# Determining number of unique hydrogens in lactupicrinal 1H NMR

I'm having trouble understanding which hydrogens are magnetically distinct in the following molecule (lactupicrinal):

I've counted all the hydrogens (there are 20) and I know that hydrogens in a methyl group are all chemically the same (I think), but I don't know about the hydrogens in the ethyl group (double bonded on the lower right side). Am I right in that all other hydrogens outside of those groups are distinct, or are the mirrored hydrogens on the benzene ring also equivalent? Please help, I'm completely at a loss and I've been trying desperately to understand hydrogen equivalence for weeks.

Your question doesn't sound like you care about exact splitting, so I'm going to ignore magnetic equivalence/inequivalence. If you do care, you need to take that into account.

For the purposes of this discussion, equivalent means chemically equivalent.

You also need to keep in mind that NMR is achiral. In other words, opposite enantiomers will produce the same spectrum.

The standard procedure is to replace the atom of interest with something that is distinguishable from the regular atom. I usually replace an $$\ce{H}$$ atom with an $$\ce{H}*$$, but anything will do, provided it's still a hydrogen but different from the others.

With this replacement you've created a new molecule.

For two specific hydrogen atoms, you can perform the replacement operation to generate two molecules.

Now, compare these two molecules.

1. If they are different molecules (e.g., geometric isomers, (E)-/(Z)- isomers, cis-/trans- isomers), then the original hydrogen atoms are heterotopic and should have different chemical shifts. Their environments are different.

2. If the molecules are diastereomers, then the two hydrogens are diastereotopic. Since we can distinguish diastereomers by achiral methods, so can NMR. That means that the two original hydrogen atoms will have different shifts.

3. If the molecules are enantiomers, then the two hydrogens are enantiotopic. They are different, but because NMR can't distinguish between enantiomers, it can't tell that these two hydrogen atoms are different. They will have the same shift.

4. If the two molecules are identical, then the two hydrogens are homotopic. Nothing can distinguish them. They are chemically equivalent, and by NMR, they will have the same chemical shift.

Try applying this test to the hydrogen atoms on the ethyl group and the hydrogen atoms on the ring.

• Thanks for you answer. I've used the H-replacement technique you're talking about in the past, but I can't always tell whether the molecules are equivalent; for example, if I replace each of the upper hydrogens on the benzene ring with H*, are those two new molecules equivalent because the single bond between the ring and the rest of the molecule rotates freely? – omzrs Oct 16 '19 at 20:31
• In the absence of a concrete example, your question would be difficult to answer because atropisomers (isomers due to restricted rotation) exist. However, free rotation seems very likely in this case, since the aryl ring is not very substituted (particularly in the ortho positions), it's not part of some larger ring, and the actual substituents are small. – Zhe Oct 16 '19 at 20:51
• @Zhe You didn't need to ping with @ - it was done automatically 'cause there was only 1 user to ping. – Mithoron Oct 17 '19 at 0:06

Your analysis is basically right. The three protons within a rapidly rotationaly averaged methyl group are magnetically and chemically equivalent, whereas the aromatic protons related by reflection symmetry are chemically but not magnetically equivalent. The ethylenic protons are not chemically equivalent and therefore not magnetically equivalent. The key to identifying magnetic equivalence is whether any proton outside of the group under consideration has identical couplings to each proton in the group. Magnetically inequivalent protons generate coupling patterns through J-couplings to each other, even though they might share a chemical shift due to chemical equivalence.