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I am currently trying to determine the structure of my unknown compound with both 1H NMR and 13C NMR spectra.

My H NMR peaks:

  • 1.16 ppm [singlet, 1H], 1.68 [singlet, 3H], 1.75 [singlet, 3H], 4.2 [doublet, 2H], 5.4 [multiplet, 1H], and 7.24 (solvent peak)

C NMR:

  • 18, 26, 59, 77 (solvent peak), 124, and 136 ppm.

I have deduced that my molecular formula is $\ce{C5H10O}$. And I am fairly certain I should have a double bond in my structure. However, I do not know intuitively where this pi bond should be.

What should I be looking for to determine the placement of this pi bond?

I unfortunately do not have characteristic IR peaks to determine whether I have an alcohol, ether, or a ketone.

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    $\begingroup$ 124 and 136 scream alkene. See how much further you get with that idea. Also, you can deduce that you have an alcohol due to only one C-O carbon (59). I'll put an answer up in two hours if you haven't posted it yourself $\endgroup$
    – Lighthart
    May 6, 2016 at 16:26
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    $\begingroup$ In fact 1.16 is unlikely low for methine, which leaves only one other possibility, according to my tables. $\endgroup$
    – Karl
    May 6, 2016 at 17:24
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    $\begingroup$ OK, my last comment was confusing, if you don't know that you should never expect coupling from OH protons. $\endgroup$
    – Karl
    May 6, 2016 at 18:40
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    $\begingroup$ I recind all comments on the OH proton. If it exchanges with chloroform to make a singlet, it should never appear at 1.16 ppm. $\endgroup$
    – Karl
    May 6, 2016 at 19:04
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    $\begingroup$ @Karl - OH will not exchange with chloroform, ever. It may exchange with water or other labile peaks IN chloroform. In the absence of fast exchange, coupling to OH peaks can be, and is frequently observed. $\endgroup$
    – long
    May 6, 2016 at 19:51

1 Answer 1

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The doublet at $4.2\ \mathrm{ppm}$ is almost assuredly an alcohol group or ether group. Since we only have one carbon that is bonded to an oxygen at $59\ \mathrm{ppm}$, we conclude this is a primary alcohol. The situation of being a doublet says smoething else is lurking around.

$5.4\ \mathrm{ppm}$ is far enough down field, we start thinking alkenes, but the multiplicity is mysterious for the moment.

$1.68\ \mathrm{ppm}$ and $1.75\ \mathrm{ppm}$ are in the range of allyl methyl groups, and the carbon signals are okay with this as well. Additionally singlets for these groups indicate a geminal dimethyl compound with a third substituent on an alkene justifying different chemical shifts.

Finally the peak at $1.16\ \mathrm{ppm}$ is difficult to explain. However, we know we have an alcohol, and they can have a fairly wide range of values. If this peak is broadened at all, I would immediately assign it to the alcohol. It being a singlet is also worrisome, but if exchange is occurring, this is not out of the question.

The multiplicity of the alkene proton indicates there are some long-range things going on, and these don't always resolve well. My guess is you are picking up some long-range coupling and is not resolving nicely in the methyl groups, which appear to be singlets.

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  • $\begingroup$ The observed multiplicity of the peaks in the question is clearly not quite right, and is important for determining the correct arrangement of substituents around the double bond. The CH2 at 4.2 should, in the least, be a doublet with coupling in the order of 5-6Hz to the CH at 5.4. This would make this alkene peak a triplet by appearance. Additionally, Coupling from the alkene through to both methyl groups should be in the order of 1-2 Hz, and these splittings should be observable in the 1H spectrum. $\endgroup$
    – long
    May 6, 2016 at 20:08
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    $\begingroup$ With a poorly shimmed spectrum 1-2 Hz might not be apparent, particularly on a 90MHz or lower. Given question, it seems the spectrum was taken on an instructional machine, so I am not very worried about the lack of resolution of the methyl peaks. And if the alkene proton really has 3 different coupling constants, multiplet makes a lot of sense here. $\endgroup$
    – Lighthart
    May 6, 2016 at 20:24
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    $\begingroup$ If the OH peak is exchanging, it cannot appear at 1ppm. The exchanging peak appears at the median of the chemical shifts involved, and this does not work with water or anything else i can think of. But if it is not exchanging, it has to make a triplett. I also came to prenol as the solution, but something is wrong here. It'd be nice to see the actual spectrum. $\endgroup$
    – Karl
    May 6, 2016 at 21:01
  • $\begingroup$ Agree. A picture would have made this much simpler. But I don't agree that it has to be a triplet. It could be smoothed out by intermoleular hydrogen bonding. I expect it is a broad singlet. $\endgroup$
    – Lighthart
    May 6, 2016 at 21:07
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    $\begingroup$ Problems with the alcohol proton aside, it matches this real well: sigmaaldrich.com/spectra/fnmr/FNMR001914.PDF, particularly if OP meant to indicate 2.16 for that peak $\endgroup$
    – Lighthart
    May 6, 2016 at 21:34

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