Give a mechanism for the following substitution and eliminatio...

Question

Give a mechanism for the following substitution and elimination reactions.

(c)

Accepted Answer

Welcome back everyone. Our next question says, what is the mechanism for the substitution reaction below? And we have just a single reactant. It's an alcohol with a long carbon chain that also has a chlorine attached to it. So we have an ho group followed by a seven carbon chain. And on carbon five of that chain, there is a chlorine and an ethyl group, it's reacting just in water and it forms a funny looking product which is a six membered ring that has an oxygen as one of its members. And the rest are carbons on the carbon adjacent to the oxygen. We have two ethyl groups and it looks like a little head with two antennae on it. So when we first glance at this, this does not look like a substitution reaction. At first, we have, it's only reacting in water. Where's our base? What's going on? There's only a single molecule and then there's water, water isn't a good Nile, but let's just go in a little closer. We started with a chain, we ended up in a ring. So we know we must have an intermolecular reaction going here. And indeed, when we look at our molecule, we have a chlorine. So we've got a leaving group and then we have this alcohol group and an alcohol is a weak Nile. In addition, when we look at our coin, it's on a tertiary carbon. So we've got a tertiary, not an alky hide, it's an alcohol, but we have a tertiary carbon. All right here. So when we have a tertiary carbon on where our leaving group is, and we have a weak nuclei that leads us right? To what type of substitution reaction? Rsn one reaction, we will not get an SN two reaction at all. This is too bulky reagent here. And you just can't get that concerted substitution reaction with a tertiary hide. And we have a weak Nile which also favors SN one. And even our solvent does, water is a polar pro solvent which favors SN one. So that's pretty clear, we don't really have any other choice. We're going to have an sn one reaction, generating a Carbocaine intermediate. And then our weak Nile, our oxygen from our alcohol group will go on and attach to that positively charged carbon forming a ring. So let's think through our steps here. Step one will be the leaving group leaves and I'll go ahead and draw that mechanism right on the structure we're given. So the electrons from the carbon chlorine bond go on to the chlorine. It departs and we now generate our Carbocaine. So we'll draw our ho and then our long chain here, it has an ethyl group on carbon five, the chlorine is gone and now we have a positive charge on that carbon five. So we've got our carbon Canion. I'm going to go ahead and number the carbons. We're going to be forming a ring and we want to keep track of where things should go on this ring. So we've got our alcohol group and then carbon 1234567. So step two, we got step one, leaving group leaves step two, here will be the attack of the Nile. So let's recall that alcohol group, that oxygen is going to have two lone pairs. So the electrons from one of those lone pairs can come on around and attached to that positively charged carbon, know that this uh intervallic reaction is possible here because we have this long chain. And since it generates a six membered ring, six membered rings are very stable because of that favorable bond angle, that's so similar to the tetrahedral angle. So I'm just going to put in parentheses, intermolecular next to attack of the nuclear fire. So now we've generated our ring. So let's draw our next step. So here's where our numbering is going to come in handy. So I will start, I'm going to, I want to make it look like the product that's already given to me. So I'll start with that oxygen sort of on the top left there and then I know the oxygen is, has a hydrogen attached to it. So we'll put that on there and then we'll go down and draw our six membered ring around that oxygen. It's a little wonky there. But you can see, hopefully there's six carbons there and we'll go ahead and number carbon one will be below the oxygen disorientation there. So we've numbered up to carbon five on my ring. So what belongs here? But we've added the H on the oxygen and then there's nothing else on the ring. So we get to carbon five, carbon five already had an ethyl group. So let's put that on there. But then don't forget, it wasn't the end of the chain, the original chain had six carbons. So there's still two more carbons bound to carbon five. So that's where that second ethyl group comes from. So we have our six membered ring oxygen with the hydrogen and two ethyl groups. One last thing we have to do, we had two lone pairs in that oxygen, one went into the co bond, but we still have remaining lone pair and that leaves oxygen with a positive charge. It has five valence electrons around it in this arrangement. So it's missing one. Well, the solution to that is pretty straightforward. We're in water in this reaction. So let's draw a little H2O with its two lone pairs. So last of all, our oxygen can get detonated by water. So just take one of those long pears from water, make it snag a proton and then the electrons from that oh bond can go on to the oxygen and that will end up giving us our final product. So our final step here, step three is de protonation and now our oxygen is neutral. It's got two lone pairs, it's happy and it's in within the six membered ring with the two ethyl groups on the adjacent carbon. So we've drawn our mechanism once again, step one, the leaving group leaves step two, an intermolecular attack of the nucleo file and step 3D protonation giving us our final product. See you in the next video.

Give a mechanism for the following substitution and elimination reactions.
(c)

Key Concepts

Nucleophilic Substitution Mechanisms

Elimination Reactions

Regioselectivity and Stereoselectivity

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