Doesn't LiAlH4 reduce carboxylic acids???

[jdpaul14]

Hey guys,

I was working on ochem and did the problem attached...the question wanted to know which compounds were the reagents used to make the product.

I dont understand how you can use LiAlH4 because I thought it would reduce the carboxylic acid to a primary alcohol? In this problem it reduces the carboxylic acid to two primary alcohols?

Hope y'all can help me

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[Kneecoal]

i think it's because LiAlH4 isn't very strong, and so won't reduce as something stronger might.

 

[krazchikin]

I thought that it was NaBH3 (or 4?) that was the weaker reducer...

 

[sixkiller]

Isn't that from the Kaplan subject test? The answer is wrong...

Lithium aluminium hydride will reduce esters and acids to a primary alcohol.

In reality, although lithium aluminium hydride will reduce a carboxylic acid, the reaction is very slow and requires higher temperatures to go to completion.

Carboxylic acids are reduced to primary alcohols in high yield under much milder conditions by using BH3 in THF and acidifying.

 

[nze82]

 

[sixkiller]

Since when does reduction of a carboxylic acid via lithium aluminium hydride generate a gem-diol?

 

[sixkiller]

In the mechanism, RCO2- Li+ reacts with AlH3....

AlH3 donates a hydrogen and coordinates with the oxygen not coordinated to the lithium ion,

AlH2O- leaves and an aldehyde intermediate results.

The aldehyde intermediate reacts with either AlH3 or AlH4- either of which is a sufficient reducing agent. Then the alkoxide coordinated to the Lithium ion is protonated to yield a primary alcohol.

 

[jdpaul14]

Yeah, the answer has to be wrong because it should only be on OH group not two, thanks a lot guys!

 

[nze82]

Since when does reduction of a carboxylic acid via lithium aluminium hydride generate a gem-diol?

Since the time they chose this to be the correct answer! And if you pay attention, the reduction doesn't produce the gem-diol. When The alcohol attacks the C=O, the ring cleaves open, and the intervening O picks up a proton to produce the intermediate product. Then LiAlH4 reduces the two C=O groups (as it always does) to generate the final product!

 

[sixkiller]

Since the time they chose this to be the correct answer! And if you pay attention, the reduction doesn't produce the gem-diol. When The alcohol attacks the C=O, the ring cleaves open, and the intervening O picks up a proton to produce the intermediate product. Then LiAlH4 reduces the two C=O groups (as it always does) to generate the final product!

I'm sorry, I'm not sure about what you're saying.

I agree that the intermediate is correct since once equivalent of alcohol is treated with maleic anhydride.

Then the reaction is simply a matter of ester/carboxylic acid reduction whereby one oxygen from each terminal functional group is displaced by hydrogen.

A gem-diol at one terminal end is impossible because one oxygen would be displaced by hydrogen. AlH20- is the leaving group that takes the oxygen group...

 

[nze82]

I'm sorry, I'm not sure about what you're saying.

I agree that the intermediate is correct since once equivalent of alcohol is treated with maleic anhydride.

Then the reaction is simply a matter of ester/carboxylic acid reduction whereby one oxygen from each terminal functional group is displaced by hydrogen.

A gem-diol at one terminal end is impossible because one oxygen would be displaced by hydrogen. AlH20- is the leaving group that takes the oxygen group...

Can't make it any more clearer than this:

 

[GiTsticker]

Have to agree with sixkiller (sorry nze82). The acid should be reduced to a primary alcohol, not a diol.
R-COOH + LiAlH4 > R-COH

 

[nze82]

Alright! I give up
If you guys can't see what I mean after all the figures I posted, I don't think any further efforts would make a difference either.

 

[sixkiller]

Can't make it any more clearer than this:

But a carboxylic acids and esters are reduced to PRIMARY alcohols.

The -OH and -OEt from the acid and ester respectively are displaced...

When the acid/ester intermediate is reduced in the presence of the hydride, first nucleophilic acyl substitution occurs and then nucleophillic addition occurs.

OH-CH2CH=CHCH2-OH is the final product.

 

[alanan84]

Alright! I give up
If you guys can't see what I mean after all the figures I posted, I don't think any further efforts would make a difference either.

It looks like you're trying to reduce just the carbonyl to an OH instead of the whole carboxy acid...

 

[192LT192]

i agree with everything up until the last step. you would never get a C attached to an OH and attached to OR after LiAlH4. the H Nuc attacks the carbonyl carbon and then when the lone pairs on the O move down to regenerate the carbonyl (bc its so stable so likes to regenerate) the OR is kicked out. so this gives us a new carbonyl that can be attacked again by the H Nuc which is why we get a primary alcohol. i dont get where the gem diol would come from at the other end of the molecule... to me, the same things happens there as i just described.

 

[GiTsticker]

nze82, you're thinking of a carboxilic acid as a ketone with an alcohol next to it. So you are reducing the ketone part while leaving the alcohol alone. This isn't the case though because -COOH is a functional group all it's own. When reduced, you don't just replace the C=O with C-OH, you simply cut the C=O and leave it be with one alcohol left; albeit this bears no resemblence to the true mechanism, but it gets you from reactants to products.

 

[sixkiller]

May as well drive the bus right over the cliff at this point........

RCO2H when treated with LiAlH4 gives

RCO2- Li+ along with hydrogen gas.

RCO2-Li+ then reacts with AlH3 to give

R-CHO2(Li+)(AlH2)

Both oxygens bear a negative charge; one oxygen is coordinated to Li+ and the other to AlH2

AlH2 "takes" the oxygen with it to give AlH2O- and an aldehyde intermediate results (with the remaining oxygen coordinated to Li+)

The aldehyde then reacts with either LiAlH4- or AlH3 to yield an alkoxide that is then protonated with water or hydronium.

The mechanism for the ester is similar but no hydrogen gas is evolved.

In both cases, the -OH and -OEt is displaced by hydrogen. In the case of the ester, the ethoxide can simply act as a leaving group in the presence of the hydride (much like saponification mechanism).

 

[UCB05]

Can't make it any more clearer than this:

I believe the point everyone is trying to make is that the "answer" shows reduction by LAH to not only reduce the ester carbonyl, but the carboxylic acid to a gem-diol (C4 having 2 -OH groups attached to it), which I am also of the opinion does not happen. LAH is a strong reducing agent and should reduce the carboxy group to a primary alcohol, not a gem-diol, first by reducing the carboxyl to an aldehyde intermediate, and then again to a primary alcohol. Unless the question implies some sort of very mild condition, i'd call this reaction bunk until someone gives a solid reason for the reduction stopping at the gem-diol step

 

[DanCanDent]

Primary Alcohol.

 

[lonhosilver]

THe point that you guys are missing is that you are reading to much into the problem and putting the knowledge (which is well respected) about reducing and carboxylic acids and everything without actually looking at the reactants and the products.... First of all you guys are all correct that LiAlH4 reduces carboxylic acids and the ester to primary alcohols, but if you step back and look at the problem long enough you start to think maybe the test writers wanted it to be ketones with and alcohol at one end and an ether at the other end and if that is the case then you would reduce both of those to the secondary alcohols that you see....

 

[Philippines03j]

Agree carboxylic acid is reduced to primary alcohol. We just pounded that in class about 2 weeks ago plus I got it correct on my test.

 

[Sublimation]

Its primary guys. I attached the mechanism to this reaction. Its too late and im sleepy, so i didnt leave any comments, or highlight any parts.

 

[vhlam]

did anyone take into consideration how the double bond might affect the reduction reaction?
http://forums.studentdoctor.net/showthread.php?t=432583

 

[sixkiller]

did anyone take into consideration how the double bond might affect the reduction reaction?
http://forums.studentdoctor.net/showthread.php?t=432583

This is a selective reduction, and the alkene does not react.

Therefore, hydride reduction is prefered to reduce a carbonyl if an alkene is present.

If an alkene and carbony are both present hydrogen gas and raney nickel will selectively reduce the alkene.

*hydrogen and Pt, Pd, or Raney Nickel will reduce a ketone or aldehyde to a secondary or primary alcohol while aryl ketones/aldehydes are reduced to aliphatic substituents.*

 

[sixkiller]

Furthermore, hydride reductions of alpha/beta unsaturated ketones and aldehydes still yield the selectively reduced product.

Look in any organic chemistry book. Most books discuss what to do when both an alkene (usually conjugated with the carbonyl) and carbonyl are present and one functional group must be selectively reduced.

 

[vhlam]

i'm actually looking through my ochem book (vollhardt) right now but can't seem to find anything on this.

there is a similar problem in the destroyer and the answer is as expected, where the carboxylic acid is reduced to a primary alcohol and the double bond is untouched.

how come the ester is still intact? aren't they supposed to be reduced also?

Edit: Ok i see where you said the ester is supposed to be replaced too. So the answer provided is completely wrong?

 

[miedvied]

nze82, you're thinking of a carboxilic acid as a ketone with an alcohol next to it. So you are reducing the ketone part while leaving the alcohol alone. This isn't the case though because -COOH is a functional group all it's own. When reduced, you don't just replace the C=O with C-OH, you simply cut the C=O and leave it be with one alcohol left; albeit this bears no resemblence to the true mechanism, but it gets you from reactants to products.

GiT nailed it on the head.

nze, you're awesome, I generally think your posts are on the money, but in this one you're mistaken about the geminal diol. When the carbonyl regenerates, the alcohol pops off - LiAlH4 always breaks carboxy acids into primary alcohols.

 

[vhlam]

can someone provide proof (preferably online) that a beta unsaturated carboxylic acid will be reduced into a primary alcohol and not a geminal diol?

 

[UCB05]

did anyone take into consideration how the double bond might affect the reduction reaction?
http://forums.studentdoctor.net/showthread.php?t=432583

To that question I would say it does nothing. After hydride attack on the carbonyl carbon, I don't see any possible resonance structures involving movement of the beta unsaturated bond short of regenerating hydride shifts, and that shouldn't happen. It's one bond too far from offering any resonance stabilization.

 

[Sublimation]

can someone provide proof (preferably online) that a beta unsaturated carboxylic acid will be reduced into a primary alcohol and not a geminal diol?

I posted the mechanism. Which is self explanatory. If you look at the mechanism u will understand why u cant make a geminal diol.

 

[vhlam]

I saw the mechanism (thank btw), but that other thread confused me, and sixkiller said most ochem books should say what to do in case of a situation like this. I'm just looking for some concrete evidence... not that I'm doubting you guys

 

[Sublimation]

I saw the mechanism (thank btw), but that other thread confused me, and sixkiller said most ochem books should say what to do in case of a situation like this. I'm just looking for some concrete evidence... not that I'm doubting you guys

I kno how you feel. Always make sure ur source is credible. Not saying we are not credible here (well technically we arent) but sometimes information here can be a tad misleading, always do ur own investigations...and personally i dont get my info online....i use wiki sometimes. however always use a text book. U should have access to them at ur local library. To be honest wat i recommend u do, is catagorize the reactions....u kno ixidations, sub, elect add, elect elimin. , nuc add, nuc sub, reductions and so on. Ull prolly have to spend a good chunk if not ur whole study time for the day on this but after that ust skim through them everynight right before bed and they will roll off ur tongue in about a week, if not less.