Question

There are two peaks, one at 1.66 ppm and another at 1.74 ppm, that integrate to 3 hydrogens each, indicating that they are the two methyl groups you identified from the DEPT 13C. However, the chemical shifts are significantly downfield of a typical methyl group, which comes at "0.9 ppm. Explain the chemical shifts of these methyl groups in terms of what functional group they are bonded to in the orange oil. Hint: tetramethylethene shows a single absorption in the 'H NMR spectrum at 1.71 ppm.

Answer 1

Typical Me proton in 1H NMR comes at 0.9 ppm around

however if you are getting peak aroun 1.7 - 2 ppm this is possibly because of presence of electron withdrawing group probably any sp2 carbon like alkene (C=C) this can be deciphered from the hind in which Me protons of tetramethylethene comes at 1.7 ppm!!

C=C is sp2 hence cause deshielding by withdrawing electron density of adjacent protons of Me