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Organic Reagents

By James Ashenhurst

Reagent Friday: LiAlH[Ot-Bu]3

Last updated: February 10th, 2016

In a blatant plug for the Reagent Guide and the Reagents App for iPhone, each Friday  I profile a different reagent that is commonly encountered in Org 1/ Org 2. 

Today’s reagent is probably a bit on the obscure side, but it solves a useful problem. Lithium tri tert-butoxy aluminum hydride is lot like lithium aluminum hydride, but with a difference. Like lithium aluminum hydride, it’s a reducing agent. As a source of hydride, it will form carbon-hydrogen bonds.

Unlike lithium aluminum hydride, which is kind of a raging beast,  reducing everything in sight, LiAlH[OC(CH3)3]3 is a lot more controlled. First of all, it only has one hydride to give, unlike LiAlH4, so it’s a lot easier to control the reaction using stoichiometry. Secondly, those big bulky tert-butoxy groups (that’s -OC(CH3)3) help to modulate (i.e. slow down) the reactivity of the reagent. They’re kind of like a fat suit around aluminum that ensure that the hydride can’t fit into tight spaces.

So what’s it used for? One big thing. It will reduce acid chlorides to aldehydes, and stop there. This is a big deal, because aldehydes are very reactive species themselves, easily reduced to alcohols. So if you use just 1 equivalent of the reagent, you’ll end up with one equivalent of the aldehyde. And aldehydes can themselves be used in all kinds of useful applications.

This serves as a way to indirectly reduce carboxylic acids to aldehydes: you can convert the carboxylic acid to an acid chloride using something like SOCl2 or PCl3, and then reduce the acid chloride to the aldehyde with LiAlH[OC(CH3)3]3 .

So how does it work? It’s pretty straightforward actually. Just like with NaBH4, the hydride from Al–H adds to the carbonyl carbon of the acid chloride, breaking the C–O π bond and forming a tetrahedral intermediate. If you’ve covered carbonyl chemistry at all, you should recognize this step as The Most Important Mechanistic Step in Carbonyl Chemistry – the 1,2-addition. Then, we’ve got this negatively charged oxygen which can then come down and re-form the C-O π bond, expelling the chloride ion (Cl-) in the process. This is the 2nd Most Important Mechanistic Step in Carbonyl Chemistry, called the 1,2-elimination. And that’s it.

Note that in the end the other products that are formed are lithium chloride and Al[OC(CH3)3]3. If you’re not careful about the number of equivalents of the reagent, it will add to the aldehyde too. But here we’re generally assuming that you’re careful.

P.S. You can read about the chemistry of CH2N2 and more than 80 other reagents in undergraduate organic chemistry in the “Organic Chemistry Reagent Guide”, available here as a downloadable PDF. The Reagents App is also available for iPhone, click on the icon below!


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9 thoughts on “Reagent Friday: LiAlH[Ot-Bu]3

  1. Regarding the mechanism, is not some sort of complexation between the alkoxide intermediate and an aluminum species the key to the overall success in obtaining pure aldehydes via this method? This chelate could not be further reduced even in the presence of excess hydride (as the reaction between a negatively charged species and a hydride is extremely disfavored), and releases the aldehyde only on quenching with water. Compare with the mechanism for the Weinreb ketone synthesis. That is how I pictured this conversion. I might be off, though.

    Proving my proposed mechanism would be fairly straightforward. Treat an aldehyde with LiAlH[OC(CH3)3]3 and see whether it is reduced down to the alcohol or not. If the aldehyde is left intact, my theory is wrong.

  2. Can be prepared by the addition of 3eqs t-butanol to a cooled suspension of 1eq LAH in THF or ether. Used in situ. Quite stable but would usually be prepared and used straight away.

  3. What happens if you add LAH to an acyl chloride? Would the acyl chloride reduce to an alcohol or would no reaction occur?

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