Double bond equivalent for
CaHbOcNdXe = a+1 - (b/2) - (d/2) + e/2
1.
Double bond equivalent for C5H9BrO2
C5H9BrO2 = 5 + 1 - (9/2) + (1/2) = 6 - 5 = 1
13C = 5 peaks
5 different types of carbon.
1H NMR
0.95 (t, 3H) - So, CH3 group and near a CH2 group present.
1.72 (m, 2H) - So, CH2 group and near more alkyl group present.
3.85 (s, 2H) - So, CH2 group with no adjacent carbon bearing hydrogen and Br group present (EWG)
4.15 (t, 2H) - So, CH2 group and near a CH2 group present. But it is at 4.15 ppm so, -COO group adjacent to it. The structure is shown in figure.
2.
C8H8O2 = 8 + 1 - (8/2) = 9 - 4 = 5
So, 5 double bond equivalent. 4 double bond and 1 ring
13C = 8 peaks
8 different types of carbon.
1H NMR
3.98 (s, 3H) - So, CH3 group no nearby no carbon bearing hydrogens.
7.60 (m, 4H) - So, 4 aromatic protrons.
10.00 (s, 1H) - So, aldehydic protron. The structure is shown in figure.
Can someone help me with these NMR questions? I have read and followed the above requirements...
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