1H NMR & IR : unknown structure determination

  1. Hi all,

    I was struggling in solving this question which is a part of a practice exam, and I need you to be patient with me and teach me step by step how to figure out the structure and how to read and analyze the spectra.

    I attached the two 1H NMR & IR spectra so you will be able to know what structure I'm talking about.

    I started with finding the saturation number which is 5 so I know that at least I have aromatic ring and double bond(s)
    I predicted the compound to have an ester but I'm not sur.

    please help me to know what is this structure.
    The molecular formula is C8H8O2

    and all the peaks and ppm readings are in the attached figures.

    http://img257.imageshack....mg257/3670/scan0004t.jpg

    http://img405.imageshack....g405/4821/scan0003aa.jpg

    http://img213.imageshack....mg213/8636/scan0005j.jpg


    thanks all.
     

    Attached Files:

  2. jcsd
  3. chemisttree

    chemisttree 3,721
    Science Advisor
    Homework Helper
    Gold Member

    Right. You note that there is the carbonyl stretch in the IR but no corresponding OH group in either the NMR or the IR spectrum.

    What might the 'R' group be in that ester? Any hint from the NMR spectrum? What group gives a singlet at ~3.7-3.8ppm that integrates to 3H? (your spectrum is integrated to peak height... it should be by integrated by area in the future)
    All of the aromatic protons are accounted for. What type of substitution do the integration and splitting patterns indicate? Mono-substitution? Di-substitution?
     
  4. I have three signals in the aromatic region that accounts for 5 hydrogens. There is only one more signal of three hydrogens. It must be a CH3 group. so this gives me
    C6H5 + CO2 + CH3?
    so I'm only thinking of C6H5-COOCH3 which is a methyle benzoate
    [​IMG]

    right?

    I don't know what type of subsitiution I have, may be a di ?
     
  5. chemisttree

    chemisttree 3,721
    Science Advisor
    Homework Helper
    Gold Member

    Right. It is a mono-substituted aromatic ring. There are three types of protons in that ring, 2 ortho, 2 meta and and a single para (think of the symmetry). The three signals in the NMR should integrate 2:2:1 (edit: or was that 2:1:2?) by area (not peak ht as they are split like crazy).
     
    Last edited: Nov 2, 2009
  6. I was actually little bit confused about this but you cleared my confussion by letting this know to me as, I will search about this little more for my further knowledge.

    Thanks!
     
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