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f324 How do you answer nmr question? URGENT PLS

Hi,

I wanted to know how do your structure you answer while answering to proton nmr questions. I know how it works but apparently my teacher said I dont get most of the marks if its not written well. This is an example of how an A* student answered nmr
http://www.ocr.org.uk/Images/67675-unit-f324-rings-polymers-and-analysis-high-band.pdf
And it kinda shocked me because I write long essays :frown: I get the structure but I got 2/7 in the mock..
Could you please share your answer for example for 5Bii

https://docs.google.com/file/d/0B9AspcGXWU4qUzBJQ0tJMGFfajA/edit

the mark scheme is here
https://docs.google.com/file/d/0B9AspcGXWU4qV0FTUlF5Qzc4X2s/edit

Thanks in advance for your help.
(edited 9 years ago)
Cant find the question?
Reply 2
Original post by ChemistryBud
Cant find the question?


i think it's the last couple (page 16 or so)

Im not sure I can help as my paper is AQA. I usually just draw the fragments
Original post by iloveitachi
Hi,

I wanted to know how do your structure you answer while answering to proton nmr questions. I know how it works but apparently my teacher said I dont get most of the marks if its not written well. This is an example of how an A* student answered nmr
http://www.ocr.org.uk/Images/67675-unit-f324-rings-polymers-and-analysis-high-band.pdf
And it kinda shocked me because I write long essays :frown: I get the structure but I got 2/7 in the mock..
Could you please share your answer for example for 5Bii

https://docs.google.com/file/d/0B9AspcGXWU4qUzBJQ0tJMGFfajA/edit

the mark scheme is here
https://docs.google.com/file/d/0B9AspcGXWU4qV0FTUlF5Qzc4X2s/edit

Thanks in advance for your help.


I normally say something like this
"the peak at delta 1.5 shows a proton as part of R-CH group. The doublet splitting pattern suggests it is bonded to an adjacent carbon with 1 hydrogen." if it shows integration say how many hydrogens it integrates for. and do it for every peak- this is the way my teacher taught us to get the most marks possible even if you don't get the final structure
Ill just go from page 16 onwards I guess.

So for question 5b)
Due to the low natural abundance of carbon (and a few other details beyond A-level), coupling is not usually observed in a 13C NMR spectrum. This sets it apart from proton (1H-NMR) spectra where scalar couplings give rise to triplets, quartets e.t.c. Apart from that and the chemical shift range (20-200ppm as opposed to 0-10ppm), there is little you need to remember, the concepts are exactly the same. That is, the number of signals represents of the number of chemical environments (7 in this case) and the peak integration gives the ratio of carbon atoms in each environment (from left to right 1:1:2:1:2:1:2).

Moving onto 1H-NMR:
6ci) reference/ standard (not sure of the course terminology)
6cii) note the key word- integration trace, it is asking how peak intensity is related to the number of hydrogen atoms within each chemical environment. 5(quintet):1(quartet):6(doublet), just as you did with the 13C spectrum.

6ciii)
the structure of C10H12O2, and an essay O_o

well the best starting point is identifying the benzene ring from the proton spectra (5 aromatics H's at roughly 7ppm, very characteristic)
relate this back to the 13C spectra (4 peaks from 120-160ppm belong to this benzene ring), as two peaks account for 2 carbons each, this tells you that the benzene ring is symmetrical and only substitited at one position.

Leaves you with a branch coming off the benzene ring with the molecular formula C4H7O2.
The peak at 180ppm in the carbon spectra is likely to be a ester (not sure if you are given a table of values in the exam?)
Given the spread of the benzene 13C peaks (ppm) you can tell that the ester group is directly joined, causing one of the beneze carbons to be deshielded (150ppm rather than the average value of 125ppm).

Left with C4H7 joined to the ester group and benzene ring.
in the carbon spectra there are two peaks unassigned- one at 40ppm and another at 20ppm (which accounts for two carbons). In the proton spectra there is a doublet (1ppm, accounting for 6 hydrogen atoms) and a
sixtet (at 3ppm). from this you should be able to deduce the remainer of the structure.

Its very hard to go through this in writing. Id recommend asking your chem teacher to show you visually with the data at hand. One of the key skills they are testing in this final question is your understanding of shielding, deshielding and how chemical shifts relate to the functional groups present; esters, carbonyls, aromatics e.t.c

Good luck, let me know if theres anything you want me to explain.
Apologies if this posts twice, the first one got filtered due to a weblink for chemical shift values
(edited 9 years ago)
Reply 5
Original post by ChemistryBud
Ill just go from page 16 onwards I guess.

So for question 5b)
Due to the low natural abundance of carbon (and a few other details beyond A-level), coupling is not usually observed in a 13C NMR spectrum. This sets it apart from proton (1H-NMR) spectra where scalar couplings give rise to triplets, quartets e.t.c. Apart from that and the chemical shift range (20-200ppm as opposed to 0-10ppm), there is little you need to remember, the concepts are exactly the same. That is, the number of signals represents of the number of chemical environments (7 in this case) and the peak integration gives the ratio of carbon atoms in each environment (from left to right 1:1:2:1:2:1:2).

Moving onto 1H-NMR:
6ci) reference/ standard (not sure of the course terminology)
6cii) note the key word- integration trace, it is asking how peak intensity is related to the number of hydrogen atoms within each chemical environment. 5(quintet):1(quartet):6(doublet), just as you did with the 13C spectrum.

6ciii)
the structure of C10H12O2, and an essay O_o

well the best starting point is identifying the benzene ring from the proton spectra (5 aromatics H's at roughly 7ppm, very characteristic)
relate this back to the 13C spectra (4 peaks from 120-160ppm belong to this benzene ring), as two peaks account for 2 carbons each, this tells you that the benzene ring is symmetrical and only substitited at one position.

Leaves you with a branch coming off the benzene ring with the molecular formula C4H7O2.
The peak at 180ppm in the carbon spectra is likely to be a ester (not sure if you are given a table of values in the exam?)
Given the spread of the benzene 13C peaks (ppm) you can tell that the ester group is directly joined, causing one of the beneze carbons to be deshielded (150ppm rather than the average value of 125ppm).

Left with C4H7 joined to the ester group and benzene ring.
in the carbon spectra there are two peaks unassigned- one at 40ppm and another at 20ppm (which accounts for two carbons). In the proton spectra there is a doublet (1ppm, accounting for 6 hydrogen atoms) and a
sixtet (at 3ppm). from this you should be able to deduce the remainer of the structure.

Its very hard to go through this in writing. Id recommend asking your chem teacher to show you visually with the data at hand. One of the key skills they are testing in this final question is your understanding of shielding, deshielding and how chemical shifts relate to the functional groups present; esters, carbonyls, aromatics e.t.c

Good luck, let me know if theres anything you want me to explain.
Apologies if this posts twice, the first one got filtered due to a weblink for chemical shift values


Thanks a lot for your answer and your time!!

Original post by Kemics
i think it's the last couple (page 16 or so)

Im not sure I can help as my paper is AQA. I usually just draw the fragments


Its fine dw

Original post by AliceCaroline
I normally say something like this
"the peak at delta 1.5 shows a proton as part of R-CH group. The doublet splitting pattern suggests it is bonded to an adjacent carbon with 1 hydrogen." if it shows integration say how many hydrogens it integrates for. and do it for every peak- this is the way my teacher taught us to get the most marks possible even if you don't get the final structure


Thanks

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