Treatment of the following alcohol was expected to give alkene...

Question

Treatment of the following alcohol was expected to give alkene A. Instead, B was produced as the major product. Suggest a mechanism by which B formed.

Chemical reaction diagram showing an alcohol reacting with H2SO4, expected product A not formed, major product B highlighted.

Accepted Answer

Hi, everyone. Let's look at our next problem. Consider the following reaction and we have a seven carbon chain with four methyl groups and an alcohol group on carbon three, it's reacting in hydro sulfuric acid with heat. And we see two possible products, product A where the alcohol group has been removed. And there's now a carbon carbon double bond between carbons three and four and product B where the alcohol group has been removed. And there's now a carbon, carbon double bond between carbons four and five under product A it says not formed and under product B it says major product, then it says the expected product is A but the major product is B what is the reaction mechanism by which B formed? So let's think about what type of reaction this is. We have the removal of an alcohol group. We're in an acidic environment with heat and we then have the formation of AC C double bond. So what we're talking about is an acidic dehydration of alcohol. So in this reaction, in the acidic environment, the alcohol group is protein which converts it into a good leaving group, it's eliminated as water. And a carbon, carbon double bond forms. We note that this is a secondary alcohol. So we can expect that this will be an E one elimination of a proton alcohol which leads the formation of a Carbocaine eliminate Carbocaine intermediate, excuse me, as part of that process. So just looking at our products, why would we have a as an expected product? When we think about elimination reactions or excuse me, dehydration reactions of alcohols, we, what we expect is alcohol is removed and adjacent or beta hydrogen is removed and then the double bond is formed. So we look here and we see our beta carbons both have hydrogens, but which side is preferred depends on which will form the most substituted alkene. And in this case, we see that carbon two has no substituents while carbon four has a methyl group. So the most more substituted alkene would be product A so more substituted al Ken. And yet, when we look at b the actual major product, we see the double bond is not from 3 to 4 or from 2 to 3, it's over from 4 to 5. Well, again, we know we have an elimination reaction. So we have the formation of that carbo Catron. And what do we recall can happen with Carbo cats? We can have a carbo cion rearrangement which can make our molecule look different. So what I would suspect here right away is a possibility of carbo cion rearrangement. So let's take a look at this possible reaction mechanism and see if this is indeed what's going on. So first, let's draw our original molecule, we know that it will be protein and the alcohol group will leave to form a Carbocaine. So I've drawn my original molecule, adding the two lone pairs on the oxygen of the alcohol group were in hydro hydro sulphuric acid. I didn't draw the whole structure of that as that's not important. I drew it as H with a bond then. So 4h, the important part is it is an acid, it will readily give away a proton. So in terms of our electron pushing here, one of the loam pairs on the oxygen will grab that proton from our acid and the electrons from the bond on the H os hso 4h go to the base there, conjugate base there. So now we've got our protonation. All right. That step one is protonation of our alcohol. And then now in step two, our alcohol group is now h2o plus which again, very good leaving group and it can leave as water. So the electrons from the co bond go on to the oxygen there. Oops, it's turned into a line and we have the leaving of water. So we go on to our next step and with water leaving our carbon three now has only three bonds to hydrogen and the two other carbons in the chain. So it is now a Carbocaine with a positive charge. So we have formation of the carbo Canion or rather, I should put that over in step two since the carbo cion is formed when water leaves. So here's where if we then moved on to detonating the beta hydrogen on carbon four, we would form that product. A. So we'd say if D proton eight, I'll put C four product, a wood form. Again, it would form from carbon 3 to 4 rather than 2 to 3 because that would produce the more substituted alkene. However, what do we remember about carbon canines? They will rearrange if you can form a more stable carbon Canion. So what do we notice with this here, this is a secondary carbo cat I am. So we need to look and say, is there a way that we could shift a methyl group or a hydride and create a more stable Carbocaine? Well, when we look at the adjacent carbons, carbon four has a methyl group. So it could form a tertiary Carbocaine by hydride, shift, tertiary Carbocaine will be more stable than a secondary Carbocaine. So that shift will be favored. So let's go on and do that mechanism and see what happens. So I'm just going to scroll up a bit to leave me room to do that. So we have this hydrogen. So I'm gonna add in the hydrogen on carbon four. Actually, that's gonna be a little tricky to fit in here. We will just, I'm going to move my dashed wedge, you have to kind of redraw these, I move my dashed wedge a bit to make room for me to draw the hydrogen. So here's our hydrogen on a front wedge and we have a hydride shift. So the hydrogen and its electrons, its a hydride shift over to that positively charged carbon. So what happens there? Now, carbon two or carbon three, excuse me is neutral as it has. It already had a hydrogen that's implied there. It now has two bonds to the chain and two hydrogens, it's neutral. So I'm now no longer gonna draw that hydrogen because again, it's implied carbon two is neutral. So we know it has four bonds, it must have two hydrogens, but carbon four is now missing a hydrogen, it has three bonds and no hydrogen. So it's become the positively charged carbon. And then one last little change, we know that when we have a carbo cion, the three bonds are sp two hybridized rather than SP three. So I'm going to change that dashed wedge of the methyl group to a solid line as we now have that trigonal planar geometry on that carbon. So now we have a tertiary carbo catt iron. So up on my previous step, all right, step three is carbo cion rearrangement. Oops I'm running out of space there. So the tertiary Carbocaine is more stable. So therefore, this arrangement is favored. So now we have our carbo caton and we can go on with the next part of the mechanism which would be the removal of the beta hydrogen by de with, by dep protonation, excuse me, with water. And we need to look at our two possible beta hydrogens. So I've just added in the hydrogens here on, there's two hydrogens available on carbon three now and there's two or one hydrogen, excuse me, on carbon five, there's also a methyl group. So there's only one hydrogen there. So now we go back to our rule about oops this don't really fit there. Let's move the betas up here. So beta and beat it. So now we go back to our rule that we talked about earlier that we would form the more substituted alkene as being the more stable alkene. And so we can see that because there's a methyl group on carbon five that forming ac four C five double bond will produce the more substitute alkene in this case. So last step here will be de protonation. I'll put this up here. Step. So step four, rather dep protonation and removal of beta hydrogen. I'm just gonna scroll up a tiny bit. I need a little bit more room and I will draw a water molecule here with its two lone pairs on the oxygen apologize. These are getting rather tiny and we see that one of these lone pairs red, we'll grab a tom and then the electrons that and that from that ch bond, we will go down to form the double bond between the two carbons, thereby eliminating the positive charge on C four. So now we have our product. So we form AC four C five, double bond. And the last thing we need to change again, we have that remaining methyl group on C five. But now that it's part, it's bonded to a carbon in a double bond. Instead of being a dashed line, it will be a solid line showing the new trigonal planar geometry around that carbon. So we see that we've now generated product B as our major product. So again, our question asked us what, what is the reaction mechanism by which B is formed as opposed to that product you might expect as a and we see we have first steps would be similar protonation of our alcohol group leaving of a water and formation of the carbo cion. But then where we would differ is that we have a carbo cion rearrangement that if we had detonated carbon four product A would form as a more substituted alkene, but we can generate a tertiary, more stable carbo car, buy that Carbocaine and rearrangement by the hydride shift. And so finally, we have a dep protonation and removal of beta hydrogen leading to the formation of product B with that double bond between carbon four and five preferred as the more substituted alchem. So finally, we've generated product B. See you in the next video.

Treatment of the following alcohol was expected to give alkene A. Instead, B was produced as the major product. Suggest a mechanism by which B formed.

Key Concepts

Elimination Reactions

Zaitsev's Rule

Carbocation Stability

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