Propose mechanisms for the following reactions.
(b)

Question

Propose mechanisms for the following reactions.

(b)

Accepted Answer

Hey everyone and welcome back in today's video. We are discussing the following problem which states show a suitable mechanism for the reaction given below. If we analyze this reaction, we noticed that in alcohol because there's an a wage group is forming two different Elkins is the presence of these double bonds and the conditions that we're using, they're still very Cassidy and heat. That's very informed for us. And let's recall. The city reaction is classified as an acid catalyzed dehydration. So we have an asset catalyzed dehydration, which as you know, follows anyone mechanism that bit of information is tremendously useful for us to predict a suitable mechanism. If we know that it's an elimination reaction, we pretty much understand what steps we have to take to derive these products. We're using heat, which basically favors elimination as well. This alcohol is B- 10 c. O. Because there are four carbon atoms in the parent chain and we are going to draw its dehydration mechanism. If we redraw B 10 T. O, We have to recall that the first step in the dehydration mechanism or an E-1 mechanism, just to be more specific. The law, the lone pair on oxygen will get protein ated by the strong acid. In this case we have sulfuric acid and sulfuric acid is a proton donor. We have two acidic protons. We will expand one of them and the remaining part of the molecule would be oso three age because the acidic hydrogen is bonded to oxygen. The lone parent oxygen gets protein ated and we have the loss of the conjugate base of sulfuric acid. This is called the protein ation staff and we are forming a protein ated intermediate. That alcohol will now have an O. H. Two group with a formal positive charge. Oxygen has three different bombs. Now this is a very good leaving group because as you know, we will form a water molecule which is neutral. So our next step in an E. One mechanism corresponds to the loss of the leaving group. We will form a carbo Karen at this position and we will remove a water molecule from here. So we may now indicate our steps. First of all, we will draw the carbo Caroline that we get which is a secondary carb. Okay, Ryan and we may also label this as the loss of the leaving group. Stop, we have formed our secondary carbo Karen, this is a secondary carbo Karen, we have to be extra careful and we have to check whether or not we can have a metal shift or a hydrogen or a hydride shift. We cannot have that because this is already the most stable secondary carbo Karen, we cannot form a tertiary one. So we're good and now we are ready to return fire beta hydrogen, the beta hydrogen czar adjacent to our positive charge. This is alpha carbon. The two beta carbons are at these positions and they have two beta hydrogen which could be removed. Now we will have two possibilities at least. But we also have to keep in mind that there's a constant bond rotation at this position or basically any position pretty much. But we are mainly interested in this position. So if we rotate our bond we will see that the metal group that was pointing up will now be pointing down. Let's draw another representation of that car bull Karen, which looks like this positive charge. But now the method group is pointing down, you will soon understand why this is super important for us. But for now just keep in mind that this is really relevant stuff we really want to keep in mind that bond rotation however, going back to our problem, we may now think about the elimination step in which a water molecule that we formed and the previous stuff will act as a weak base. So first of all, if we think about this hygiene in particular, a water molecule, H. T. O. Will use its lone parent oxygen to D protein, eight hydrogen. And essentially my apologies, I should have expanded that bond to make the mechanism clear what I'm going to do. I'm just going to expand this C. H bond so that the mechanism is consistent. Now the water molecule uses as Sloan pair to attack that hydrogen and these bonding electrons and that's the reason why I should have expanded that hydrogen. These bonding electrons will form a pi bond in this position and this is how we form our first product. We may now draw it As the following Elkin which corresponds to structure one. This is our structure number one. Now if we remove the green hydrogen, our other beta hydrogen, we will show that a what a molecule with the lumpy in oxygen at tax hydrogen. And we are forming another double bond at a different position. This is how we get our second product. We have a double bond at position number two. So this is indeed our second product which is transit its arrangement. Right. The two al kiel groups are pointing in opposite directions. We can say that trans is our second product which is already thrown over here. We will label it as structure # two. And now we are going back to a more complex part. We call that previously said that there was a significant importance in that bond rotation because if you look at this representation of that same car, broken iron. And if you visualize that beta hydrogen and disposition. And if you draw the mechanism for the for the D protein nation step, the lone pair would attack it. And we would form a double bond in a different in a different arrangement which would be sis because now the two al kiel groups are pointing down they're on the same side of the double bond. And this is how we get our sister I Zimmer. So we may call this structure number three, we got a total of three products. And in addition, all of these steps are the D protein nation steps well done. So we have our three products. We got product number one corresponding to the second alkaline products two and three corresponding to CIS and trans eyes MERS. For the first structure drawn. We may now label our complete mechanism for this problem. We started off here and we got our products. We may summarize that the correct answer to this multiple choice question is C. And if you have any questions, don't hesitate to leave them down below in the comments section, Thank you for watching this video and I'll see you once again soon.

Propose mechanisms for the following reactions.
(b)

Key Concepts

Elimination Reactions

E1 Mechanism

Regioselectivity and Stereochemistry

Watch next