There are 4 major methods to create ethers. Most of these should look familiar in some way since they are variations of reactions we have used before.
The name given for the SN2 substitution of an oxide with an alkyl halide.
Concept: The Mechanism of Williamson Ether Synthesis.4m
Now I want to talk about ethers. It turns out that in organic chemistry one, you're going to be responsible to know four different methods to synthesize them. Some of these methods are going to be stuff that you already know from prior chapters. Some of it is going to be really related to other reactions you've learned, it's just going to be tweaked a little bit. Overall, this isn't so hard, you just have to keep track of the four reactions to make ethers.
Let's start off with the simplest one, which is called Williamson Ether Synthesis, which sounds complicated. You're thinking, “Oh man, another name that I have to memorize.” But really, this is an easy reaction. All this is going to be is an SN2 reaction of a primary or a methyl alkyl halide with an oxide base.
Now this is just a typical SN2 that we would use the flow chart for. You might be wondering, “Well, Johnny, why does it have that funky name?” Well, Williamson Ether Synthesis is just the name for that specific route that you take on the flow chart. Cool? So even if you forgot the name of this reaction, you would still know what to do just by using the flow chart.
Let's go ahead and talk about SN2 for a second. Remember that what does SN2 really require. A good backside. That's why primary and methyl alkyl halides are awesome at Williamson Ether Synthesis because they have a really good backside.
Well, what happens if I try to use a secondary or tertiary alkyl halide instead? Let's say that I'm like screw the primary alkyl halide. I want to start with my tertiary. Can we do that? No, we can't. Because remember that secondary and tertiary alkyl halides are sterically hindered in the back. What that means is that they're actually going to favor E2, not SN2.
So now that I've kind of explained the regents, let's just draw the mechanism. As you can see, what kind of alkyl halide am I starting off with? This would be primary. What kind of nucleophile do I have? Is it neutral or is it negatively charged? Well, in this case, this is negatively charged because it's going to dissociate into OEt negative.
Even if you didn't know what this was, we could just use the flow chart. According to the flow chart, the first question is is my nucleophile negative or neutral? It's negative. Two, is it one of my bulky bases? No. Three, what kind of alkyl halide do I have? Primary alkyl halide. Does that always favor a certain reaction? Yes, it favors SN2. Even if you didn't remember that this is Williamson Ether Synthesis, it's fine because you can just use the flow chart to figure it out.
Now we just have to draw the mechanism. The mechanism would be a backside attack. My OEt would kick out the Br and look what I get. I wind up getting a carbon with an O and then an ethyl group on the other side. Now, I'm just replacing the ethyl group – this is the thing that was Et before. I'm just drawing it out. But notice that look what functional group I have at the end. I have an ether. I was able to use an SN2 reaction just from the flow chart to make an ether. This is your first and probably most common form to make ethers in this chapter.
Good so far? Let's go ahead and move on to the second way to make ethers.
Condensation reactions join two smaller molecules together to form a single, larger molecule.
Concept: The Mechanism of Alcohol Condensation.5m
So another way to make ethers is through a reaction called Acid-Catalyzed Alcohol Condensation. I know this sounds really complicated, but it's not that bad. As you guys will learn later in orgo two, a condensation reaction is simply a reaction that takes two molecules and makes them into one bigger molecule. I'm just going to say it's a reaction that takes two smaller molecules and then it turns them into one bigger molecule. That's the definition of condensation.
What we're going to be doing here is we're going to be taking two alcohols, it's an alcohol condensation, so we're going to take two alcohols, we're going to put them together. We're going to condense them and they're going to turn into one ether.
How does this work? Let me just go ahead and just draw the mechanism for you. The way this works is you have alcohol in the presence of acid and heat. What's going to wind up happening is that the acid's going to protonate one of the alcohols.
Let's go ahead and just draw this part really quick. I've got my H3O+ that I'm going to write like this because it's easier to deprotonate that way. Same thing as H3O+, I'm just writing it a little bit different. So my OH is going to grab an H from the acid and what I'm going to wind up getting is something that looks like this. I have a protonated alcohol now.
Now what's going to happen is that that protonated alcohol just turned into a good leaving group. Water is a good leaving group. So my other equivalent of alcohol, the one that did not get protonated is going to do a backside attack on this good leaving group. We're basically going to get an SN2 reaction where I get this attacking that carbon and kicking out the good leaving group.
So now what we're going to wind up getting is – let me just draw it in the same colors that I used. The black alcohol that still has an H on it, but now that's going to be attached to the two-carbon chain from the red alcohol. On top of that, there's going to be a water that just left by itself. Does that make sense so far?
So we've got the black one attacking the red one. This looks like an ether, but we've got a problem. There's a formal charge. So what can we do about that formal charge? Remember, this is called acid catalyzed for a reason. That means that you always have to end up with the same amount of acid that you started off with because it's a catalyst. It can't be consumed or destroyed in the reaction.
What that means is that I use the water to pick up the proton. And what I'm going to wind up getting at the end is I'm going to get an ether plus the same H3O+ that I started off with. There you have it. We just condensed an ether out of alcohol.
Now there is going to be a significant limitation for this synthesis. Can anyone tell me? It's only going to yield a certain type of ether. Actually, there's a typo here that I will correct in your notes. This should not say alcohols. It should say ethers. But it's going to form only symmetrical ethers.
The reason is because we're always going to be reacting acid and alcohol and you're going to have an abundance of alcohol. What that means is that one molecule is going to react with another molecule of the same alcohol and you're going to wind up getting the same R groups on both sides. That's why I'm saying that it's symmetrical because you're always going to get the same R groups on both sides.
Sometimes you want that. For example, you wanted an asymmetrical ether. It has to be asymmetrical, maybe Williamson Ether Synthesis would be a better choice because that one it doesn't matter. You can just add R groups as you want.
So let's move on and keep talking about ethers.
Same reagents as oxymercuration, except with alcohol as the nucleophile instead of water.
Concept: The Mechanism of Alkoxymercuation.5m
Another way that we can synthesize Ether is through a reaction called alkoxymercuration and you might immediately notice that the sounds a lot like another reaction that you're supposed to know by now and that's oxymercuration so it turns out this is going to be the same exact mechanism as oxymercuration or Oxymerc except that in the top reagent in the oxymercuration step we're going to use alcohol as our nucleophile instead of water, OK? So what you can see is that the biggest difference here is this right here, I have alcohol in place of the water and that's going to make an ether instead of alcohol, OK? So let me just walk you guys through this mechanism really quick, so remember that we get is HG-OAC-OAC......Opps it came out weird OAC and in the first step what we do is the double bond attacks the mercury kicks out one of the OACs and the Mercury attacks back so what we wind up making is a bridge, an Ion bridge so what that's going to look like is like this where now I have a dotted partial bond to HG it's only attached to 1 OAC now, dotted partial and a positive chart, OK? And there should be actually a dotted partial bond here too and a positive charge, OK? So now we've got our intermediate, notice that we can't get any shifts because there's no carbocation and now this is the part that typically water would come in and attack which side do you remember? In this case I have a Markovnikov site here and I have an anti-Markovnikov site there which one would it attack? Markovnikov because that's going to be the one with the most positive character but instead of water what's actually going to attack is in this case ethanol, OK? So my water attacks......I'm sorry my ethanol attacks kicks out the HG And what I wind up getting is something looks like this and I'm going to try to fit in here where now let's say that the alcohol attacked from the bottom then I would draw that the HG the mercury attacks from top or not attacks but is now placed in the top, OK? Cool? Awesome stuff the end of the Oxymerc step, OK? In this case alkoxymerc because I used ethanol, OK? Then remember that we have a reduction step, the reduction step is just NaBH4 and OH negative and you don't need to know the mechanism for this so I'm just going to say don't need to know mechanism, OK? But what we do need to know that happens is that well number there is a positive charge in that O actually, OK? And what's going to happen is that I'm going to wind up reducing the mercury just to an H so this is going to become H, OK? And I'm going to wind up deprotonating my oxygen so now look what that turns into, it turns into ether, OK? Because now what I have is RoR, it's a weird looking ether but it is an ether so I get an ether as my product and that's it's the same as OxyMerc except that you have to just now you have to be aware every time you see an Oxymerc make sure that you look to see whether it's water or alcohol that's attacking because that will make a world of difference in terms of what your product looks like, alright? So I hope that made sense same as I did before we're just using alcohol instead, alright? So let's go ahead and move on.
Same reagents as acid-catalyzed hydration, except with alcohol as the nucleophile instead of water.
Concept: The Mechanism of Alkoxylation.5m
Another method for making ethers is called acid catalyzed Alkylation and it's going to be very similar to another reaction we've learned before which is acid catalyzed hydration, OK? Really this is the same exact reaction as acid catalyzed hydration on a double bond except for our nucleophile we're going to use alcohol instead of water, OK? So let's just go through this mechanism really quick I know it might have been a while so I'm just going to go ahead and start from scratch, I've got a double bond remember that double bonds are good nucleophiles and I've got a strong acid, OK? So here I've got my acid I'm writing it a little bit funny but that's because can be easier to deprotonate that way and remember that the first arrow or the first part of this mechanism would be that my nucleophile attacks the H in my strong acid, OK? My sulphuric acid what that's going to give me is that's going to wind up giving me a Carbocation that looks like this with the H attaching over here now this H I'm going to go ahead and ignore it for the rest of this reaction because you don't need to draw Hs, right? But I just want to show you where it attached to, so now I've got that carbocation there do I go ahead and nucleophilic attack it? What should I do? I need to shift it, right? So at this point I would do a carbocation shift because remember that carbocations will shift any time it's possible for them to move to a more stable location, in this case would this be a methyl shift or a hydrate shift? It will be a methyl shift because I only have methyl groups adjacent to this carbocation so I'm just going to write here it's a 1-2 CH3 shift and what that's now going to give me is really a new looking molecule that looks like this with a carbocation there, OK? Notice that really all that happened is that the CH3 from the top moved to the bottom and now my carbocation is on the left, it's tertiary it's a lot more stable so now this is the part where my nucleophile attacks, now for acid catalyzed hydration water would have attacked and I would have gotten an alcohol easily but now I'm going to use an alcohol instead what that means is that I'm going to wind up getting an OR group instead of an OH group so I'm going to get here H, CH3 there we go, now I did notice that I've been making an error here does anyone see what the error is? OK what the error is that I had an H right here because I wanted to draw it in and I've continued to draw that stick in both of these that should not be there if you did draw the stick just put the H there just so you guys will know that that's supposed to be an H but also you could just erase that stick completely, that would also be fine, OK? Now this oxygen needs a formal charge so I'm just going to give it a formal charge, what do you think is the last step here? How can I deprotonate that? Well I could just use the conjugate base of my sulfuric acid so I could use OSO3H negative and I can deprotonate and what that's going to do is it's going to give me my ether plus it's going to give you my original sulfuric acid, alright? I know that I'm right over that so I'm just going to.... Ops so I'm just going to disappear really quick, there you go oh gosh OK We're having some technical difficulties I'll move to the side just so you guys can see that there is an ether ROR and I have sulfuric acid present, OK? So this is another way that we can use to make ethers, let me know if you guys have any questions but if not let's do some practice problems.
Problem: Predict the product of the following reaction.4m
Problem: Predict the product of the following reaction.6m
Show how you would make the following ethers, using only simple alcohols and any needed reagents as your starting materials.
(c) benzyl cyclopentyl ether
(e) the TIPS ether of (d)
(f) cyclohexyl cyclopentyl ether
Predict the organic product(s) for the following reaction. Be sure to indicate stereochemistry when appropriate, if stereoisomers are produced draw one and state the relationship between the other stereoisomer formed. (enantiomers, diastereomers, etc.).
Predict the organic product(s) of the following reaction. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product, if stereoisomers are produced in the reaction be sure to indicate the relationship between them. Draw all answers in skeletal form.
Draw the ether of molecular formula C 5H10O from BrCH2CH2CH2CH2CH2Br + NaOH in CH3OH.
For the reaction shown below, draw the structure of the major product(s). Where appropriate, write "No Reaction". Assume all chiral starting materials are optically pure. Show stereochemistry where appropriate. Indicate whether the product exists as a single enantiomer or a racemic mixture.
Predict the product of the following reaction.
Compare the following reactions and decide which reaction occurs fastest.
Device a synthetic pathway to form the following product.
Starting with an appropriate alkyl halide and using any other needed reagents, outline syntheses of each of the following. When alternative possibilities exist for a synthesis, you should be careful to choose the one that gives the better yield. (d) Methyl phenyl ether
Starting with an appropriate alkyl halide and using any other needed reagents, outline syntheses of each of the following. When alternative possibilities exist for a synthesis, you should be careful to choose the one that gives the better yield. (c) Methyl neopentyl ether
Starting with an appropriate alkyl halide and using any other needed reagents, outline syntheses of each of the following. When alternative possibilities exist for a synthesis, you should be careful to choose the one that gives the better yield. (a) Butyl sec-butyl ether
With methyl, ethyl, or cyclopentyl halides as your organic starting materials and using any needed solvents or inorganic reagents, outline syntheses of each of the following. More than one step may be necessary and you need not repeat steps carried out in earlier parts of this problem.
With methyl, ethyl, or cyclopentyl halides as your organic starting materials and using any needed solvents or inorganic reagents, outline syntheses of each of the following. More than one step may be necessary and you need not repeat steps carried out in earlier parts of this problem. (i) CH3OCH3
Show how you might use a nucleophilic substitution reaction of 1-bromopropane to synthesize each of the following compounds. (You may use any other compounds that are necessary.) (d) CH3CH2CH2—S—CH3
Show how you might use a nucleophilic substitution reaction of 1-bromopropane to synthesize each of the following compounds. (You may use any other compounds that are necessary.)
Predict the product(s). If no reaction would occur, say so. Show stereochemistry where relevant. If more than one isomer or product would form, show all the ones which would form.
Which reaction would produce phenyl propyl ether?
Which compound will be the organic product after the following two steps?