/sci/ - Science & Math

[Anonymous No.6386369]

Sup guys, i'm preparing for my finals, but i just can't solve this one, if anybody can help that would be great.
I don't see how i can have a CH surrounded by 5 hydrogens without a ch2, and since there's no ch2 ( C13 NMR spectrum). Doesn't seem to be Nh2 and OH aswell so..... yeah :/
Maybe i'm wrong with the isopropyl substituent, i don't know.
Hope the french won't be an issue, words are pretty much the same anyway.

[Anonymous No.6386370]

second part

[Anonymous No.6386371]

>>6386369
Sorry, I don't speak Spanish.

[Anonymous No.6386373]

You don't need to understand the language to guess the chemical formula from the graphic

[Anonymous No.6386376]

>>6386371
That's French

[Anonymous No.6386798]

I'm guessing nitro is in the 3-position.

Para would give only two distinct aromatic protons.

In the aromatic region, the small splitting is from vicinal protons, the big splitting is from neighboring protons.

So the doublet is the most deshielded, F, is in the 2 position between the nitro and the isopropyl (which is in the 1 position). Because it has no neighbors, it only exhibits vicinal coupling (JCCCH << JCCH << JCH), you can only see JCCH and JCH at this resolution. E is double doublet, and is probably in the 4 position. D is double triplet, and probably para to the nitro.
C is probably in the meta position. Problem is you'd expect it to be a triplet. It could be, it looks like a triplet but JCCH is making it looks like a double doublet?

Also, 5 looks really high for chemical shift of benzylic isopropyl, even if it is nitro substituted...

[Anonymous No.6386912]

i tried
but im not sure

[Anonymous No.6386944]

Oh i get it now,so it's the long range coupling giving those strange results, Wasn't making any sens without that. Thanks a lot !

[Anonymous No.6386972]

>>6386369

this book helped me solve aromatic NMR problems


https://dl.dropboxusercontent.com/u/10787669/Techniques%20in%20Organic%20Chemistry%20-%20Mohrig.pdf

[Anonymous No.6387046]

Merci bien ;)

[Anonymous No.6387070]

>>6386972
Wow. Thanks, anon!

[Anonymous No.6387150]

there is no way they would give you a problem like this on a final.


that being said:

in general:

a) your integration suggests that A corresponds to at least 3-6 protons, whereas the other peaks correspond to 1-2 protons each.

b) the expanded region is 4 signals, not 5 or 6.

they are split because the Nitro group rotates on the timeframe of NMR (NMR is "slow" because the decay of an ensemble of excited nuclear spin states occurs over the course of milliseconds to seconds; this is why you see broad peaks for alcohols and amines, or no peak at all, because the protons exchange with the solvent environment much faster than the spectrometer can differentiate the signals in the FID)

c) B is almost certainly a benzylic proton, and given the integration, I would agree that its probably 1 proton.


D) thus we have 4 aromatic protons, 1 benzylic proton, and 6 methyl protons, correctly adding up to H11.

6 aromatic carbons, 1 benzylic carbon, and 2 methyl carbons adds up to 9.

this is consistent with your guess.


E) in assigning the relative bonding to the aromatic ring for the Nitro and alkyl group, remember:

because you have 4 signals, you know that its not Para (para would create a mirror plane that would cut the number of signals in half)


you can use any logic you want here.


you could even step out of the straight de-shielded/rational lewis structure and consider that:

alkyl groups are electron donating to aromatic systems, and nitro groups are electron withdrawing.

eg: alkyl gropus are othro-para directors, and Nitro groups are meta directors.

thus the bonding is probably either ortho or meta (because we already know its not para).


to assign them, further, you can use lewis structures to identify which carbons have partial positive charges on them due to the nitro group

[Anonymous No.6387171]

>>6386912


I agree with the overall structure, but not the assignments.


I would say that the E and F assignments that you have made are actually the C and D signals from the spectrum.

C is probably assigned to the location that you have labelled for E/L.

D is probably assigned to the location that you have labelled for F/M.


F is probably assigned to the location that you have labelled for I/C

E is probably assigned to the location that you have labelled for K/D


Rationale:

a) aromatic groups are de-shielding because of the ring current generated in the molecular orbital above and below the ring plane.

b) nitro groups are strongly electron withdrawing and create a region of partial relative positive charge density at the carbons that are ortho and para to the nitro group.

c) by *decreasing* the electron density at these positions, you decrease the de-shielding, which actually INCREASES their PPM.

d) alkyl groups are weakly electron donating, with the greatest affect being nearest to the alkyl group.

[matsci !!invJra9x72G No.6387450]

>>6387150

Yeah, which is exactly what I wrote here, although I can't remember if I used geminal/vicinal correctly:

>>6386798
here.

And it's probably meta because one of the aromatic protons doesn't show the correct coupling -> one of the aromatic protons has no adjacent protons aromatic protons.

Only problem though even though nitro is deactivating, a chemical shift of 5 for a benzyl is way too high. For 4-nitrocumene, it's only at 3... 4 seems more legit, 5 just seems way too high, that's just gut instinct, though

>>6387171

You're going more on the chemical shift, and not looking at the multiplicity, though.

F is most likely right because no strong coupling -> no adjacent protons.

C is probably right because meta to nitro should have lowest ppm.

D and E might be switched, not sure.

Whatever, it's been years since I've run NMR.

[Anonymous No.6387471]

>>6387450


coupling on aromatic protons is not common, nor is it usual to use aromatic proton coupling as a method for structural elucidation.


>meta to nitro should have lowest PPM

again:

nitro is electron withdrawing

aromatic group deshields protons because of electron density above and below ring plane

electron density is lowest at carbons that are ortho and para to nitro group (next to the nitro group, and opposite the nitro group, see your lewis resonance structures).

carbon between nitro and alkyl group receives electron density from alkyl group, with electron density removed by nitro group.

carbon "on the other side" of nitro group (para to alkyl group) is *only* experiencing electron withdrawal by nitro


electron density is lowest at that carbon, so de-shielding is lowest at that carbon.


consequently, THAT carbon (and the hydrogen bonded to it) has the smallest amount of de-shielding, and consequently has the lowest PPM of the aromatic protons

[Anonymous No.6387474]

>>6387450


geminal and vicinal are for substition on a saturated alkane.

example: geminal and vicinal dihalide


E, Z, cis/trans, and ortho, meta, para are used for constrained systems (Such as those that are constrained by aromatic or unsaturated bonds)

[matsci !!invJra9x72G No.6387492]

>>6387471

You're just looking at the ppm, and I'd agree with you if that were all to consider, but you are completely neglecting the multiplicity. F should be ortho to isopropyl and nitro.

Yes, electron density is lowest at carbons that are ortho and para to nitro. Low e- density = higher ppm. So ortho and para to the nitro should be highest ppm, i.e. these are D, E, and F. Meta is least deshielded, has highest e- density of all the aromatic protons, so meta should be at lowest ppm, i.e. C.

>>6387474

Hrm, I know there's nomenclature for the aromatic ones, can't remember what it is, though. Oh well.

[Anonymous No.6387498]

>>6387492


no.

you really need to refresh your memory about shielding.

de-shielding is an effect in aromatic systems because of ring current. it is not an electrostatic effect.

the opposite occurs in, as an example, acetylenic systems.

in other words, its not a measure of the simple existence of electron density.


if we restrict ourselves to aromatic systems, where the effect is "de shielding," which is the result of ring current, then the greater de shielding (higher ppm) occurs when there is greater electron density: electron donating (ortho para director).

the smaller de-shielding (lower PPM) occurs when there is electron withdrawing going on (meta director).

[Anonymous No.6387500]

>>6387492


the nomenclature is ortho, meta, and para, or 1,2,3,4 (all cyclic systems)

[matsci !!invJra9x72G No.6387526]

>>6387498

Deshielding is more than aromatic systems. Deshielding simply means lower electron density, which means a higher ppm. Google "deshielding aldehyde."

Like I said, greater deshielding -> higher ppm occurs with lower electron density.

[matsci !!invJra9x72G No.6387543]

>>6387498

And yes, I agree (at least, vaguely remember) de-shielding occurs as the result of ring current in aromatic systems.

The only thing we disagree on is the effect of electron density upon deshielding. I say that electron withdrawing groups show greater de-shielding. SDBS assignments of molecules like 4-nitro-o-xylene will confirm what I have said.

[matsci !!invJra9x72G No.6387556]

>>6387498

And yes, it makes sense that increased electron density in the ring would increase ring current -> deshield all of them.

However, inductive/resonance effects would win out. Thus, ortho and para to nitro have highest ppms.