Timeline for Why does the rate law of Cannizzaro reaction include a term for base?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 6 at 18:08 | comment | added | mechanist | @Maurice, but I've always been told by instructors that rate laws follow only from the rds, which is why for eg. e2 & e1cb have a base term but not e1 as e1 has cation formation as rds. Sam202, that could explain it, I will try to find more info, but for now this is the only paper, that too in a predatory one, showing that as the rds. | |
| May 6 at 17:42 | comment | added | Sam202 | This source (page 2) shows reaction between $\ce{RCHO}$ and $\ce{OH-}$ as the rate limiting step. | |
| May 6 at 16:08 | comment | added | Maurice | If the ion $\ce{[OH-]}$ reacts in step $1$, the reaction rate must reasonably be proportional to $\ce{[OH-]}$, or to some power of this concentration. | |
| May 6 at 9:43 | comment | added | mechanist | @Maurice, I've looked at the mechanism of the reaction here and I see that the base attacks in step 1, while step 2 is the actual hydride shift rds, so the base term still doesn't make sense to me. | |
| May 6 at 9:34 | history | edited | andselisk♦ | CC BY-SA 4.0 |
added 142 characters in body; edited title
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| May 6 at 9:30 | comment | converted from answer | Maurice | Please look at Cannizzaro Mechanism through Google. You'll see plenty of reports saying that the Cannizzaro reaction is due to the action of $\ce{OH-}$ on the aldehyde. | |
| May 6 at 8:21 | history | asked | mechanist | CC BY-SA 4.0 |