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I'm stumped on this question, could anyone propose a reasonable mechanism for the attached reaction?
How might I yield this product? (Ochem)
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A mechanism or a synthesis? For mechanism, it's just an "oxidation" of an alcohol. One could envision a pathway where you simply pluck off a proton from the alcohol and kick off the methoxy as methanol. It's not a particularly good leaving group, but it looks like it's gonna have to leave if this reaction works.
Edit: "Oxidation" placed in quotes because it's not really an oxidation - alcohol does go to an aldehyde, but the carbon does not change oxidation state.
Edit: "Oxidation" placed in quotes because it's not really an oxidation - alcohol does go to an aldehyde, but the carbon does not change oxidation state.
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Yea this looks like an unusual case of alcohol oxidation with a leaving group. Using one of the weaker oxidizing agents (e.g. PCC) will prevent the aldehyde from being further oxidized to carboxylic acid.
Sorry, I misspoke earlier! It's not an actual oxidation because the carbon atom does not change oxidation state. It started with two bonds to oxygen atoms and ended up with two bonds to an oxygen atom - the oxidation state is the same. The mechanism is basically just collapse of the tetrahedral form - it's the formal reverse of hemiacetal formation.
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So the second intermediate is not quote right. When the base plucks off the proton and collapses the tetrahedral hemiacetal, it becomes the aldehyde - there's no charge buildup on the oxygen. Other than that, as long as this is in aqueous solution, I don't see a problem with it, although it would be better to protonate the methoxy group first as you have it drawn. That makes it into a better leaving group.
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Both are wrong.
Your first one could work but you're going to need LiAlH4 after forming carbonyl to have hydride to attack
The hydrogen on the product you see is from the hydride
There makes no sense whatsoever. There is no reduction going on here. Check the formal charges. Moreover, there is no gain or loss of mass in the reaction. Count your atoms. LAH is not needed.
The first alternative you gave is correct (except there should be a positive charge on the oxygen of the methoxy group after you protonate it - conservation of charge). Protonate the methoxy group first to give the oxygen a positive charge and make it a good leaving group. Then pluck off the proton from the hydroxyl and collapse the tetrahedral form to form the aldehyde.
The problem with your second alternative is you have a very unstable carbocation. A good rule of thumb to remember is that carbocations are good only with tertiary carbons.
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The formation of hemiacetal is a reversible process under acidic conditions, as seen below
No redox reaction necessary
EDIT: here's the general mechanism
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No worries! There actually is a carbonyl equivalent - instead of having two C-O bonds to one oxygen, the substrate has two C-O bonds to different oxygens. These are carbonyl equivalents because it's just the hemiacetal/acetal form of the carbonyl.
Not a huge fan of that mechanism. I guess since we are on SDN we just need to know mcat level material, but there is no need to keep doing things in concerted steps. If you are going to have the base deprotonate and create an alkoxide, then why have an acid step at the same time?
What would really happen is a Base deprotonates the alcohol. Then in a new step, the alcohol expels the methoxide as a leaving group. Here is where you guys are going wrong and need to remember your organic 2. The leaving group in a tetrahedral intermediate is not important, its not in the rate determining step. Unless it is prohibitively bad (H- or R-) then it is fair game to leave. What prevents certain groups from leaving in a 1-2 addition to a carbonyl (NOT a substitution!!!) is the initial addition step of the nucleophile.
Sorry to go on a tangent.
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I agree. I just took a picture from the interwebz since i got lazy writing all that down 😛
The mechanism is definitely condensed because that would imply a ternary transition state. How the OP has it is fine. Protonate the methoxy first, then collapse the carbonyl.
Before you reply, I do want to note that in general, you are correct in that additions normally favor nucleophilic addition as the RDS. However, this is not an addition reaction. That's important. Nothing is being added to the substrate - in fact, you're losing things, namely a proton and methoxide.
Furthermore, you are not correct in that addition of nucleophile is RDS invariably. Think about a peptide bond. Studies (I can find them if you're interested) have shown that if you label the carbonyl oxygen, it gets scrambled in aqueous solution. That means that the first step of addition is reversible and the leaving of the LG is RDS. That's because a basic amine is a terrible leaving group. So is an alkoxide.
So it's good to have organic rules in the back of your mind, but in the real world, things aren't so cut and dry as in your organic chemistry class. When we teach organic chemistry, we have to teach it as a set of rules - but you can't take that completely to heart, or else you'll be a terrible chemist. These rules are more like guidelines. In general, these rules apply but once you start accepting them as dogma, then we start to worry. The reason we have to teach it this way though, is that undergraduate organic chemistry courses aren't data-driven. That is, we can't just give you a rate experiment and expect you to figure out the RDS from that - that would take a lot of time and we would have to teach you elementary kinetics to do that. Better saved for another course. So instead, you get a set of guidelines and every time you see something on the exam, it will follow those guidelines because again, we can't include rate or Hammett plots, etc. So in sum, you have to weigh many factors at once and in the end, it's faster to just do the kinetics experiment.
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As an organometallic grad student, I resent that. Jk. Nobody likes organic and these classical organic methods are outdated in labs anyway - at least in organometallics labs 😛
Of course you aren't adding things. This is the 2nd step of the addition, the E1cb mechanism.
Thats not what that means.
https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/crbacid2.htm
That situation is a result of the amide causing a different reactivity of the initial attack at the E+ carbon. But that doesnt affect the rate diagram enough to justify switching the steps and their RDS status. I dont have time to get more data until late this evening.
Of course you run the kinetics experiment. Of course you "weigh all the factors". No one here is saying otherwise.
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Thank you for enlightening me as to the mechanism. I also took graduate advanced organic chemistry and phys org, so I think we both know the mechanism you're referring to. Here's the problem. Look at the substrate. Just look in OP's original post. What's being added here? Presumably you think that the hydroxyl was the carbonyl prior to the addition. So then you have one of three choices. You either added methoxy to an aliphatic aldehyde (methyl formate) or you added an ethyl group via a Grignard or nucleophilic carbene or you reduced an ester with a strong reducing agent. If you argue the former, the mechanism OP wants is just the formal reverse of the addition, governed by microscopic reversibility. That means that the nature of the leaving group must matter. If you argue either of the latter, we can proceed. Now, the point is that either the first or second step could be rate-determining. I know you were taught that nucleophilic addition is rate-determining in these mechanisms. But we teach that because it's a rule of thumb, not incontrovertible fact. If you have a terrible leaving group, the second step could be rate-determining. Which brings me to my second point.
I'm not talking about the amide causing any special reactivity in the first step. What we usually teach in introductory organic chemistry is that amide bonds are not very reactive because of amide bond resonance. This is true when you're comparing amides to, say, acyl chlorides. But the problem with that is that amide bond resonance isn't that strong. See: http://pubs.acs.org/doi/abs/10.1021/ja0663024?src=recsys. Amide resonance is situation-specific. Still, they mainly lie in the region of 15-20 kcals. Now see the problem? An amide stabilization of 20 kcal/mol is within the room-temperature limit for spontaneous reaction. That means that amide bond resonance is not as strong as you think. Now, couple that with a poor leaving group and you could easily have nucleophilic addition being the easy step. That's exactly why the 18-O label scrambles - addition is fast, leaving is slow. I should also clarify the lingo - what I mean by "scramble" is that when you label the carbonyl oxygen, you recover starting material after reaction that has no 18-O label at that site, meaning that water attacked, broke the amide bond, and then instead of going forward and hydrolyzing the peptide bond as you would expect if that was the RDS, it reverted to the substrate and kicked out water - the water is now labeled. Again, your reactivity tables we taught you in Orgo 2 are very good for comparing across different groups, e.g. esters, amides, etc., but 1) should not be taken as dogma and 2) should not be used to compare reaction profiles. Otherwise, you're missing the point.
Your quote: "Here is where you guys are going wrong and need to remember your organic 2. The leaving group in a tetrahedral intermediate is not important, its not in the rate determining step."
That all being said, I do feel satisfied that what we teach in organic chemistry is sticking - at least with one student! Just try not to over-apply rules and instead look at each case separately.
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