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Please help, NMR question!

The aldehyde has the molecular formula C5H10O. The 1H NMR spectrum of the aldehyde contains a doublet at δ = 0.9 ppm with a relative peak area of six compared with the aldehyde proton. Analyse this information to deduce the structure of the aldehyde. Explain your reasoning.

http://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Past-Papers/OCR-Old/Unit-4/June%202010%20MS%20-%20Unit%204%20OCR%20Chemistry%20A-level.pdf
Markscheme question 3c

In the markscheme it says
(relative) peak area indicates 2 x CH3 (in the same environment)

what does that mean?
I'm so confused about NMR
(edited 5 years ago)
Reply 1
Anyone??
I only learnt this today during school, but I'll try my best to share my limited knowledge.
I believe the markscheme is referring to the peak on the NMR graph, and the '2 CH3' is referring to the "environments" in which the hydrogens exist in the molecule, being that there are two of them, existing in the environment with CH3 surrounding it. You should watch some YouTube videos about it, it's a bit difficult to explain over text but lots of the practice questions I've been completing for NMR have a set structure to them. I suggest learning the 'core' knowledge and understanding for NMR off YouTube and then applying it to past paper Qs.
Original post by chem222
The aldehyde has the molecular formula C5H10O. The 1H NMR spectrum of the aldehyde contains a doublet at δ = 0.9 ppm with a relative peak area of six compared with the aldehyde proton. Analyse this information to deduce the structure of the aldehyde. Explain your reasoning.

http://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Past-Papers/OCR-Old/Unit-4/June%202010%20MS%20-%20Unit%204%20OCR%20Chemistry%20A-level.pdf
Markscheme question 3c

In the markscheme it says
(relative) peak area indicates 2 x CH3 (in the same environment)

what does that mean?
I'm so confused about NMR


Relative peak area means that there are 6 equivalent protons. The most likely arrangement of this is (CH3)2CH-

The CH causes the splitting of the six equivalent protons into a doublet.

OK so now you just have to place C2H3O

You know that there is a CHO group at the end (it's an aldehyde)

That leaves CH2 ...

(CH3)2CH-CH2CHO
Reply 4
Original post by charco
Relative peak area means that there are 6 equivalent protons. The most likely arrangement of this is (CH3)2CH-

The CH causes the splitting of the six equivalent protons into a doublet.

OK so now you just have to place C2H3O

You know that there is a CHO group at the end (it's an aldehyde)

That leaves CH2 ...

(CH3)2CH-CH2CHO


Why can't we say R-CH6?
0.9 ppm means R-CH..
Area of 6 so 6 protons?
so R-CH6?

And when it says compared to the aldehyde proton what do they mean? I thought they are talking about the H in the CHO group
(edited 5 years ago)
Original post by chem222
Why can't we say R-CH6?
0.9 ppm means R-CH..
Area of 6 so 6 protons?
so R-CH6?

And when it says compared to the aldehyde proton what do they mean? I thought they are talking about the H in the CHO group


1. Carbon can only form four bonds, so CH6 is impossible
2. They ARE talking about the CHO proton. This is only 1 hydrogen atom so if another signal is six times bigger it must contain 6 hydrogen atoms
Reply 6
Original post by charco
1. Carbon can only form four bonds, so CH6 is impossible
2. They ARE talking about the CHO proton. This is only 1 hydrogen atom so if another signal is six times bigger it must contain 6 hydrogen atoms


Is that why we have 2 CH3's?
From what I have understood:
So aldehyde proton is the H in CHO, however, just because the area ration is 1:6 doesn't mean that the groups are right next to each other in the molecule? Area ratio just means number of protons.
So the hydrogen attached to carbon in CHO is not necessarily next to the group that has a doublet peak with 0.9ppm, is this correct?
But what we can say is that it has 6 hydrogen atoms so 2 CH3's and is a doublet so it has a neighbouring carbon with one H attached - which is not the CHO

Sorry I keep asking, just want to make sure I understand it properly
Original post by chem222
Is that why we have 2 CH3's?
From what I have understood:
1. So aldehyde proton is the H in CHO, however, just because the area ration is 1:6 doesn't mean that the groups are right next to each other in the molecule? Area ratio just means number of protons.
2. So the hydrogen attached to carbon in CHO is not necessarily next to the group that has a doublet peak with 0.9ppm, is this correct?
3. But what we can say is that it has 6 hydrogen atoms so 2 CH3's and is a doublet so it has a neighbouring carbon with one H attached - which is not the CHO

Sorry I keep asking, just want to make sure I understand it properly


1. True. But in reality it is difficult to have six hydrogen atoms in the same environment unless they are in the form of (CH3)2C-. There are exceptions to this, such as dimethylbenzene (3 isomers), benzene itself, and propanone etc. BUT you are told that the molecule is an aldehyde, mening that there is a CHO at one end. In this case the only possibility is (CH3)2C-
2. Correct
3. Correct
Reply 8
Original post by charco
1. True. But in reality it is difficult to have six hydrogen atoms in the same environment unless they are in the form of (CH3)2C-. There are exceptions to this, such as dimethylbenzene (3 isomers), benzene itself, and propanone etc. BUT you are told that the molecule is an aldehyde, mening that there is a CHO at one end. In this case the only possibility is (CH3)2C-
2. Correct
3. Correct


Thanks a lot!!!
Can I please ask one more question?
I really appreciate it!

http://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Past-Papers/OCR-Old/Unit-4/January%202012%20QP%20-%20Unit%204%20OCR%20Chemistry%20A-level.pdf

For question 4c
I was able to identify the molecular formula from the mass spectrum- C8H16O2
Looking at the NMR spectrum, there are 4 peaks so 4 hydrogen environments
Area under the peaks is in the ratio 9:3:2:2
First peak is 0.9 so R-CH
Second peak 1.4 is also R-CH
Peak 2.3 is HC-CO
And peak 4.3 is O-CH

According to the area under the peaks ratio it will be
First peak R-(CH3)3 singlet so no neighbouring hydrogens
Second peak R-CH3 2 hydrogens on neighbouring carbon
Third H2C-CO no neighbouring hydrogens
Fourth O-CH2 3 hydrogens on neighbouring carbon

I have no idea how to work out the structure now?
I am confused on the last two peaks for the structure
Original post by chem222
Thanks a lot!!!
Can I please ask one more question?
I really appreciate it!

http://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Past-Papers/OCR-Old/Unit-4/January%202012%20QP%20-%20Unit%204%20OCR%20Chemistry%20A-level.pdf

For question 4c
I was able to identify the molecular formula from the mass spectrum- C8H16O2
Looking at the NMR spectrum, there are 4 peaks so 4 hydrogen environments
Area under the peaks is in the ratio 9:3:2:2
First peak is 0.9 so R-CH
Second peak 1.4 is also R-CH
Peak 2.3 is HC-CO
And peak 4.3 is O-CH

According to the area under the peaks ratio it will be
First peak R-(CH3)3 singlet so no neighbouring hydrogens
Second peak R-CH3 2 hydrogens on neighbouring carbon
Third H2C-CO no neighbouring hydrogens
Fourth O-CH2 3 hydrogens on neighbouring carbon

I have no idea how to work out the structure now?
I am confused on the last two peaks for the structure


A 9-H singlet signal is tertiary carbon, (CH3)3C- this has to be a termination.

You are told that it's an ester, so it contains a -COO- group
The relative mass is 144.
Subtract COO(44) leaves 100, the nmr suggests 16 H atoms hence the carbons add up to 84 = seven carbon atoms

Molecular formula C8H16O2

IHD = 1, accounted for in the ester, hence the rest of the molecule is saturated.

(CH3)3C- accounts for four carbon atoms and 9 H atoms, leaving three carbons and 7 hydrogen atoms to be located.

NMR
quartet at 4.2 (2H)
singlet 2.2 (2H)
triplet 1.3 (3H)

The triplet and quartet are clearly linked CH3CH2- and the high shift of the quartet suggests adjacent to a strong electron withdrawing atom, such as oxygen.
This gives

CH3CH2-OCO-

This just leaves a CH2 singlet shifted to typical adjacent to a carbonyl group 2.2

Hence:

(CH3)3CCH2COOCH2CH3
Reply 10
Original post by charco
A 9-H singlet signal is tertiary carbon, (CH3)3C- this has to be a termination.

You are told that it's an ester, so it contains a -COO- group
The relative mass is 144.
Subtract COO(44) leaves 100, the nmr suggests 16 H atoms hence the carbons add up to 84 = seven carbon atoms

Molecular formula C8H16O2

IHD = 1, accounted for in the ester, hence the rest of the molecule is saturated.

(CH3)3C- accounts for four carbon atoms and 9 H atoms, leaving three carbons and 7 hydrogen atoms to be located.

NMR
quartet at 4.2 (2H)
singlet 2.2 (2H)
triplet 1.3 (3H)

The triplet and quartet are clearly linked CH3CH2- and the high shift of the quartet suggests adjacent to a strong electron withdrawing atom, such as oxygen.
This gives

CH3CH2-OCO-

This just leaves a CH2 singlet shifted to typical adjacent to a carbonyl group 2.2

Hence:

(CH3)3CCH2COOCH2CH3


Thanks a lot
1. But one thing that confuses me is the peak at 0.9 that is a C(CH3)3 is attached to CH2 which has 2 hydrogens attached to it? but the peak at 0.9 is a singlet?
And from what I have understood the CH2 at 2.2 is a singlet because it's attached to a C(CH3)3 which isn't attached to any hydrogens directly?
(edited 5 years ago)
Original post by chem222
Thanks a lot
1. But one thing that confuses me is the peak at 0.9 that is a C(CH3)3 is attached to CH2 which has 2 hydrogens attached to it? but the peak at 0.9 is a singlet?
And from what I have understood the CH2 at 2.2 is a singlet because it's attached to a C(CH3)3 which isn't attached to any hydrogens directly?


The tertiary butyl group is attached to a carbon that has NO hydrogen atoms and therefore is a singlet.

The next carbon then has two hydrogen atoms, with no H atoms on the adjacent carbon and so is also a singlet.

(CH3)3-C-CH2-COO-
Reply 12
Original post by charco
The tertiary butyl group is attached to a carbon that has NO hydrogen atoms and therefore is a singlet.

The next carbon then has two hydrogen atoms, with no H atoms on the adjacent carbon and so is also a singlet.

(CH3)3-C-CH2-COO-


Thanks a lot
Your explanation has really helped!
I have been doing questions but then I came across another one which I found confusing

http://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Past-Papers/OCR-Old/Unit-4/June%202011%20QP%20-%20Unit%204%20OCR%20Chemistry%20A-level.pdf

for question 5c
In the question they say The peak centred at δ = 3.4 ppm would normally be expected at a chemical shift value about
1 ppm to the right, i.e. 2.4 ppm.

What do they mean by this? And I am confused about the structure, why is there 2 COOH
Original post by chem222
Thanks a lot
Your explanation has really helped!
I have been doing questions but then I came across another one which I found confusing

http://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Past-Papers/OCR-Old/Unit-4/June%202011%20QP%20-%20Unit%204%20OCR%20Chemistry%20A-level.pdf

for question 5c
In the question they say The peak centred at δ = 3.4 ppm would normally be expected at a chemical shift value about
1 ppm to the right, i.e. 2.4 ppm.

What do they mean by this? And I am confused about the structure, why is there 2 COOH


Compound F
You have the molecular formula C4H6O4

NMR
0.9 doublet
3.4 quartet
11.0 singlet

The 0.9 signal is typical of an unshifted alkyl group - it is split into a doublet suggesting that it is adjacent to a CH grouping.
The 11.0 signal is typical of COOH and as it is a singlet both COOH groups must be in the same environment. (there are two because we have to account for four oxygen atoms)
The higher shift of the CH is due to it being adjacent to not one but two COOH groups

HOOC-X-COOH

where "X" is C2H4

HOOC-CH(CH3)-COOH

fits the bill.
Reply 14
Original post by charco
Compound F
You have the molecular formula C4H6O4

NMR
0.9 doublet
3.4 quartet
11.0 singlet

The 0.9 signal is typical of an unshifted alkyl group - it is split into a doublet suggesting that it is adjacent to a CH grouping.
The 11.0 signal is typical of COOH and as it is a singlet both COOH groups must be in the same environment. (there are two because we have to account for four oxygen atoms)
The higher shift of the CH is due to it being adjacent to not one but two COOH groups

HOOC-X-COOH

where "X" is C2H4

HOOC-CH(CH3)-COOH

fits the bill.


Thanks a lot
But what about the peak at 3.4? That's an O-CH group? but not present in the molecule?
Original post by chem222
Thanks a lot
But what about the peak at 3.4? That's an O-CH group? but not present in the molecule?


No, its the proton that is sandwiched between two carboxylate groups.

HOOC-CH-COOH
Reply 16
Original post by charco
No, its the proton that is sandwiched between two carboxylate groups.

HOOC-CH-COOH


Is 3.4 not an O-CH group?
in the question it says The peak centred at δ = 3.4 ppm would normally be expected at a chemical shift value about
1 ppm to the right, i.e. 2.4 ppm.
do we use 2.4 or 3.4
(edited 5 years ago)
Original post by chem222
Is 3.4 not an O-CH group?
in the question it says The peak centred at δ = 3.4 ppm would normally be expected at a chemical shift value about
1 ppm to the right, i.e. 2.4 ppm.
do we use 2.4 or 3.4


The CH-COOH proton would normally be expected at 2.4

BUT it is between two COOH groups, shifting the signal further downfield
Reply 18
Original post by charco
The CH-COOH proton would normally be expected at 2.4

BUT it is between two COOH groups, shifting the signal further downfield


Thanks a lot
http://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Past-Papers/OCR-Old/Unit-4/June%202012%20QP%20-%20Unit%204%20OCR%20Chemistry%20A-level.pdf

for question 5b the last peak at 7.5 is a benzene ring because.. there are 6 peaks so attached to carbon with 5 hydrogens, is that the hydrogens in the benzene ring?
also how can you work out the molecular formula using m/z without looking at the nmr spectrum first?
(edited 5 years ago)
Original post by chem222
Thanks a lot
http://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Past-Papers/OCR-Old/Unit-4/June%202012%20QP%20-%20Unit%204%20OCR%20Chemistry%20A-level.pdf

for question 5b the last peak at 7.5 is a benzene ring because.. there are 6 peaks so attached to carbon with 5 hydrogens, is that the hydrogens in the benzene ring?
also how can you work out the molecular formula using m/z without looking at the nmr spectrum first?


It's not the number of peaks, that is due to splitting within the aromatic environments, it is simply the fact that the signal is due to five protons.

You can't work out the molecular formula from the m/z unless it is a high resolution MS, in which case to 8 decimal places a database search will be able to give you the exact molecular formula. This is because the actual mass of isotopes is not integral and every molecular formula has a unique relative mass.
(edited 5 years ago)

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