What product do you expect to obtain from each of the followin...

Question

What product do you expect to obtain from each of the following reactions?

a. Chemical structure showing a reaction involving hydroxyl groups and HCl, prompting expected product identification.

Accepted Answer

All right. Hi, everyone. So for this question, let's determine the product of the following reaction. And here we have four hydroxy five methyl Heino acid reacting with HCO. Now recall really quickly that this is an example of an in intermolecular fisher ification, right? Because of fisher ification reaction involves a carboxylic acid reacting with an alcohol in the presence of acid to yield. And but notice how our starting material contains both a carboxylic acid functional group and a hydroxy group in the same molecule. So therefore, right, instead of reacting two separate organic compounds, the same organic compound is going to react with itself to yield a lactone or a cyclic ester. So let's go ahead and see how that plays out. So, on the bottom of the screen, I'm going to go ahead and redraw my starting material so we can go ahead and proceed through the mechanism. So here I just want to make sure I draw everything and I also want to go ahead and actually label my carbon tier because I find that it's easier to keep track of them that way. So what I'm going to do now is I'm going to label the six carbons of my parent chain with my first one being the carbon of the carboxylic acid. So given the fact that we have acid in our system, right, recall that the first step is going to be the protonation of our carbon with the proton of HCL. So my carbonyl oxygen is going to take the proton of HCL and recall that Fisher certification is an equilibrium. So here I'm going to draw equilibrium arrows. But now in my next step, I'm going to have a proton car bottle. So she is my oxygen with an extra hydrogen, which is going to give my oxygen a formal charge of plus one. So now I'm going to number my carbons once again and now that my carbon has been proton, right. Our next step, if you recall is going to be the in intermolecular nucleic attack, buy my hydroxy oxygen. So in this case, a lone pair of electrons on my hydroxy oxygen, here is going to go ahead and attack my carbonyl carbon and that is going to result in the pi electrons of my carbon moving towards oxygen. So my next step, I'm drawing another equilibrium arrow. And this time we formed a ring. So before I draw the ring, notice how we have four carbons between the carbona itself and carbon four where my hydroxy group is. Now, I have four carbons and my oxygen from the hydroxy group is also going to be involved in the ring. So therefore, right, my end result is going to be a five member ring because I have four carbons in between. Right. And my additional hydroxy oxygen. Now, speaking of the hydroxy oxygen, because it still had a hydrogen attached when it attacked the carbonnel, it's going to have a positive charge in the ring. So now I have two additional hydroxy groups on position one due to the fact that my car bottle was lost. So here I'm labeling my carbons once again. And I also have to take into account, not only carbons five and six, right? But also my methyl substituent outside. So now our issue stems from the fact that our oxygen here on carbon four doesn't appreciate having a positive charge, right? So in this case, our chloride ion which is the conjugate base of HCL that we used in the first step is going to come in and detonate that oxygen. And here we are left with a neutral ring structure overall. So here are my two hydroxy groups and here are the other carbons of my starting material. Now, now we're almost there, right? But in this case, in order to form a Caronna, once again, on position one, here, we have to go ahead and remove one of the hydroxy groups and reform a carbon once again. So to kickstart that process, another molecule of HCL is going to come in and proton one of the two hydroxy groups because recall that pronated hydroxy groups make fantastic leaving groups in acidic conditions. So here I can take either hydroxy group in this case and have it be poin by hydrochloric acid. So when I scroll down I met equilibrium arrows once again, and that's going to give me a proton hydroxy group. So here I have one and here I have the other now because the other hydroxy group has been pronated, it's going to have a formal charge of plus one, right? So at this point, recall it, it's a fantastic leaving group because it can go ahead and detach itself like soap. Right? First and foremost, the neutral hydroxy group on carbon one, here is going to donate a lone pair of electrons between itself and carbon to form a carbonyl. And that is going to kick out, so to speak, the pronated hydroxy group and have it depart as a molecule of water. So here is our five member ring, our substituent and now we have a proton carbon here we have a proton carbon giving my oxygen here on carbon one, a formal charge of plus one. So now our final step right is the deep protonation of my carbonyl group to yield my final lactone product. So again, recall that we have chloride in solution from our second equivalent of HCL. So here HCLD protons, our carbon, I should have drawn this bond in black. So I'm going to do that. But regardless the bond between oxygen and hydrogen. Those electrons are going to move towards oxygen to yield a neutral carbonnel at the end. And I'm actually running out of space. So I'm going to try and move this to the side, see if that helps it does. OK. So here I have my lactone at the end and here is my final product. So this structure, right, our final lactone product is going to correspond with option A in the multiple choice. So there you have it, our product is going to be option A in the multiple choice. So now let's go ahead and recap all of our steps, right? This was an example of an intra molecular fissure ification. The first step involved proton the car bottle at position one to yield a pronated car bottle followed by the nucleic attack of our hydroxy group onto the carbonyl carbon. Now that given the fact that it's intra molecular resulted in the formation of a ring structure. So after the derotation of the oxygen inside the ring one hydroxy group on position one was pronated once again. So the neutral hydroxy group on position one regenerated a carbonyl, whereas the pronated hydroxy group detach itself from carbon one entirely. And after deportation of the resulting carbon, we ended up with a neutral lacto which is a cyclic ester. So with that being said, thank you so much for watching if you stuck around to the end. And I hope you found this helpful.

What product do you expect to obtain from each of the following reactions?
a.

Key Concepts

Functional Groups

Acid-Base Reactions

Reaction Mechanisms

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