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Show how to convert hexanoic acid to each amine in good yield:a)N-CH: CH3b)...

Question

Show how to convert hexanoic acid to each amine in good yield:a)N-CH: CH3b)

Show how to convert hexanoic acid to each amine in good yield: a) N-CH: CH3 b)



Answers

Write a structural formula for each amine. (a) $2-$ Aminoethanol (b) Diphenylamine (c) Diisopropylamine

Hi in this problem, we are given some of the structures of our mites. Using these structures, we have to draw the condensed formulas for the carb oxalic acid and I mean, and we're used to form these um it's the first one of the structure in this compound. This part of the molecule would have come from the Tarbox liquor, sending this would have come from that. I mean, and carbolic acid loses high Doxil part. I mean loses hydrogen to form the um it So the car box liquor said used for this molecule has the structure, Well, we have this kind of arrangement. This is the car box lickers it part and each of this carbon item has a hydrogen wanted to it and double bond. And the um in part here it is NH two. This would have come by the loss of hydrogen from Ammonia. So the I mean, he used his actually ammonia. The next molecule is this and we can draw its structure as shown here. So this molecule would have come from the car box, like I said, which has a structure shown here. And I mean used is this in the next molecule the car box leaders had used for the preparation of this amid. Has the structure shown here? And uh I mean, has the structure.

All right. And this problem were given four different Amin's and we need to draw them and write their formulas. So the first is going to this Ethel Amin. So you know that you will have an end touch A to H is plus in fo Group, and this will have the Formula C two each five and H two. So then you'll have a Die Purple Amin's. He'll have an end with two propane groups as waas in each. So then counting up, you'll have C six each 14 and h If you wanted to, you could put the hydrogen Ah, that's a touch the end, Like you could say C six h 15 on. But I like to keep the house, Uh, just for my own No, I can see where the Amin is basically. So then secondly, you're gonna have our third. You're gonna have beautiful dime. Ethel Amin, you have an end with a beauty in group on it was to metals So counting up again, you'll have C six each 15 and and as you can see, this one and the one before have you could almost have the same formula, but by separating them you can tell that it is a secondary hemming versus a tertiary Amin sewn on this. Try Ethel Ami nearing him and 1 to 1 to that story. Yeah. 1212 And then this will be C six each 15 n as well.

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

Hey, guys. So in this question were given three reactions and were asked to find whether they are acid base or condensation and were asked to also find the products for each of these reactions. So, in part A, we have the reaction between die metal ah, mean and hydrochloric acid, and right away we can know that this is an asset based reaction since we know on acid based reaction is, of course, between a Nassib and a base. So we can all write that down, that this is an asset based reaction and our product for this is going to be ammonium salt or a version of an ammonium self. So we have the base, Ah, parent chain that we have forgiven. This is going to have a positive charge because of the amount of bonds it has, it has four bonds and the chlorine, um, molecule from the hydrochloric acid is going to be bonded or attracted to this. A positive charge from the ammonium. All right, in part B, we have a reaction between Ethel Amine and propane OIC acid, so we know that Ah, condensation reaction is a reaction that forms, uh to ah forms a bond between two different molecules and it forms the loss of a small molecule. And we also know that a means usually undergo condensation with, ah, car carb oxalic acids to form a meid. So here we have on a mean and we also have a car box cilic acid. So I'm going to go ahead and say, This is going to be a condensation reaction and our product for this is going to be and Ethel pra propia in pro pie on a mydd excuse, my partner pronunciations. So we're going to have ah, part of each chain from the amine and then apart from the A meid as well there's going to be a double bond on this carbon right here, and Ross even be forming water. So this is our n Ethel pro pie on a mydd. And in part C, we have what looks like another amine so methyl amine and sulfuric acid. So right away. We can also say this is an acid base reaction and our product for this is going to be, um, and ammonium salt again. So we have aversion from Arm Ethel chain and our ammonium, and we'll have our salt forming from these two ions and this is our final


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