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Create a pathway for making Butyne from ButanoicAcid...

Question

Create a pathway for making Butyne from ButanoicAcid

Create a pathway for making Butyne from Butanoic Acid



Answers

How could you synthesize the following compounds starting with a carboxylic acid?

This is the answer to Chapter 20. Problem number 33 from the McMurray Organic chemistry textbook. Ah, this problem asks us to convert butin OIC acid into five different compounds. Um, okay, So the first, uh, product that we're asked to make is one beautiful, um, and so we can get there from beauty malic acid. Actually, just in a single step eso we can use. Ah, nice, strong reducing agent. So, in this case, lithium aluminum hydride. Ah, And then that, um, will use an acidic work up. So h 30 plus general acidic work up on, and that is going to get us to warn Beautiful. Okay, us? That's part a for part B. Um, we're actually gonna start from that one Beautiful that we made in part A So starting from one beautiful, um, and we're gonna we're gonna start from there because we want to make one bruma butane in part B. And so we can get there from one butin all again in a single step. And in this case, it's going to be PBR three that will use. Remember, PVR three basically replaces an alcohol with a roaming. And so from one uh, butin all Teoh one bruma butane. Um, pretty sure you forward. Okay, so that's part B for part C. We are trying to make Penton OIC acid. Um, and so again, we're going to start. Uh, well, not again. I guess eso this time we'll start from the one Burma beauty in that we made in part B. So there's are one form of beauty mean, um and eso this time there are, I guess, a couple different ways that we could do this that were discussed in this chapter. I'm so I am going to used sodium cyanide on, make the night trial and then hide relies the no trial and so we can write that with a single arrow. So Step one, ah, will be sodium cyanide. Ah, and then we can just hide relies that night trial with some acid. So h 30 plus, um And so that sequence will get us, um, to Penton OIC acid. So we we had to add a carbon there. Um, which is why making the night trial and then hide realizing it worked so well because we needed that Fifth carbon eso There's our Penton OIC acid. It's then in part D. Um, we are going to make one beauty mean, and so we will start from our product from part B again. So one broom of butane Onda we can get from here to one beauty mean again in just a single step. And so in order to do that, we're going to use Ah, strong hindered base. So potassium turkey tac side is sort of the, uh, standard here. Um, so potassium turkey tac side. Ah, I write it that way. There are different ways to write it. Um, but so that is going to get us to one beauty, which looks like that. All right. And so then, lastly for part E, we are hoping to make, um, octane. So this octane is Ah, hydrocarbon, obviously a carbons. Um, and so this is actually going to take several steps again? We'll start from one Burma Butane, which we made in part B. And so we're actually gonna form the Gilman re agents of the lithium Cooper eight. Um, and so the first thing to do here is going to be to treat our alcohol hail I'd with two equivalents of lithium metal, and so that is going to get us to the alkalinity. Um Okay, um, and from the alkalinity, um, Then we will use copper. I died, and so that's gonna actually form our gilman re agent. Um, and so remember, for the human re agent, it's too. Ah, equivalents of the Al Qaeda group. Um, so to, uh, beauty chains attached to the copper. Um, and then we can use that gilman re agent directly on inappropriate alcohol. Hell, I'd, um and we will make a carbon chain. And so, since we have four carbons here, and I are, uh, Gilman, every agent and we want eight carbons total. In the end, um, we can use another equivalent of Roma butane or one Burma butane, rather. And so that reaction is going to get us toe octane. So there we go. Um, so part e are required a couple more steps than the previous parts, but still very straightforward. Just formation of the gilman. Every agent on then using that in conjunction with the appropriate alcohol Allied. Let's make octane. Okay. Ah. And so that is the answer to Chapter 20. Problem number 33

Okay, so we want to we want to write a try try a cell cholesterol. It can be formed from the following acids. So glycerol is a three carbon alcohol with alcohol is on each carbon. I think this and the first one is paul merrick acid, which is a 16 carbon copies of acid. So I'm sure to put and our group right here, subsequent 15 for 15 were carbons. And this is really just between a carbonic acid and alcohol or I die cross passage and a dial. So if we react a carbonic acid with an alcohol, we'll get asked her. I'm going to do is drop it. So age group this hydrogen and combine these two structures. Just give me the structure the trickiest cholesterol. And if we have reacted all the alcohols with all the diplomatic acids, we get these esters. It's typically drawn like this. These carbons are attached to each Esther to a 15 carbon june. I would just have um to hodgins over here, but for the middle of just one hydrogen. Okay, so for the next one, all we need to do is just replace the number 15 with however long chain is in the following acids. Next one is always like acid and this is an 18 carbon carbolic acid and assist confirmation. So just drawing this three carbon chain first, you know, it's gonna form an ester because between a corrosive acid and the alcohol glycerol. So we just react all these alcohols. I'm gonna draw this one out because we are. They are. Yeah, keane just 12 3456789 10, 11, 12, 13, 14, 15, 16, 17, 18. Now, kean's on carbon nine relative to yester 123456789 So we should draw the rest of the chain down here because it's cysts 10 11, 12, 13, 14, 15, 16, 17, 18. Okay, so, we have two options over here. I have one here to have four bonds and to have their and the last one is with But no, look acid. And this is another 18 carbon Car myself acid. We're gonna have to It's a dye into all two double bonds and it's also in the cIS confirmation. Okay, so just drawing the through carbon chain first, your cholesterol reacting. Each of the alcohol is with carbolic acid, which is the 18 carbon of linoleic acid. Okay, these esters and each ester attached to 18 carbon chain with two Wilkins in the chain. 12 3456789 Is your keen? It's this 10, 11, 12. There are keen. 13, 14, 15, 16, 17, 18. I just came here to 123456789 10, 11, 12, 13, 14, 15, 16, 17, 18. Okay, They're just out of the Hodgins to their one here. That's you there. Remember the structure of a try using glycerol is just glycerol attached to any carbon carbolic acid in this case for each one. Well, in the last week, 18 carbon carbon acid, but it could be any number of carbon atoms. It's just between lists are all, which is a three carbon alcohol with an alcohol in each carbon attached to a car bus of acid, too, generating Esther.

They were just looking at the structure of the product using beauty Nike acid on missile. Amy as our starting materials. So we generate this structure here where our circle, What was I would be to note acid this potion waas from on me Thought aiming I don't We've done is removed a water molecule in order to generate are a might. So that answers The second point that we needed to cover was what functional groups are presidents, that we haven't a my present because we have a carbon doubly bound to an oxygen where that carbon is also bound to a 90 Jim. So the last thing I've covered here is just writing out the balanced chemical equation for this reaction where, as you can see, we remove awards up in order to join these two molecules that we have a condensation reaction here

This is the answer to Chapter 20. Problem number 34 from the McMurray organic chemistry textbook. This problem is sort of the opposite of the problem before it. Number 33 in that, uh, in number 33 we were asked to start from butin OIC acid on and convert that into five different things. But here, number 34 were asked to start from five different starting materials and convert each of them into beauty malic acid. Okay. Eso looking at a, um we have one butin all here. Ah, and we want to get to be, you know, like acid. And so we can do this simply by using a strong oxidizing agent. So, um, chromium try oxide is good one on. And then we would do an acidic work up there, and that is going to get us to our butin OIC acid. Okay, um and then for B and C were actually going Teoh, try to get toe one beautiful and then again, use chromium try oxide to oxidize that too, you know, gas it. Okay, so for B um, as I said, we need to start by getting from Burma. Butane one, Burma. Butin to, ah, one. Beautiful. And so we can do that just by using sodium hydroxide. Um, And so that is going to get us 21 Beautiful. And then again, just like in a chromium trioxide and some acid gets us tribute No gas and okay, And then for C, um, we will ah, start again by getting the one beauty home. And to do that and C, we need to do an anti Markov niqab hydration. So we need to add the elements of water, Um, in an anti Markov knockoff fashion on and so the way to do that eyes to use a boron re agents of bh three works. Uh, and we always right that we include. You know, that we do that in th f. Um, and then we need to follow that with hydrogen peroxide and some base. So h two of two. Ah, and then just hydroxide Ion is usually how that's written. And so that is again going to get us to one beautiful. And then from there, justus in a and B chromium try oxide and asked, Okay. Ah, so that's a B and c. So then here in D, um, we are starting from one bro mo propane. So we need Teoh not only get from the alcohol. Hey, lied to the carb oxalic acid. We actually also need to add a carbon. Um, and so there are two ways that we could do this We could, ah, follow ah green yeard pathway and use carbon dioxide. Or we could use sodium cyanide and then hide relies that no trial that that we make. And so that's what I'm gonna dio. So I'm gonna use sodium cyanide, as I said in a CNN Ah, and then follow that with some acid and that is going to get us to where we need to be. So we've added a carbon. Ah, and we've gotten to the carb oxalic acid here. All right, And then So lastly, we need to go from four octane to, um do you know a gas it, um And so basically, when we look at four octane, there are four carbons on each side of this double bond eso we need to Cleveland's double bond in such a way that we end up with carb oxalic acid. Ah, and so we could do this. Ah, via those analysis so we could do, you know, ozone followed by a knock sedate of work up, Um, or I think, the answer that the book prefers eyes going to be, um, potassium permanganate. So OK, I m in 04 and then again, some acid in there, and so that is going to get us to equivalence of Butte. Knock acid. Okay. Ah, and so, um, that's part E. Ah. And that is the answer to Chapter 20. Problem number 34.


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