyl group? Both of these things happen better if the carbonyl group(s) involved are more electron deficient!

Hope that helps... good luck!