Hey, I've recently started to revise for a spectroscopy test and had a random retardation moment where I asked myself "Why dont protons on a methyl group couple to eachover?" The answer seemed obvious, there all in the same chemical environment and all sigma bonded to whatever substituent you can imagine and therefore their spin states wont interact. Being stubborn that a slightly hand waveing argument for why a methyl would show as a singlet (supposing no vicinal coupling takes place) I checked on the internet and found a pretty knowledgeable blog post explaining common myths and misconceptions in NMR: http://nmr-analysis.blogspot.com/2008/01/1h-nmr-analysis-common-myths-and.html Most of the point's in it I understood (I'm studying university level chemistry, so strong coupling and all the usual suspects for spectroscopy have been taught) but the end of the post mentions that protons within a methyl would couple to eachover, but the effect would be even? Can someone experienced perhaps explain it to me? My perception on coupling at the moment is that if you have a proton in a magnetic field, it has two states [tex]\alpha[/tex] and [tex]\beta[/tex] and these two states are distroted by the nearby states of other atoms in the molecule, causing more than two possible states (and therefore one transition) to turn into multiple states allowing for multiple transitions. Thanks for the help, no urgency as with most exams, understanding < just knowing what to write (Gotta love the system).