I am little confused on this question. On one hand, I remember my teacher saying that $\ce{LiAlH4}$ is like $\ce{BH3}$ of hydroboration oxidation of alkene, where you one mole of $\ce{BH3}$ can oxidize 3 mole of alkene molecule. but on the other hand my textbook keeps on mentioning excess LAH when reducing ester? Can't one mole of LAH reduce four mole of ester molecules as well? Thanks.

  • $\begingroup$ A single reduction of a single gives a aldehyde, which is a more reactive carbonyl than an ester is. $\endgroup$ – Zhe Feb 24 '17 at 2:33
  • $\begingroup$ I know that. I am asking why can't one mole of LAH reduce ester twice since it havs 4 hydrogens attacked to Aluminum. $\endgroup$ – TLo Feb 24 '17 at 2:48
  • 1
    $\begingroup$ LAH is a bit more fickle than borane. Some of the hydrides might not add. Depends on the binding of the reduced components to the central aluminum. They form an oxide, and the oxide can reduce the nucleophiliticity of the hydrides and increase steric crowding. $\endgroup$ – Zhe Feb 24 '17 at 3:07

Lithium aluminium hydride or LAH is the more everyday form of $\ce{LiAlH4}$. When originally produced, most of it is indeed $\ce{LiAlH4}$ but that quickly partially hydrolyses giving the grey powder you may or may not know from the lab. As such, $\ce{LiAlH4}$ is practically never entered into reactions in pure form. Since it is impure, excess reagent is added to make sure the reduction job is performed as desired.

Also, remember to consider the reaction meachnism:

$$\ce{R-COOR' ->[LAH][- HOR'] R-CHO ->[LAH] R-CH2OH}\tag{1}$$

Note that the second step, the reduction of an aldehyde, is more rapid than the first as aldehydes are more reactive towards nucleophiles than esters. Thus, if you attempted to bargain, hoping to arrive at the aldehyde selectively, you would end up with under half of your reactant reduced to the alcohol and half of it still as an ester. Another reason to always add excess just to be sure.

  • For one, the action of $\ce{BH3}$ (or a borane in general) on an alkene in the course of a hydroboration is not an oxidation, rather than an reduction. The second step, the action of hydrogenperoxide on the product of this addition (an alkane) is an oxidation.
  • For the reduction of an ester by LAH, there are two things to retain:
    • general rule: LAH reacts with a carbonyl compound, initially forming a lithium alcoholate and $\ce{AlH3}$, and subsequently they form the lithium alanate (decomposed during mildly acidic aqueous workup)
    • The first H of $\ce{LiAlH4}$ leads only to the aldehyde, and the second one to the alcohol you are able to isolate.

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