Identify the peaks of the H-NMR of C10BrH7. There is no additional information. Problem 9 - 'H NMR spectrum (CDC13, 500 MHz)1H1H2H1HJ= 8 H21H1HClick on the highlighted area to zoom. Click again to zoom back out.

Answered: Problem 9 - 'H NMR spectrum (CDC13, 500…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Identify the peaks of the H-NMR of C10BrH7. There is no additional information.

Problem 9 - 'H NMR spectrum (CDC13, 500 MHz)
1H
1H
2H
1H
J= 8 H2
1H
1H
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Expert Solution

Step 1

According to the given organic compound formula, the double bond equivalent (DBE) = 10 - (8/2) + 1 = 11 - 4 = 7

So, this means that the compound should have present unsaturation. As all the ¹H NMR peaks come at the $δ$ 7 to 8 ppm region, the compound should have aromatic benzene ring. Now There are number of unsaturation is 7 and all the peaks are in the deshielded region, so the compound should contain 2 fused benzene ring.

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