Why are some spectra more complicated than others?

Oct 3, 2015

I presume you mean NMR spectra. Some spectra are more complicated than others because some molecules are less symmetric than others.

Explanation:

Symmetry is a fundamental, underlying principle of all spectroscopy, not just the proton NMR spectrum. The more symmetric a molecule, the less signals that will be observed in the spectrum.

If we take a simple formula, say ${C}_{5} {H}_{12}$, proton NMR spectroscopy would directly identify the particular isomer. The most symmetric pentane, $C {\left(C {H}_{3}\right)}_{4}$, neopentane, would give rise to precisely 1 absorption in its proton NMR spectrum, as all its methyl groups can be interchanged by symmetry, and are thus, therefore, equivalent. The straight chain pentane still has some symmetry, i.e. ${H}_{3} C - C {H}_{2} C {H}_{2} C {H}_{2} C {H}_{3}$ and would give rise to 3 absorptions (i.e. the terminal methyl groups, the 1st methylene, and the second central methylene group.) On the other hand, ${H}_{3} C - C {H}_{2} C H \left(C {H}_{3}\right) C {H}_{3}$, its isomer, would give rise to 4 absorptions. Why? Because symmetry dictates that there are 4 types of proton environment. Which are they?

Not only is a consideration of symmetry important in spectroscopy, but also the number of absorbing species. The intensity of the observed signal directly correlates with the number of absorbing protons. Proton NMR spectra can thus be integrated, and integration can routinely allow assignment of the signals to particular protons. If we go back to the example, ${H}_{3} C - C {H}_{2} C H \left(C {H}_{3}\right) C {H}_{3}$, we would thus have 4 absorptions in a 3:2:1:6 ratio. Such integration is the best method of assigning particular absorptions, because numbers do not mislead.

When you consider a molecule and predict its likely NMR spectrum it is always advized to draw out the molecule as symmetrically as possible, and decide which protons are in the same environment, or are equivalent, and that, therefore, will give rise to the same absorptions. This also operates in carbon NMR spectroscopy. n -pentane would also give rise to 3 absorptions only in its carbon spectrum: $C 1$, $C 2$, and $C 3$. What about the carbon NMR spectrum of neopentane?