Timeline for 1H (proton) NMR spectra for alkanes
Current License: CC BY-SA 3.0
11 events
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| Feb 16, 2019 at 5:22 | comment | added | Tan Yong Boon | Your explanation using the conformational analysis seems to suggest that each of the 2 enantiontopic hydrogens of any one of the 2 methylene groups of the butane molecule are non-equivalent and hence give rise to 2 different peaks in the NMR spectra. That means to say H2 and H2' are not equivalent? Does your explanation also suggest that H2' is magnetically equivalent to H3' and H2 is magnetically equivalent to H3? | |
| Feb 16, 2019 at 4:20 | comment | added | Zhe | @TanYongBoon You're 100% correct. | |
| Feb 16, 2019 at 4:14 | comment | added | Tan Yong Boon | @Zhe Are you sure about that? Aren't the hydrogens of the methylene group in propane homotopic and the hydrogens on the methylene in butane enantiotopic? | |
| Nov 24, 2016 at 23:09 | comment | added | Zhe | Also important to note that NMR is slow compared to the time scale of molecular rotations, so you're going to be time averaging the various signals together... | |
| Aug 19, 2015 at 12:27 | comment | added | user258521 | Also, I tried using the simplified spectra predictor from 'bon' and came up with this for pentane. [nmrdb.org/new_predictor/index.shtml?v=v2.17.5] Why does it show 6 peaks for the CH2 group attatched to the CH3's when we just said that equivalent groups do not couple, surely the CH2 groups attatched to the CH3's should produce a quartet if we use the simplified way of looking at the problem? I also noticed the central CH2 group had singlets only, which is what would be predicted as equivalent hydrogens do not couple. Have I gone wrong here? | |
| Aug 19, 2015 at 12:21 | comment | added | user258521 | I know that this may be a little off topic from what you have just said but what would a multiplet be and how would that form? | |
| Aug 19, 2015 at 9:44 | comment | added | bon | Very interesting - I too was never taught this. | |
| Aug 19, 2015 at 1:10 | comment | added | long | For the anti conformer (centre), J2,3 != J2',3, so H2 and H2' are not equivalent in this environment. For the gauche conformers (left and right), the couple pairs J2,3 != J2',3 OR J2,3' != J2',3', again introducing non-degeneracy. Anti/gauche equilibrium is temperature dependent for these molecules, and the observed J2,3, J2,3' couplings can be used to calculate the equilibrium populations of conformers. At 300K, J2,2=-13.3, J2,3=6.12, J2,3'=8.66Hz | |
| Aug 19, 2015 at 0:33 | comment | added | ron | long wrote, "At no time is the coupling between H2 and H3 the same as H2' and H3" (that last H3 should be H3' right?). Aren't those couplings equivalent (or enantiotopic) in the anti-conformation pictured in the center of your figure? | |
| Aug 19, 2015 at 0:19 | vote | accept | user258521 | ||
| Aug 19, 2015 at 0:14 | history | answered | long | CC BY-SA 3.0 |