Overlapping signal sets on HNMR
I've spent the past day trying to figure out the structure for the given formula C6H12O from the given HNMR and for the life of me, I don't know if I'm doing something wrong or I just don't have a deep grasp of HNMR, but I can't figure out the structure. I think it's mainly the overlapped set that's confusing me. I read online somewhere that the 3H overlap could be a CH2 overlapping CH or OH but I didn't learn anything like that so I'm uncertain about this, so if anyone could explain what this means, I'd really appreciate it. Thanks.
What I have so far:
There should be a double bond in there somewhere since the IHD is 1.
The 3H set around 1 should be a -CH2-CH3
The 3H set around 1.5 should be -CH3 with 1 neighboring H so either a -CH(-C)-CH3 with no H's on the (-C) or -CH=CH3.
The 1H set around 3.3 should be a CH with 3 H neighbors and alpha to O atom so CH3-CH(-C)-O- with CH's fourth substituent being a C with no H's or it could be -CH2-CH(-CH)-O- maybe?
The 2H set around 2ppm is either a quartet or a pentet with an overlapping 3H singlet? I'm already confused about it overlapping so I hope someone could explain what this could mean.
What I have so far:
There should be a double bond in there somewhere since the IHD is 1.
The 3H set around 1 should be a -CH2-CH3
The 3H set around 1.5 should be -CH3 with 1 neighboring H so either a -CH(-C)-CH3 with no H's on the (-C) or -CH=CH3.
The 1H set around 3.3 should be a CH with 3 H neighbors and alpha to O atom so CH3-CH(-C)-O- with CH's fourth substituent being a C with no H's or it could be -CH2-CH(-CH)-O- maybe?
The 2H set around 2ppm is either a quartet or a pentet with an overlapping 3H singlet? I'm already confused about it overlapping so I hope someone could explain what this could mean.
Hi, not sure if this is right or not, I do not have any chemical shift data on hand.
1. 5 signals, so 5 proton environments. The overlapping signals at 2ppm are just a singlet and a quartet, I think. They overlap because they have the same chemical shift. The nmr is run on a machine with a low frequency, probably a 60MHz machine, if you run on a machine at say 300 Mhz then the signals might separate. Basically the higher the frequency the longer the piece of paper the spectrum is printed on. Its like the scale from 1-8 instead on being on a piece of A4 paper for example, its on a A3 piece, same spectrum, just spread out more, so overlapping signals can be separated.
2. CH3CH=C(CH3)OCH2CH3 gives the correct splitting pattern if the chemical shifts are right!
3. CH3- CH will give the doublet intensity 3 (shift 1.5) and the quartet intensity 1(shift 3.2)
4. 0-CH2-CH3 will give the quartet intensity 2 (shift 2) and the triple intensity 3 (shift 1.1)
5. C=C(CH3) would give the big singlet at shift 2 intensity 3.
Hope this helps a bit, but not sure if its right!
1. 5 signals, so 5 proton environments. The overlapping signals at 2ppm are just a singlet and a quartet, I think. They overlap because they have the same chemical shift. The nmr is run on a machine with a low frequency, probably a 60MHz machine, if you run on a machine at say 300 Mhz then the signals might separate. Basically the higher the frequency the longer the piece of paper the spectrum is printed on. Its like the scale from 1-8 instead on being on a piece of A4 paper for example, its on a A3 piece, same spectrum, just spread out more, so overlapping signals can be separated.
2. CH3CH=C(CH3)OCH2CH3 gives the correct splitting pattern if the chemical shifts are right!
3. CH3- CH will give the doublet intensity 3 (shift 1.5) and the quartet intensity 1(shift 3.2)
4. 0-CH2-CH3 will give the quartet intensity 2 (shift 2) and the triple intensity 3 (shift 1.1)
5. C=C(CH3) would give the big singlet at shift 2 intensity 3.
Hope this helps a bit, but not sure if its right!
Original post by scimus63
Hi, not sure if this is right or not, I do not have any chemical shift data on hand.
1. 5 signals, so 5 proton environments. The overlapping signals at 2ppm are just a singlet and a quartet, I think. They overlap because they have the same chemical shift. The nmr is run on a machine with a low frequency, probably a 60MHz machine, if you run on a machine at say 300 Mhz then the signals might separate. Basically the higher the frequency the longer the piece of paper the spectrum is printed on. Its like the scale from 1-8 instead on being on a piece of A4 paper for example, its on a A3 piece, same spectrum, just spread out more, so overlapping signals can be separated.
2. CH3CH=C(CH3)OCH2CH3 gives the correct splitting pattern if the chemical shifts are right!
3. CH3- CH will give the doublet intensity 3 (shift 1.5) and the quartet intensity 1(shift 3.2)
4. 0-CH2-CH3 will give the quartet intensity 2 (shift 2) and the triple intensity 3 (shift 1.1)
5. C=C(CH3) would give the big singlet at shift 2 intensity 3.
Hope this helps a bit, but not sure if its right!
1. 5 signals, so 5 proton environments. The overlapping signals at 2ppm are just a singlet and a quartet, I think. They overlap because they have the same chemical shift. The nmr is run on a machine with a low frequency, probably a 60MHz machine, if you run on a machine at say 300 Mhz then the signals might separate. Basically the higher the frequency the longer the piece of paper the spectrum is printed on. Its like the scale from 1-8 instead on being on a piece of A4 paper for example, its on a A3 piece, same spectrum, just spread out more, so overlapping signals can be separated.
2. CH3CH=C(CH3)OCH2CH3 gives the correct splitting pattern if the chemical shifts are right!
3. CH3- CH will give the doublet intensity 3 (shift 1.5) and the quartet intensity 1(shift 3.2)
4. 0-CH2-CH3 will give the quartet intensity 2 (shift 2) and the triple intensity 3 (shift 1.1)
5. C=C(CH3) would give the big singlet at shift 2 intensity 3.
Hope this helps a bit, but not sure if its right!
Thank you so much for your help! Your explanation really helped me out.
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