Organic Chemistry 9th Edition

Published by Brooks Cole
ISBN 10: 1305080483
ISBN 13: 978-1-30508-048-5

Chapter 13 - Structure Determination: Nuclear Magnetic Resonance Spectroscopy - Exercises - Page 419g: 57


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Work Step by Step

Step 1: C8H9Br molecule has the same unsaturation degree as 'C8H10', which is 4. And it has 8 carbons, so most likely it contains one aromatic ring. One can verify the presence of an aromatic ring because the peaks at 120 ppm, 129 ppm, 131 ppm and 143 pm in C-13 NMR and the peaks at 7.0 and 7.4 ppm in H-1 NMR are peaks at the aromatic ring region. Step 2: Based on the symmetry analysis, the presence of four peaks at the aromatic ring region in C-NMR indicates there is one symmetry plane (shown by the blue dash line in the molecular diagram above) and two equivalent zones in the aromatic ring. Step 3: The molecule C8H9Br has nine H atoms with peak area ratio is 2:2:2:3 (from left to right) in H-NMR. So the numbers of H atoms from the four peaks are 2 2, 2 & 3 respectively. Among them, two peaks are at 7.0 ppm and two are at 7.4 ppm at the aromatic ring region. This shows there are two H atoms in each equivalent zone occupying positions 2 and 3 in the aromatic ring as shown by the molecular diagram above. So the molecule should have two groups (R1 and R2) substituting positions 1 and 4 in the aromatic ring as shown by the diagram. Step 4: All four unsaturation degrees of the molecule come from the aromatic ring, so R1 and R2 groups only contain saturated alkyl groups from those two aliphatic carbons and one Br atom. According to the calculation of number of H atoms for each peak in H-NMR from step 3 (2,2,2 & 3), the peak at 2.6 ppm has two H atoms from CH2 group and the one at 1.2 ppm has three H atoms from CH3 group. And from the split patterns of the two peaks, the CH2 group has three neighbor H atoms and the CH3 group has two neighbor H atoms. Finally, we get the structure as shown by the molecular diagram above.
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