A student was studying terpene synthesis, and she wanted to ma...

Question

A student was studying terpene synthesis, and she wanted to make the compound shown here. First she converted 3-bromo-6-methylcyclohexene to alcohol A. She heated alcohol A with sulfuric acid and purified one of the components (compound B) from the resulting mixture. Compound B has the correct molecular formula for the desired product.
(c) Propose a mechanism for the dehydration of alcohol A to compound B.

(c) Propose a mechanism for the dehydration of alcohol A to compound B.

Accepted Answer

Hi everyone and welcome back in today's video, we are looking at this problem which reads a graduate student performing a complex organic synthesis required one ethel cyclo Penta won three, dying as one of her compounds of interest. She started with three chloral, five ethyl cyclops and one in and was able to convert it into four methyl cyclo pants two in one oh. She then performed an acid catalyst dehydration on four ethyl cyclo pence two in 10. To convert it into her required compound writing mechanism for the dehydration reaction. Now, I know that you may think that there's a lot of information here, but let's just condense our problem. Notice that if you carefully read it, it's not really important to consider that chloral compound because it already had been converted into four ethyl cyclops and two in one oh. And we simply wish to write the dehydration mechanism for that compound to produce the dying. Let's do that. Let's just draw our structure that we're interested in as our starting material and we will start by drawing cyclo Fencing Ring. So that would be five member ring. And we understand that this is an alcoholic position number one, it has an alcohol group. So we may simply add an a wage group at position number one and at position number two we have, in which means we have a double bonds. We may add a double bond at this position and position number four has an ethyl substation. That's 1234 in this position. We may add sl that's our starting material for ethel cycle pan two in one oh and we wish to dehydrate it. An acid catalyzed dehydration. As you recall implies that we are going to use a strong asset in the general case because the acid is not given. We may indicate the presence of that asset by the presence of hydro ni um itself. And now also we need to use water which will act as a weak base recall. Also that these conditions imply an E. One reaction. So we will be dealing with an E. One mechanism because we wish to represent an asset capitalized dehydration in which the base is a weak base. The product that we wish to draw is the Dyin and it says we have one ethyl cycle Penta won three die in so we are going to draw another ring of cyclo fencing. We understand that position number one has an ethyl group. So why don't we why don't we just add it at the exact same position as we saw it previously. So there will be an ethnic group at that position that be position number one and it also states that dying contains one comma three, dying basically that means there are two double bonds at positions one and three. So between one and two we have one double bond and that's position number three from 3 to 4. We have another double bond. So that's our final product. That's how we wish to perform our reaction. And essentially going back to the original problem, we wish to write a mechanism for it. Let's go ahead the first step And the dehydration mechanism is actually two protein A the alcohol group. So what we can do here, we may recall that oxygen has a lone pair and can easily be protein ated due to the presence of a strong asset which we represented as hydro ni. Um we will show our protein nation stop in the protein nation stop the lone parent oxygen will attack one of the hydrogen and we will have a loss of water. So we'll be forming water in the step. However, the most and more part is to understand that we will form a protein ated intermediate. And we may call this step as protein Nation. During the protein nation step, we are forming that protein eight intermediate. So let's simply redraw our ring with all of the substitutions looking pretty much the same. Except from the fact that now the alcohol group of wage will be O. H two with a formal positive charge, which is a good leaving group because it would form a water molecule after we lose it. Now, essentially, this is exactly what we want to do because this is a good leaving group. It has a formal positive charge. We would have a loss of the leaving group in this step during the loss of the leaving group. We are forming a carbo Karen, but let's not worry about that yet. Let's just simply break it down step by step and write down that this step is called the loss of the leaving group. And as you recall, the last of the leaving group gives us a carbo cast iron. Let's draw that carbo cast iron. There is still that double bond at this position, there are still an ethyl group. But after the loss of the leaving group, we are forming a positive charge at this carbon. Now, what you may notice is that this is a secondary carbo Karen that is resident stabilized because there's a presence of a double bond adjacent to it. So we may simply draw our residence forms. What if that double bond shift over here and our pi electrons shift towards the carbo Karen to stabilize it. We would form another version of that car Bocaranda disposition. So we may draw our resonance forms. This is the first one and the second one would essentially have a positive charge here because there wouldn't be no double bond anymore on the left, let's draw that structure. In this case there is no more double bondable bonus on the right. The positive charges. Now in this position because we have removed the valve on the ethyl group is still present. This is another secondary carbo carrion. They are both secondary carbo can answer their resonance stabilized and now we have to think about it a bit. Can we make this cardboard? Can I'm more stable, we actually can do that. If you recall one comment to hydrate shift. If there is a source of hydrogen that we may shift towards that car bo katan, we may definitely do that as as long as we form a more stable carbo Carolan and notice that at this position in particular, there is a source of hydrogen which may essentially shift over here to form a more stable tertiary carbon cast iron. So we are definitely going to do that. We are going to indicate this step as my apologies, let me fix my arrows. We are going to indicate this step As 1:02 hydrate shift because that's exactly what we are observing. One comment see hydride shift as the arrow suggests, we are shifting this hygiene onto that carbon to form a tertiary carbo can iron. We are ready to draw that new car, broken iron. So let's do that, redrawing our ring, getting our double bond at the right position. Now there's a hygiene so there's no more positive charge, There's a positive charge at this position where we have an ethyl group which indeed makes it a tertiary carbon Karen that's perfect for us. We have stabilized it as much as possible and now we are ready to think about our final step because we have our carbon Caroline ready for the reaction, we think about the D protein nation step and we wish to find our beta hydra jin's there are two beta hydrogen that these positions because there are adjacent to the carbo can iron and notice that there's actually that line of symmetry. This compound is symmetric with respect to that vertical line, meaning the two beta hydrants are equivalent to cheddar. It doesn't really matter which one you choose to eliminate in your mechanism. You will essentially get the same product. The base that we're using is water because we're seeing an asset catalyzed dehydration mechanism. This is a weak base that we're using in E one reactions. And in this final step we are deep protein ating it, removing the proton simultaneously forming a double bond out of these bonding electrons. And this would be called the deep protein nation stuff because we are removing a proton. And during the D the D protein Nation step we are forming an additional double bond and disposition. So let's draw our final product. The original double bond, the ethyl group that we had previously and the newly formed double bond we have formed our final product. If we compare our product to the previously drawn structure in the overall reaction. This is indeed the same compound. We may notice that this dying that we had to synthesize is seen in the final step of our mechanism. So we are good. It verifies that our mechanism makes complete sense. We may now summarize that we have successfully proposed a mechanism for this reaction and the correct answer choice to this multiple choice question is D. However, if you have any questions, don't hesitate to leave your comment down below in the comment section, Thank you for watching this video.

Key Concepts

Dehydration Reaction

E1 and E2 Mechanisms

Terpene Synthesis

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