Propose a mechanism for each of the following reactions:

Question

a. Chemical reaction diagram illustrating dehydration of a diol using sulfuric acid, resulting in two carbonyl compounds.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to suggest a suitable mechanism for the following reactions. Now look at the two reactions, we can essentially start with a and we will propose a suitable mechanism. So first of all we notice that we are starting with a one comma two dial right? We have two alcohol groups. We are going to produce a key tone. The ring gets smaller. We understand that the starting material has seven carbon atoms in the ring. Now we are forming a six member ring, so there might be an L kill shift involved. So let's see how that looks like If we essentially start by drawing our starting material. Okay, so that's our seven members ring. Now we are going to add a method group, an alcohol group and another alcohol group in metal group. The first part, as you notice there's only one region that says acidic hydrogen. We can essentially say that there will be proton capture because as you recall, oxygen has a lone pair. So there might be proton capture. It doesn't really matter which one you choose. Let's suppose that we go with this one. We're capturing that proton and we are forming our protein ated intermediate. In the first step. So we can say that here we are forming the protein ated intermediate. So let's destroy it. Okay, and now we are going to add our method group alcohol. This would be the protein ated form O. H two with a formal positive charge. And this is our protein ated intermediate. So we got our protein ated intermediate. And I say no, this is a very good leading group because getting rid of it will form neutral water and we will form a stable tertiary carbo cast iron. So why not? We can do that. Yeah, So let's see what we get. We have the loss of the leaving group. That's also called the first step the protein nation stop just in case in here we have the loss of the leaving group which forms our carbo can iron. So once again let's throw our seven member ring. Okay, and now we understand that we still have a metal group over here, an alcohol group and specifically here we are only left with that metal group and a positive charge, that's our tertiary carbo carrion. And now this is where we have to start thinking because we're getting a smaller ring, we must have an al kill shift. So to show the formation of that product specifically, we will think about that al kill shift. And if we notice we can essentially transfer these electrons onto that carbon. So to do that, we are just going to indicate that all kill shift. And we are going to show that we are breaking that carbon carbon bond and instead we are essentially transferring these electrons to that carbon. So we are making our ring smaller right from seven carbon atoms, we will have six carbon atoms. Now this is not necessarily favorable. As you will see because we may form a less stable carbo Karen, but the ring will be smaller and therefore if it's a six member ring that might compensate the stability of the carbo karan to some extent. Even though this carbo Karen is really stable, this is still possible and favorable because the result in kabul Caroline will be our secondary carbo Karen and it's still relatively stable. Right? So let's see what we get when we perform that alcohol shift, we can control our arrow, we can call this al kill shift. Now we understand that the ring will be our sixth member ring because once again we are breaking this carbon carbon bond, right? We are essentially opening it at this position and we are just making this, making a new bond between this carbon and that carbon. So we are skipping one carbon atom and therefore the rain here would be a six member drink. So we can draw a six member drink and now we end up with several substitutions. So first of all, at this position we have a metal group. So let's add that metal group. Okay, and now the remaining substitutions are C C3 And a weight. Right? So we are going to draw that carbon atom was a muscle group and an alcohol group and we understand that because we had a loss of electrons at this position, there will be positive charge of this carbon and this is exactly what we were referring to. Notice how this is a tertiary carbon. Karen, This is a secondary one because it has to neighboring carbon atoms. Right, So this cardboard Karen is less stable but it's still relatively stable. Right? Even though it's less stable than the previous one, it's relatively stable. So it's still possible to achieve a reaction in such a manner. Okay, so we got our cardboard Karen. Now let's think about resonance, specifically this carbo Karen is resident stabilized. So it appears that even though it's secondary because its residents stabilized, it's even more stable than just a secondary carbo Karen. And the reason why it's resident stabilizes, because if you look at it auction has a lone pair. So we can use that lone pair and we can essentially de localize that lone pair into the formation of a pi bond. So we can outdraw resonance, we can indicate that here, we will have our resonance forms. Let's draw our ring. Once again our methyl group. Now here we are forming our o age with a formal positive charge and auction because now it has three bonds. And we also need to make sure that we are adding that method group. Okay, so now we got our residence forms and now we are going to remove that acidic proton because if you look at the product, we are nearly done. And to remove that acidic proton, we are going to use water. The reason why we're using water is because if you think about the general form of H plus it's actually hydro ni. Um Right. And the leaving group is water. So we have plenty of water. The molecule of water will essentially capture that proton and restore acidic conditions. That's why it's called an acid catalyzed reaction. So we are removing the proton and forming our final product This way we can now show that the final product or we can also call this deep rotation stop. Okay, so during the deeper termination step we get our found product. Okay, so let's see if that's correct. We got our metal group, we got our kids on. Yeah, that looks good. So this is our mechanism for reaction. A and now let's think about reaction be for reaction be we will have a bit of an opposite process. We are going to form a bigger ring out of a smaller one. So this is this will be our reaction be. So let's label it properly. Let's draw the starting material that's our six member ring. We have an O. H. Substitute print. Then we have two final groups and another alcohol group. Okay and then the world first step what we're going to do here, we are going to treat it with an asset. So just as we said before, we are going to use an asset, we can indicate the presence of hydro ni um in in its simplest form. H Plus we have two alcohol groups. The question is now, which one are we going to protein eight? And because we are going to perform a ring expansion any possible al kill shift? We definitely want to form our carbo khurana disposition. So simply looking at our product. If we want to have an L kill shift, we want to use the protein nation of this a wage even though the loss of each weight which would produce us with a thirst theory. Carb okay. Ryan we are going to protein it. This one based on the identity of the product because the protein nation of this would allow us to get rid of it form a positive charge here and increase the size of the ring by an al kill shift. So that's why we're going with this group. Just based on the outcome of the mechanism. The lone pair on the oxygen will get protein ated. So we can draw our protein nation staff label our protein ated intermediate. That would be O. H. Two with a formal boss of charge because oxygen now has three bonds. Now this is followed by the loss of the leaving group and the formation of that stable tertiary carbo Karen. Besides, we can also say that this car bo Karin is stabilized by two final groups. Right? And due to that stabilization by or resonance stabilization is more likely that the carbo Karen will be formed here. So that's another reason why we are protein ating that wage instead of that one. Right? Because considering that a wage, it's bonded to a different carbon atoms. So it's not residents stabilized. It's a bit further from fennel rings. And therefore this is just another way to verify why your protein eating this. So wages basically because the loss of that water would form a positive charge here and it would be residents stabilized by two final rings. That's really favorable. Right? So now we are just ready to draw the loss of the leaving group. Let's do our six member ring the original alcohol group. And now we have our two final groups remaining with a positive charge. So we got our tertiary car Bocaranda. This resident stabilized from both sides. And now we are ready to think about the al kill shift because we're increasing the size of the ring. So we are going to open that carbon carbon bond, right, we are essentially going to open that ring and we are going to shift these electrons towards this carbon atom. So in other words, you can think of it as increasing your size of the ring by one unit. It will have seven carbon atoms because you're breaking this bond over here and you're transferring it over here. So you will have a bond between this carbon and that carbon instead of this carbon and that carbon. Right? So we will have our seven member ring. We can call this step as an al kill shift. Now let's draw the seven member ring that we are expected to have. Okay, and now if you look at the substations that we have, we are going to add our oh age substitution because we are losing electrons from here. That's where the positive charge will be. Right. And at the adjacent position over here we have our two final substitutions, so one final and then another. And now similarly to what we saw before. Let's think if residents as possible and indeed it's definitely likely because we have a positive charge on the carbon atom that has oxygen. There is a lone pair on oxygen which will form a carbon will bond. So we are essentially de localizing our positive charge and oxygen. We can essentially say that here we will have resonance forms. Let's draw that our seven member dring. Okay, and now here we are forming double bond oxygen, hydrogen that we are positive charge and oxygen to final groups. We can clearly indicate that these are our residence forms. And they were looking at the second form, we are going to d protein it to form our final product. And just as before we are going to use water because when we used that acidic proton from hydro nian, we formed water and water will act as that protein except er to restore the conditions deep rotation step over here and we are ready to draw the found product, let's throw bond angles a bit more differently so we can get our seven members ring properly. Okay, and now we understand that here we get our Carbonell group and two final groups adjacent said, okay, so that's our mechanism for reaction be. We will label this is our final answer. We understood how reactions take place for A and B. And looking at the outcomes that we got. We can conclude that the correct answer choice the small cultures question is A. However, if you have any questions, don't hesitate to leave your comment down below. Thank you for watching.

Propose a mechanism for each of the following reactions:
a.

Key Concepts

Reaction Mechanisms

Nucleophiles and Electrophiles

Transition States and Intermediates

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