Propose mechanisms for the following reactions. In most cases,...

Question

Propose mechanisms for the following reactions. In most cases, more products are formed than are shown here. You only need to explain the formation of the products shown, however.

(c) Chemical reaction diagram showing dehydration of a compound with two hydroxyl groups, producing multiple products.

Accepted Answer

Welcome back, everyone propose the mechanism of the given reaction. Note, more products may be formed only show how the given products are produced in the reaction. So we're given a dial and it reacts with sulfuric acid in the presence of heat to produce a mixture of products. And most of them are die ins. One of them is a ketone. Let's understand that whenever sulfuric acid is used in the presence of heat and an alcohol, we're going to observe an acid catalyzed dehydration. So we're going to use that logic in this problem. First of all, let's redraw the starting material. And let's understand that in the welfare step of an acid catalyze dehydration, we're using sulfuric acid to proton oxygen. So we're going to expand our hydrogen oxygen bond H OS 03 H, that's sulfuric acid. And we're going to illustrate the protonation step at oxygen in the protonation step. We're going to draw the protein at intermediate or basically Oxonian. This allows us to get a perfect living group because now we are producing 02 plus, there's a formal positive charge on water and we are going to remove it. Now we are going to draw the result on carbo cum. So we're getting a positive charge, we have a stable tertiary carbo cion. And from here, we can just remove the beta hygen. The reason why we are removing it is because we can see that we have one of our Dynes formed that way. So we're going to draw the beta hygen and we're going to perform an elimination reaction. E one, right? So what are, acts as a base oxygen has a long pair of electrons and it attacks the beta hydrogen which allows us to warm our first pi bond. Well done. This product is not a product in any of our three products, right? So we essentially need to consider what is the next step while we still have hydroxyl presence. So we can perform the protonation step. Once again, let's draw the acid H OS 03 H and let's perform that proton capture. Once again, we are forming another Oxon ion. Oht oxygen gets a formal positive charge. Now, from here, we can remove the living group, right? We already know that we have produced a perfect living group and we are getting a carbo C now that we have our carbo carion produced. It is really important to understand that it is re and stabilized. So what we're going to do from here is basically perform the pi bond shift to indicate the second possible carbo cat I am. It is analytic carbo carry on So we're just performing that pipe on shift and we can notice that we end up with a positive charge at a different position. Now, we have two different Carbo car ons, right. We can easily get our products from here. Those would be products number one and number two. If we take a look at the first Carbo Caron formed, we can just remove the beta hygen with water. So we're performing elimination once again. And this allows us to get our first product, one of the two dy ins well done. As you understand, for the second product. Once again, we're going to take one of our resins forms, that will be the second carbo Caron. And we're going to remove the beta hygen. And this gives us our second conjugated die in. So we have two out of three products already well done. So we have products number one and number two. Now let's consider the formation of the third product for the third product. We basically want to consider the formation of our initial carbo carry on. So let's go ahead and illustrate it once again. So let's start over and let's recall what we did. Initially, we had two alcohol groups, we performed protonation, we had the loss of the living group and we formed a carbo cion. Now, this carbo cion can actually undergo alky shift, right? And we understand that we in our final product, we are forming 25 membered rings. So what we're going to do is essentially perform an alky shift and decrease the size of the six membered ring into a five membered ring. So simply speaking, we're performing an alky shift. And from here, this allows us to get 25 member ranks. So one of them and now the other and we are simply going to draw do group. Now where is the positive charge? Our carbon that is bonded to oxygen gets a positive charge because now it lost e bond. So it is electron deficient. Now that we have our new intermediate form, we have to understand that it is re stabilized because oxygen has a long pair of electrons. So we're going to draw once again, 25 member drinks and car bono, but that oxygen still is bonded to hydrogen. So oxygen gets a formal positive charge. And of course, from here, our final step is the deep rotation step using water, which acts as a bass. We're removing that acidic hydrogen in forming the final ket zone, which is product number three. Well, then we have our mechanisms. Thank you for watching.

Propose mechanisms for the following reactions. In most cases, more products are formed than are shown here. You only need to explain the formation of the products shown, however.
(c)

Key Concepts

Reaction Mechanisms

Types of Organic Reactions

Intermediates and Transition States

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