LiAlH4 also known as lithium aluminum hydride or LAH, is a strong reducing agent meaning it will create less bonds to oxygen (more bonds to hydrogen). Commonly seen is its use on carbonyls such as carboxylic acid, esters, and amides.
For now all you really need to know is that LiAlH4 is going to add H’s to many types of molecules.
For some of LiAlH4’s properties here is the full rundown. We will start by looking at the structure of LiAlH4, or lithium aluminum hydride.
For most reactions in Organic Chemistry you will not need to memorize this structure unless drawing a mechanism. You will also see that LiAlH4 is sometimes abbreviated as simply LAH. Therefore, feel free to use either in reactions.
A better question is: What can’t LiAlH4 do?!
You will soon see that LiAlH4 can be used in a variety of reactions in Organic Chemistry. Some of the most common places you will see it are the following:
(Notice that the carbon of the carbonyl contains an invisible H not drawn in)
Yes, most commonly seen is the use of LiAlH4 to reduce an aldehyde to a primary (1˚) alcohol. This occurs through a nucleophilic addition of H (hydrogen) from our reducing agent, followed by a protonation step.
What this means is we take an H from LiAlH4 and add it to our carbonyl. We then have an oxygen with a negative charge that obtains an H in the last step. (More on the mechanism later)
This will occur the same as aldehydes except that we will obtain a 2˚ (secondary) alcohol due to the fact that ketones contain an R group instead of the H that aldehydes have.
Here, we see lithium aluminum hydride + our acid workup reducing a carboxylic acid to a 1˚ alcohol. Note, this will require 2 successive reduction steps.
Just to make sure you are staying awake, here’s a little challenge for you.
We know that LAH will reduce the blue carboxylic acid, but will it reduce the red alkene? Hmm… I will let you guys think that one over for a bit until we get to the end. (Stay tuned!)
Ester to Alcohols via LiAlH4
Now, we get into esters being reduced to alcohols. We will obtain 2 products for a non-cyclic ester:
- one being the alcohol that forms from the carbonyl which will be primary (1˚).
- the other will be the alcohol that forms from the carboxylate portion, or simply the O + the 3 carbons labeled in red. This alcohol does not have to be primary.
If a cyclic ester is our starting material, we will yield a diol as seen below:
For an epoxide, the situation changes slightly because there is no carbonyl group present. After forming the epoxide through Epoxidation it simply becomes an Epoxide Ring Opening through the base-catalyzed mechanism.
Pop quiz: Can you draw in the expected product? Remember epoxides get reduced to alcohols in the presence of LiAlH4.
The reactions below feature an amide using LiAlH4 to yield an amine, and a nitrile or cyanohydrin being reduced via lithium aluminum hydride to a primary (1˚) amine.
Click here to learn more about the reduction of amides via LiAlH4, as well as other reduction reactions to form amines.
More on Cyanohydrins (Nitriles).
Acid Chloride with LiAlH4
The last situation I want you to be aware of involving LiAlH4 is its use to reduce acid chlorides to primary (1˚) alcohols. This will occur via an aldehyde intermediate similar to carboxylic acid and require a protonation step after the second reduction with LAH.
(*Note: There may be other less common reactions not listed)
Still confused? Check out Clutch’s videos on reduction here
Answer to pop quiz:
About the Mechanism of LiAlH4:
The type of mechanism used will be that of Nucleophilic Addition. The main steps involved are:
1) Attack carbonyl carbon (E+) with your nucleophile (Nu-)
2) Form the tetrahedral intermediate
3) Protonate to from your alcohol
Here is an example of what the mechanism would look like if we used LiAlH4 to reduce a carboxylic acid to an alcohol, by first reducing it to an aldehyde.
See how far you can get with the above mechanism (Scroll to the end for answer)
NaBH4 or sodium borohydride is another reducing agent that is not as strong as LiAlH4 therefore we will use it for only aldehydes and ketones where 1 equivalent of hydrogen is added. It can also be used for reduction of acid chlorides seen above.
Quick note: You may also see a reagent by the name of DIBAL-H (Diisobutylaluminium hydride). If you see it this is the molecule you should think of:
It is another reducing agent out there, kind of like LiAl(OtBu)3 (Lithium tri-tert-butoxy aluminum hydride).
Link to videos on each of the 3 situations mentioned above.
What LiAlH4 (LiAlH4) can’t do:
One big area where LiAlH4 fails is when we want to stop at aldehydes because it is so strong. This is also the case with alkenes and alkynes since they are non-polar groups. LiAlH4 will also not reduce such groups as well.
Uniquely, NaBH4 is able to reduce one special type of double bond. Sodium borohydride or NaBH4, unlike lithium aluminum hydride (LiAlH4), can reduce conjugated carbonyl compounds.
Here’s an example:
NaBH4 vs LiAlH4
If you would like a quick refresher on non-polar groups & polarity here are 2 links you can check out!
A Quick Derivative of LiAlH4:
Just in case you see D’s instead of H’s in LiAlH4 don’t be alarmed. Deuterium is a heavy isotope of hydrogen and can be treated in a similar fashion as seen above!
So remember, these reagents such as LiAlH4, NaBH4, DIBAL-H and Lithium tri-tert-butoxy aluminum hydride are all reducing agents, which means they will add H’s to a molecule. (There may be a few others not mentioned as well but these are the most common)
They will add H (hydrogen) via a Nucleophilic addition and primarily yield alcohols except in a few cases.
In all reactions you can assume acid workup as your final step, however most cases involving these reagents will not require a mechanism.
In the case that you do need to know the mechanism for reduction, here it is!
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