Here are two different reactions in which lithium aluminium hydride, $\ce{LiAlH4}$, reduces a carbonyl group:

Reduction reactions

Why does $\ce{LiAlH4}$ completely remove the oxygen from the amide, but only reduces the carbonyl group of a ketone/aldehyde to a hydroxyl group? Is the difference due to the fact that $\ce{H2O}$ is added in the first reaction (i.e. if the aldehyde/ketone reduction had a second step where water was added, would the carbonyl group be reduced to a methylene group)?


1 Answer 1


You've got it correct in your comment with the $\ce{N}$ lone pair assisting. The reason why the $\ce{N}$ lone pair kicks oxygen out, and not the other way round, is because oxygen-based groups are better leaving groups than nitrogen-based ones under the reaction conditions. (Just as a rough guide, you could compare $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 15.7$ and $\mathrm{p}K_\mathrm{a}(\ce{NH3}) = 38$.)


(Don't worry about how I've drawn the metals coordinating to oxygen - the exact details aren't important and sometimes aren't even fully understood.)

In the aldehyde reduction, there is no lone pair to assist in the expulsion of oxygen. So the intermediate alkoxide just sits there, until you add water or acid:

Aldehyde reduction mechanism

The aqueous workup serves two purposes: firstly to reprotonate the product (for example, with the aldehyde reduction, your final product is an alkoxide ion and adding water gets you the alcohol), and secondly to destroy any unreacted $\ce{LiAlH4}$. In fact, the aqueous workup is also carried out in the reduction of the ketone or aldehyde - we often just omit it because it's so common and it's tiresome to have to write it on every reaction we do. So even though it's not there, it's still implied.

If you wanted to reduce the carbonyl group of a ketone/aldehyde to a methylene group, the three most well-known methods available are the Wolff-Kishner reaction, the Clemmensen reduction, and the Mozingo reduction.

  • $\begingroup$ Why then, does an ester not undergo assisted reduction with the help of the neighbouring O's lone pairs? $\endgroup$ Apr 3, 2019 at 4:13
  • $\begingroup$ I don’t get the point. Esters are fully reduced to alcohols. @YUSUFHASAN $\endgroup$ Apr 3, 2019 at 9:11
  • $\begingroup$ No, I mean..I did exactly what you did up there with amide, just in place of N, used the ester's oxygen to remove lithium oxide...and then added a hydride at the carbon moiety of the doubly bonded oxonium ion so formed..But I ended up with an ether..So why can't assisted removal occur here? $\endgroup$ Apr 6, 2019 at 10:23
  • 1
    $\begingroup$ @YUSUFHASAN because OR is a better leaving group than O[M], whereas NR2 is worse so N kicks out O[M] $\endgroup$ Apr 6, 2019 at 10:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.