Master Organic Chemistry Reaction Guide

Reduction of carboxylic acids to primary alcohols using LiAlH4

Description: Addition of lithium aluminum hydride [LiAlH4]  to carboxylic acids leads to the formation of primary alcohols (after addition of acid)
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15 thoughts on “Reduction of carboxylic acids to primary alcohols using LiAlH4

    1. Do you have any specific suggestions for how to make it better? I’d love to hear them! Send by email or the anonymous feedback button above.

    1. What’s probably leaving here is the aluminum salt, Li(+) (-)O-AlH2 . You could draw a resonance form as Li(+) O=AlH2 (-) ; the strong Al-O bond stabilizes the negative charge that will be formed here. I should probably adjust the drawing, thanks.

  1. Most of the reactions I’ve done in lab I don’t do an acidic workup. This tends to produce low yields. It is best to use a NaOH(aq) soln between 15%-25% this causes much of the salts to precipitate out and you can almost just pour off your organic layer and have some close to pure product. This is pretty standard via Fieser and Fieser for metal hydride workups…
    For every mole of carboxylic acid use 3-5 molar equivalents of LiAlH4. Typically run with a THF solvent under inert atmosphere/ dry conditions, in which the carboxylic acid soln is added dropwise to the LiAlH4 suspended in THF at 0*C. Workup is for every Xg of LiAlH4 add XmL of H2O slowly, XmL of 15-25% NaOH(aq) soln, 3XmL of H2O, then allowing to warm to room temperature and stirred for variable amounts of time, then separated, precipitated salts washed , and combined organic layer and extractions dried with MgSO4, filtered and your done. Through it in a column and you get even analytically pure product.
    Anyways I personally love the detailed mechanism, but could you maybe do one for a basic workup as well. I’m finding organic mechanisms and reactions in my textbooks are not the whole story or are not conducive to what really goes on in lab.

    1. You are absolutely correct. The Fieser workup that you mention is the most practical way to work up this reaction. “Acid” could refer here to water, which serves as a proton source for the aluminum hydride.

      I had to make a choice when starting this site: cater to students just starting out, or to aim matters at a more advanced type of reader. Due to the relatively scattered resources available for students in the first category, I chose the former.

      You are clearly in the second category. There are more resources available for someone in your situation, such as March’s Advanced Organic chemistry, Not Voodoo, Chemical Forums, Reich’s site at Wisconsin, and a lot of other blogs and websites.

      I wish I could easily please both groups of people but I had to make a choice. Thanks for understanding. James

      1. Yeah I use Not Voodoo all the time it is an amazingly helpful website. I’m personally having trouble with purification/isolation of my product (nicotinic acid –> 3-pyridinemethanol) due to functional groups that like to make Zwitterion’s which is why I asked. So I’ll definitively have to check out these other sources. Thanks alot!

          1. Hi, I don’t know how old this entry is, but I was having a similar problem with reducing a carboxylic acid which is ortho to a methoxy and para to an amine using LAH. However, I converted the acid into a methyl ester and then it worked! I just can’t find a reasonable explanation as to why it would work on an ester and not the acid. I thought that Li might be making a complex between the carboxylate on the methoxy oxygen?

          2. – additional comment:

            Also when i tried reducing the acid, then very little reacted and could be extracted into the organic phase – meaning that pprox. 80% of the acid stayed, unreacted in the water phase… this didn’t happen with the ester

          3. The issue is that reducing carboxylic acids requires considerably more forcing conditions, since the first thing that will happen is that LiAlH4 will react with the acidic OH to make the carboxylate, and the carboxylate is even harder to reduce than the acid (O- is a better pi donor into the carbonyl).Esters don’t have that problem.
            LiAlH4 likely would have done the job, you just would have to heat it more and for longer.

    1. You can, in theory, it just requires very forcing conditions (high catalyst loading, high concentration of H2).

  2. If heat is introduced to the product in the presence of acid another reaction may occur. What reaction would occur and what is the product that would form?

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