Reaction of ethers with strong acids to form an alcohol and an alkyl halide. Created by Jay.
Want to join the conversation?
- At6:32you show adding a bromine group to a tertiary carbon and again at6:58. Are these carbons not sterically hindered? Is this hindrance different than the hindrance in the next example with benzene ring which prohibits the Sn2 mechanism?(9 votes)
- Those carbons appear to be sterically hindered but I think the mechanism in this case is SN1 with the leaving group leaving first (the C - O bond is broken) and then the Br attacks the resulting carbocation(9 votes)
- In the video, he shows the X- attacking the R' attached to the oxygen, and the ROH group leaving, to ultimately give the products of ROH and R'X. Was the carbon of choice for attack arbitrary? (Because it seems that the X- could attack R (not R'), and form R'OH and RX instead)
Thank you!(7 votes)
- The choice was arbitrary in this case, because the structures of the R groups weren't specified.
With real molecules, both R and R' are attacked. The R group that is more easily attacked will react faster. Thus if in R' the C next to the O is 1° and in R the C next to the O is 3°, the product is mostly R'X, but with a smsll amount of RX,(6 votes)
- at8:40ish, if one were to make a grignard reagent with the dibromide molecule shown, which Br would bond with Mg?(4 votes)
- Grignard reagents are made from alkyl halides and magnesium, not diatomic halogen molecules. even if you did make a Grignard reagant from diatomic bromine, which Br bonds to Mg is irrelevant because the two Br atoms are indistinguishable (unless you want to talk about different isotopes)(4 votes)
- Why doesn't the nucleophile attack at the positively charged oxygen ?(2 votes)
- It probably does, but that would give the O atom 10 valence electrons. The attack will be unsuccessful.(7 votes)
- It is wrong for Jay to say that there are no resonance stabilization of the phenyl carbocation . You can put the left pi electrons move to the carbocation so you can have some resonance stabilization.(4 votes)
- That doesn't really help though, because the positive charge would be localised to an orbital in the same plane as the ring and so at a right angle to the pi system. So the pi system cannot stabilise it through resonance.(2 votes)
- 11:25Why is benzene sterically hindered, when we know that it is a planar molecule? Also carbon which is bounded to the oxygen is sp2 hybridised.(4 votes)
- SN2 reactions require an sp3 Carbon. The Carbon in question is, like you said, sp2 hybridized. Therefore it can't undergo an SN2 reaction.(0 votes)
- When he writes Hbr, delta - does the delta mean that the temperature is raised?(3 votes)
- The benzene ring is an exception to the formation of two alkyl halide groups, are there any other molecules that as soon as we see them,, we should conclude are going to give an alcohol and an alkyl halide?(2 votes)
- What if you only had H3O to work with? Instead of H-Br & heat.(2 votes)
- Ethers are relatively inert, so it requires a strong acid and heat to react faster. If you only had H3O, some reactions would occur, but the K constant would not be favorable.(2 votes)
- In the second example, for a cyclic ether, how do we determine which bromine is added first to the original molecule?(2 votes)
Normally ethers are very unreactive, which is what makes them good organic solvents. However, if you react them with strong acids, you get acidic cleavage of the ether. So if we start with our ether over here on the left, and we add excess hydrogen halide, and we heat things up, the ether gets cleaved to form an alcohol and an alkyl halide. Normally, the reaction conditions will then convert that alcohol into yet another alkyl halide. So you will usually end up with two alkyl halides as your product. However, not always. In terms of what hydrogen halide do you use? HBr and HI work the best. And that's because when you have the bromide or the iodide anion, the electrons are further away from the nucleus, which increases the nucleophilic strength of those anions compared to, say, something like chlorine, where the electrons are closer to the nucleus since it's a smaller size. Let's look at the mechanism for the acidic cleavage of ethers. And we start with our ether like this. So this is our ether, which we are going to react with our hydrogen halide. So we start with an acid base reaction. So a strong acid like hydrabromic acid will donate protons. The ether is going to act as a base. Lone pair of electrons are going to pick up the proton from our hydrogen halide. The electrons will kick off onto the halogen like that. So we will end up with-- we're going to protonate our ether. So we have a hydrogen there. We have a lone pair of electrons on this oxygen, which gives this oxygen a plus 1 formal charge. So we also have our halogen. It used to have three lone pairs of electrons. It just picked up one more lone pair of electrons to form a halide anion, which is going to function as our nucleophile. So our nucleophile is going to attack the carbon bonded to the oxygen. And then the electrons between the carbon and the oxygen are going to kick off onto the oxygen. So this is an SN2 type mechanism where your halide anion is going to function as your nucleophile. OK. So this is why bromide anions and iodide anions work better than chloride anions. So the result of that nucleophilic attack-- we now have are alcohol. So we got an extra lone pair of electrons on our oxygen there. And our R-prime group is now bonded to our halogen. So we've made our first alkyl halide like that. And sometimes the reaction will stop here. And sometimes the reaction will keep going. So if the reaction keeps going, we can get another hydrogen halide molecule in here. And so the lone pair of electrons on our alcohol are now going to function as a base and pick up a proton, which kicks these electrons off onto our halogen. So if we draw the result of that acid-base reaction, we're going to protonate our alcohol. So our alcohol now has two hydrogens, a lone pair of electrons on our oxygen, plus 1 formal charge. And our halogen now has an extra lone pair of electrons, giving it a negative charge, making it a halide anion. So in the next step of the mechanism, the halide anion will function as our nucleophile and attack the carbon bonded to the oxygen, kicks these electrons off onto the oxygen. And that's how we make our second alkyl halide. So we'd form R-X as our other product. And also water would be produced in this as well-- so H2O, like that. Now, I drew this second part of the mechanism like it's an SN2 mechanism. And it would be an SN2 mechanism if we were starting with a primary alcohol. So if this guy over here is a primary alcohol, and after it gets protonated, a primary alcohol would work the best for an SN2 mechanism because that would be decreased steric hindrance. However, if we were dealing with something like a tertiary alcohol at this point, things would likely proceed via an SN1 type mechanism. So it's important to look at the structure of the alkyl halide. Let's do an example of the acidic cleavage of ethers. And we'll start with an ether that looks like this. All right. So we're going to react this ether with excess hydrobromic acid. And we're going to heat things up. And when we think about our products, we know that the ether's going to go away. And we know that we're going to get two alkyl halides out of this. So we just need to find our alkyl groups. So if I look over here, here's one of my alkyl groups. And if I look over here, here's my other alkyl group. So all I have to do is turn those alkyl groups into alkyl halides. And they're going to be alkyl bromides, since we're using hydrobromic acid here. So I'm going to draw one of my alkyl halides like that. So it would be bromine attached to the ring. And then my other alkyl halide will be this methyl group over here. So I take a methyl group and I attach it to bromine. So methyl bromide would be my other product. And we'd also produce water, as well. But your major organic products would be these two alkyl halides. Let's do another one. This one will make it a little bit tricky. So in three dimensions-- so we're going to have to think about acidic cleavage of ethers in three dimensions. So it makes it much harder. So if I look at an ether, which looks like this, and I add, once again, excess hydrobromic acid, and I heat things up, I'm going to get acidic cleavage of that ether. And when I'm trying to figure out the products, I know that oxygen's going to go away. And I know that the carbons that are bonded to that oxygen are the ones that are going to form my alkyl halides. So I look at this carbon that's bonded to my oxygen. That's going to be bonded to a halogen. And if I look at this carbon on the other side of my ether, that carbon is also going to bonded to a halogen. So when I'm trying to draw the products-- so this ring here is going to stay the same. So I'm just going to go ahead and draw my ring like that. All right. And let's first look at this carbon right here, this one in red, which is this one over here. So it was already bonded to a methyl group right here. And the bond between the carbon and the oxygen is going to break. And we're going to form a new bond with our halogen, which happens to be bromine. So we're going to put bromine there like that. And the opposite side, when we look at this carbon over here-- so let's go ahead and draw out that carbon in. So we're going to get like that. So it's the carbon in blue. The bond between the carbon in blue and the oxygen is going to break. And there's going to be a new bond formed to our bromine, which is our halogen. So we can go ahead and draw our bromine in there like that. And so that will be our product. So we get two alkyl halides in the same molecule this time. We could draw the product like that, or we could kind of flatten it out a little bit. And let's think about that ring system there. So what kind of a ring system do we have? We have one, two, three, four, five, six carbons present. So if I'm going to draw this molecule in a different way, I would have a six carbon ring-- so cyclohexane. And if I take a look at-- let's do this carbon right here. So that's this carbon. What is attached to that carbon? Well, this carbon is attached to another carbon, there are two methyl groups, and then a bromine like that. It doesn't really matter the order that you put your methyl groups and your bromine. And this is because this carbon right here is not a chirality center. So you don't have to worry too much about that. Let's look at this carbon down here-- so in green. So I'll make this one down here in green. There's a methyl group attached to that carbon. So if I make a methyl group attach to that carbon right here. And there's also a bromine attached to that carbon. So these are just two different ways to represent the exact same molecule for your product after the acidic cleavage of the original ether. So kind of a tough one because your thinking in three dimensions here. Let's do one more example for the acidic cleavage ethers. And we'll start with ethyl. Let's do ethyl phenol ether here. So I'll draw a phenol group. So here is our phenol group. And then we'll have an ethyl group on this side. So what will be the product of ethyl phenol ether reacted with excess hydrobromic acid, and everything is heated up? All right, so let's think about the mechanism. We know that in our mechanism the first step is to protonate the ether. So the ether's going to function as a base, pick up a proton from hydrobromic acid. So if we go ahead and draw the first step, the acid base reaction there. So we're going to protonate our ether. And that means there's going to be a hydrogen attached to our oxygen now. Lone pair of electrons still left on our oxygen, plus 1 formal charge on our oxygen like that. Well, we have the bromide anion left over after HBr donates a proton. Br negative is left. The bromide anion is going to function as my nucleophile. And it's going to proceed via an SN2 mechanism. Therefore, it's going to attack the least stericly hindered alkyl group, which is, of course, the alkyl group on the right. So lone pair of electrons is going to attack this carbon right in here. And these electrons would then kick off onto my oxygen. So if I draw the result of that nucleophilic attack, I now have my ring over here. And I just turned everything into an OH group. Because now I have an oxygen with two lone pairs of electrons on it like that. So I make phenol as one of my products here. And the the bromide anion just added on to an ethyl group. So we're going to form methyl bromide as our other product. So ethyl bromide would be the other product here. And let me just highlight those carbons. So these two carbons are these two carbons right there for my product. And it turns out that this reaction will stop at this point. So these are your two products. Phenol and ethyl bromide. And the reaction doesn't really continue on. Let's pretend like the reaction does continue on. And we'll see why it actually stops here. So if we were to continue on in our mechanism, we would next protonate the phenol. So we have HBr here again. The addition of hydrobromic acid, we would protonate the phenol. So if we protonated that phenol right there, we would get something that looks like this, with a plus 1 formal charge on our oxygen. And we would also form the bromide to anion. And if this mechanism were to continue like usual, the bromide anion would function as our nucleophile. And it would attack this carbon right here on our ring. But that's going to be a problem. That's going to be a problem if you're talking about an SN2 mechanism. Because the benzene ring has all the stuff that gets in the way to stericly hinder that bromide anion as a nucleophile. So because of that steric hindrance, the bromide anion cannot attack. So there's no SN2 mechanism here to form another alkyl halide. So that won't work. What about an SN1 type mechanism? So if this is an SN1 type mechanism, we would need to form a carbocation. And so these electrons in here would kick off onto the oxygen. And we could form a carbocation that way. So it would be plus 1 charge on that carbon. The problem with this carbocation is there's no resonance stabilization. We can't draw any resonant structures to help stabilize this. And so therefore, an SN1 mechanism won't work either. So no SN1 mechanism to turn everything into a second alkyl halide. No SN2 mechanism because of steric hindrance. And for those reasons we are stuck with our phenol as our product, and ethyl bromide as our other products. So only one alkyl halide was produced in this reaction.