The following reaction was used in the synthesis of aculeatin ...

Question

The following reaction was used in the synthesis of aculeatin A, a natural product that is active against KB cell lines. Although it only worked under acidic conditions, a mechanism can be drawn where the reaction might proceed under basic conditions. Suggest this mechanism (J. Org. Chem. 2014, 79, 1498–1504). [The most acidic proton is indicated . . . and number your carbons!]

Chemical reaction diagram illustrating the Michael Addition mechanism for synthesizing aculeatin A, highlighting the most acidic proton.

Accepted Answer

Hi, everyone and welcome. Our next problem says the following reaction occurs in acidic conditions, but a mechanism can be proposed for when it occurs in basic conditions. And we have a reaction shown a really complicated molecule with several double bonds, hexen ring, triple bond that's conjugated with a ketone. We've got two alcohol groups and it's reacting in a base. We have ko lowercase TB U and TB uoh tt beetle alcohol. And our reactant product ends up with three different rings. The cyclohexane ring with a ketone group on it that the original molecule had. And then a five atom ring with an oxygen incorporated in it and then a six atom ring with an oxygen incorporated. And another ketone in the proposed mechanism for the basic or is the proposed mechanism for the basic conditions shown below possible. Note, the most acidic proton is below is shown. So let's work our way through just looking at each step in this mechanism and determining whether overall this is a possible mechanism for this reaction of basic conditions. So let's start by numbering the sort of tail. Our mo almost looks like a kite with a tail and our reaction is going to be taking place mainly in this tail that's going to end up as these rings. So I have that cycle heene ring, two carbon, carbon double bonds within it. It has a kone group on it. And I'm going to number the Carmens coming down that tail. So starting from the ring, the first bond is to carbon one. There's also an alcohol group on that ring carbon that's attached to. And we have carbon two, carbon three and four are connected by a triple bond. Carbon five has a carbonyl group on it. Carbon six, carbon seven has another alcohol group as a substituent and then carbon eight. So now we have sort of numerical terms by which to refer to the different carbons here. So in the first step, it was indicated that our most acidic hydrogen is from the oh group that's attached to the ring. So along with carbon one, and we're in basic conditions. So our base molecules shown with the one of its lone pairs coming and grabbing the proton from that oh group. So the hydrogen without its electrons and the electrons from the ho bond going on to the oxygen and leaving it with a negative charge. So the first thing we want to check when checking the mechanisms is the electron pushing, correct. Does it make sense? Is it going in the right direction? So our electron pushing is OK. And we have now generated this negative oxygen. So that was going to be our Nile. So generation of Nile and that Nile was generated by de protonation. So now we go to our next step where we're going to see a nucleic attack. Now let's look carefully at what's happening here. The oxy negatively charged oxygen nucleo file is attacking. What if we look closely is the beta carbon of an alpha beta unsaturated carbon. So we have a nucleo nucleic attack because carbon five in my tail here is a carbon carbon and then it has an alpha carbon and carbon four and a beta carbon and carbon three, they're connected by a triple bond rather than what we see more regularly, a double bond, but that's still unsaturated. So we have essentially an alpha beta unsaturated carbonyl. So we're talking here about an ox michael reaction and these oxy reactions or when an oxygen based Nile such as our alcohol attacks, an alpha beta unsaturated carbon. And this involves the formation of a new carbon carbon bond. So let's look at the next step in the mechanism we have that negatively charged oxygen, its lone pairs are attacking the beta carbon of our triple blonde there, thereby forming a new bond. And now with this new bond forming. And the other thing that happens, we have our oxygen, the lone pair attacking that being a carbonell. This is a conjugated addition. So we have multiple things happening at once. Then a pair of electrons from our triple bond jumps over and forms a new double bond from the alpha carbon to the carbon of carbon. Em in a pair of electrons from the double bond in the carbonel group goes on to that oxygen atom giving it a negative charge. And essentially the series of electron movements means that the negative charge from the nuclei is delocalized in the alcide anion. So I'm going to number these carbons again. So we can keep track of where things went. So we have this newly formed five atom ring. The oxygen in that ring was the oxygen from our Nile. So that means it has attached itself to the beta carbon from that alky which was carbon three. Well, number three there. And that means as we trace back in this ring to the left of carbon three will be carbon two and then carbon one which was attached to the psycho heine ring. So our new five atom ring here has a carbon from cyclohexane ring and then come 123. And that's attached to the oxygen, which was the alcohol group. Carbon three. Now has a different arrangement. It had been part of a triple bond. It now has a double bond to carbon four instead of a triple bond as a pair of those electrons jumped over to the next bond between carbons four and five. So put number five there. So carbon four now has double bonds on either side So carbon five was the carbon carbon. Now, we know that in our previous step, the electrons from the double bond to the oxygen went on to the oxygen giving it a negative chart. You see that's there. And then we go on to carbon six, which is the alpha carbon on the other side. And then carbon seven which has an alcohol group and then carbon eight is there at the end of the chain. So we've kind of found ourselves again here. And then in this step, what's happening is we have two things, the double bond between carbon five and carbon, excuse me, carbon four and carbon five, a pair of electrons from that double bond grab a proton. So we're essentially adding a proton to that double bond, bringing it down to a single bond. We see it's grabbing a proton from the molecule we generated when our base detonated the alcohol group in the first step. So we've grabbed a proton with those pi electrons which would leave a positive charge on one of the carbons, except that one of the lone pairs smart negatively charged oxygen has come down to reform that carbon bond. So we've reformed co bond c double bond. Oh I should say now, we're focusing just on sort of the lower part of this molecule. So I won't number all of the carbons again. But we have, I'll start with our five atom rings of carbon three that's within that ring, carbon four. So we have that double bond between three and four, carbon five, which is our carbon, carbon, carbon six, carbon seven had an alcohol in it and then carbon eight. So now we're going to sort of repeat step number one, we have that other alcohol group down off of carbon seven. And we're going to deproteinate that alcohol group with the base in our solution. So de protonation of another alcohol, now we'll be able to do another micro reaction because we are alpha beta unsaturated bond was a triple bond. So we still have a double bond there. We still have can do this conjugated addition. So we deproteinate that alcohol and we're going to do another one of these micro reactions. So when we look at our next step and we'll go ahead and label those carbons again. 34, 678 that negatively starts oxygen is acting as a nuclear file. It's attacking carbon three, which is are beta, carbon beta in an alpha beta unsaturated carbonel. So again, we have that series of reactions. After as part of this conjugate addition, the after the lone pair electrons make that bond from oxygen to carbon three, then the pie electrons from the double bond between three and four jump over to form a new double bond between carbons four and five, five is the carbon carbon. So the electrons from that co double bond hop up and go to the oxygen there So again, we see that the negative charge to the nuclei is delocalized in the alcide anion. So that was a second scroll up a little bit Michael reaction. And since this is a reaction that's happening internally, instead of with another reagent, we've made another ring with this reaction. So now we have this additional six carbon ring on the bottom incorporating in oxygen. So we still have carbon three at the bottom of that five atom ring. But then that goes right into a six atom ring. So carbon four now has a single bond between three and four due to our reactions and then a double bond between four and five, a single bond to a negatively charged oxygen there. Then we have carbon six, carbon seven has a methyl group that is attached to the oxygen in the ring. And now we reached our final step where we're going to regenerate our carbonel group. And we're going to reduce our carbon carbon double bond that's was between carbon four and five by adding a proton again from the molecule we generated when we dep proin the dep proton, the alcohol. So numbering our carbons one more time, 34, 5678. Notice we now have a ring that we formed in our last step. Carbon seven, which had an alcohol group that is now bonded to it has its oxygen bonded back into carbon three. So forming this six atom ring. So we've now made an ether from that alcohol. And we have our net loan pair from our negatively charged oxygen has moved over to reform the carbon co double bond. Meanwhile, that the pi electrons from that double bond between carbons four and five, I've grabbed a proton off of again, that molecule that we generated when we detonated the alcohol with our base. So we've got our final product here where we've generated two new ether rings. Let me perform two Michael editions. So back to our original question, if we remember that is the proposed mechanism for the basic conditions down below possible. And the simple answer is yes, it is possible. All the steps make sense. Our electron pushing is correct and we have two Michael editions taking place. See you in the next video.

Key Concepts

Acid-Base Chemistry

Reaction Mechanisms

Natural Product Synthesis

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