Question

: How would you use 'H NMR spectroscopy to distinguish between the following compounds based on the benzene ring protons? O II 0 OA) Benzene protons for I and II both show up between 7.5 - 8.5 ppm OB) Benzene protons for I show up between 6.5-7.5 ppm and II between 7.5 - 8.5 ppm OC) Benzene protons for I and II both show up between 6.5 - 7.5 ppm OD) Benzene protons for I show up between 7.5 - 8.5 ppm and II between 6.5-7.5 ppm Predict the product(s) for the following reaction: SO Bry/FeBr3 OCH3 ОА) Br o Br OCH3 OCH OB) Br O Br Br OCH3 OC) Br O Br OCH3 OCH3 OD) Br o O Br Br OCH3 OCH3 OCH3

Answer 1

Based on the given options, the correct answer for the first question is OB) Benzene protons for I show up between 6.5-7.5 ppm and II between 7.5 - 8.5 ppm. This is because in compound I, the benzene protons exhibit a chemical shift between 6.5-7.5 ppm, while in compound II, the benzene protons show a chemical shift between 7.5-8.5 ppm. This difference in chemical shifts can be used to distinguish between these two compounds using H NMR spectroscopy.

For the second question, based on the reaction scheme provided, it appears to be an electrophilic aromatic substitution reaction where a bromine atom is being substituted with a nucleophile. The correct product would be OD) Br o O Br Br OCH3 OCH3 OCH3, where the bromine atom is replaced by a methoxy (OCH3) group, resulting in the formation of the product with three methoxy groups attached to the benzene ring.

Step-by-step explanation: