This is the answer to Chapter 14. Problem number 65 Fromthe Smith Organic chemistry textbook. Ah, and this problem asks us to propose to structures. Ah, that are consistent with two sets of data that we've been given eso for each structure that we're supposed to propose. We've been given a molecular ion. Um, and I are spectrum in an enemy war spectrum. Okay? And so, starting with J, we're told that Jay has, um, its molecular ion at 72. Um and so from there, we can start to think about some formulas. S 06 carbons equal 72 s 05 carbons on 12 hydrogen ZX could get us to 72. Um, but I'm going to jump around a little bit. Ah, And in light of the fact that J has an i r. Peek at 17. 10 wave number, which is consistent with a carbon oxygen double bond, I'm going to say that there is anoxia gin in Jae. And so when I see the mtz at 72 I think this is going to be C four each aids. Oh, um and I think that because again, this I r tells me that there has to be a carbon oxygen double bond in here. Um, and so if the formula is see for H 80 the degrees of on saturation for that formula are going to be, uh, two times four plus two minus eight. All of that over too. And so that's going thio be one degree oven saturation on dhe. So I think this is, ah, the correct formula. And I think that we have a carbon oxygen double bond in this molecule s. So then we need to look at our anymore. Signals s 1.0 ppm. Triplet. That integrates to three. That's going to be a method group. Um, and I'm gonna skip the next signal on Look at this 2.4 ppm quartet that integrates to two. And that's gonna be a methylene group. Ah, And so these two groups are splitting each other. Uh, right, because we have a quartet on the methylene signal on, So that means it has three neighboring protons on the signal for the metal group split into a triple it, which means that it has to neighboring protons on. So these two are next to each other. They're splitting each other. Um, And then the last signal of the second signal here, the 2.1 ppm single it that integrates to three. Uh, that's going to be a method group with no neighboring Proton because it's a single it. Ah. And so if we put all of this information together, um, what we get looks like this. So here's our Carbonell. And then there's another method group on just to sort of spell this out S o. This signal is going to be this metal group. This signal is going to be this method group, and this signal is going to be this math linger. Okay, so there's, uh there's j and now we just need to do the exact same thing for Kay. So looking uk, we have ah molecular ions at 88. Um, and since looking at the i r for j proof so helpful, let's start with the i r. Here. So this time on the i r. Is a peak between 3200 and 3600 wave numbers. That's a broad peak. Ah, and so this is consistent with an O. H. So we have an alcohol in this molecule Um, And so again, we need to account for the presence of an oxygen. Ah, and so C five, um, case that 60. And then oxygen is gonna be 16. So that's 76. So let's go see five each. 12. Oh, okay. See? Five each 12 0 And so if we do the hydrogen deficiency index for that formula, what we get is two times five plus two minus 12 all of that over to equal zero. So there are no degrees of on saturation in this molecule us and no double bonds, no rings. Just a, uh, chain of carbons with an alcohol in it. Okay. Um, and so looking at our anymore the fourth signal, the 1.6 p p. M. Single it jumps out to me. This is probably, uh, the proton of our hijack see group. Um, Okay, So the 0.9 p p. M. Triplett are that integrates to three. Uh, that's going to be a method group. The next signal, the 1.2 p p. M. Single it that integrates to six is going to be to methyl groups. So there two metal groups in the same environment and they're not split by anything. Um, and so I'm gonna go ahead and guess that they are on our carbon with the alcohol on it. And then, lastly, we have 1.5 ppm. Quartet are that ingrates to two. So this is going to be a methylene group. And so again, notice we have We have a triplet are that integrates to three and a quartet that integrates to two. And so these two signals are splitting one another. Um, okay. And so when we put all of that together, we can put it together like this. Um, so let's do it like this, Okay? And I think actually, drawing it out with letters in this case maybe makes it a little easier to see. So obviously, uh, single for the O h. Eyes is this? Oh, age. Um, we have these two methyl groups that are equivalent. So that's gonna be these two method groups. Uh, and then we have, um, as I said, the methylene and the metal group that are splitting one another. Well, let's be consistent, I suppose. Oh, dear. There we go. Okay. So sorry. All right, so here we go. So this black arrow Obviously this is the methylene group, and then the blue arrow eyes referring to this method, Bert. And so this is a good structure for this molecule that fulfills all of the requirements that we were given on. And so that's the the answer for K. And then the answer again for J looked like this. Um and that's the answer to Chapter 14 problem number 65.