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    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/...%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
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    Anyone??
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    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.
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    (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/...%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
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    (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
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    (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
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    (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
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    (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
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    (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/...%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
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    (Original post by chem222)
    Thanks a lot!!!
    Can I please ask one more question?
    I really appreciate it!

    http://pmt.physicsandmathstutor.com/...%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
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    (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?
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    (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-
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    (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/...%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
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    (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/...%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.
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    (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?
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    (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
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    (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
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    (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
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    (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/...%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?
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    (Original post by chem222)
    Thanks a lot
    http://pmt.physicsandmathstutor.com/...%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.
 
 
 
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