IR/NMR question! Easy 10 points!!!?

Provide a structure for the following compound: C6H13N; IR: 3280, 1653, 898 cm–1; 1H

Answer 1

Is there any way you can post a picture? With the quartet – are you sure it’s a quartet, and not a hextet (ie 6 lines)?

EDIT: The thing you describe as “3H doublet” is actually a triplet. I think this compound might be CH3CH2NHCH2C(=CH2)CH3. In terms of what’s wrong with your suggestion – your double bond has three protons on it, and there are only two protons in the alkene region of the spectrum!

Answer 2

No chemical shifts on the NMR spectrum??

Your formula C6H13N has one degree of unsaturation, which could be either a double bond or a cycle. But 1653 cm-1 is the IR resonance for a C=C double bond, so I’ll conclude that the molecule contains 1 double bond and no rings.

The IR resonance at 3280 cm-1 suggests a secondary amine group. That N-H group is probably the 1H singlet.

The NMR 3H singlet suggests an isolated methyl group. The ethyl group is wrong as it should be a 2H quartet and a 3H triplet. Instead, the 3H doublet would be a methyl group attached to a CH group (but then the CH group would be a quartet, and I don’t see a 1H quartet there).

Is the 2H quartet a normal quartet (1:3:3:1) or is it a doublet-of-doublets quartet (1:1:1:1)? The latter would indicate splitting by two different hydrogens.

I can’t figure out your compound either. “Easy 10 points” my #@$.

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Edit: Given your chemical shifts I’ll agree with you about its being a primary amine.

Thus, given your NMR integrals, you have one NH2, two CH3’s, two CH2’s, one CH, and one C without hydrogens in your compound. I’ve tried writing several structures, but I haven’t found one consistent with your data. I’m still thrown off by the 3H doublet, which would suggest that there is a 1H quartet (or multiplet if it’s close to another hydrogen) somewhere.

Answer 3

It looks like it is unsaturated. If you were to remove the NH2, you’d end up with C6H11, an unsaturated alkyl group. Not sure whether it is an alkene or a cycloalkane, though.

You also don’t have a CH3-CH2 group in the molecule, judging by the NMR, since the CH3 would be split into a triplet, not a doublet.

Have you any chemical shift values for the NMR? I can’t make heads nor tails of these figures.

Edit: you have a triplet 3H, not a doublet. This would point towards an ethyl group. The broad singlet could be a secondary amine proton, which tend to occur as singlets due to proton-deuteron exchange. This could explain the doublet for the 2H.

Answer 4

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