This source says that DIBAL-H reduces esters to aldehydes. Further, it says that, like NaBH4, it reduces aldehydes too.
Why in reduction of ester, the aldehyde so formed is not converted to alcohol, if it can reduce both ?
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It depends on the reduction conditions.
1 eq of Dibal-H takes esters to aldehydes at low temperature. Use multiple eqs at a higher temperature and you get further reduction. This is because of the formation of a hemi-acetal aluminium complex intermediate that is stable at low temperature and resistant to further reduction.
DIBAL is an “electrophilic” reductant. That is, the first step in the reaction is coordination of a lone pair from the carbonyl oxygen (a nucleophile) to the aluminum (electrophile). It is only after coordinating to its carbonyl host that DIBAL delivers its hydride to the carbonyl carbon, resulting in formation of a neutral hemiacetal intermediate that is stable at low temperatures.
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$\begingroup$ Thanks sir but when we say 1 eq. , we theoretically make product that would be formed if only one molecule react. But in the solvent, there are many molecules and practically we should get more reduction products even if 1 equivalent is used? $\endgroup$ Sep 18, 2022 at 5:48
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$\begingroup$ @Answer_Me if you control the temperature of the reaction the intermediate (top right of the diagram) is stable and does not react with more reagent. Of course, if you add reactant too fast and get local heating then it can react and reduce your yield. In practice, run at -75C and with slow addition, this is a good high yielding reaction. $\endgroup$ Sep 18, 2022 at 5:59