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I have been given the following data and am asked to determine what the structure is. So far based on the IR spectra given I (think) I have identified a $\ce{C-O}$ bond, $\ce{sp^3}$ $\ce{C-H}$ bond however, I am really thrown off by the 5 hydrogens in a multiplet when there seems to be only one big peak. Any chance I can get help with this?

problem

This is my proposed structure so far (note I am still missing 2 Hydrogens, there's 11 total and my structure contains only 9) however, I am not too sure about it's correctness this is the closest I can get with the information which I think I have interpreted correctly.

what i think i have

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    $\begingroup$ Ignoring the multiplet for now, draw out what you have and share it with us. $\endgroup$
    – Melanie Shebel
    Commented Apr 7, 2017 at 21:02
  • $\begingroup$ I went ahead and added it for you but I question how close I am to achieving the actual structure $\endgroup$
    – Basel Anani
    Commented Apr 7, 2017 at 21:14
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    $\begingroup$ Two thoughts: 1. Whoever wrote this question couldn't be bothered to actually look up the NMR or run the sample themselves?! They're using a computer generated spectrum... Lazy... 2. If they had given you a real spectrum, you might notice that the multiplet at 7.2 is actually 3 sets of peaks (1H, 2H, 2H), all multiplets. Two of these happen to have very similar shifts and look like 3H but they're not equivalent. Hope this helps. $\endgroup$
    – Zhe
    Commented Apr 7, 2017 at 21:30
  • $\begingroup$ The 1H and 13C NMR data disagree very strongly with your proposed structure. The fine structure of the multiplet is important in its own right, but you should probably start with the basics: look at the chemical shifts. What does a proton chemical shift at ~7 ppm mean? The IR data should not usually be your starting point unless there is something very characteristic, e.g. carbonyl. But even if you were starting with the IR data, your proposed structure has a carbonyl group, and there's no IR absorption anywhere near 1700 cm-1. $\endgroup$
    – orthocresol
    Commented Apr 8, 2017 at 0:25
  • $\begingroup$ The organic experts here may be able to work with just the NMR, but if you were given an IR spectrum, too, you should also have included it in the question. $\endgroup$
    – hBy2Py
    Commented Apr 8, 2017 at 15:24

1 Answer 1

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In a step wise approach, based on the NMR data provided:

  • The broadband decoupled 13C NMR spectrum accounts for six magnetically different nuclei. (2-Pentanone drawn by you is not consistent with that, its carbon atoms would yield at maximum five signals.) The four signals in the interval of about 120 to 140 ppm could be carbon atoms of an aromatic ring; the two between 40 to 60 could be by an alkyl chain adjacent to some electron withdrawing group.

  • The 1H NMR spectrum displays a multiplet in the "aromatic region" worth 5 H. There is a broad singlet around 4.5 ppm; shape and to some extend location of this signal may refer to an exchangable proton by an OH group (this again is not explained by your first attempted attribution).

  • Still refering to the 1H NMR spectrum, there are two signals like for an alkyl chain adjacent to an electron withdrawing group; yet their chemical displacement ($\delta$) is a bit too high to be just a normal alkyl chain. This is equally consistent with the two signals in the alkyl region of the 13C NMR spectrum, too.

  • Mass spectroscopy mentions a molecular ion, correlated to a molecular mass of 122 g/mol.

Hence

enter image description here

How is this consistent with the IR data provided? An absorption in the range of 3600 to about 3200 1/cm is typical for free OH, like in the instance of alcohols. In this particular example, with (perhaps arbitrary selected) absorptions mentioned by their position (only), there is little of information you may extract. Maybe for educational purpose in tune of "not every single information adds to knowlegde promptly useable".

Why – on purpose – a different compound like the isomeric p-ethyl phenol does not represent a valid answer?

enter image description here

The 1H NMR spectrum for this compound would be different. The pattern in the alkyl region were a quartett (accounting for two protons) and a triplett (accounting for three protons). Very often, the para-subsitution pattern would split the signals in the aromatic region into two dublets of equal intensity, accounting each for two magnetically equivalent protons.

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