[lutesium]

While reading the synthesis of migraine medications something has puzzled my mind! How can LiALH4 reduce the two =O groups of Oxalyl Chloride to a hydrocarbon???

Thanks!!!

[spirochete]

This is a special case.  Draw the mechanism up to the diol intermediate.  It's easiest to see with the geminal alcohol/amine side.  The lone pair on the amine allows for the generation of an intermediate that can be reduced by LAH. 

The same general rule applies for functionality adjacent to the other alcohol.

[lutesium]

This is a special case.  Draw the mechanism up to the diol intermediate.  It's easiest to see with the geminal alcohol/amine side.  The lone pair on the amine allows for the generation of an intermediate that can be reduced by LAH. 

How can the amine alllow the generation of an intermediate? And what's the structure of this intermediate???

The same general rule applies for functionality adjacent to the other alcohol.

What do you mean by this sentence? Where's the alcohol??? Or is he the "will be formed" alcohol???

Thanks!!!

[spirochete]

Yes everything I'm saying is in reference to the alcohol intermediate that that results from LAH adding one equivalent of hydride to each carbonyl group.  I referred to this as a vicinal diol.  Vicinal just means "adjacent."

Mechanistically, it's not favored to reduce a plain alcohol with hydride.  That would be an SN2 on an alcohol and OH- is too strong a base to be kicked out in SN2.

But what if OH- were to leave before attack by hydride, creating a carbocation intermediate?  Then it could be attacked by hydride.  A secondary alcohol normally does not just ionize under basic conditions.  Normally you would need acid which is not present here.  But it's more reasonable for OH- to leave if it's going to create a carbocation that is strongly stabilized by resonance.

I'd suggest drawing the carbocation intermediates that result from OH- leaving spontaneously from the alcohol intermediate.  Do each side individually or it will be needlessly confusing.  In both cases loss of OH- creates resonance stabilize intermediates where positive charge is shared by nitrogen.

[english]

I don't agree with the existence of a sufficiently long lived carbocation intermediate, but I do agree with what is being said overall.

here

[spirochete]

I don't this is a case of the neighboring group effect because the nitrogen lone pairs are both in conjugation with the reaction centers.

As for the existence of a discreet carbocation intermediate, I'm not sure.  Are you saying that lone pair donation into the antibonding orbital of the C-OH bond weakens the bond and allows for SN2 to happen?  You could test it by isolating one stereoconfiguation of the diol and treating it with lithium aluminum deuteride. 

[orgopete]

I don't agree with the proposed mechanisms at all. LAH doesn't reduce alcohol because they react with LAH to give the alkoxide. There needs to be a group that can enhance the loss of that oxygen.

The mechanism has to be similar to a LAH reduction of a carboxylic acid. The first step of its reduction is the formation of the carboxylate and hydrogen. Amides react in the same way. So if an amide can be reduced to an amine, that same mechanism can be applied here on this vinylogous amide.

[spirochete]

Orgopete, I should have just looked this link up for the OP right away.  It describes reduction of an amide to an amine by LAH. 

http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch22/ch22-2-2-4.html

The main point I missed is aluminum complexing with the deprotonated alcohol, allowing oxygen to be expelled as part of a leaving group.  Clearly any alcohol created will be deprotonated first by LAH but I glossed over that part.  I assume this is how carboxylic acid reduction works as well, although in that case I know often people people make an ester first to avoid mixing acid and LAH.

[lutesium]

Orgopete, I should have just looked this link up for the OP right away.  It describes reduction of an amide to an amine by LAH. 

http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch22/ch22-2-2-4.html

The main point I missed is aluminum complexing with the deprotonated alcohol, allowing oxygen to be expelled as part of a leaving group.  Clearly any alcohol created will be deprotonated first by LAH but I glossed over that part.  I assume this is how carboxylic acid reduction works as well, although in that case I know often people people make an ester first to avoid mixing acid and LAH.

Great find but there lies a question! How can the second -OH group be reduced as he is far away from the amine or doesn't it make sense???

[spirochete]

It makes sense.  Start drawing out the mechanism for reduction of the vinylogous amide (the one on the right).  Stop at the carbocation intermediate and think about all the resonance structures you can draw.  Hint:  one of them puts a positive charge on nitrogen.  This has a stabilizing effect that is about equivalent to the lone pair being directly adjacent to the carbocation. 

[orgopete]

Orgopete, I should have just looked this link up for the OP right away.  It describes reduction of an amide to an amine by LAH. 

http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch22/ch22-2-2-4.html

The main point I missed is aluminum complexing with the deprotonated alcohol, allowing oxygen to be expelled as part of a leaving group.  Clearly any alcohol created will be deprotonated first by LAH but I glossed over that part.  I assume this is how carboxylic acid reduction works as well, although in that case I know often people people make an ester first to avoid mixing acid and LAH.

I don't mean to break the thread on this topic, but I had just answered another post on how organic chemistry "ought" to be taught. In that post, I argued, "… that if students learn the reaction mechanisms (the logical steps showing how the products form), they can solve any problem with that same mechanism."

I had disagreed originally as the proposed mechanisms were not making chemical sense. (I concede to selling a reaction mechanism book in which one of my main objectives was to make every step logically complete.) My rational for doing so is to encourage students to think mechanistically. By that, I mean that every step must make logical chemical sense.

The problem that I have with this solution or link, is that it fails on those grounds. It does not make logical chemical sense. The distinction is that if only partial steps are required and we are allowed to guess at any missing steps, logical or not, then we are encouraging students to skip writing a reaction mechanism. I argue that doing so begs the issue in this case as it is the absence of a reaction mechanism that initiated this question.

While I am not answering the OP's question in this reply (I am breaking the thread), I am offering up a mechanism for amide reduction from A Handbook of Organic Chemistry Mechanisms. I cannot guarantee the accuracy of this mechanism, however I believe they meet the concept of writing complete logical chemical steps (and improve learning). If one reviews the prior posts, I think they extended this discussion for precisely this reason (for example, no ^-OH or ROH groups are present during the reduction). Had all posters written a complete mechanism as I have done, they may not have offered some of the suggestions.

Disclaimer, it is difficult to refrain from being a self serving advertisement for my books. However, my books are simply an extension of my organic chemistry courses of the 1960s. I knew my professors wrote the reaction mechanisms that were to be repeated on a new problem on an exam. If I wished to write the mechanism for the exam problem, I had to be able to write at least one example. In learning and writing the mechanisms, each step had to pass the test of chemical logic.