Predict the major products of the following crossed Claisen co...

Question

Predict the major products of the following crossed Claisen condensations.

(c)

Accepted Answer

All right. Hi everyone. So for this question we must determine the product formed in the cross clouds and condensation reaction of the given molecules. So let's begin by first noticing that in our starting material here, we have an ester on the right side and we have a key tone on the left right. So we have a car bottle. Now because we have both an ester and another car bottle in the same molecule. This is going to be an intra molecular collision condensation, intra molecular because it's happening within the same molecule. So if we have an intra molecular clouds and condensation, we're going to end up with a ring structure for me. Now, what's interesting about this molecule as well is that if we look at our key tone on the left side here, we actually have two alpha positions, right? We have two alpha positions adjacent to the car bottle and they each have um hydrogen and they're not the same. So that's going to mean that we have two different um products that can be formed. So if we go ahead and we count our carbon starting from the carbon of the ester, You'll notice that we have seven carbons. So carbon number five is one possibility for a kleasen condensation, which means that we're going to form either a five or a seven carbon ring, depending on um the alpha carbon that we're talking about. So what I did on the bottom here was that I just redrew our starting material twice so that we can um discuss both possibilities for this class and condensation. So let me just erase my labels and now let's start with um the alpha position on carbon five. So I'm going to point out one of the hydrogen in our alpha position and our base here, sorry, a fox side, R O E t minus, that's going to dip rotator alpha hydrogen. And in the next step, the pair of electrons from the hydrogen is going to jump in between the alpha carbon and the carbonnel, therefore displacing the pi electrons of the car bottle towards oxygen. So this first step is going to generate our um or in a li ion. So our Esther is going to stay untouched. And now, instead of a car bottle, we have an O minus with an all keen in between um itself and the alpha carbon. So this is going to be our all keen. And now let's go ahead and actually relabel, starting from the carbon over Esther. So we have 1234 and five. The fifth carbon is going to go ahead and attack the car bottle of our Esther. So, using the pi electrons In between carbon five and 6, we're going to see a nuclear attack of the car bottle from the ester and that is going to displace its electrons towards oxygen. Now. Meanwhile, right, R O minus on carbon six is going to jump towards becoming a car bottle again. So like I said before, we are going to end up with a five member ring, this is our five member ring and I'm just gonna go ahead and label our carbon so that we're not confused. Now, carbon one is going to have our own minus from the car bottle and it's also going to have this with oxy group for my Esther. Meanwhile, attached to carbon five is going to be our ketone. So now in the final step of the class in condensation, we're going to see a um a displacement, so to speak. Or rather elimination. The Toxie group from our Esther is going to be eliminated after the car bottle or rather the o minus on position one becomes a car bottle once again, So this is going to be our five member ring and now we have car bottles on Both Carbons one and 6. Alright, so this is the first possibility and this is going to be our first product, which I'm going to write on the top here. So I'm just gonna read write there. So this is our first possibility. So now let's go ahead and draw out the second. So I am again, this is the same starting material, but now we're going to highlight um the alpha hydrogen on position seven relative to the carbon over ester. So once again, our dockside is going to make an appearance And it's going to deep protein eight, that alpha hydrogen causing the pair of electrons from hydrogen to jump in between the carbonnel and the alpha carbon Therefore our car bottle is going to become an oh minus. So we have our, you know, late in a different position. This time, our Esther stays untouched for now. So now we have our own minus and are in a late now between carbon six and 7. So because carbon seven of our kitchen is the alpha hydrogen, it's going to go ahead and attack the carbonnel of our Esther, so it's gonna attack like that and the carbon of our Esther is therefore going to become an O minus. Meanwhile, right, The O - on carbon six is going to become a carbonnel once again after it donates a pair of electrons towards that carbon. So this time we're going to end up with a seven member ring. So this is a seven member ring and it's always a good idea to go ahead and label your carbons. Just that you're keeping yourself in check. Right? So if I go ahead and I label my chain of carbons here, I can see that we have an O minus on position six or sorry card bottle And the minus is actually on carbon one. So there we go. And now r O minus on carbon one is going to become a car bottle once again after it donates a pair of electrons to carbon one and this is going to eliminate the Foxy group from our Esther. So we're going to end up with our seven carbon ring, which I'm not too good at drawing, but and now we have two car bottles on positions one in six, we have two car bottles on positions one in six, and this is going to be our second product. Alright, so these are the two products that are going to be formed after the inter molecular collision condensation reaction. And the reason we have two products is because there are two different um alpha hydrogen. So with that being said, if you stuck around to the end, thank you very much for watching and I hope you found this helpful.

Predict the major products of the following crossed Claisen condensations.
(c)

Key Concepts

Claisen Condensation

Enolate Ion Formation

Crossed Claisen Condensation

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