{"id":1839,"date":"2012-04-24T19:13:06","date_gmt":"2012-04-24T23:13:06","guid":{"rendered":"http:\/\/blogs.vassar.edu\/magnes\/?p=1839"},"modified":"2013-07-11T10:29:36","modified_gmt":"2013-07-11T14:29:36","slug":"interpreting-a-c-13-nmr-spectrum","status":"publish","type":"post","link":"https:\/\/pages.vassar.edu\/magnes\/2012\/04\/24\/interpreting-a-c-13-nmr-spectrum\/","title":{"rendered":"Interpreting a C-13 NMR spectrum"},"content":{"rendered":"

In Vassar’s 300 MHz NMR, the Larmor frequency of an unaltered hydrogen nucleus with shielding factor \"\sigma=0\" is 300 MHz. \u00a0\"^{13}\"C nuclei are different than \"^{1}\"H nuclei, and have a different Larmor frequency in the spectrometer’s 7.046 \"T\"magnetic field. \u00a0According to Jacobsen in “NMR Spectroscopy Explained,” the Larmor frequency of unaltered\u00a0\"^{13}\"C nuclei with \"\sigma=0\" is 75.43 MHz in this magnetic field. \u00a0The magnetogyric ratio \"\gamma\", is 672.650\"*10^{5}\frac{rad}{Ts}\" \u00a0We bring back the equation for Larmor Frequency:<\/p>\n

(1) <\/span>   <\/span>\"\begin{equation*}<\/p>\n

You might ask why \"^{13}\"C nuclei are the focus of carbon NMR instead of the much more common and ordinary \"^{12}\"C nuclei. \u00a0It comes down to the composition of the nucleus. \u00a0Only nuclei with nonzero spin are magnetically active. \u00a0 Spin is a type of angular momentum intrinsic to subatomic particles. \u00a0Particles with nonzero spin can have magnetic moments that can be influenced by a magnetic field. \u00a0Any nucleus with an even number of protons and an even number of neutrons will not be magnetically active because its spin is 0 and it has no magnetic moment. \u00a0Carbon-12 has 6 protons and 6 neutrons, so it can’t be studied with NMR. \u00a0Carbon-13 has 6 protons and\u00a07<\/strong>\u00a0neutrons, so we can<\/strong>\u00a0study it with NMR. \u00a0The structure and \"^{13}\"C-NMR spectrum of 3,3-dimethyl-2-butanol is shown below.<\/p>\n

\"\"<\/a><\/p>\n

\"\"<\/a><\/p>\n

At first glance, the spectrum looks almost the same as the H-NMR spectrum. \u00a0However, there is little indication of how many carbons a single peak is referring to. \u00a0Each peak corresponds to a type of carbon in the molecule. \u00a0The triplet peak at 77 ppm represents the deuterated chloroform, CDCl\"_{3}\", used to dissolve the sample. \u00a0Excluding that triplet, the molecule has four different kinds of carbon. \u00a0To help us find what carbons are represented here, we turn to the other two spectra of interest. \u00a0They are called the DEPT-90, and the DEPT 135. \u00a0DEPT stands for Distortionless Enhancement by Polarization Transfer. \u00a0The technique hits the sample with five successive radio pulses designed to excite either hydrogen or \"^{13}\"C nuclei. \u00a0The sequence of pulses is:<\/p>\n