Aug 22, 2013
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Pre-Dental
Hey guys I've watched chad several times on proton NMR.. I know how to indicate how many there are in a molecule, but the triplet, quartet, doublet singlet junk.. I do not understand.. I YouTube it too! :( can someone please explain to me a simple way to identify them? Here are example problems you can dumb down for me
1. CH3C=OCH2CH3
2. CH3CH2C=OCH2CH3
3. Cyclohexanol
Thank you guys!
 
Apr 7, 2012
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It all depends on what is attached to you. If ch2 is next to you, you are a triplet. If ch3 is next to you, you are a quartet. If ch is next to you, the signal is a doublet.
 

Thanhn

5+ Year Member
Nov 17, 2012
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Splitting is caused by protons bonded to adjacent atoms.
The general rule is that it's considered neighboring if it's 3 bonds away.

1. CH3C=OCH2CH3
You would see 3 total peaks because there are 3 different types of protons.
The methyl group next to the Carbonyl will be a singlet, that's because the protons of said methyl group are not next to any hydrogens.
So, why is it a singlet? This is based on the n+1 rule.
"n" represents the number of hydrogens in the adjacent atom(s). "n+1" tells us the number of signals (singlet, doublet, triplet, quartet, etc.)
So if you know a peak is a quartet, that means that it's next to 3 H's.

Next, there would be a -CH2- peak that's split as a quartet. This is due to the protons on the terminal -CH3.
Lastly, the peak at ~1.1ppm would represent the terminal -CH3- but have triplet splitting due to the -CH2-.
2. CH3CH2C=OCH2CH3
There are only 2 types of hydrogens here due to the symmetry in the molecule.
You would have a peak that integrated to 6H's, but it would be triplet splitting, because each of the -CH3- are next to a -CH2-.
You would also have a peak that integrates to 4H's but would be quartet splitted, because of the 3H's in the -CH3- next to each -CH2-.

3. Cyclohexanol
This one is a little more difficult, but let's think about it. You would have 5 different peaks.
Peak A - H on the -OH-
Peak B - H on the -CH- attached to the -OH-
Peak C - 4H from the two "equivalent" -CH2- immediately adjacent to the -CH-
Peak D - 4H from the two "equivalent" -CH2- attached the Peak C
Peak E - 2H from the -CH2- opposite of the -OH-

Peak A - I would say that this would be singlet splitting because it's attached to a very electronegative atom so it wouldn't split.
Peak B - multiplet splitting
Peak C - quartet splitting because each -CH2- is next to 3 total H's. n = 3, n+1 = 4 (quartet)
Peak D - quintet splitting because each -CH2- is next to 4 total H's. n = 4, n+1 = 5 (quintet)
Peak E - quintet splitting because the -CH2- is next to 4 total H's. n = 4, n+1 = 5 (quintet)
 
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OP
kiki23
Aug 22, 2013
56
5
Status
Pre-Dental
Splitting is caused by protons bonded to adjacent atoms.
The general rule is that it's considered neighboring if it's 3 bonds away.

1. CH3C=OCH2CH3
You would see 3 total peaks because there are 3 different types of protons.
The methyl group next to the Carbonyl will be a singlet, that's because the protons of said methyl group are not next to any hydrogens.
So, why is it a singlet? This is based on the n+1 rule.
"n" represents the number of hydrogens in the adjacent atom(s). "n+1" tells us the number of signals (singlet, doublet, triplet, quartet, etc.)
So if you know a peak is a quartet, that means that it's next to 3 H's.

Next, there would be a -CH2- peak that's split as a quartet. This is due to the protons on the terminal -CH3.
Lastly, the peak at ~1.1ppm would represent the terminal -CH3- but have triplet splitting due to the -CH2-.
2. CH3CH2C=OCH2CH3
There are only 2 types of hydrogens here due to the symmetry in the molecule.
You would have a peak that integrated to 6H's, but it would be triplet splitting, because each of the -CH3- are next to a -CH2-.
You would also have a peak that integrates to 4H's but would be quartet splitted, because of the 3H's in the -CH3- next to each -CH2-.

3. Cyclohexanol
This one is a little more difficult, but let's think about it. You would have 5 different peaks.
Peak A - H on the -OH-
Peak B - H on the -CH- attached to the -OH-
Peak C - 4H from the two "equivalent" -CH2- immediately adjacent to the -CH-
Peak D - 4H from the two "equivalent" -CH2- attached the Peak C
Peak E - 2H from the -CH2- opposite of the -OH-

Peak A - I would say that this would be singlet splitting because it's attached to a very electronegative atom so it wouldn't split.
Peak B - multiplet splitting
Peak C - quartet splitting because each -CH2- is next to 3 total H's. n = 3, n+1 = 4 (quartet)
Peak D - quintet splitting because each -CH2- is next to 4 total H's. n = 4, n+1 = 5 (quintet)
Peak E - quintet splitting because the -CH2- is next to 4 total H's. n = 4, n+1 = 5 (quintet)
Omgosh!!! Thank you so much for your help!!!!
 
OP
kiki23
Aug 22, 2013
56
5
Status
Pre-Dental
It all depends on what is attached to you. If ch2 is next to you, you are a triplet. If ch3 is next to you, you are a quartet. If ch is next to you, the signal is a doublet.
Thank you so much for your help!