How would you use 1H NMR spectroscopy to distinguish between these two compounds? Select all that apply. chool B The methyl labeled F will be a 3H singlet at about 1.9 ppm, while the methyl labeled C will be a 3H triplet at about 0.9 ppm. The methylene labeled B will be a 2H sextet at about 1.4 ppm, while the methylene labeled E will be a 2H triplet at about 2.2 ppm. There will be 5 aromatic H signals in the left compound, and only 3 in the right compound. There will be 6 proton signals in the left compound, and only 5 proton signals in the right compound. The proton labeled A will be a 2H triplet at about 2.2 ppm, while the proton labeled D will be a 21 triplet at about 2.5 ppm The aromatic protons in the right molecule will be shifted further to the left compared to the same protons in the left molecule due to the electronegative carbonyl group being closer to the aromatic ring in the right molecule