Prasad's+Analysis+of+the+CNMR+and+HNMR+spectra+of+Taurine

=__**Prasad Krishnakurup**__=

**Structure of Taurine (2-aminoethanesulfonic acid)**
Spectra obtained from: http://www.chemspider.com/Chemical-Structure.1091.html

This is an H-NMR of Taurine depicting all peaks.

This is a magnified view of the H-NMR of Taurine. This is a set of two triplet of peaks at 3.38 to 3.42 ppm which corresponds to the “Ha” protons attached to the carbon at the “1” position and at 3.23 to 3.28 ppm which corresponds to the “Hb” protons attached to the carbon at the “2” position. These peaks correspond to the sp3 hybridized hydrogen atoms bonded to the first and second carbon in the compound. There is splitting of the signal in the form of three peaks as the hydrogens are affected by two equavalent adjacent protons present bonded to the first and second carbon; from the N+1 rule, N would then equal 2 for each proton, so N+1, (2+1 = 3), 3 peaks would be observed. The three peaks corresponding to “Ha” deshielded by the electron-withdrawing group SO3H, as well causing a chemical shift to the left. This is so because normally without deshielding an sp3 hydrogen attached to a carbon is around 1.4 ppm. For “Hb” is deshielded by an amine group, which is less electronegative, and therefore less electron withdrawing than the effect seen on “Ha” so it is moved to the left less. The triplet peaks exhibit similar heights because the height of peaks corresponds to the ratio of protons between carbons; since both carbons have two H’s, the ratio is 1:1, so their heights are similar.

This is the C-NMR spectra of Taurine depicting all peaks.

This is a magnified view of the C-NMR of Taurine. It depicts a single peak from 50.1-50.23ppm and corresponding to the first carbon bonded to both the sulfur as well as the second carbon. The reason why the C-NMR would correspond to 50.1-50.23 is because the H-NMR corresponding to the protons attached to Carbon #1 is around 3.4 ppm. Since C-NMR peaks represent 15-20 times the amount of ppm of the protons attached to it, 3.4 x 15 is approximately 51 ppm. (See above explanation for H-NMR peak at 3.4 ppm)



This is a magnified view of the C-NMR of Taurine. There is a single peak at approximately 38.20 ppm, which corresponds to the second carbon bonded to the nitrogen group. The reason why the C-NMR would correspond to 38.2 is because the H-NMR corresponding to the protons attached to Carbon #2 is around 3.2 ppm. Since C-NMR peaks represent approximately 15-20 times the amount of ppm of the protons attached to it, 3.2 x 15 is approximately 48 ppm. (See above explanation for H-NMR peak at 3.2 ppm). Although the peak doesn’t exactly match with the H-NMR prediction, it can be deduced that this C-NMR peak correlates with Carbon #2 because it is the lesser ppm of the two C-NMR peaks. This is the structure of 2-ethylaniline This is the overall H-NMR of 2-ethylaniline

This is an H-NMR of the compound 2-ethylaniline. This particular magnified view depicts a quartet at 3.12-3.18 ppm. This quartet refers to the hydrogens present on the second carbon in the alkyl chain whcih is next to the amine group. Furthermore, these peaks are split by the last carbon present on the alkyl chain which has three quavalent hydrogens causing due to the N+1 rule the formation of a quartet. Also, these valued correspond to the hydrogens attatched to the second to last carbon, Hc, as the hydrogens on the last carbon, Hd, would produce a triplet with values around .9 ppm. Also, the hydrogens attatched to the benzene ring would be around 7.2 ppm. Furthermore, the fact that the Hc hydrogens have values around 3.12-3.18 ppm vindicate the assignment of very similar values to the hydrogens on the "2" position of 2-aminoethanesulfonic acid. Thus, if one of the two sets of peaks can be comparitively determined to correspond to the "Ha" hydrogens on carbon with position "1" then the other must correspond to the 'Hb' Hydrogens on carbon "2".
 * [You can't really use those arguments to assign the different hydrogens and carbons. One approach you could take is to find similar compounds that only have one of the groups and compare their H and C NMRs - for example methanesulfonic acid vs. methylamine. This can require some digging because not all spectra can be found for all molecules. JCB]**

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