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1H NMR splitting pattern question

I dont understand what this splitting pattern is, I think the molecule it belongs to is an alkene and I thought this was a doublet of doublets, but none of the differences between the peaks are the same, whats going on? the integration is 1H
Since there are four peaks (quartet), surely that means that the hydrogen atom(s) which this environment represents is adjacent to carbon atom(s) with 3 hydrogen atoms in total following the n+1 rule?
Reply 2
Original post by Don Pedro K.
Since there are four peaks (quartet), surely that means that the hydrogen atom(s) which this environment represents is adjacent to carbon atom(s) with 3 hydrogen atoms in total following the n+1 rule?


surely the J values between the peaks should be constant if it is a quartet? And I don't think it would be possible looking at all the other data - C4H6O2 I think its an alkene with an ester group
I'm pretty sure you're right - looks like a doublet of doublets - its roughly the same separation, could just be due to the resolution of the spectrometer?

You should be able to see the other roofed environment with the slope pointing the other way, which gives you another clue as to whats going on.
Reply 4
Original post by KombatWombat
I'm pretty sure you're right - looks like a doublet of doublets - its roughly the same separation, could just be due to the resolution of the spectrometer?

You should be able to see the other roofed environment with the slope pointing the other way, which gives you another clue as to whats going on.


roofing? I don't know what that is :frown: also, what would I report the J value as?
Original post by jacksonmeg
roofing? I don't know what that is :frown: also, what would I report the J value as?


Actually why roofing happens occurs is a bit complicated, but it basically gives that slope in intensities you see. It happens when you've got two environments with similar shifts that couple - with two doublets, its always the inner peak thats higher so the other roofed peak should point in the other direction.

I've just remembered, what I said about the resolution is probably rubbish! Another effect of roofing is actually that the peaks are shifted slightly, so the gaps do end up being different. Because its a doublet of doublets it gets even more complicated so I'd just report it as roughly 1.6 ppm (or convert it to Hz if you know the field strength).
Reply 6
Original post by KombatWombat
Actually why roofing happens occurs is a bit complicated, but it basically gives that slope in intensities you see. It happens when you've got two environments with similar shifts that couple - with two doublets, its always the inner peak thats higher so the other roofed peak should point in the other direction.

I've just remembered, what I said about the resolution is probably rubbish! Another effect of roofing is actually that the peaks are shifted slightly, so the gaps do end up being different. Because its a doublet of doublets it gets even more complicated so I'd just report it as roughly 1.6 ppm (or convert it to Hz if you know the field strength).


Im really not sure about this question, the 13C NMR makes no sense
Original post by jacksonmeg
Im really not sure about this question, the 13C NMR makes no sense


Do you want to post the full spectra?

The DEPT thing is basically because the phase is arbitrary - in DEPT 135 if you know CH is positive, then CH2 must be negative and CH3 positive. If CH is negative then CH2 is positive, CH3 negative. Its obviously the convention that CH is positive, so usually if it comes out negative the spectra will be flipped but I guess they're just trying to test you!
Reply 8
Original post by KombatWombat
Do you want to post the full spectra?

The DEPT thing is basically because the phase is arbitrary - in DEPT 135 if you know CH is positive, then CH2 must be negative and CH3 positive. If CH is negative then CH2 is positive, CH3 negative. Its obviously the convention that CH is positive, so usually if it comes out negative the spectra will be flipped but I guess they're just trying to test you!


that's a trick question :angry:
Reply 9
Original post by KombatWombat
Do you want to post the full spectra?

The DEPT thing is basically because the phase is arbitrary - in DEPT 135 if you know CH is positive, then CH2 must be negative and CH3 positive. If CH is negative then CH2 is positive, CH3 negative. Its obviously the convention that CH is positive, so usually if it comes out negative the spectra will be flipped but I guess they're just trying to test you!


sending it in a sec, I think the molecule is CH2=CH-OCO-CH3
Original post by KombatWombat
Do you want to post the full spectra?

The DEPT thing is basically because the phase is arbitrary - in DEPT 135 if you know CH is positive, then CH2 must be negative and CH3 positive. If CH is negative then CH2 is positive, CH3 negative. Its obviously the convention that CH is positive, so usually if it comes out negative the spectra will be flipped but I guess they're just trying to test you!


There's also a question

There is some suggestion of second order effects in the 1H spectrum, how is this apparent and perform a calculation to show if this is expected, i dont know how to do this
Original post by jacksonmeg
I dont understand what this splitting pattern is, I think the molecule it belongs to is an alkene and I thought this was a doublet of doublets, but none of the differences between the peaks are the same, whats going on? the integration is 1H


It certainly looks like a doublet of doublets.

The J values of the two obvious doublets look the same and the other does not matter.

It looks like splitting from two adjacent H atoms on an alkene. The shift of the signals is about right also.
There also seems to be an aromatic unit. It looks more like C6H5CH=CHCOOCH3 or something like
Original post by charco
There also seems to be an aromatic unit. It looks more like C6H5CH=CHCOOCH3 or something like

molecular formula is C4H6O2
Original post by jacksonmeg
molecular formula is C4H6O2


That little gem of information would have helped a little! -)
Original post by jacksonmeg
molecular formula is C4H6O2


OK

H2C=CH-OCOCH3

The hydrogens on carbon #1 are non-equivalent and split each other. They are also split with a different J value by the H on carbon#2. These are the two signals in the region of delta 5. It is a doublet of doublets.

The two non-equivalent H atoms on #C1 both split the H on C#2 giving the messy signal at delta 7(ish) shifted downfield by the adjacent ester group.

Finally the CH3 of the ester gives the singlet at delta 2 (ish)
Original post by charco
OK

H2C=CH-OCOCH3

The hydrogens on carbon #1 are non-equivalent and split each other. They are also split with a different J value by the H on carbon#2. These are the two signals in the region of delta 5. It is a doublet of doublets.

The two non-equivalent H atoms on #C1 both split the H on C#2 giving the messy signal at delta 7(ish) shifted downfield by the adjacent ester group.

Finally the CH3 of the ester gives the singlet at delta 2 (ish)

thanks :smile:
Original post by KombatWombat
Do you want to post the full spectra?

The DEPT thing is basically because the phase is arbitrary - in DEPT 135 if you know CH is positive, then CH2 must be negative and CH3 positive. If CH is negative then CH2 is positive, CH3 negative. Its obviously the convention that CH is positive, so usually if it comes out negative the spectra will be flipped but I guess they're just trying to test you!


thanks :smile:

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