I've zoomed in and taken more pictures of the 4 H NMR spectrum. I would appreciate any explaination or work. Thank you so much!!
SOLN 6
The structure is Isobutylbenzene
The signal 'A' is due to 6H of two Methyls that appears at upfield around 1 ppm. it split into double to 1 neighboring CH protons
The signal 'B' 1 H septet is due to CH protons adjacent to two CH3 groups
then signal 'C' is due to 2H of CH2 adjacent to CH hence give doublet at 3 ppm
the final sinal at 7-8 ppm is due to aromatic protons which gives multiplet due to complex splitting
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I've zoomed in and taken more pictures of the 4 H NMR spectrum. I would appreciate...
Any work or explanation would be much appreciated. Thank you! 6. (Bpts) For each example below, draw a structure that is consistent with the formula and H NMR spectrum provided. In some cases, additional IR spectral information is provided. Assign each peak to the appropriate hydrogen atoms in the structure by using the labeling convention discussed in class U n o u. U. JH, L. 2H, D. TH, M. On CH1002 IR: 1676 cm-1 Breton NMR 7 2 PPM 11.15...
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just the circled ones only 8.10 Suggest structures consistent with the following 'H NMR data: a. Сно triplet (1.0 ppm, 3H) singlet (2.1 ppm, 3H) quartet (2.4 ppm, 2H) b. С Н.о, triplet (1.2 ppm, 3H) singlet (2.1 ppm, 3H) quartet (4.1 ppm, 2H) усно singlet (2.4 ppm, 3H) multiplet (7.5 ppm, 5H) singlet (1.6 ppm, 6H) singlet (3.1 ppm, 2H) multiplet (7.3 ppm, 5H) 1. Сно, triplet (1.2 ppm, 3H) singlet (2.4 ppm, 3H quartet (4.2 ppm, 2H) ....
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