(••••) LOOKING AHEAD In Chapter 19, we will learn about the hy...

Question

(••••) LOOKING AHEAD In Chapter 19, we will learn about the hydrolysis of t-butyl esters. In the reaction below, the hydrolysis is coupled to the decarboxylation reaction learned in this chapter. Suggest a mechanism for this reaction. [Hint: The formation of t-butanol proceeds by an SN1 reaction.]

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Accepted Answer

Welcome back. Everyone in the reaction below the hydrolysis is coupled to the decarboxylate reaction. Suggesting mechanism for this reaction ignores stereo chemistry. As we can see in the given problem, we're given a carbonel compound. It reacts with an aqueous solution of sulfuric acid and then it is heated to produce a ketone carbon dioxide and an alcohol. So let's consider this mechanism. And first of all, we're going to perform a protonation step because we are using sulfuric acid, which is a strong acid. So if we expand the acidic hydrogen, we end up with H OS 03 H. The question is which oxygen should we proton? And the answer is the car bals oxygen at the ester because if we think about the resonance structure, specifically alcoy oxygen is an electron donor, right. So we are going to have a formal negative charge on that oxygen and it would be the oxygen atom that is most likely proton, that we have a proton capture. And we are producing a proton at intermediate, meaning we simply want to add the oxygen hydrogen bond. And of course, oxygen gets a formal positive charge. As we understand it is resin stabilized. So we can shift our p electrons on to oxygen and obtain a formal positive charge on carbon, which tells us that the electrophysi of that carbon has increased. There's no necessity to worry about resonance forms. We're simply going to draw the nucleic attack step, which is the next step in the reaction and the Nile is the water molecule. So we are attacking the electro pilic carbon and breaking the pi bond. This is how we're going to draw our next intermediate. And to draw the next intermediate, we are simply going to add water to our structure, remove the formal positive charge. This gives us a single bond between carbon and oxygen. And now we are simply adding the water molecule 02 or basically oh and another h right? We're only interested in that specific hygiene because it provides oxygen wet the formal positive charge. So this is our second step. Now, let's consider the next step in the reaction. Of course, we're going to remove that hydrogen because we have plenty of water in our solution, right. So water can act as a base or a proton acceptor which gives us our next structure and we can simply remove that hydrogen. Let's go ahead and tap and we have two hydroxyl groups bonded to the same carbon atom well done. Now, let's consider the next step in the reaction. And the next step would be the protonation step. Now, the alcoy oxygen is going to be proton because it's bonded to an alkyl group and an alkyl group is an electron donating group, right. So essentially, we're going to consider our protonation step, we have previously formed hydro num. So let's illustrate the protein capture from hydro num oxygen gets proton that which essentially means that it gets a formal positive charge. We have got a good living group. As we can see, we are forming a molecule of ethanol. And this is possible if we consider our two hydroxyl groups, if we take one of them, we can form car bono pie bond and remove ethanol from the structure. Let's draw the resultant intermediate to draw the resultant intermediate. We're going to remove ethanol. And as we can see, we have oxygen, one of our oxygen atoms is double bonded and it gets a formal positive charge because it has a total of three bonds. Now, once again, we have to understand that this structure is resonant stabilized. Specifically, we can shift our pi electrons on the oxygen which tells us that carbon will have a formal positive charge, right. So now from here, if we consider our products, one of our products is a key zone. So what we're going to do is essentially the proton, it oxygen to begin with, this will give us the carboxyl group. We can use water once again, acting as a base, we're illustrating the proton capture. And this step allows us to get two car bals, one of them is a ketone carbonel and the other one is the carboxy well done. Now that we have our intermediate product, we can consider bond rotation specifically. If we rotate our bond for the Decor box station stop, we can draw oxygen with a single bond pointing up, we can expand the oxygen hydrogen bond. And now we can shift coronal downwards. This is the same coo a tribe. And from here, we want to draw the dep protonation step actually decoration. My apologies, which is achieved in the presence of heat. Now, how does it happen? Well, essential. We want to remove co two from the structure. So we are going to need a total of three arrows. We're forming a carbon oxygen double bond, right? We have another oxygen already double bond. So this gives us co2. The next step is the cleavage of the carbon carbon bond. And of course, the pipe bond of carbona is broken. We're capturing high hydrogen. Now, what do we get? Well, essentially, we are removing co two from the structure and we get an EOL as we know eos are less stable than carbon compounds that correspond to those ens. So we're going to observe a keto iny tote ket, which allows us to get the final coronal compound. In this case, it is a heat on. And this is how we have got our mechanism. Let's make sure it is completely visible. We have already broken all of the steps down. Now, we can just label our mechanism as the final answer to this problem. Well done. We have our final answer. Thank you for watching.

Key Concepts

Hydrolysis of Esters

SN1 Reaction Mechanism

Decarboxylation

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