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The rate law of Cannizzaro reaction in mild base is

$$\mathrm{rate_1} = k[\ce{RCHO}]^2 [\ce{OH^-}],\tag{1}$$

while in concentrated base the rate becomes

$$\mathrm{rate_2} = k[\ce{RCHO}]^2[\ce{OH^-}]^2\tag{2}$$

according to most sources I've read. But in both cases in the RDS hydride shift there is no base molecule among reactants. It's just the anion or dianion of the hydrate and the aldehyde reacting. So, why does the rate law include a base term?

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    $\begingroup$ 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. $\endgroup$
    – Maurice
    Commented May 6 at 9:30
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    $\begingroup$ @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. $\endgroup$
    – mechanist
    Commented May 6 at 9:43
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    $\begingroup$ 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. $\endgroup$
    – Maurice
    Commented May 6 at 16:08
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    $\begingroup$ This source (page 2) shows reaction between $\ce{RCHO}$ and $\ce{OH-}$ as the rate limiting step. $\endgroup$
    – Sam202
    Commented May 6 at 17:42
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    $\begingroup$ @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. $\endgroup$
    – mechanist
    Commented May 6 at 18:08

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