Another method for converting alcohols to chloroalkanes makes ...

Question

Another method for converting alcohols to chloroalkanes makes use of chlorotrimethylsilane (TMSCl) and DMSO. Suggest a mechanism for this reaction to form (a) a 1° chloroalkane and (b) a 3° chloroalkane. [The reaction begins by the reaction of DMSO and TMSCl and is analogous to the Swern oxidation.]

Diagram illustrating the reaction mechanism for converting alcohols to chloroalkanes using TMSCl and DMSO.

Accepted Answer

Hi, everyone. Let's look at our next problem. It says alcohols can be converted to Chloro Alkanes using T MS Cl, Chloro tri methyl Cyle and D MS O dimethyl sulfoxide propose a mechanism for the transformation shown below. And then there's a hint, the mechanism starts with the reaction of T MS CL and D MS O similar to sworn oxidation. And then in our reactions, we see we have a, in each case, we have a molecule of D MS O plus TMS L. And in our first reaction, our alcohol, we have cyclohexane ring which has Gemel substituents and ethyl group and an alcohol group. So it's a tertiary alcohol. And in the product, the alcohol group has been replaced by a chloride. I have regenerated my D MS O and then my other product, my chlorine in my T MS L has been replaced with that oxygen group. Then same uh process for the other reaction except the alcohol. In that case is a primary alcohol. We have a 1234 carbon chain with the oxygen, excuse me, the alcohol group at one end and then two methyl groups on the third carbon and again, similar reaction, the alcohol group is replaced by a chloride, regenerate the D MS O. And on the T MS L, the chlorine's been replaced by alcohol. So this one looks a little intimidating. So let's start by thinking about this reaction of D MS O and our T MS LTM SS CL. And we know that we have to generate a reactive intermediate. The sworn oxidation involves D MS L also and it forms a complex with ox aal chloride. And in the same way, it forms this reactive intermediate, you sort of call it uh activated DMSO binds to the oxygen and alcohol and it oxidizes it to an a carbonel group. So let's think about what our reactive intermediate would be. So we'll start with our dmso, we have that positively charge sulfur and two methyl groups. And then we have that negatively charged oxygen. So it's got three lone pairs around it and it's reacting with our tri methyl cy Cyle. So I've got a silicon with three methyl groups and it has a chloride on it. Well, our negatively charged oxygen there on D MS O is a good N file and we have that um very electro cilic on there. So our, one of our loan pairs from the oxygen can go over and attach to the silicon and then the electrons from the silicon chlorine bond will go over to that the electron chlorine and that will generate our intermediate. So let's draw that and in that reactive intermediate, we've maintained the positive charge on the sulfur, but we now have a new bond from the oxygen to the silicon. And we've also generated a chloride ion, which we'll want because we want to perform that substitution reaction to substitute the chloride for the alcohol group. So I'm going to label this reaction as essentially the first part of my mechanism for both reactions. So this would be the formation of the reactive intermediate. So now let's start looking at these alcohols. Well, we know that the alcohol group is a poor leaving group. Uh We usually think of proton it with an acid and acid catalyze reaction to convert it into a good lead leaving group. But sometimes you might want an alternate way where you don't need to have an acid present to convert that alcohol into a good leaving group. So I'll start by my first alcohol. So I've drawn my reactive intermediate and I've drawn the alcohol here and I'm going to call this step two A and we'll just call that top reaction reaction a to keep the two of them straight. So we recall the first step we need to take is convert that alcohol into a good leaving group. So it's going to react the oxygen of the alcohol group, the lone pair there, we'll go ahead and attack the electro pilic silicon in my reactive intermediate. I have a good leaving group in this what was the D MS O? So the electrons that have bonded the oxygen, the silicon are happy to hop back on this oxygen. Now, this step is a little bit confusing because you have this positively charged sulfur and the silicon, they're both electro pilic. And so you might question, why is my, why is my lone pair going to attack the silicon that has a partial positive charge rather than the sulfur with its complete positive charge? And the answer is essentially, our reaction tells us when we look at the top that we're going to regenerate our D MS. So we know that needs to happen and that we're going to end up with that alcohol attached to the silicon. So we just need to trust that those products are more stable that that reaction will be favored. Basically because it's given to us in the problem. That is what will happen. But it can be confusing, especially because in the sworn oxidation, the al the alcohol group does those that lumbar does attack the positively charged sulfur. But of course, there's no silicon atom involved there. So just trust that that silicon is electro enough that the oxygen will attack. And again, we know from our problem up here that we have that product of the silicon bonded to the alcohol group. So we have this going on, we've now converted our alcohol into a good leaving group because the oxygen now has a positive charge which it will not like. And so we end up with the next step where we have our good leaving group here because our oxygen has a positive charge. And we need to think about what kind of substitution reaction are we going to have? Well, we have a tertiary alcohol so it will make a pretty stable Carbocaine. So we can guess that we'll go through or we can say that we'll have an sn one mechanism happening here. N one there. Notice also that we have regenerated the D MS O as we knew we wanted to do. And so we'll have the electrons from me C bod go to that positively charged oxygen. And again, it's an sn one reaction. So we'll have the leaving group leave first generating a Carbocaine intermediate. So we now have our Carbocaine. It has the positive charge on that tertiary carbon that has the ethyl group on it. And then notice that we've generated the second of our products that tri methyl silicon with an alcohol group on it. So we've regenerated our D MS O. We've generated our second product. And don't forget now we have this Carbocaine, we generated a chloride ion in that initial step, one formation of our reactive intermediate since the formation of the intermediate kicked off the chloride. So we have chloride ions around to act as a Nile. So let's give them their, give it its lone pairs negative charge. It's a strong Nile though, there's not that much of it and we can go and attach it to that positively charged carbon. And the end result will be the product shown in our problem above. I'm not going to write that all out because we have it already given to us. So I'll just write an arrow and say that that produces product A. So that's the reaction for our tertiary alcohol. Now, let's look at our second reaction in which we have a primary alcohol. So I'm going to scroll up. So we'll call this step two B that we'll start with. And again, we start exactly the same way converting our alcohol into a good leaving group. So we have our reactive intermediate, the same one and we have our 33 dimethyl butanol, which is our alcohol there. So again, the lone pairs on the alcohol come on over to attach two, that's silicon atom which will regenerate our D MS O and give us our good leaving group. So that step is the same as with our tertiary alcohol. So we now have this oxygen which has the positive charge since it now has the silicon attached to it. But now we have to do it a little differently because this is a primary alcohol. So it cannot generate a stable Carbocaine intermediate. So instead we'll have an sn two reaction. So again, that will be a concerted reaction of substitution. So we have our chlorine, which is again, it's a strong nucleo file so it can do that. Put all its lone pears on. So the lone pair on that chlorine will perform a backside attack coming around and oops, sorry, I drew my arrow the wrong way there getting a little tangled up in all my lines here. The low power on the chlorine will form a backside attack coming over to the carbon, bring that over there. And at the same time as the electrons in ac O bond come over to the oxygen. So the chlorine comes in as this oxygen silicon complex is leaving. But again, we generate the same type of product, we generate our alkyl halides with our silicon alcohol there. And in this case, I had a little room. So I drew that alkyl Halide there, the oxy the alcohol replaced by the chloride. The we had that D MS O regenerated and the silicon with an alcohol group on it. So again, that was reaction B. So we have our mechanism for using DMSO and TMS CL to generate a reactive intermediate that will help me convert an alcohol to an alkyl Halide. In this case, it works for both a tertiary alcohol and a primary alcohol, but using two different mechanisms. So we have step one being formation of the reactive intermediate with the D MS O and the T MS CL that also generates our chloride ion that will be our nucleo file for the substitution reaction. And then for our tertiary alcohol, which I called reaction A when my leaving group leaves, after my reactive intermediate complexes with it show that there, then I form a Carbocaine intermediate which undergoes an SN one substitution as chlorine comes into the carbo Cron. And then in reaction B with my primary alcohol, I start out the same way by converting my alcohol into a good leaving group and then finish up with an SN two concerted reaction where the chlorine comes in as my complex leaves. But again, generating that alkyl Halide in the end, regenerating my SDM SS O and that silicon alcohol product. So there we go. There's our reaction mechanisms. I hope to see you in the next video.

Key Concepts

Swern Oxidation

Mechanism of Nucleophilic Substitution

Primary vs. Tertiary Chloroalkanes

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