# Matching the NMR to a compound

I was given a molecule that I had to identify the NMR of. The molecule is 2-methylpent-1-ene (shown below):

The correct NMR spectra for this molecule is:

My question is, why is this the right spectra for this molecule. Particularly, how and why did the single peak come up? As far as I know, the only two single peaks are the two that are downfield (aka closest to the $\ce{sp2}$ hybridized carbon).

• The singlet peak is obviously due to methyl group which is closer to the double bond. See, it is attached to carbon with no hydrogens on it, hence no splitting. – Ivan Neretin Feb 23 '18 at 8:49
• This spectrum is too perfect (simulated?). See what it looks like at 400 and 90 MHz. [sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi] – user55119 Feb 23 '18 at 16:27

Due to the double bond, there is a small $^4$J$_H$$_H$ coupling between the CH3 and olefinic CH2. If you look carefully, you can see in your spectrum that although it says singlet it is actually a triplet due to the tiny shoulders that are barely visible about a bit more than half way on the peak! However, for a weak spectrometer (200-300 MHz) and depending on the solvent and conditions the measurement takes place, it can appear as a slightly broad singlet. Similarly, the olefinic peaks that are stated as doublets are not real doublets but multiplets due to multiple small coulings with the aliphatic CH3 and CH2. They can appear as slightly broad doublets though if resolution is not great or even a broad singlet. You would be able to see all these finer details clearly with a high filed NMR instrument (e.g. 700 MHz) and zooming in the aquired spectrum.