NMR for Molecule S1:

Explanations for S1:
a: This is an aromatic singlet, that would be very downfield due to electron withdrawing groups, so the only value to match would be 8.46.
b: This is another aromatic but with one neighbor, c, it is near an electron withdrawing group so it would be downfield, at around 8.45
c: Another aromatic doublet, but not as close to an ewg as the other two, so it would be a little more upfield, and couples with b.
d: Aromatic hydrogens show up near 7, this hydrogen is not as close to a large deshielding group so is would show up more downfield, but not by much.
e: Since e is aromatic and closer to a larger deshielding group, it would show up somewhat upfield.
f: This hydrogen has one neighbor and is attached to a double bond, so it would appear quite downfield. Since it is on a double bond, it would not share the share the exact same coupling constant as its neighbor so it would appear with more complex peaking—a dd.
g: This hydrogen has three neighbors and would normally show up around a quintet, but they do not share the same coupling values. So three neighbors would become a ddd, the only value to have this is 5.83, which also makes sense since it is a hydrogen on a double bond. We can also tell from the coupling value that it needs to be near whichever hydrogen had the value of 6.02.
h: A hydrogen close to two oxygens would be very downfield, and shows up at a triplet with only two neighbors, the only peak to match is 4.53.
i: This hydrogen only has one neighbor, so it would show up as a doublet. It's very close to oxygen as well as a carbon bonded to a hydroxyl group, so it would need to be pretty downfield.
j: This is a 3H singlet attached to an oxygen and should show up around 3.8, since it is attached to a benzene ring, we would expect it to be even more downfield at 4.37.
k: This would be very downfield because it is directly attached to a hydrogen so it was assigned 4.37, which is only labeled as 1H. This is a little confusing since it seems like it would have neighbors, however.
l: This is a 3H singlet attached to an oxygen and should show up around 3.8.
m: This hydrogen is attached to a ring and very close to a hydroxyl, it would be very downfield, it also has several neighbors that are coupled different, so it was assigned to 3.35 as a multiplet.
n: This is a 3H singlet attached to an oxygen and should show up around 3.8.
p: This is a 3H singlet attached to an nitrogen, so it would show up a little more upfield, near 3.
q: This hydrogen was labeled at 2.86 because it is near h and they both have J values of 4.5, so much participate in coupling.
s: These two hydrogens are sp3 and not extremely close to any electronegative atoms so they were assigned at 2.04-2.16.
As for the remaining multiplets, o, r, t assigning these were difficult. o was assigned the most downfield value because it is attached to a nitrogen, r is close to a deshielding group so it would be a little more upfield near 2-2.5, and the hydrogens assigned to t are the most like a normal sp3 group that is not near many deshielding groups, so it could be seen as the most upfield of all the values.

NMR for Molecule 37:

Explanations:
a: These hydrogens are attached to a carbon near a –OH group, so would be very downfield, they also have only 2 neighbors which would give a triplet.
b: These hydrogens are relatively downfield and show up as a broad singlet, so these should be attached to the oxygens.
c: These were assigned 1.80 because they would be a bit upfield, being farther from and cis to the –OH on the ring.
d: The 10H, multiplet is difficult to assign because it means the peaks on the spectra were not very clear, so it was assigned to the remaining hydrogens that weren't as close to the –OH groups as the others.


Return to the top!