# CNMR vs HNMR and Intramolecular H-Bonding¶

When the protons are put into a magnetic field ($$B_\circ$$) and align themselves with or against it, these is a splitting in energy that occurs.

This excitation follows the EM relationship:

$\Delta E=h\nu=\gamma \hbar B_\circ\\ \nu=\frac{\gamma B_\circ}{2\pi}$

Where:

• $$B_\circ=$$ the strength of the magnetic field
• $$\hbar=\frac{h}{2\pi}$$
• $$\Delta E=$$ the difference in energy between the two spin states
• $$\gamma=$$ a constant that is specific to the atom type. This gives us atomic specificity.

We can use a bigger magnetic field to increase the sensitivity of the NMR spectrometer, as $$\nu\propto B_\circ$$. This results in higher resolution spectra:

When Shielding or deshielding occurs, what’s happening is that we get a change in the effective $$B_\circ$$, that results in a change in $$\nu$$

## $$\cnmr$$ vs $$\hnmr$$¶

Both provide information about the number of chemically nonequivalent nuclei in the sample and both give us information about the electronic environment that those nuclei exist, however:

• A $$\cnmr$$ spectra is $$\e{-4}$$ times weaker than a $$\hnmr$$ spectra, as a $$\ce{^13C}$$ nucleus is only about $$1\%$$ as intense as the $$\ce{^1H}$$ nucleus and is significantly less abundant ($$\sim1.1\%$$) of a sample of carbon (most are $$\ce{^12C}$$).
• The $$\cnmr$$ spectra is more spread out over a larger range than the $$\hnmr$$ spectra, making it easier to identify, count and categorise the nuclei.
• Since the abundance of $$\ce{^13C}$$ is so low, the likelihood of having two $$\ce{^13C}$$ atoms next to each other in a molecules is icredibly low, and accounts ofr why we don’t see splitting in $$\cnmr$$

## When to use $$\cnmr$$ Spectroscopy¶

When the molecule is heavily unsaturated, the proton information may not give you much clarification to the structure. It may also be used when the $$\hnmr$$ spectra are just too complex to be analysed alone