4
$\begingroup$

I was working the following spectroscopy problem from Clayden 2e that asks us to identify the structure of a compound from information about its carbon-13 NMR peaks.

Problem 3.5: The solvent formerly used in some correcting fluids is a single compound $\ce{C2H3Cl3}$, having 13C NMR peaks at 45.1 and 95.0 ppm. What is its structure? How would you confirm it spectroscopically?

I initially thought that the structure would be 1,1,1-trichloroethane (shown below) since there will be two different carbon chemical shifts: one bonded to three $\ce{H}$ atoms and one $\ce{C}$ atom in the 0-50ppm region (corresponding to the 45.1 ppm peak) and another peak for a carbon bonded to three $\ce{Cl}$ atoms in the 50-100ppm region (corresponding to the 95.0ppm peak).

enter image description here

Although it considers 1,1,1-trichloroethane, the solutions manual to the book concludes that the correct structure is 1,1,2-trichloroethane (shown below), the other possible structure with formula $\ce{C2H3Cl3}$.

enter image description here

However, this compound would have both peaks close to eachother in the 50-100ppm range.

Wouldn't the correct molecule be 1,1,1-trichloroethane, and wouldn't the solutions manual be incorrect? The NMR specs below that I found online corroborate my thinking.

enter image description here

enter image description here

References

Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

https://www.chemicalbook.com/SpectrumEN_71-55-6_13CNMR.htm

https://spectrabase.com/spectrum/5WUteTf4M4u

$\endgroup$
5
  • 2
    $\begingroup$ I don't have a copy of the textbook so can't confirm, but the spectra predictions you post are accurate. The wikipedia (infinite in its wisdom) writes of 1,1,1-trichloroethane as a solvent, but not of 1,1,2-TCE. A blooper? $\endgroup$
    – Buck Thorn
    Dec 15, 2018 at 16:09
  • $\begingroup$ Most likely was a blooper on the part of the authors. I just wanted to make sure that my explanations for the two NMRs I found online were correct. $\endgroup$
    – Ethiopius
    Dec 15, 2018 at 16:21
  • $\begingroup$ So probably you should write a very short answer so the Q can be closed $\endgroup$
    – Alchimista
    Dec 15, 2018 at 17:43
  • 1
    $\begingroup$ You can find another reliable 13C-NMR on this site: hmdb.ca/spectra/nmr_one_d/3289, which also confirms you are right about 1,1,1-trichloroethane (can find in: pubchem.ncbi.nlm.nih.gov/compound/…). $\endgroup$ Dec 15, 2018 at 22:44
  • $\begingroup$ SDBS: sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi confirms your conclusion. $\endgroup$
    – user55119
    Dec 16, 2018 at 2:17

1 Answer 1

1
$\begingroup$

I fully agree with your assessment and assume that this is a mistake in the book. I wonder what exactly the authors suggested as spectroscopic verification. I personally would suggest using 1H-NMR as a verification method. In 1,1,1-trichloroethane, only one hydrogen peak should be observed as the three protons of a methyl group are always homotopic. In 1,1,2-trichloroethane, two signals in a $1:2$ ratio should be observed as a triplet and dublet, respectively. Depending on what the book suggests as verification, it should be easy to confirm the nomenclature error.

$\endgroup$
2
  • $\begingroup$ The authors did suggest using 1H-NMR as the method to "confirm it spectroscopically". This is what they provided as the suggested method of spectroscopic confirmation: "The second structure (1,1,2-trichloroethane) looks better but it would be easily confirmed by proton NMR as the first structure (1,1,1-trichloroethane) would have one peak only but the second would have two peaks for different CHs. The solvent is indeed the second structure 1,1,2-trichloroethane." $\endgroup$
    – Ethiopius
    Dec 16, 2018 at 18:10
  • 1
    $\begingroup$ @Ethiopius Yup, that sounds like the authors were having a major derp moment. $\endgroup$
    – Jan
    Dec 17, 2018 at 3:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.