Why is $C D C l_{3}$ $CDCl_3$ a triplet in $C^{13} N M R$ $C^13 NMR$ ?

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Answer 1 · 2015-01-09T20:34:13

The deuterium in the CDCl₃ splits the $^{13} C$ $""^13"C"$ signal into a triplet.

Whenever you run a $^{13} C$ $""^13"C"$ spectrum in CDCl₃, you always get a triplet solvent peak at 77.5 ppm.

For example, here's a spectrum of t-butyl chloride.

www.oc-praktikum.de

You see the characteristic CDCl₃ triplet at 77.5 ppm.

The reason is that $^{2} H$ $""^2"H"$ has a spin quantum number $I$ $I$ = 1.

A nucleus with $I = 1$ $I = 1$ has $2 I + 1 = 3$ $2I + 1 = 3$ possible orientations.

These orientations have the same energy in the absence of a magnetic field.

In the presence of a magnetic field, these energy levels split. Each level is labelled with a magnetic quantum number $m$ $m$ = +1, 0, -1.

So a $^{13} C$ $""^13"C"$ nucleus will be split into three different energy levels depending on the +1, 0, or -1 orientation of the deuterium atom.

Each orientation has the same probability, so the signal is a 1:1:1 triplet.

Why is CDCl3CDCl_3 a triplet in C13NMRC^13 NMR?