Provide an arrow-pushing mechanism that rationalizes the outco...

Question

Provide an arrow-pushing mechanism that rationalizes the outcome of the reaction shown.

Chemical reaction diagram showing an alcohol converting to a chlorinated compound with an arrow-pushing mechanism.

Accepted Answer

Hi, everybody. Welcome back. Let's look at our next problem. It says, consider the given reaction and we have a reagent, single reagent. That's a secondary alcohol. It's our groups are methyl and then a rather bulky group here. That's a cyclo butane ring with two methyl substituents. It's reacting in hydrochloric acid and it makes a product that's a cyclops ring with three methyl groups and sharing a carbon with one of those methyl groups is chlorine. It says, what is the arrow pushing mechanism that explains the product of the reaction? So this one's a little bit intimidating. First, we've got a group leaving a group joining, but it's on a different carbon atom. The oh is off on a side chain, the chlorine is on the ring. We've got a four carbon ring changing to a five carbon ring. So it's a little intimidating at first. But let's just kind of look at this overview and think about what's going on. What kind of reaction is this likely to be? Well, obviously, the most striking element of this reaction is the cyclone butane going to a cyclops, four carbon ring to a five carbon ring. Well, right away. This should make us think about ring expansion, psycho butane is under a lot of ring strain. It's very happy to become a more stable cop pentane. And of course, that's a very common method of Carbocaine rearrangement. So that should make us think about Carbocaine rearrangement. And that would also lead us to a substitution where the atom coming in joins the molecule at a different place than the atom leaving or the group leaving I should say. And of course, Carbocaine rearrangement requires a Carbocaine intermediate, which leads us to the idea that this is most likely an sn one reaction. Let's look a little more closely and see. Does that make sense? Well, again, we have the possibility of Carbocaine rearrangement sn One reactions use Carbocaine intermediates. We are having uh we do know we have to have an SN one, we have to have our leaving group leave first. So that might give us a little pause because alcohol groups are not good leaving groups. However, note that we have a strong acid present were in HCL. So a strong acid means that the alcohol can be proton, it can then become 02 plus and become an excellent leaving group. It can leave as water. So there we go in that acidic environment, we do have the potential of our leaving group leaving as a first step. So if we proton that alcohol it can leave, we generate, we have a secondary alcohol here So when that alcohol leaves as water that will generate a secondary carbo cion, it's adjacent to a cyclo butane ring. So ring expansion definitely favorable in that, that will generate a tertiary Carbocaine. So yes, we have these conditions for our sn one reaction. In addition, with an SN two reaction, those take place in one step. There's no way you can have all this happening in a single step. So we're set on our sn one substitution reaction. But as we know, our first step here must be protonation of the alcohol group. So we'll draw our reagent here. So I've drawn it out and I've added the two lone pairs to the oxygen of the alcohol group. Because again, we are looking for our electron pushing mechanism need to see where those electrons are. And I've drawn the HCL showing the bond between the H and the CL. So one of those lone pairs from the oxygen are going to come over and snag a proton from the HCL, it's a strong acid, it will readily give up a proton and then the electrons from the HCL bond go to the chlorine. So there we go with step one. And we have now converted our reagent and given ourselves a really good leaving group here. So our former oh is now 02 plus oxygen does not like having a positive charge. So the electrons from the co bond here go on to the oxygen causing it to leave as water. Of course, in the first step, we've also generated or that first step, we've also generated chloride ion. So now we move on to our next step. We've eliminated a water molecule. And now we've generated our carbo cat iron. So again, we have our cyclo butane ring, it has its two methyl groups. And then now we have this essentially an ethyl group CC and then a positive charge on that first carbon where the alcohol was attached. So there's our Carbocaine step two again was water leaves. And now we'll have step three. When we look at our Carbocaine, this is a secondary Carbocaine again adjacent to that ring. So sips three will be ring expansion are carbo cion rearrangement. So now here's where things get a little tricky to keep straight, put a little arrow down and I'm going to scroll up a bit to make more room. So this step is one where it can be really tricky to keep straight when the new ring forms, where do the substituents go? And where does that positive charge go? So I like the method that Johnny used in the Pearson channels. Videos of using different colors to help me keep track otherwise you could number them. Um If you don't have the chance you're on a test or something like that. But for sake, here of doing this here, I will use different colors. So I'm going to put a green dot on the carbon that has the positive charge out on that chain there. And I will put a blue dot on the carbon in the ring. And I will number the carbons in the ring for the sake of keeping our substituents straight. So that will be carbon number one. And then we'll just go down to carbon two, then the other bottom left to carbon three and then top left carbon four. So I've numbered the carbons in the ring. So carbon one in the ring is blue has the methyl group and then the chain that has the positive charge and a red dot on carbon two in the ring, which has no substituents. So what is going on in terms of electron pushing in a ring expansion? Well, keep in mind this is a symmetrical ring. So it doesn't matter whether we're using carbon three or, or carbon two, excuse me or carbon four in the ring here to do this. So I'm just using carbon, the bond between carbon one and two, those electrons are going to come up and join the positively charged carbon on the side chain. So now we will have a new bond. We've broken the bond between carbons one and two in the ring, we have a new bond between our red carbon two and the green carbon that had the positive charge. And meanwhile, our blue carbon is still bonded to the green carbon. So our ring has been expanded by one and made into a cycling. So to draw that, let's draw our five carbon ring here, accidentally drew it in red, switched to blue. So we now have a five carbon ring and now we've just got to make sure we're correctly placing all our substituents in our positive charge. So I'm just going to dedicate not the top carbon in the ring, but the first one to the right as my blue carbon that was carbon one in the ring, it has a methyl group. So let's put that on there. Oops and note that in the past, it had no hydrogens, right? It had four bonds on it were no hydrogen, the original ring. So it's still, it didn't get any extra hydrogens, but it now only has three bonds. It has the methyl group, it has the bond to what was carbon four in the ring and then it has a bond to the green carbon. So now the green carbon is adjacent to it in the ring instead of being a substituent. So it only has three bonds, no hydrogens. So this is the site of our new positive charge. So our new Carbocaine, let's make that a little less messy. Has that positive charge on the blue carbon that was carbon one. So again, it could be a little tricky to keep straight with everything moving around. We formed a new shape of ring, where are things going? Where does the positive charge? Now go. But by looking at the hydrogens, you can find it. So now let's move on to our green carbon. It was originally part of that side chain. It's now in the ring, it has a bond to the blue carbon on one side, on the other side, it has a bond that newly formed bond to the red carbon. So that's now adjacent to it in the ring. But it also has a methyl group. Remember when we look back at the original chain, so it will now have a methyl group coming off the ring at the green carbon. So now let's move on that. We had that new bond to the red carbon, which was carbon two in the original ring, no substituents on carbon two. And again, we can double check that positive charge. Our green carbon originally had two bonds and a hydrogen that hydrogen applied and a positive charge. So it still has one hydrogen and three bonds to another carbon. It's neutral. Now and then our red carbon originally in a ring, no substituents. So two hydrogens, it's still in a ring with no substituents having two hydrogen, it's also neutral. So we already would expect that to be the case since we located our positive charge. But again, you can use the number of hydrogens to double check. Then the last thing we need to do is place that last methyl group, it was on carbon three in the butane ring. So we're on the red carbon, which is number two. So adjacent to that red carbon, we have our third methyl group. So now we have our carbo Canion. And so we just need one more step. Now, we're going to have nucleic attack, buy a chloride ion. That chloride of course has four lone pairs. It's got lots of electrons. So our final step step four will be nucleic attack. That final step of the sn one react reaction. So now we form our product. Oops gotta draw my electron pushing here one of the lone pairs from the chlorine. We'll go on up and perform that nuclear attack on our positively charged carbon. So let's draw our product. I'm just going to redraw that cyclops as it looks pretty bad straw five carbon ring, we now have a methyl group. Again, that's our blue carbon. It's not as hard to keep track now, but just for the sake of consistency and as well as the methyl group, it now has a chlorine. And then we've got our two other methyl groups, one on the green carbon adjacent to this one and then one over on the carbon that was carbon three in the original ring. So this ring, of course, number one would be the blue carbon. So we've got methyl groups on carbon one carbon two and carbon four. So again, I'll just put our green and red on here to keep straight. But this is so similar to the Carbocaine. It's this step is pretty easy to keep straight. So we double check, look at our product here. Does it match? Yes, it does. So we have the arrow pushing mechanism that explains the product of the reaction. We have our first protonation of the alcohol. Step two. So I've got the electrons moving there. Step two, the water leaving step three, the ring expansion, step four nucleic attack leading to our product. See you in the next video.

Provide an arrow-pushing mechanism that rationalizes the outcome of the reaction shown.

Key Concepts

Arrow-Pushing Mechanism

Nucleophiles and Electrophiles

Reaction Mechanisms

Watch next