Question

6. (Bpts) For each example below, draw a structure that is consistent with the formula and H NMR spectrum provided. In some

some cases, additional IR spectral information is provided. Assign each peak to the appropriate nydiyye structure by using th

46.6 = 5.0 19,6 = 2.1 24 = 1.0 a = 6.2 9.4 Adjusted Integration Values: D:5H, C: 2H, B: 1H, A: 6H CH1002 IR: 1676 cm-1 TTTTTT

TTT 8 Proton NMR 2 PPM L 15 Intergal Values 17.46 16.62 C5H1002 IR: 1741 cm-1 4.5 Proton NMR 4.0 0.5 PPM 3.0 Intregral Values

00 4.5 P TTTTTTT un NMR 4.0 3.5 3.0 TTTTT 1.5 1.0 0.5 PPM 26.6 Intregral Values 17.3 27.1 C5H1002 IR: 1743 cm-1 TTTT 3.0 4.5

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!!

Chemical Shift 0.5-6 ppm 0.9-2 ppm 1-5 ppm Type of H C-O-H (alcohol) aliphatic C-H C-N-H C=C-C-H O=C-C-H Ar-C-H CEC-C-H CEC-H

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Answer #1

SOLN 6

The structure is Isobutylbenzene

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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