[SOLVED] IR/NMR question! Easy 10 points!!!?

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

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

NMR spectrum: 2H in a doublet, 2H in a singlet, 2H in a quartet, 3H in a singlet, 3H in a doublet, 1H in a singlet

I’m think the 2H in a quartet and the 3H in a doublet would have to be Ch2-Ch3 on a central carbon with no hydrogens attached to that central carbon; I am also thinking this is a primary amine with (R-NH2); what i’m stuck on is how you can have a split 2H doublet and no splitting on the 1H singlet?

Any help would be appreciated; this question is making me angry!


2H in a doublet, 2H in a singlet, 2H in a quartet, 3H in a singlet, 3H in a doublet, 1H in a singlet

chemical shifts respectively: 4.8, 3.2, 2.6, 1.8, 1.2, 1 delta values in ppm

Update 2:

that’s the question and my attempt at answering

Update 3:

i feel like my response satisfies all of the criteria. maybe i need to brush up on my 1H NMR…..


This is the Best answer for this question

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!

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 #@$. 🙂


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.

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.

i think this is the correct answer

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