How to determine the regiochemical product of a Diels-Alder reaction.
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- Ok for finding resonance for the dienophile, in the beginning of the video, couldn't we bring down the oxygen's double bond to the carbon on the left and kick out the bond of the carbon to make it have two electrons Which makes it electron Rich, instead if how you did it?(5 votes)
- That gives you a higher-energy resonance contributor, because it puts a positive charge on the electronegative O atom.
The contributor with the positive charge on the C atom is more important, because the C atom is less electronegative than O.
So the reaction will go the way he did it.(9 votes)
- in resonance structure ,the oxygen in diene, which is electron withdrawing group due to its high electronegativity than carbon, why its giving away its lone pair of electron instead of pulling carbon's pi electron like you did for dienophile(4 votes)
- Oxygen can't have more than an octet of valence electrons, so the only way to do that would be to break a carbon-oxygen bond ...(3 votes)
- How are you able to recognize that you have to use resonance structures to determine that that specific product is the major product in the example given?(4 votes)
- The diene is asymmetric with an electron-donating group on it.(1 vote)
- if two differnt corbonyl group present at dinophile terminal then how we decide exo an endo product?
i think that corbonyl which has more tandancy to donating electron density. m right? suggest a example(2 votes)
- Dear sir,
I conducted the Diels-Alder reaction using 3-hydroxy-2-pyridone(Diene) with N-phenylmaleimide(Dienophile) in the presence of organocatalyst. The recent reports shows only the formation of Endo adducts. But, I would like to cross-check the my afforded products weather endo or exo using H1NMR coupling constants (J Values). could you please help me for solving this problem....?(2 votes)
- But do you get any of the other product? Say this was on the exam, should we just draw the favored one or do we draw both and say one is favored over the other? Also will there be an enantiomer, because I remember that when the dienophile is trans, you get plus enantiomer. Thanks!(1 vote)
- Depending on the substituents, you can get either enantiomers or diastereomers.
Unless you are asked to give both major and minor products, you should be safe in giving only the major product.
Clarify this with your instructor before the exam!(2 votes)
- if we rotate the dienophile 180 degree, can't we get another product?(1 vote)
- Yes. One orientation gives the "para" product, and the other gives the "meta" product.(1 vote)
- Why is this the diene which turns on itself and not the dienophile ?(1 vote)
- You will want the dienophile's double bond to face the diene.(1 vote)
- What is the stereochemistry in this product? Would the ketone be down (dashed bond) to the cyclohexene?(1 vote)
- i think ketone would be outward (wedge) as it's right to and outside the DA bonding(1 vote)
- why do yo draw the resonance?
I think your purpose are to show different product, is it right?(1 vote)
- Resonance was drawn to reveal which carbon on the diene was electron-rich (partial negative) and which carbon on the dienophile was electron-poor (partial positive). This was necessary as these opposite partial charges attract, resulting in the regioselective product forming.(1 vote)
- [Voiceover] For this Diels-Alder reaction, I've added on an electron donating group to the diene. So here's the diene, and notice there is a methoxy group attached to this carbon. That means there are two possible regiochemical outcomes for this Diels-Alder reaction. So we could form this product, or we could form this product. So which product is favored? To figure that out, we need to draw resonance structures for both the diene and the dienophile, and let's start with the diene. We could take this lone pair of electrons on this oxygen and move them into here, and then push these electrons off onto this carbon. So let's draw that resonance structure. So now this oxygen would have a double bond to this carbon, and then this carbon would have a lone pair of electrons on it. Let me go ahead and finish drawing in the rest of these bonds, and this oxygen still has one lone pair of electrons on it which gives this oxygen a plus one formal charge, and this carbon would get a negative one formal charge. Let me highlight that carbon, so this carbon right here gets a negative one formal charge. Let me draw in my resonance brackets here. Let's look at the dienophile next. So we know that this oxygen is electronegative, so electron density is going to flow towards that oxygen. I could take these electrons and move them into here and these electrons come off onto the oxygen, so let's draw that resonance structure. We would have a double bond here, and our oxygen would have three lone pairs of electrons around it, which gives that oxygen a negative one formal charge. Let me put in my resonance brackets, and notice we took a bond away from this carbon, so this carbon gets a plus one formal charge, so our diene has a carbon that is electron rich, and our dienophile has a carbon that is electron poor, and we know that opposite charges attract, so we could just line up these two carbons, and that allows us to predict the regiochemistry for this reaction. Technically the Diels-Alder reaction is not an ionic reaction. It's a pericyclic reaction. But this trick does allow you to predict the product, so let's go ahead and use it. Let's get some more space down here, and let's redraw our dienophile first. So let me draw this in, so we had our double bonds, and then we had our carbonyl. We know that this carbon down here is the one that is electron poor, so I'm saying that's partially positive, so we need to line up that electron poor carbon with the electron rich carbon on the diene, so we want to make this carbon the electron rich one, and that means the methoxy group must come off of this carbon, and now we can see the regiochemistry for our Diels-Alder reaction. Remember the Diels-Alder reaction is a concerted movement of six pi electrons. So these pi electrons gonna move into here to form a bond between these two carbons. These pi electrons move into here to form a bond, and these pi electrons move down. So we form our cyclohexene ring, and then we would have our methoxy group coming off of this carbon, and we would have our ketone coming off of this carbon, so following our electrons, so these electrons in red formed this bond, and then our electrons in blue formed this bond, and our electrons in magenta formed this bond. So this turns out to be the product of our Diels-Alder reaction, and let me go back up to here, so that's our product as opposed to this one.