Master Organic Chemistry Reaction Guide

Addition of LiAlH4 to ketones to give secondary alcohols

Description: Addition of lithium aluminum hydride to ketones leads to formation of secondary alcohols (after addition of acid) Content available for Reactionguide members only. Not a member? Get access for about 30 cents / day!

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12 thoughts on “Addition of LiAlH4 to ketones to give secondary alcohols

  1. LiAlH4 seems to function in the same alcohol reduction reactions as the reagent, NaBH4. If LiAlH4 achieves the same products as NaBH4, what is the point of having to remember NaBH4 as a reagent?

    1. NaBH4 is a milder reducing agent and is able to be used in protic solvents. It will only reduce the aldehyde or ketone. LAH is a very strong reducing agent and cannot be used in protic solvents due to a violent deprotonation releasing H2 gas. Therefore it must be used separate to the aqueous work up.

    2. NaBH4 reacts only with aldeydes and ketones. LiAlH4 reacts with aldehyde, ketone, carboxylic acid, acid chloride and esters

    3. NaBH4 is for selective reduction. For example, if one had a molecule in which a ketone and a carboxylic acid were present and the carboxylic acid was wanted in the final product, one would use NaBH4 because it cannot reduce carboxylic acids (or any derivatives). LiAlH4 is a much stronger reducing agent and would simply convert both of those groups to secondary alcohols.

      I hope this helps!

      -Brian

  2. On stereochemistry – Since this is a flat structure before AlH3, does this produce equal amounts of enantiomers if there is a stereocenter at the Carbon?

  3. I’m curious about something. In the protonation step, where does the (seemingly) H3O come from that protonates the O^-? The diagram says we are using H2SO4/H20 so does the H3O happen from the H2SO4/H2O interaction? I’m sorry if this is a stupid question.

    1. The Al starts in the -5 oxidation state with [Ar] valence. It prefers +3 [Ne] valence configuration. It pushes off an H- first. That extra electron ends up breaking one of the bonds on the protonated oxygen giving O- which then immediately bonds with the proton (H+). From a charge accounting perspective we have a leftover Li+. The Li+ goes with the HS04- I *think* because the sulfate anion is going to more eletronegative and thus LiSO4 is a lower energy state than if it displaced the hydrogen on the newly formed OH group.

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