What hydrocarbons that contain a six-membered ring will have a...

Question

What hydrocarbons that contain a six-membered ring will have a molecular ion peak at m/z = 112?

Accepted Answer

All right. Hi, everyone. So for this question, let's identify the six member ring or rings with a significant peak at master charge of 100 and 11 and a molecular ion peak at master charge of 100 and 26. So option A is N propol cyclohexane. Option B is isopropyl cyclohexane. Option C is 114 triethyl cyclohexane. Option D is both A and B. Option E is both B and C and option F is all three. So A B and C, so before we begin, what I want to bring to your attention is this comparison right? Between our master charge of 1 11 and our master charge of 1 26. Now, when I subtract uh 1 11 from my 1 26 I get a difference of 15, right? And recall that 15 corresponds to the loss of a methyl radical. So here our first step is to determine our molecular formula using the molecular ion value. Now recall that when we're given the molecular ion, we can determine the molecular formula using the rule of 13. And the rule of 13 means that when we divide our molecular ion or our M plus by 13, we get a number, right. We get our quotient, which I'm going to represent with the letter N and sometimes a remainder, which I'm going to represent with the letter R. Now using this information, right, using these variables, we can figure out our molecular formula, right? Because the number of carbons according to the rule of 13 equals our quotient to our end value. But our hydrogens, the number of hydrogens corresponds to the sum of our quotient and our remainder. So N plus R. So this is our molecular formula in general, right? And then following the calculation of our molecular formula, we should always figure out the I H D of that formula just to make sure that it makes sense, right? So I H T is equal to two times the number of carbons in your molecular formula plus two plus any nitrogens you may have minus any hydrogens minus any halogens All of which you're going to divide by two. Now, with this information in mind, right, when we apply our molecular formula or sorry, our molecular ion peak, I should say to our rule of 13. Here We calculate 126 divided by 13, which is going to give us an end value of nine with a remainder of nine. So our corresponding molecular formula is going to be C and age nine plus nine. So at 18, right. So none of the answer choices have any hetero atoms. So we know for a fact that our molecular formula corresponds only to carbon and hydrogen. So from there, we can calculate our molecular formula which is two multiplied by nine plus two, There are no nitrogen. So I'm not going to write that, but there are 18 hydrogens. So that's -18 there are no halogens either All of this is going to be divided by two. And so that's going to give me an I H D E equal to one. Now, an I H D equal to one for all three of our answer choices here is going to correspond to our one cyclo heine ring. All right. So now that we've verified the molecular formula for each of these compounds. Let's go ahead and discuss their individual fragmentation patterns Because it's important to recognize that the question is asking for 600 rings that only form A significant fragment at master charge of 111, which we know corresponds to a method radical. So for our reference, right, I've gone ahead and I've drawn each of our options on the bottom here. So I have N propol hexane isopro cyclo heine And 1, 1 4 tri methyl cyclo heine. So from this, let's go ahead and focus our attention on our end Propol cycle heine. So we know that when we're looking at our end propal cyclo heine, our methyl radical is going to detach itself from our propal substituent here Right. So one carbon at the end of our substituent is going to separate itself and that's going to leave us with this remaining carbo ion. Now, what I wanna bring to your attention is that this carbo cat ion is primary, but it can rearrange itself readily to a secondary curb cat, right. So a secondary carbel cat ion is of course more stable than a primary one, but the method radical itself is very unstable, right. So because a meth radical is unstable, our end propo cyclohexane is not really going to favor the formation of it, right. It's not going to favor formation of this unstable methyl radical here. So because of that, it's not likely that a meth radical is going to dissociate. Oh did not mean to do that. And instead an E O or a propal radical is more likely to dissociate instead. So here what I'm going to write on the bottom is that an E O or a propal radical is favored, right. So because N propo heine can form more stable radicals, the formation of this methyl radical is not favored. And therefore, we do not expect to see A significant peak at massive charge of 111. So from here, we can move on to our isopropyl cyclohexane and our methyl radical is going to stem from our isopropyl substituent. All right. So that's going to give us the following secondary car cut out, right. And so this secondary carbo ion just like the first one, it can rearrange itself to a more stable form, right, which in this case is actually tertiary. So here the carbo ion can rearrange itself to become tertiary instead. But here's the problem, right, like I mentioned in our discussion of end propol heine this method radical is not stable, right. So when a more stable radical can form instead of the methyl radical, then that is what's going to be favored, right. So here what's more likely to occur is that a secondary propal radical is going to form instead. So our secondary propal radic was favored over our unstable methyl radical. So we are not going to see a significant peak at Massive charge of and last but not least, right. If we look at our 114 tri methyl cyclohexane, what I want to reach your attention is that all of our substituent on the cyclohexane ring are methyl groups, right? So that means that 114 tri methyl cyclo heine can only lose sorry a methyl radical. So either the methyl radical can leave or detach itself from carbon one or it can detach itself from carbon two. So in this case, we have two carbo ion fragments, right? We have a tertiary one. I wasn't happy with the way I drew that in my bed. Here we go, we have this tertiary one from carbon one, but we also have the secondary carbo ion from carbon four. So given this information, right, our answer choice here is going to be choice C 114 tri methyl cyclo heine because it can only form or rather it can only lose a methyl radical when we look at the fragmentation patterns of the molecule compared to the other answer choices, right. So for options A and B, because more stable radicals could form instead of the methyl radical, Then a significant peak was not likely to form at massive charge of 111. So that's going to be your final answer if you stuck around to the end. Thank you so much for watching and I hope you found this helpful.

What hydrocarbons that contain a six-membered ring will have a molecular ion peak at m/z = 112?

Key Concepts

Hydrocarbons

Mass Spectrometry

Cyclohexane Derivatives

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