Which of the following ethers can be formed in good yield by c...

Question

Which of the following ethers can be formed in good yield by condensation of the corresponding alcohols? For those that cannot be formed by condensation, suggest an alternative method that will work.

(a) dibutyl ether

(b) ethyl n-propyl ether

(c) di-sec-butyl ether

Accepted Answer

Hey everyone and welcome back in today's video, we're going to learn how to solve the following problem, determine which of the following ethers can be synthesized via condensation of the corresponding alcohols and good yield, proposing synthesis plan for ethers that cannot be made via condensation. And here we have three different eaters Di ethyl ether, methyl ethyl ether and diane propio ether. We missed probably throwing these three structures. Number one, Number two, Number three, starting with the ethyl ether. It essentially implies that there are two ethnic groups bonded to the central oxygen atoms. So that would be di ethyl ether. Looking at methyl ethyl eater, one muscle group is wanted to oxygen followed by another ethyl group. And finally die. And properly throw means die normal propel ether. So to propel chains are bonded to that central oxygen atom. So these are our ethers and we once understand what condensation means. First of all, let's analyze the mechanism of a condensation reaction for alcohols, essentially you want to recall that you want to take to exactly same molecules of the same alcohol. So I suppose we have an alcohol chain, our siege two bonded to the alcoholic group. What we're going to do, we're going to draw these two molecules and we are going to analyze the mechanism essentially what we want to do here. We want to understand that we definitely need to use a strong asset, which would protein 81 of the alcohols. We are going to say that we're adding a strong asset in the presence of heat. And the reason why that's needed is because we want to protein eight, that oxygen which attacks hydro knee. Um So we may say that it attacks the acidic hydrogen within hydro ni um fold by the loss of the water molecule. And in the first step we get that protein ated alcohol now because it's protein ated. And because water is a good leaving group, another molecule of that same alcohol can attack that center because this other molecule can act as a nuclear file now. Right? So if we draw that nuclear file, that second molecule of exactly same alcohol now, the lone pair on oxygen essentially attacks the electro fill a carbon atom followed by the loss of the leaving group. And we are forming our protein ated intermediate. Right? If we look at it, we have our cease to now it's bonded to oxygen which still has that hydrogen. And that's why it's protein ated. And then the remaining part of the molecule would be seized to our after the deep protestations that we're going to remove this Haijun by another molecule of water. So the molecule of what attacks Let's just expand this auction hydrogen bond. And then these bonding electrons are placed in the form of lone parent oxygen to form our ether. Our cease to oh cease to are. Now the reason why this is extremely important is because if you look at this condensation mechanism, you understand that if your ether is symmetrical, you can use condensation and we may immediately say that if you look at our three ethers, the two that are symmetrical, our number one and number two. So we may essentially say that number one and number two, they might be formed using the condensation mechanism. Right? Because they are symmetrical. We noticed that we may draw a vertical line passing through oxygen that will divide the molecule into two equal parts, which are symmetrical to each other. So diethylene ether and die. And properly ther they can be synthesized using the condensation mechanism in good yield. Now regarding number two, it cannot be synthesized using that mechanism because it's just not symmetrical. Right? We essentially understand that it's not symmetrical because it has a metal group and an ethnic group. So these two groups are not equivalent to each other, meaning it cannot be synthesized using their condensation mechanism. So as the problem suggests, we are going to propose a synthesis plan for methyl ethyl ether. And let's see how that looks like for number two to synthesize methyl ethyl ether. We are going to think about two alcohols that might form that molecule in particular. If we split it apart, we will notice that we have a methyl groups and an ethyl group. So we essentially need methanol CH 30. H. And ethanol. Right? Because there are two carbon atoms. So methanol and ethanol are required for our synthesis. However, we need to think about how to synthesize it from these two alcohols. So recall that alcohol is a really bad living groups, but we can still make them good living groups if we use Tosel chloride. Right? So our goal is to think about the nuclear Felix substitution reaction SNC which will help us form the required either. Why don't we start with this alcohol, ethanol and turn it into to oscillate ester? Right, so what we can do here, we can essentially say we're creating it with castle chloride in Carradine, which essentially will allow us to replace this hydrant with the Tosel group and we will get a much better leaving group. So that would be our puzzle Esther. Now this is a very good leaving group and this is a very common reaction whenever you want to turn the alcohol group into a much better leaving group. So we have formed our tassel Esther and now, because it's a good leaving group and we have a primary substrate, you may essentially see that this carbon has only one neighbor. So this is a primary substrate. We may make use of an essence to reaction. But the problem is that methanol is not charged, right? It's a weak nuclear file. That's how you think about it because it's a weak nuclear file. We want to d protein ate it and the way to do that is just too displays that hydrogen with sodium metal. So if we add sodium we will form the meth oxide and iron. And here we may say that we are forming meth oxide with a negative charge, sodium possibly charged. But sodium is an ion spectator, it's not really bonded directly. Right? There's an ionic interaction between the two ions. So we understand that we have meth oxide, our strong basis now charged we have our primary substrate and we may perform a lesson to reaction. So if we do that, the negatively charged with oxide, we may essentially um it sodium. Right? So the negatively charged with ox that attacks the electra Felix center, followed by the loss of the leaving group. And we are forming methyl group that oxygen atom Now bonded to two carbon atoms. And this is indeed methyl ethyl ether exactly what we wanted to synthesize. So, we have proposed a synthesis plan for Eastern Number two. Right. We saw that we went to start with two alcohols, methanol and ethanol for methanol. We're treating it with sodium to produce meth oxide and for ethanol, we're turning the alcohol group into docile ester group. That is a perfect leaving group for us using fossil chloride and parading. And then between the two intermediates, we noticed that if we apply an sN two mechanism, we are forming our required ether. So that'd be pretty much it. If we now look at our answers and summarize our findings, we may say that di ethyl ether and diane propel ether can be synthesized using the condensation mechanism because they're both symmetrical, however methylene ethyl ether cannot be synthesized that way because it's not symmetrical, but we have seen a way to synthesize it without condensation. Now the correct answer choices. This multiple choice question is C However, if you have any questions, don't have states. Leave a comment down below in the comment section, thank you for watching.

Which of the following ethers can be formed in good yield by condensation of the corresponding alcohols? For those that cannot be formed by condensation, suggest an alternative method that will work.
(a) dibutyl ether
(b) ethyl n-propyl ether
(c) di-sec-butyl ether

Key Concepts

Ethers and Their Formation

Condensation Reactions

Steric Hindrance

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