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# Homework help please? Identifying spectrums Watch

1. Hi guys, I have a homework on spectroscopy and I have to identify the structure shown by each spectrum. I am struggling grasping the ideas behind it and it's due in tomorrow.
https://imgur.com/aKVHjHy
Help me please 😩
2. What have you done so far?
3. (Original post by alow)
What have you done so far?
By the looks of it, I think it's Xylene C8H10 but I need help with actually explaining each spectrum as to how I concluded with C8H10. I only know a few bits but it's not specific enough so if you could help me specify like which peak shows what part of the structure etc.
4. (Original post by Rnddlcrz)
By the looks of it, I think it's Xylene C8H10 but I need help with actually explaining each spectrum as to how I concluded with C8H10. I only know a few bits but it's not specific enough so if you could help me specify like which peak shows what part of the structure etc.
A good start would be showing how you found the emperical formula from percentage composition data

Then I'd explain how the Mass spec molecular ion peak allowed you to work out the formula of the compound. Knowing the formula is always a useful place to start
5. (Original post by MexicanKeith)
A good start would be showing how you found the emperical formula from percentage composition data

Then I'd explain how the Mass spec molecular ion peak allowed you to work out the formula of the compound. Knowing the formula is always a useful place to start
Well, by looking at the mass spec I know that the Mr is 107. So to get the number of carbon I did (90.6/100)x107 which gave me 96.942 then I did 96.942/12 to get the number of carbons which I found to be 8.0785 or 8. I did the same with the hydrogen (9.4/100)x107 = 10.058 or 10. That's how I came up with C8H10. I now need to explain the structure using the the four spectrum and thats what I'm struggling with 😬 help me? please? haha thanks
6. (Original post by Rnddlcrz)
Well, by looking at the mass spec I know that the Mr is 107. So to get the number of carbon I did (90.6/100)x107 which gave me 96.942 then I did 96.942/12 to get the number of carbons which I found to be 8.0785 or 8. I did the same with the hydrogen (9.4/100)x107 = 10.058 or 10. That's how I came up with C8H10. I now need to explain the structure using the the four spectrum and thats what I'm struggling with 😬 help me? please? haha thanks
That's a neat way to find the formula!

What exactly do you think the product is (ie which isomer?)
7. (Original post by MexicanKeith)
That's a neat way to find the formula!

What exactly do you think the product is (ie which isomer?)
I'm not really sure, though from the data sheet and from what I know I think there are a few c=c etc. I think I may be able to analyse most of the Mass spec and IR spectrum but Im not sure how to interpret the 1H and 13C spectrum. I just asked my friend too he isnt sure either but he said that there are three isomers and im not sure how to specify which is the right one according to the spectra
8. (Original post by Rnddlcrz)
I'm not really sure, though from the data sheet and from what I know I think there are a few c=c etc. I think I may be able to analyse most of the Mass spec and IR spectrum but Im not sure how to interpret the 1H and 13C spectrum. I just asked my friend too he isnt sure either but he said that there are three isomers and im not sure how to specify which is the right one according to the spectra
Ahh, so earlier when you said it was Xylene, was that conclusion based purely on the formula?
9. (Original post by MexicanKeith)
Ahh, so earlier when you said it was Xylene, was that conclusion based purely on the formula?
It was a hunch yes haha. I remember seeing from a chemistry magazine in the library and for some reason I cant forget it 😬 so when I got C8H10 I just thought "Oh Xylene" 😁 most likely wrong but I cant tell if I cant properly interpret the spectra
10. (Original post by Rnddlcrz)
It was a hunch yes haha. I remember seeing from a chemistry magazine in the library and for some reason I cant forget it 😬 so when I got C8H10 I just thought "Oh Xylene" 😁 most likely wrong but I cant tell if I cant properly interpret the spectra
Okay, so, the IR and the Mass Spec won't give you much info about the isomer (NMR is a god send if you can interpret the spectra )

The IR in fact doesn't show you a lot, the compound is a hydrocarbon so the spectra naturally has C-H stretches and bends, but not a lot that is structurally diagnostic!

The mass spec, you've already used the molecular ion peak to find the formula. Next you might want to try to identify some of the other peaks, the obvious fragment peak to think about is the large peak at 91 (15 below the molecular ion peak) what fragment could break off to drop the fragment mass by 15?
11. (Original post by MexicanKeith)
Okay, so, the IR and the Mass Spec won't give you much info about the isomer (NMR is a god send if you can interpret the spectra )

The IR in fact doesn't show you a lot, the compound is a hydrocarbon so the spectra naturally has C-H stretches and bends, but not a lot that is structurally diagnostic!

The mass spec, you've already used the molecular ion peak to find the formula. Next you might want to try to identify some of the other peaks, the obvious fragment peak to think about is the large peak at 91 (15 below the molecular ion peak) what fragment could break off to drop the fragment mass by 15?
I'm trying to learn it but this week has been busy so I didnt have time but I will be learning it over the weekend. Just an inconvenient time to not know much about it as clearly it's causing problems in homeworks haha.

Anyway, I think CH3 mightve caused the drop
12. (Original post by Rnddlcrz)
I'm trying to learn it but this week has been busy so I didnt have time but I will be learning it over the weekend. Just an inconvenient time to not know much about it as clearly it's causing problems in homeworks haha.

Anyway, I think CH3 mightve caused the drop
Yep so you know your compound has -CH3 groups, so of the compounds with the formula C8H10 you know it can only be one with a methyl group you could gain more from the mass spec but that information can also be gained from the NMR's

Start with the carbon-13 NMR, we tend to do something called decoupling when doing C-13 nmr so we see no splitting in the peaks, this makes them easier to interpret. Hopefully you have data sheets and can tell that the peaks on the left are aromatic or alkene peaks

That pretty much tells you that your product should be a Xylene isomer, or ethylbenzene!

Just a note, you should be careful with Carbon-13 NMR when looking at peak integration (area) because it doesn't necessarily equate to the number of carbons very well, this is because decoupling more or less messes about with the peak integrations.

Hopefully you can follow where I've gone so far, if not, tell me and i'll explain things more carefully.

After that, do you know how to count environments in a molecule?
13. (Original post by MexicanKeith)
Yep so you know your compound has -CH3 groups, so of the compounds with the formula C8H10 you know it can only be one with a methyl group you could gain more from the mass spec but that information can also be gained from the NMR's

Start with the carbon-13 NMR, we tend to do something called decoupling when doing C-13 nmr so we see no splitting in the peaks, this makes them easier to interpret. Hopefully you have data sheets and can tell that the peaks on the left are aromatic or alkene peaks

That pretty much tells you that your product should be a Xylene isomer, or ethylbenzene!

Just a note, you should be careful with Carbon-13 NMR when looking at peak integration (area) because it doesn't necessarily equate to the number of carbons very well, this is because decoupling more or less messes about with the peak integrations.

Hopefully you can follow where I've gone so far, if not, tell me and i'll explain things more carefully.

After that, do you know how to count environments in a molecule?
Sorry for asking a bit too much but could you please explain a bit more on which peaks on the H and C nmr shows which part of the structure? You kind of lost me when you talked about alkenes and I wasnt sure which specturm youre referring to. Thanks 😅

Yes, I can identify hydrogen and carbon environments
14. (Original post by Rnddlcrz)
Sorry for asking a bit too much but could you please explain a bit more on which peaks on the H and C nmr shows which part of the structure? You kind of lost me when you talked about alkenes and I wasnt sure which specturm youre referring to. Thanks 😅

Yes, I can identify hydrogen and carbon environments
So far I've only discussed the C-13 NMR and the fact that the peaks on the left are in the aromatic/alkene region, that combined with the fact you know it must have -CH3 present narrows it down so, just from the C-13 NMR it must be a Xylene or ethylbenzene.

Hope that's clear

Next, draw out the molecules you've narrowed it down to and see how many carbon environments they have (still not even thinking about the proton NMR)!
15. (Original post by MexicanKeith)
So far I've only discussed the C-13 NMR and the fact that the peaks on the left are in the aromatic/alkene region, that combined with the fact you know it must have -CH3 present narrows it down so, just from the C-13 NMR it must be a Xylene or ethylbenzene.

Hope that's clear

Next, draw out the molecules you've narrowed it down to and see how many carbon environments they have (still not even thinking about the proton NMR)!
Yep thats better. I kind of cheated here and looked up the structure of ethylbenzene but I think there's 5 carbon environments or 6? CH3 CH2 then the C that links the benzene with the c2h5 the two CH and another two CH. Do you count the lone CH at the end as another environment? If you do then I guess there would be 6
16. (Original post by Rnddlcrz)
Yep thats better. I kind of cheated here and looked up the structure of ethylbenzene but I think there's 5 carbon environments or 6? CH3 CH2 then the C that links the benzene with the c2h5 the two CH and another two CH. Do you count the lone CH at the end as another environment? If you do then I guess there would be 6
Yep Ethylbenzene has 6 environments
Now think about the different types of xylene (the systematic names are 1,2-dimethylbenzene, 1,3-dimethylbenzene, and 1,4-dimethylbenzene) try to draw them without looking them up using the systematic names! then count the Carbon environments!

You're getting close to the answer
17. (Original post by MexicanKeith)
Yep Ethylbenzene has 6 environments
Now think about the different types of xylene (the systematic names are 1,2-dimethylbenzene, 1,3-dimethylbenzene, and 1,4-dimethylbenzene) try to draw them without looking them up using the systematic names! then count the Carbon environments!

You're getting close to the answer
Okay I think that 1,2 has 4 environments and 1,3 has 5 environments and 1,4 has 3 environments. Is the answer 1,3? There's 5 peaks on the C NMR
18. (Original post by Rnddlcrz)
Okay I think that 1,2 has 4 environments and 1,3 has 5 environments and 1,4 has 3 environments. Is the answer 1,3? There's 5 peaks on the C NMR
Exactly! 1 peak per environment, so you've got to the point where you're pretty sure its 1,3 dimethylbenzene (meta-xylene) without even needing to look at the proton NMR!

Now, hoping you know how to predict splitting patterns, look at the proton environments on your compound and see if you can match up the peaks in the proton NMR to the protons!
19. (Original post by MexicanKeith)
Exactly! 1 peak per environment, so you've got to the point where you're pretty sure its 1,3 dimethylbenzene (meta-xylene) without even needing to look at the proton NMR!

Now, hoping you know how to predict splitting patterns, look at the proton environments on your compound and see if you can match up the peaks in the proton NMR to the protons!
It's n+1 rule right? I havent done H nmr since before the summer holidays haha so I need to brush up on it a little
20. (Original post by Rnddlcrz)
It's n+1 rule right? I havent done H nmr since before the summer holidays haha so I need to brush up on it a little
Yep so each proton adjacent to the ones you're considering couples and splits the peak. If there are n adjacent protons your peak is split into n+1.

This molecule isn't too bad, and remember your data sheet should tell you where abouts different protons should appear on the spectrum

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