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Wikipedia says that esters can be reduced to aldehydes using DIBAl-H. Can anyone please explain how this reaction occurs?

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DIBAL exists as a bridged dimer, and it becomes a reducing agent only after it has formed a Lewis acid–base complex, so it reduces electron-rich carbonyl groups most rapidly.

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DIBAL will reduce esters even at –70 °C, and at this temperature the tetrahedral intermediate that forms during the reaction may be stable. Only in the aqueous acidic work-up does it collapse to the aldehyde (by expelling EtOH in the example reaction below) when excess DIBAL has been destroyed so that no further reduction to the alcohol is possible.

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A stable tetrahedral intermediate is more likely in the reduction of lactones, and DIBAL is most reliable in the reduction of lactones to lactols.

Edit:

As requested by the OP here are some more details about the collapse of the tetrahedral intermediate: During the aqueous acidic work-up the alcohol group is turned into a good leaving group via protonation. Thus, it is easily expelled from the molecule, when a water molecule removes the $\ce{Al^{$i$}Bu_2}$ group and the $\ce{C=O}$ double bond is reformed.

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A different route, where the former carbonyl-oxygen is protonated twice - thus being turned into a good leaving group - and afterwards expelled from the molecule to give an oxonium ion is possible too. But such an oxonium ion is rather unstable and highly reactive, so that the reverse reaction is very fast and the oxonium ion is not produced in any appreciable amount.

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  • $\begingroup$ Why is that tetrahedral intermediate so stable? In other words, why is (O-AliBu2)- a bad leaving group? If LiAlH4 was used as the reducing agent, then (O-AlH4)2- would come off. So what is the difference here? $\endgroup$ – Joshua Meyers Mar 8 '16 at 12:20
  • $\begingroup$ @JoshuaMeyers I would guess that steric hindrance is the reason. The $\ce{{}^{$i$}Bu}$ groups make it much harder for, say, another ester molecule to attack the $\ce{Al}$ in the tetrahedral intermediate and kick out the original (now reduced) ester leading to its collapse. This is not the case for the small hydride groups. $\endgroup$ – Philipp Mar 8 '16 at 16:48

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