4
$\begingroup$

Image of a cannizzaro reaction mechanism

In the given mechanism, is it correct to prefer hydrogen as a better leaving group than the formyl group?

The negative charge on an sp2 hybridized carbon, with the carbonyl group attached to it, seems to be more stabilized than a negative charge on hydrogen. (I have starred the step I am referring to)

Improve this question
$\endgroup$
9
  • 2
    $\begingroup$ Hydride is a terrible leaving group and a terrible nucleophile. $\endgroup$
    – Lighthart
    Commented Apr 13, 2016 at 14:23
  • 1
    $\begingroup$ @Lighthart That's true, but a hydride transfer is apparently not impossible (in this reaction). I just think it needs an acceptor here. $\endgroup$
    – SendersReagent
    Commented Apr 13, 2016 at 14:35
  • 7
    $\begingroup$ The mechanism you have drawn is wrong. The hydride is transferred to the other carbonyl directly, it's not going to exist in solution. $\endgroup$
    – orthocresol
    Commented Apr 13, 2016 at 14:40
  • 3
    $\begingroup$ Your mechanism is fine, just don't show the hydride going into solution. The hydride is transferred directly to the electrophilic carbon. Hydride transfer is common enough, but the hydride never just goes into solution like that. Your mechanism makes sense, just combine steps two and steps three and show an intramolecular hydride transfer so it is more representative of the real chemistry. $\endgroup$
    – gannex
    Commented Apr 13, 2016 at 15:28
  • 1
    $\begingroup$ I don't know what you mean by migration. Do you mean rearrangement? The reason you can get this hydride transfer is because the hydride has a very electrophilic place to go -- the aldehyde carbonyl. $\endgroup$
    – gannex
    Commented Apr 13, 2016 at 22:42

1 Answer 1

Reset to default
1
$\begingroup$

If you'd like to read more about this mechanism, you should out this paper:

Swain, C. G.; Powell, A. L.; Sheppard, W. A.; Morgan, C. R. Mechanism of the Cinazzaro Reaction, 1979, Am. Chem. Soc. 3576

They did a kinetic study with deuterium oxide. They gave two equivalent mechanisms:

enter image description here

and

enter image description here

I think that this mechanism, which is essentially what you have shown except with intramolecular hydride shift (it cannot shift through solution) is the way I would represent it. It makes sense that this can happen if you think of the transition state. In my mind, it makes sense that this hydride transfer can occur given the very electrophilic nature of the aldehyde.

This is the transition state: enter image description here

Improve this answer
$\endgroup$
3
  • 3
    $\begingroup$ One does not 'prove' transition states 'kinetically'. $\endgroup$
    – Lighthart
    Commented Apr 13, 2016 at 23:15
  • $\begingroup$ Sorry I must be conflating things. I was thinking that their kinetic study was involved because that is how they knew that the hydroxide reactant was already fully bonded before the rate determining step, so their study of the reaction's kinetics would have implications about the structure of the transition state. I guess what I'm trying to say is that this is how they know that the hydride is transferred intramolecularly from carbon to carbon, rather than through some other mechanism. $\endgroup$
    – gannex
    Commented Apr 14, 2016 at 1:41
  • $\begingroup$ What they are doing is proposing a transition state. The closest you can do to 'proving' a transition state is QM calcs to verify it is saddle point. And this is not really proof, just supporting evidence that using this transition state to explain a mechanism is 'reasonable' . $\endgroup$
    – Lighthart
    Commented Apr 14, 2016 at 20:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.