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    hello

    so i'm kinda stuck on this question and was wondering if someone could explain this to me. thanks in advance

    Q. C,D,E and F are isomers with molecular formula C3H6O. Their low resolution n.m.r spectra show the following integration patterns, C 1:2:3, D 1:2:3, E 1:2:1:2. F shows only one absorption. D does not react with Tollen's reagent. Identify the compounds.
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    First I would think about the possible kinds of functional groups you can have with that formula. Since you only have C, H, and O atoms, you are limited to a few different types, the most obvious ones would be alkene, alcohol, and carbonyl (aldehyde/ketone). Since there is only 1 oxygen atom, carboxyllic acids are not possible.

    A good thing to do if you're not sure where to go after that in these kinds of questions is to draw out/imagine a bunch of different molecules that fit the molecular formula, and see if they also fit the other conditions.

    F:
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    For F, the 6 hydrogens must all be in exactly the same environment, so there must be symmetry in the molecule. The easiest way to do this is with a 3 carbon chain, with a ketone group in the middle and 6 hydrogens on either end which are then all in exactly the same environments as each other (propanone).



    C and D:
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    As C and D both have an integration trace of 3:2:1, they must both have 3 different environments, with 3 hydrogen bound to one carbon, 2 to another, and 1 to the last. One molecule like this is propanal, or another possibility is methoxyethene (this one is probably the hardest to recognise since ether groups are relatively uncommon). Since aldehydes react with tollen's reagent and ethers/alkenes do not, you should be able to identify which is C and which is D.



    E:
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    Finally, E must have 4 differnent environments, and the only way this can happen is if one hydrogen is bonded to an oxygen instead of a carbon (3 carbons would normally only mean 3 environments). Propanol would have too many hydrogens, but prop-2-en-1-ol would work.

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    (Original post by lizardlizard)
    First I would think about the possible kinds of functional groups you can have with that formula. Since you only have C, H, and O atoms, you are limited to a few different types, the most obvious ones would be alkene, alcohol, and carbonyl (aldehyde/ketone). Since there is only 1 oxygen atom, carboxyllic acids are not possible.

    A good thing to do if you're not sure where to go after that in these kinds of questions is to draw out/imagine a bunch of different molecules that fit the molecular formula, and see if they also fit the other conditions.

    F:
    Spoiler:
    Show



    For F, the 6 hydrogens must all be in exactly the same environment, so there must be symmetry in the molecule. The easiest way to do this is with a 3 carbon chain, with a ketone group in the middle and 6 hydrogens on either end which are then all in exactly the same environments as each other (propanone).




    C and D:
    Spoiler:
    Show



    As C and D both have an integration trace of 3:2:1, they must both have 3 different environments, with 3 hydrogen bound to one carbon, 2 to another, and 1 to the last. One molecule like this is propanal, or another possibility is methoxyethene (this one is probably the hardest to recognise since ether groups are relatively uncommon). Since aldehydes react with tollen's reagent and ethers/alkenes do not, you should be able to identify which is C and which is D.




    E:
    Spoiler:
    Show




    Finally, E must have 4 differnent environments, and the only way this can happen is if one hydrogen is bonded to an oxygen instead of a carbon (3 carbons would normally only mean 3 environments). Propanol would have too many hydrogens, but prop-2-en-1-ol would work.


    thank you very much and I was just wondering for compounds C and D the integration patterns are 1:2:3 so does it matter if the ratios are not exactly in the right order so for example when I drew out the ester for D the first carbon is bonded to two hydrogens rather than 1
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    Yes, the actual order on the carbons is not relevant as long as you have one with 3, one with 2, and one with 1. Also be careful with naming ester vs ether. It must also have a C=O group on one of the C-O-C carbons to be an ester.

    An ester, ethyl ethanoate:


    An ether, ethoxyethane:
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    (Original post by lizardlizard)
    Yes, the actual order on the carbons is not relevant as long as you have one with 3, one with 2, and one with 1. Also be careful with naming ester vs ether. It must also have a C=O group on one of the C-O-C carbons to be an ester.

    An ester, ethyl ethanoate:


    An ether, ethoxyethane:
    Ahh okay thank you for clearing that up, I always seem to get them muddled up and thanks for helping out
 
 
 
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