Propose a mechanism for the following reactions:
a.

Question

Propose a mechanism for the following reactions:

a.

Accepted Answer

Welcome back everyone. Illustrate the reaction mechanism for the following. We're given a 2 comma 5 diethylfuran which reacts with hydrobromic acid in the presence of water or simply an aqueous solution of hydrobromic acid to produce a diketone. And if we want to think about the mechanism of this reaction, first of all we have to recall that the lone pair within furan is delocalized. Right? So apparently we are going to observe a protonation step, but we are going to consider it not from oxygen itself with its lone pair, but we're going to illustrate the delocalization of this lone pair into our conjugated system and use the py bond from the ring where the protonation stop. So let's illustrate that. Now we have our first arrow showing the formation of a pibond, and the second arrow where the proton capture from hydrobromic acid. We're going to draw the resultant intermediate. As we can notice, carbon gets hydrogen, and now oxygen has a total of 3 bonds and therefore it gets a formal positive charge. This makes carbon bonded to oxygen more electrophilic. And we're given water, right? Because this is an aqueous solution of an acid and water is a good nucleophile. It would attack the electrophilic carbon. When we attack the electrophilic carbon, we're simply breaking the pi bond between oxygen and carbon. What we're going to do is essentially introduce an oxonium ion oxygen is now bonded to 2 hydrogens. Right? Because it comes from water, that same oxygen is bonded to carbon, so it gets a formal positive charge. Now from here, our next step is the deprotonation step. We have the bromide anion formed previously, and it can deprotonate oxonium. Well then, we have our first intermediate formed and now we can repeat the process. Now we're going to take oxygen from the ring. Right? Because eventually we are opening the ring. And illustrate the proton capture using hydrobromic acid. Now our system is not aromatic anymore. Right? We don't have a planar ring since one of our carbon atoms is apparently s p 3 hybridized. That's efficient. Right? So we can illustrate the protonation step directly. Oxygen is going to capture hydrogen. And therefore, oxygen gets a formal positive charge. Now from here, we can essentially understand that oxygen adjacent to it from the alcohol group has a lone pair of electrons, so this is how we can open the ring. We can use that lone pair of electrons and delocalize that lone pair of electrons into the formation of a pi bond, eventually opening the ring, transferring the electrons onto oxygen within furan. Right? So here, what we're going to do is just illustrate the formation of a cyclic structure. Now, I want to pay our attention that the reaction has reversible arrows so for complete accuracy, we're just going to replace all of our arrows with reversible arrows. Well done. Now we want to draw our acyclic intermediate. And if we look at the number of carbon atoms, we have 1, 2, 3, 4, 5, 6, 7, 8. Right? So we want to draw an 8 member chain. And what we notice is that carbon number 3 from the left is going to have that carbon oxygen single bond. Right? So we're going to draw carbon oxygen single bond. Oxygen is going to have hydrogen bonded to it. So we're drawing an alcohol group and there's also a pi bond. Right? So we're going to introduce that pi bond, which essentially tells us that we have an eno. And now if we look at the right part of the ring, carbon number 3, 123, is going to have carbonyl. In addition to that, oxygen is bonded to hydrogen so oxygen is going to have a formal positive charge. So we have our next intermediate and we can essentially deprotonate it. Right? Because oxygen has a formal positive charge. So we want to illustrate a deprotonation step using bromide. Right? It can act as a base. We're capturing a proton and forming carbonyl. Well done. Our next step is going to be a protonation step once again because we have an enol and there's plenty of acid available to us. So we have a lone pair of electrons on oxygen, we can form that carbonyl, and now the ribon will essentially attack the acid, so we are going to observe the protonation step. Specifically we're attacking partially positive hydrogen, and this is how we're illustrating the proton capture. Let's draw the next intermediate form. And essentially what we want to do is draw the carbon oxygen double bond. Oxygen is going to get a formal positive charge. And from here of course, our final step is the deprotonation step, because as we can see we are very close to the final product, so we can once again use bromide as a base for the final proton capture which yields final product. Well then, we have our mechanism for this reaction. Let's label it and thank you for watching.

Propose a mechanism for the following reactions:
a.

Key Concepts

Reaction Mechanism

Nucleophiles and Electrophiles

Transition States and Intermediates

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Propose a mechanism for the following reactions:a.

Key Concepts

Reaction Mechanism

Nucleophiles and Electrophiles

Transition States and Intermediates

Watch next

Propose a mechanism for the following reactions:
a.