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    I really need talking through this question, Ive done the theory in class in my past couple of days, and I attempted this last question in a past paper but Im struggling too much with it.
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    Bump, please help.

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    Yea, this is a nasty one.

    The septet tells you that there's 1 hydrogen atom in that environment which has 6 adjacent hydrogen atoms. Then there's the peak at around 1.1ppm which has a doublet and an area of 6. Those two show you that you've got a (CH3)2CH- in the molecule. The peak at 7.4-ish shows that you have a benzene ring in there too. The relative area is 5 so it only has 1 substituent. The septet is around 2.6ppm so it shows that you've got a HC-C=O environment. Using this you can then obtain the fragment: (CH3)2CHCO. That's about all you can get from the protium nmr.

    From the carbon-13 nmr you get a peak at 185-ish which as you wrote, corresponds to an ester group. This can only be attached to the fragment from above so you get (CH3)2CHCOO. It's not going to be a carboxylic acid group as a carboxylic acid peak wasn't in the protium nmr. The peaks between 120 and 160 show that you've got a benzene ring and as you wrote there's two c-c peaks.

    That identifies all of the peaks but then you've still got your fragment and the benzene ring to stick together. You can see from the missing hydrogen on the benzene ring and the unused bond in the oxygen that the join must be an O-C6H5 bond. That gives you (CH3)2CHCOOC6H5, which also satisfies the 7 carbon environments.
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    (Original post by Peroxidation)
    Yea, this is a nasty one.

    The septet tells you that there's 1 hydrogen atom in that environment which has 6 adjacent hydrogen atoms. Then there's the peak at around 1.1ppm which has a doublet and an area of 6. Those two show you that you've got a (CH3)2CH- in the molecule. The peak at 7.4-ish shows that you have a benzene ring in there too. The relative area is 5 so it only has 1 substituent. The septet is around 2.6ppm so it shows that you've got a HC-C=O environment. Using this you can then obtain the fragment: (CH3)2CHCO. That's about all you can get from the protium nmr.

    From the carbon-13 nmr you get a peak at 185-ish which as you wrote, corresponds to an ester group. This can only be attached to the fragment from above so you get (CH3)2CHCOO. It's not going to be a carboxylic acid group as a carboxylic acid peak wasn't in the protium nmr. The peaks between 120 and 160 show that you've got a benzene ring and as you wrote there's two c-c peaks.

    That identifies all of the peaks but then you've still got your fragment and the benzene ring to stick together. You can see from the missing hydrogen on the benzene ring and the unused bond in the oxygen that the join must be an O-C6H5 bond. That gives you (CH3)2CHCOOC6H5, which also satisfies the 7 carbon environments.
    This is fantastic, thank you. I got stuck on another, if you wouldn't mind having a look. I'll post again in about 10 minutes, with the question and my proposed structure.
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    Name:  1456856721173.jpg
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Size:  267.4 KBI've attached the question, along with what I think it is so far. Working across the spectrum, there's an aldehyde proton whose peak is split into a triplet, meaning there must be two protons on the adjacent carbon. There is a peak for two HC-C=O protons which is split into two, so the adjacent carbon has one proton so these groups must be coupled together:
    H2CHO
    The peak at 7.1 ppm must be for a benzene ring, as it cannot be phenol. The fact it only has four protons suggests there is another non hydrogen group on the ring.
    Which is where the triplet and quartet come in, these indicate an ethyl group.
    The result has the right molecular formula, but the benzene peak is split into four, so the adjacent carbons must only have 3 protons in total. For my structure to be right, the benzene ring should be a quintet. Lastly, I don't think I have the right number of carbon environments.
    It'd be helpful if you could point me in the right direction and let me figure it out

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    (Original post by pineneedles)
    Name:  1456856721173.jpg
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Size:  267.4 KBI've attached the question, along with what I think it is so far. Working across the spectrum, there's an aldehyde proton whose peak is split into a triplet, meaning there must be two protons on the adjacent carbon. There is a peak for two HC-C=O protons which is split into two, so the adjacent carbon has one proton so these groups must be coupled together:
    H2CHO
    The peak at 7.1 ppm must be for a benzene ring, as it cannot be phenol. The fact it only has four protons suggests there is another non hydrogen group on the ring.
    There are two sets of non equivalent protons splitting each other. This tells you that the benzene is 1,2 or 1,4-disubstituted

    Which is where the triplet and quartet come in, these indicate an ethyl group.
    The result has the right molecular formula, but the benzene peak is split into four, so the adjacent carbons must only have 3 protons in total.
    Yes, ethyl


    For my structure to be right, the benzene ring should be a quintet. Lastly, I don't think I have the right number of carbon environments.
    It'd be helpful if you could point me in the right direction and let me figure it out

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    so, you have accounted for eight carbon atoms ...

    ... just two to go!

    I'll let you continue
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    Sorry I got distracted, I'll get right onto it now
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    Huh... I was working through this one earlier.

    I got that the peak at around ten was an aldehyde group with 2 adjacent hydrogens and that the benzene ring has 2 substituents due to it's relative area being 4. It's a doublet of doublets so you've got 2 different coupling constants. That means that the benzene ring's substituents have to be opposite each other in order for it to be a symmetrical ring and give the same coupling constant to 2 hydrogens instead of just 1.

    The peaks at 3.7ppm and 2.6ppm both correspond to hydrogen atoms on a carbon which is adjacent to a benzene ring. Since one is shifted you can tell it's pretty close to the aldehyde group. That's backed up by the fact that it also has a doublet, so it's clearly attached to the aldehyde group. So far this gives you OHCCH2C6H4. You can see from the peaks at 1.3ppm and 2.6ppm that you've then got an ethyl group as the other substituent. Overall that gives OHCCH2C6H4CH2CH3. That also satisfies the requirement for 8 carbon environments.
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    (Original post by Peroxidation)
    Huh... I was working through this one earlier.

    I got that the peak at around ten was an aldehyde group with 2 adjacent hydrogens and that the benzene ring has 2 substituents due to it's relative area being 4. It's a doublet of doublets so you've got 2 different coupling constants. That means that the benzene ring's substituents have to be opposite each other in order for it to be a symmetrical ring and give the same coupling constant to 2 hydrogens instead of just 1.

    The peaks at 3.7ppm and 2.6ppm both correspond to hydrogen atoms on a carbon which is adjacent to a benzene ring. Since one is shifted you can tell it's pretty close to the aldehyde group. That's backed up by the fact that it also has a doublet, so it's clearly attached to the aldehyde group. So far this gives you OHCCH2C6H4. You can see from the peaks at 1.3ppm and 2.6ppm that you've then got an ethyl group as the other substituent. Overall that gives OHCCH2C6H4CH2CH3. That also satisfies the requirement for 8 carbon environments.
    Thanks, does this mean my structure above is almost correct?
    I'm not familiar with a point both you and charco have raised, about the doublet of doublets? Could either of you explain what you mean in more detail? We've only discussed peaks in terms of chemical shift, different environments, and splitting so far in class 😊

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    (Original post by pineneedles)
    Thanks, does this mean my structure above is almost correct?
    I'm not familiar with a point both you and charco have raised, about the doublet of doublets? Could either of you explain what you mean in more detail? We've only discussed peaks in terms of chemical shift, different environments, and splitting so far in class 😊

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    There are two identical environments on the benzene ring, each of which has two protons.

    But they are physically separated so they behave like a double integral single signal and split each other into a doublet.

    I'm referring to the protons at positions 2 and 6 (both identical but physically separated)

    ... and the two protons at 3 and 5 on the ring (once again both identical but physically separated).

    If there were only one on 2 and the other on 3 they would have an integral of 1 each and split each other into doublets.

    But this is doubled (in terms of integrals) but not in terms of splitting.

    To explain it more thoroughly you would have to understand that the coupling constant between position 2 & 3 MUST be identical to the coupling constant between 5 & 6 on the ring.

    This means that you only see a doublet due to protons 2 & 6 (integral 2) and a doublet due to protons 3 & 5 (integral 2)
 
 
 
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