Lithium Aluminum Hydride as a Reducing Agent

[JLeBling]

I came across this in EK 30 minute orgo exam, and it left me scratching my head:

1 - I didn't know that Lithium Aluminum Anhydride could reduce a ketone down to and alkane.

2 - Assuming that it cannot reduce a secondary alcohol, why wouldn't the -OH group on the carboxylic acid get reduced as well? (That don't look like no secondary alcohol that I've ever seen).

Errata, or am I missing something?

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[JLeBling]

 

[DrDashInd]

* No errata, LiAlH4 can reduce -COOH (http://en.wikipedia.org/wiki/Lithium_aluminium_hydride)

* You can't reduce a -OH any more than it is already reduced.

Hope that helps!

 

[JLeBling]

Why is one equivalent of Lithium Aluminum Hydride required to reduce the carboxylic acid to a primary alcohol, and another equivalent of hydride required to "completely reduce it"?

This kind of makes it seem as if first the Carbonyl group of the Carbacid is reduced to an alcohol, which is then "completely reduced" (to an alkane?)

But that is wrong, because the Carbacid gets treated as a single group, not disparate carbonyl and alcohol groups... So what is the second equivalent of hydride doing exactly?

 

[Endure]

I came across this in EK 30 minute orgo exam, and it left me scratching my head:

1 - I didn't know that Lithium Aluminum Anhydride could reduce a ketone down to and alkane.

2 - Assuming that it cannot reduce a secondary alcohol, why wouldn't the -OH group on the carboxylic acid get reduced as well? (That don't look like no secondary alcohol that I've ever seen).

Errata, or am I missing something?

You should really look at the mechanism for this. It's the only way it'll make sense. This is a form of nucleophilic acyl substitution that very closely resembles the reduction of an ester, only two steps differ: (a), a hydrogen of LAH abstracts a proton from the carboxylic acid forming a charged carboxylate, and (b) the oxygen on the newly formed carboxylate attacks aluminum. The remainder of the reaction proceeds via NAS.

Here's the mechanism:

(Pulled from Chemgaroo)


Plug and chug questions become much more straightforward once you get the mechanism down.

 

[JLeBling]

I've been looking all over for an illustrated reaction like that!

Many thanks!!!

 

[JLeBling]

If that mechanism is correct, than EK's explanation is faulty.

It seems to me that the hydride dehydroxylates the carboxylic acid first, and then reduces the aldehyde that is formed in the second step.

 

[Endure]

You need two equivalents of LAH because the first equivalent deprotonates the hydrogen on the carboxylic acid, activating the negatively charged carboxylate and forming an O–AlH3 bond. After a hydrogen from the AlH3 reduces the carbonyl, nucleophilic acyl substitution takes over with the second equivalent of LAH

The mechanism never lies.