19
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ $\ce{-OCH3}$ usually shows a $+M$ (or $+R$ in your terminology) effect … $\endgroup$ – Jan Jun 4 '16 at 11:58
20
$\begingroup$

The mechanism of the Cannizzaro reaction is illustrated below.

enter image description here

The first step involves attack by the nucleophilic hydroxide ion on the positively polarized carbonyl carbon to form a tetrahedral intermediate. Once the tetrahedral intermediate is formed, substituents on the aromatic ring can have little resonance interaction with the former carbonyl carbon because it is now $\ce{sp^3}$ hybridized. So we need to look at the step leading to the tetrahedral intermediate - the first step.

However, if we examine some of the resonance structures for the starting carbonyl compounds (where the substituents on the aromatic ring can interact with the carbonyl through resonance) we see that the nitro compound places positive charge adjacent to the already positive polarized carbonyl carbon - a destabilizing situation (remember, destabilizing something means making it higher energy, more reactive). On the other hand, the methoxy compound places negative charge adjacent to the already positive polarized carbonyl carbon - a stabilizing situation.

The destabilized carbonyl in the nitro compound will be more reactive towards nucleophilic attack.

enter image description here

$\endgroup$
  • $\begingroup$ Im really sorry about asking doubts on closed questions, but i just have a wee bit trouble, you have said that "..nitro compound will be more reactive.." Are you implying that $\ce{H-}$ is easily leaving in the nitro compound? $\endgroup$ – Sujith Sizon Dec 30 '15 at 14:56
  • 1
    $\begingroup$ @SujithSizon No, I'm saying that since the carbonyl in the nitrobenzaldehyde is destabilized, it will form more of the tetrahedral intermediate in the first step (which is an equilibrium). More tetrahedral intermediate leads to faster reaction. $\endgroup$ – ron Dec 30 '15 at 15:12
  • $\begingroup$ So far so good, after more of the tetrahedral intermediate is formed (of the nitro compound) what is going to happen, i believe the reaction slows down (for nitro compound) after that? And doesn't more tetrahedral product only mean more final product formed? $\endgroup$ – Sujith Sizon Dec 30 '15 at 16:13
  • $\begingroup$ @ron whereas the methoxy group though lesser amount of intermediate formed but the H- transfer step happens fast (and in overall could be faster than the latter). $\endgroup$ – Sujith Sizon Dec 30 '15 at 16:45
  • 1
    $\begingroup$ No, that carbon is sp3 hybridized, you can't have resonance with the ring $\endgroup$ – ron Dec 30 '15 at 17:14
3
$\begingroup$

Looking at the reaction after the fact (always easier!): Two factors favoring compound a are (1) the equilibrium between the aldehyde and its adduct with hydroxide ion will more strongly favor the adduct when it has an electron-withdrawing p-nitro group; (2) once that intermediate is formed, it will be faster to undergo hydride transfer to another molecule of the aldehyde, when the latter has an electron-withdrawing group (nitro).

$\endgroup$
  • $\begingroup$ Why will the equilibrium between the aldehyde and its adduct with hydroxide ion more strongly favor the adduct ? $\endgroup$ – Shubham Jun 18 '15 at 13:48
  • $\begingroup$ An inductively electron-withdrawing group (like nitro) on the benzaldehyde will make the aldehyde group more electrophilic, and hence more reactive with hydroxide. On the other hand, the p-methoxy group donates electrons across the ring into the aldehyde group, making it less reactive. $\endgroup$ – iad22agp Jun 18 '15 at 14:20
0
$\begingroup$

You haven't considered that once the tetrahedral intermediate is formed, the nitro group will also compete for the hydride. When I realized the Cannizzaro reaction with o-nitrobenzaldehyde, the only product I obtained (in high yield) was the o-aminobenzoic acid (i.e. an intramolecular Cannizzaro).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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