The Student Room Group

NMR spectrum

image-fc1aed16-6e24-40af-bd2b-5d576d2957be7609384645939494377-compressed.jpg.jpeg

x 7
y 8
z 2

structure?
Reply 1
Original post by googie3
image-fc1aed16-6e24-40af-bd2b-5d576d2957be7609384645939494377-compressed.jpg.jpeg

x 7
y 8
z 2

structure?


image-ace4bcba-36b2-4487-a619-b8f4d002c5334442476716678600371-compressed.jpg.jpeg
Reply 2
Original post by _NMcC_
Mass spec

I multiplied that M:M+1 ratio by 4 to get 8 for the M+1 value (100 for M). You can that divide the M+1 value by 1.1% (the natural abundance of C13) to get just over 7, or you can divide 100 into 12.01 to get around 8. So we know the number of carbons is one of 7 or 8.

So yes X = 7 would be fine.

NMR
Now the number of protons must be 8 (sum of the peaks.) So roughly speaking, looking at the ratios of C;H we can say that it's a very unsaturated molecule with loads of double bonds. You can confirm this if you wish using the 'Degree of Unsaturation equation'.

Y = 8

Looking at the NMR, I find it's easiest to look in the Alkyl region first. A singlet with 3 Hydrogens @ 3.8ppm in this region must be a -O-CH3. This means it's likely an ester group (but could also be an ether). The homotopic hydrogen coupling averages out to a singlet.

The singlet @ 5.5ppm I would interpret as a C=CH- due to it's region, the lack of splitting indicates that it could be either attached to a C=O carbon (indicating ester) or directly attached to a oxygen (ether). Which is where the interpretation that there could be 1 or 2 oxygens comes in.

The last singlet at 6.7 perhaps the trickiest and requires some creative thought. Normally this is the aromatic region however simply as this is a singlet with 4 hydrogens suggests to me that it isn't a phenyl group as normally you get a lot of coupling with substituted aromatics and the total number of carbons would be too high. The number '4' hydrogens seems strange also. However going back to the deduction that it's a highly unsaturated compound. There are a couple of initial ideas;

A cyclopentadiene ring (with a double bond C attached to the 'top' carbon). This would however mean 2 sets of hydrogens in slightly different environments, therefore we would expect to see more coupling between alkene hydrogens.

So to avoid coupling somehow, perhaps look for Magnetic Symmetry...

So the other possibility is that it's a Cyclopentanyl ring. With the C=C of the 'chain' attached to the top carbon. This would make more sense as a Carbon-Carbon triple bond at the 'bottom' would hinder the amount of coupling. If you had 2 symmetrical CH2 groups both bonded to the same C triple bond C, and also attached to the same Carbon double bond - Chain. This would create a singlet with 4 hydrogens as the the two CH2 groups would be magnetically equivalent and couldn't couple to anything else.

The downfield shift of these two CH2 groups may be due to a C=O being present along with the Alkyne (due to increased resonance)

If you take the number of carbons as 8, therefore the complete molecule may look something like the attached structure.

C8H8O2

There is room for ambiguity though in my opinion and there may be other structures possible, which is why it's a tricky 'suggest' type question.

Another perhaps more accurate structure would be an ether instead of the C=O, which removes a carbon+oxygen = C7H8O. But I think either could be correct.

If you had IR data you could differentiate properly between Ester or Ether, (a distinct carbonyl peak would appear if it was an Ester).


*Update; I think the M:M+ ratio would also work better with Ether, so yes, ether is more accurate in my opinion. Ester is just a bit high.


mark scheme says only 4 methoxylphenol. i dont know why, i do t know how. thanks for your time but if possible, give a reason for actual 4 methoxylphenol answer
Reply 3
Original post by googie3
mark scheme says only 4 methoxylphenol. i dont know why, i do t know how. thanks for your time but if possible, give a reason for actual 4 methoxylphenol answer

Ah interesting. They must have averaged out the aromatic signals. Which alters the structure determination...

The -O-Me group is fine.

They seem to be taking the aromatic signals as 1 peak, which is unusual but I suppose possible if you have a hydroxyl group para to the -O-Me group which could then allow the 4 aromatic hydrogens to be treated as a magnetically similar and thus any slight multiplet would overlap in a lower resolution spectrometer, to produce a singlet.

So strictly speaking a multiplet should be expected for the aromatic hydrogens as -OH and O-Me as they are technically different groups. Which is why I initially thought it wouldn't be an aromatic compound but it appears this multiplet is overlapping, perhaps they are only considering the -O atoms and not the -OH or -O-Me

The -OH group therefore explains the last singlet, which without it, wouldn't explain the aromatic 'magnetic equivalence'.
(edited 5 years ago)
Reply 4
But a direct 4 H peak makes no sense. They should have told about overlapping. also is it 4 different environments for benzene ring with OCH3 group and OH group in meta? if we just have to tell the number of peaks expected, total would be 6 right?
Original post by _NMcC_
Ah interesting. They must have averaged out the aromatic signals. Which alters the structure determination...

The -O-Me group is fine.

They seem to be taking the aromatic signals as 1 peak, which is unusual but I suppose possible if you have a hydroxyl group para to the -O-Me group which could then allow the 4 aromatic hydrogens to be treated as a magnetically similar and thus any slight multiplet would overlap in a lower resolution spectrometer, to produce a singlet.

So strictly speaking a multiplet should be expected for the aromatic hydrogens as -OH and O-Me as they are technically different groups. Which is why I initially thought it wouldn't be an aromatic compound but it appears this multiplet is overlapping, perhaps they are only considering the -O atoms and not the -OH or -O-Me

The -OH group therefore explains the last singlet, which without it, wouldn't explain the aromatic 'magnetic equivalence'.
Original post by googie3
But a direct 4 H peak makes no sense. They should have told about overlapping. also is it 4 different environments for benzene ring with OCH3 group and OH group in meta? if we just have to tell the number of peaks expected, total would be 6 right?


4-methoxyphenol would have two different aromatic environments. It is quite possible that they are close together as both 'para' attachments are oxygen linked.
Reply 6
Original post by charco
4-methoxyphenol would have two different aromatic environments. It is quite possible that they are close together as both 'para' attachments are oxygen linked.


why two environments? i think there will be 4.... please confirm?
Original post by googie3
why two environments? i think there will be 4.... please confirm?


Both the 2 and 6 positions on the ring are in identical environments, as are the 3 and 5.

As I suspected, the shift for each environment is virtually identical:

4-methoxyphenol.png

https://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi
Reply 8
Original post by charco
Both the 2 and 6 positions on the ring are in identical environments, as are the 3 and 5.

As I suspected, the shift for each environment is virtually identical:

4-methoxyphenol.png

https://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi


you marked 4 positions as A, the same environment, why that so?
Reply 9
Original post by charco
Both the 2 and 6 positions on the ring are in identical environments, as are the 3 and 5.

As I suspected, the shift for each environment is virtually identical:

4-methoxyphenol.png

https://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi


whg are 2,6. and 3,5. not different environments? How do you see their values in data booklet?
Thanks a lot
Original post by googie3
you marked 4 positions as A, the same environment, why that so?


I didn't mark anything!

The spectrum comes from the spectral database referenced. You will see that all four protons have a shift of 6.77 ppm.

Quick Reply