Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Altered State on September 24, 2013, 05:46:06 PM
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This is the part of a synthetic route where I'm stucked now...
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi.imgur.com%2FO6ej3pV.png&hash=8f6070bb157eeb7557f3bb10b4ea22df692967fa)
In first step, the only thing that can occur is that LDA substracts an acid proton α to carbonyl, so we get the enolate, [P]
I think that next step, MeI, will do an alkylation on that alpha postion to carbonyl. So Q and R might be the same but with different stereochemistry, (One R and the other one S). But if this were true, I don't see the purpose of H3O+ adding as third step. May this protonate ketone? or nitrogen? I still don't see why if that were the case.
Then I've been told that the compound that we add the peroxyacid to is a bicyclic compound, and that PhCO3H "will attack preferably one of the sides of the bicycle", and to know which one would be and why.
I guess it can be Baeyer-villiger on ketone or epoxide formation on alkene. I think Baeyer-villiger would be more favorable, but I'm not sure.
But the real problem anyway is that I can't figure out what happens exactly in the [P]--->Q+R steps. Hope you can enlighten me
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But if this were true, I don't see the purpose of H3O+ adding as third step. May this protonate ketone? or nitrogen? I still don't see why if that were the case.
It's probably just a workup step. I can't see why it was included in the question, but I might be missing something.
There is no ketone present. Revise your functional groups!
I guess it can be Baeyer-villiger on ketone
There is no ketone present. Revise your functional groups!
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As Dan mentioned there is not a ketone present. There is a carbonyl, but no ketone. Do you know what a nitrogen next to a carbonyl in cyclic form is called?
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Oh, yes, it's a lactam so.
Btw, what's in your opinion Q + R? Both lactames the same as the starting compound but with methyl group (one case R and the other one S) alpha to carbonyl?
And then... Would peroxyacid be able to attack lactame? I think not... So the only posibility is epoxide formation? And S+T are both S and R products again? I'm not sure of this...
What's you opinion? I don't usually ask for fast help, but this time I need this for tomorrow, I'd really appreciate help..
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In this case I believe that you are correct. The conditions are what need to be more specific. Under kinetic control the mono alkylation is going to be acheived. Though there is regioselectivity, steroselectivity will not be acheived in this case. If the conditions were more thermodynamically favorable polyalkylation could occur. Also there is slow intermolecular aldol condensation that can occur, but again, in this case I think it is just asking that you recognize there is no steroselectivity. The epoxidation is the only viable process I see happening, but If I remember correctly it will be exo and endo product along with any stereoisomerism.
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Ok, if no one has anything else to say and you think I'm correct, I'll take it like that.
Anyway, this exercise being so easy compared to the other ones I've been told to do still smells bad for me...
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What does that last part mean?
Still smells bad for me??
Pardon?
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What does that last part mean?
Still smells bad for me??
Pardon?
Rofl, I misexpresed myself so bad :D (used an spanish expresion)
I mean, seems bad/strange to me.
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Yes I understand what you are saying. I have been looking at it and it seems straight forward. Like I said, I think the poly alkylation is the only other product in that first reaction. It is an amide and its not like you are going to make the ammonium salt and have Hoffman elimination. Thats a no for sure. I think we came to the right conclusion. Please let me know if I was wrong.
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Yes I understand what you are saying. I have been looking at it and it seems straight forward. Like I said, I think the poly alkylation is the only other product in that first reaction. It is an amide and its not like you are going to make the ammonium salt and have Hoffman elimination. Thats a no for sure. I think we came to the right conclusion. Please let me know if I was wrong.
Sure. I'll do tomorrow as soon as I come back home
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I'd guess two different diastereomeric monoalkylations, followed by two different diasteroemeric mono-epoxidations. I'm not 100% sure about amide reactivity with peroxyacids, though. I'm guessing the H3O+ is there to quench excess LDA?
Extra info you don't need: The favored diastereromeric product is pretty easily predicted also. Cis fused bicycles have two distinct faces, called exo and endo faces. One of the two faces is much more sterically accessible. This is a good time for model building. If I'm remembering the terminology correctly it's the exo face that is much easier to hit.
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It was right this way.
Thank you.
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First of all, I don't know the answer. I am a little skeptical that H3O+ is neutralization, especially as the product(s) is neutral. If it were not, then the product would have reacted. This could lead to a cascade of different products.