Propose mechanisms for the following reactions.
(d)

Question

Propose mechanisms for the following reactions.

(d)

Accepted Answer

Welcome back everyone. Draw a plausible mechanism of the reaction given below. We're given an amine, which reacts with an alpha beta unsaturated ketone followed by an acidic workup to produce an alpha beta unsaturated ketone, but essentially we are forming a 6 membered ring. So we have to recognize that this reaction is Robinson Annulation. Having recognized the reaction side, we can begin thinking about the mechanism itself. So what we're going to do is essentially redraw the starting material. We have a 6 membered ring bonded to another 6 membered ring. Let's redraw the structure. Let's add the methyl groups. And now, let's consider the nucleophilic attack step. Now, since we have nitrogen with a lone pair of electrons, the first step is essentially the formation of a pi bond and the cleavage of the adjacent pi bond. And that specific pi bond is going to attack the alpha beta unsaturated ketone. We're going to draw the structure of the ketone, And perform the attack followed by the formation of a carbanion, right? So let's draw our first intermediate. We're going to introduce a side chain. Let's begin with our first arrow and it represents the formation of a pi bond. Nitrogen is going to get the formal positive charge, and now we are introducing a side chain. Specifically, we are adding 123 carbon atoms. So 1, 2, and 3. Carbano, and then we have a methyl group. We also have to understand that there is a negative charge on carbon and here we are going to consider the addition of an acid, right? To protonate the carbon atom. So we want to draw hydronium And we are illustrating the proton capture. This allows us to form a neutral intermediate. So we're simply removing the negative charge from carbon. Well done. So we have formed our first intermediate, right? Or basically the second intermediate if we consider the previous one. From here, we have increased the electrophilicity of carbon because notice that carbon is double bonded to nitrogen. Right? And essentially this intermediate is resonance stabilized. Since it is resonance stabilized and we have increased the electrophilicity of carbon, we're going to perform the nucleophilic attack step by water, and break the pipe bond between carbon and nitrogen. Nitrogen is going to lose the formal positive charge. And we have formed a carbon oxygen bond. Oxygen is still bonded to 2 hydrogens, and therefore, oxygen gets a formal positive charge. Now from here, another molecule of water is going to deprotonate the resultant oxonium intermediate. Let's draw it out. And we can simply introduce our o h group. Right. Additionally, we are going to observe a proton transfer, so we can just state that we are observing a proton transfer onto nitrogen, so we can just say plus a plus. And this allows us to protonate nitrogen. Nitrogen is going to be bonded to hydrogen. It gets a formal positive charge, and we have formed a good leaving group. So we are demonstrating the loss of the leaving group in the next step. So let's draw the resultant intermediate by removing the 6 membered ring from the structure. There's no necessity to draw the side products. We can just remove it entirely and leave the alcohol group. Well done. And now since we had a loss of the leaving group, we're going to form a positively charged carbon, right? We have a carbocation. From here, we're going to consider the hydrogen atom at the alpha position relative to carbono, and we're going to use water to essentially remove that hydrogen, form a pi bond, and break the adjacent pi bond from carbonyl, which will allow us to form an enol. Right? Because we have acidic conditions. So we can introduce hydronium once again, perform the proton capture, and this allows us to draw the enol form of the original ketone. So we have simply introduced tautomerization, and now we can draw an enol form. Well done. We have our enal form, and now we are going to perform the nucleophilic attack at the positively charged carbon. Which allows us to form a 6 membered ring. So in this step, we are going to introduce a 6 membered ring. We're also going to introduce carbanyl. Oxygen is still bonded to hydrogen, so oxygen gets a formal positive charge. We're going to lose that positive charge at the the original secondary position, but we still have the o h group present. In this step, we want to consider the proton transfer, so we want to remove hydrogen using a base which is simply water. This allows us to form carbonyl. And we have formed a neutral intermediate. From here, we also want to understand that we are going to protonate the o h group, so we can just say that the second step is simply the addition of hydronium. Right? This allows us to form a perfect leaving group. We're going to form water, and we're going to illustrate the loss of water from the given intermediate to form a carbocation. Now that we have a carbocation, the final step is to introduce a pi bond. We have alpha hydrogens available. We have water acting as a base, so we can just remove the hydrogen atom and form a pie bond. This is how we get the final product for this problem. And that's it. We have the final mechanism. Thank you for watching.

Propose mechanisms for the following reactions.
(d)

Key Concepts

Reaction Mechanisms

Types of Organic Reactions

Catalysts and Reaction Conditions

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Propose mechanisms for the following reactions.(d)

Key Concepts

Reaction Mechanisms

Types of Organic Reactions

Catalysts and Reaction Conditions

Watch next

Propose mechanisms for the following reactions.
(d)