/0's Chem Questions

[TrueTears]

There was a question like this in checkpoints (2 actually) and it was this sought of spin-spin coupling question.

[cns1511]

The answers are wrong for the first question. I remember dekoyl asking about this in the holidays as well.

Should be 12 peaks. (3+1)(2+1) = 12

Answers probs left out the 2.

I've been taught the multiplicity for peaks as well. I assumed it was just something we were supposed to know.

[dekoyl]

Okay I better learn it here, then (VN > School, huh.)

What if we have:

1. H  (butan-1-ol)
2. H

How many peaks in proton NMR would those bolded hydrogens have? Would multiplicity apply for the bolded hydrogens? (cases 1 and 2)

Thanks heaps

[/0]

I'm guessing
1. (3+1)(2+1)=12
2. 4+1-1+1=5
amirite?

[shinny]

Don't worry about multiplying to work out peaks. My brother who was in 2nd year uni for Pharmacy hadn't been taught this yet, and it's definitely not in the VCE syllabus. However if you want to learn it for self-interest (as I did), and particularly in regards to why there's multiplying, and why the n+1 rule exists in the first place, then go for it. Understanding all of NMR just makes the stuff you need to know for VCE level piss easy. However since it's getting close to midyears, I'd say you've got higher priorities.

[shinny]

The answers are wrong for the first question. I remember dekoyl asking about this in the holidays as well.

Should be 12 peaks. (3+1)(2+1) = 12

Answers probs left out the 2.

On another note, the H on the hydroxyl group shouldn't be counted should it? If so, it'd be 8 peaks.

EDIT: Whoops, drew it wrong

[cns1511]

The answers are wrong for the first question. I remember dekoyl asking about this in the holidays as well.

Should be 12 peaks. (3+1)(2+1) = 12

Answers probs left out the 2.

On another note, the H on the hydroxyl group shouldn't be counted should it? If so, it'd be 8 peaks.

From the bold H it has CH3 on the left and CH2 on the right. So its (3+1)(2+1) = 12 still. Not 8.

[chem-nerd]

hmmm wouldn't case 2 be 9?

[shinny]

hmmm wouldn't case 2 be 9?

They're identical environments on both sides, so it's just back to good ol' n+1 rule.

Sorry, yep 9

[chem-nerd]

not identical on both sides
the Hs on C1 are next to an OH whilst the Hs on C3 are next to a CH₃

[dekoyl]

So my case 2 is 5 peaks?

Because for case 2, would I got 6 peaks (haha =[ )

[shinny]

Ok that's the last time I eat an entire large pizza for dinner. Just misread two questions. And yeh, that'll make it 9.

[dekoyl]

I was using the n+1 rule.

So for butan-1-ol case 1 and 2, multiplicity is taken into account? (I did (2+1 from the -OH side) times (2 + 1) on the left of the CH2).

[shinny]

Yep, different hydrogen environments on both sides. But remember what I said in my first post though, this isn't really that necessary =\

[Mao]

For the interested:

The peak multiplicity thing can be kinda 'simplified' with n+1 rule. To understand this, first recognise that under 'normal' conditions for n+1 splitting, the chemical shift between split'd peaks are the same. Let's call this constant J.

For

Br-CH2CH2CH2-OD
[D stand for Deuterium, heavy hydrogen, which cannot respond to 1H NMR because it has a proton and a neutron]

The bolded hydrogen would have coupling effects with H in the Br-CH2- and the -CH2-OD. The couping constants J_ab and J_bc are similar (but slightly different), and you get something like this:



as you can see, that sorts itself into n+1 = 5 'groups' of peaks, even though the multiplicity is 9. The peaks kinda merge into each other and are kinda hard to distinguish.

So at the end of the day, the advice is to

FORGET SCIENTIFIC TRUTH AND STICK TO VCE LAND

sauce: http://www.chem.uic.edu/web1/ocol/spec/NMR3.htm