Propose mechanisms for the following reactions.
(c)

Question

Propose mechanisms for the following reactions.

(c)

Accepted Answer

everyone and welcome back in today's video, We are looking at this problem which asks us to draw a suitable mechanism for the reaction given below. Now, if we look at this reaction, we have an alcohol and the conditions that were using their sulfuric acid and heat, which implies that we have an acid catalyzed alcohol dehydration. We have an asset. We have an alcohol. Were forming two different Elkins. So it implies that we have and asset catalyzed dehydration. Recall that this reaction undergoes an e. one mechanism because we'll be using a weak base, which is water. In particular, we are forming two different Elkins. We wish to look at the mechanism and how they are formed. We are going to draw our substrate the substrate here is a secondary alcohol. We may essentially add the alcohol group. The method groups shown we are using sulfuric acid and heat. And recall that in the world first stop of the reaction. We are protein eating the alcohol. The lone pair on oxygen will attack the acidic proton of the acid and in particular, they're too acidic protons in sulfuric acid, we may expand. One of them is directly bonded to oxygen and then we would have the remaining atoms. We may label them as S. 03 H H two, S. 04, the lone pair at tax. We have the loss of the conjugate base. Now this forms a protein ated intermediate. So we may draw the result on protein it and intermediate that would be O. H. T. With a formal positive charge because now oxygen contains three different bonds. We may also label this as a proto nation stop for reference and going back at this step, we have to think carefully, we call that this is a very good leaving group because it would form a neutral water molecule. So we may have the loss of the leaving group and we will form a cardboard Karen. So we are going to indicate this is our next step. The laws of the leaving group during the loss of the leaving group will form a secondary carb. Okay, Ryan, let's draw that carb. Okay, Ryan, that's our secondary carb. Okay, Ryan. But we may also notice that this carbo Karen maybe may undergo one comma two hydride shift. There is a source of hydrogen disposition to produce a more stable tertiary carbon canton. So he may indicate that after the one comma two hydride shift, we are forming a more stable tertiary carbo Karen. Let's try it Now, essentially, what we're going to do, we're going to expand that implicit hydrogen and we will show how that shift takes place essentially because these two are adjacent positions. This hydrogen shift towards the carbo Karen, we have a loss of hygiene and disposition forming a carbo Karen at that same position that would be a tertiary carbon. Karen that is more stable. Now we have two different carbo karan's formed and we will analyze each of them separately to see what our next step could be In the next step recall. This is an E. one mechanism. We are going to use our weak base which is water. We have we have the loss of the water molecule in this step and that water molecule will act as a weak base. So if you look at the two cardboard Karen's formed, we have multiple possibilities here and for complete clarity. We will just label them. This will be our cardboard karan number one in our cardboard camera number two, lets label it in green. Now for the blue one, if we look at that secondary carbo cast iron, we have a positive charge here and a metal group. We need to find the positions of the beta hydrogen and they are adjacent to the shower carbo can iron or the positive charge. So we may eliminate these two beta hydrogen. We will have two different paths. If we eliminate this one in particular, water will act as a protein except er in this case that's our base, it will attack that hydrogen and we will form a double bond out of these bonding electrons. That's our first possibility. So we may indicate that this would be one of our products. The double bond goes here, we have a method group and indeed that corresponds to our product # one. In addition, we may also eliminate this hydrogen atom on the left. This is a beta hydrogen. If a water molecule attacks it D protein, it's that and we would form a double bond and disposition. This step would be the deportation stop. We have already shown how the deeper termination step takes place for this beta hygiene for this one. We are forming another product. The double bond would be at this position and then we have a method group and this product doesn't even appear in our in our on on our product side. So we're not really worried about this one, We only care about this first one. So we're not really worried about chewing the step for the attack at the left beta hydrogen simply take into account this hydrogen that forms the first product and you're good. Now we have shown the cheap possibilities for the secondary carbo Karen. Now let's look at the tertiary. If you look at the tertiary one in particular, we can throw it in blue actually let's draw it in green because that's how we labeled it. So we draw it. It's our cyclone fencing ring the method group and the positive charge. We again have to identify our beta hydrogen due to the plane of symmetry. The two beta hydrogen is that these positions are equivalent. They're exactly the same. So we may notice that these two hydrogen are exactly the same. It doesn't matter which one you decide to remove and there is an additional beta hygiene there are actually three but it's sufficient to expand at least one of them. So we again have two possibilities for this deep throaty nation step water may attack this hydrogen atom forming an L keen and this gives us the exact same structure as we had previously we made through that structure. And if we remove that hydrogen atom, Another molecule of what are my d protein eight, this hydrogen and then we would form a double bond at this position. Net foot form an additional product which would have a double bond. This position destroyed clearly. So this is indeed our other product. Now does that if we look at the reaction scheme, we see that there to Elkins one of them is the one that we drew here which is exactly the same as this one. So let's suppose that if we label them, that's our structure one, That's our structure too. So we have shown two different paths which allow us to form product number one and regarding product number two, we see that we got it just over here, we may now label this as our complete mechanism because as the problem suggested, we wanted to draw a complete mechanism indicating the formation of these two products. And we may say that this is our mechanism for this problem. Now we may also stay that comparing the alc in stability in particular because this is a tri substituted alkaline is more stable. So we may say that this would be the major product and the second one would be the minor one because it's less substituted, its tri substituted. it has three al kiel groups around it and this one only has two alcohol groups, so it's di substituted, It's less stable. And according to the days of school, the days of school states that the more substituted alcohol would be the major product because it's more stable and the less substituted al King would be the minor one because it's less stable, Well done. The correct answer to this multiple choice question is D and if you have any questions, feel free to leave them down below in the comments section, thank you for watching and I'll see you in the next video.

Propose mechanisms for the following reactions.
(c)

Key Concepts

Elimination Reactions

Regioselectivity

Carbocation Stability

Watch next