Timeline for Which compound reacts faster in the Cannizzaro Reaction?
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14 events
| when toggle format | what | by | license | comment | |
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| Dec 30, 2015 at 17:14 | comment | added | ron | No, that carbon is sp3 hybridized, you can't have resonance with the ring | |
| Dec 30, 2015 at 17:13 | comment | added | Sujith Sizon | @ron in the case of methoxy compound there is resonance interaction between C (from where H is removed) and ring right? | |
| Dec 30, 2015 at 17:06 | comment | added | ron | The carbon the H is being removed from is sp3 hybridized, there is no carbonyl and because the C is sp3 there is no resonance interaction between it and the aromatic ring. | |
| Dec 30, 2015 at 17:03 | comment | added | Sujith Sizon | @ron i agree, but for removing H- shouldn't the carbonyl carbon be more electron rich so that hydrogen can take some extra electrons? | |
| Dec 30, 2015 at 16:56 | comment | added | ron | @SujithSizon 1) No, more intermediate can also mean faster rate of reaction (e.g. rate is dependent upon concentration); 2) Even in the H transfer step there is more positive charge on the nitro aldehyde carbonyl than the methoxy case (draw resonance structures) | |
| Dec 30, 2015 at 16:45 | comment | added | Sujith Sizon | @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). | |
| Dec 30, 2015 at 16:13 | comment | added | Sujith Sizon | 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? | |
| Dec 30, 2015 at 15:12 | comment | added | ron | @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. | |
| Dec 30, 2015 at 14:56 | comment | added | Sujith Sizon | 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? | |
| Aug 3, 2015 at 17:04 | vote | accept | Shubham | ||
| Jun 18, 2015 at 23:24 | history | edited | ron | CC BY-SA 3.0 |
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| Jun 18, 2015 at 19:19 | history | edited | ron | CC BY-SA 3.0 |
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| Jun 18, 2015 at 14:50 | history | edited | ron | CC BY-SA 3.0 |
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| Jun 18, 2015 at 14:45 | history | answered | ron | CC BY-SA 3.0 |