[gibbeenrgy]

I copied this problem from my friend and tried to solved it. However, after 3 days , I still have no idea. So, I decided to post it here. Any helps are always appreciated .


Edit: I have uploaded it again. You can double click on the image to make it bigger. Sorry for the small size image.
BTW, this problem is really really interesting. I spend so much time on drawing tree diagram, trying almost possible structure but still can't obtain the result. I think the formula will have 3 C-N .

[Ψ×Ψ]

Sorry, but I totally can't see it in a decent enough size for me to offer any suggestions.  (Apart from "larger image, please.")

[movies]

There is no way those expansions came from that NMR spectrum.

[Dolphinsiu]

By Rule of Thirteen, MWt = 121/13 = 9 + 4/13

Base formula = C9H13

As MWt is odd, then implies at least one N's, Molecular formula becomes C8H11N
DBE (Double bond equivalent) = [2( 8 ) + 2 - 11 + 1] / 2 = 4
As peak > 3000, this confirms it contains unsaturated =C-H STR
As no peak around 3500-3200cm-1 in IR, tertiary amine is possible! -N(CH3)2

Molecular formula left = C6H5

Obviously, I will predict it is Ph - N(CH3)2

[movies]

The mass would be too high and you would have too many peaks in the ¹³C NMR.

[movies]

Okay, you chaged it so the mass is close, but you would still see more than three peaks in the ¹³C NMR if it was N,N-dimethylaniline.

[gibbeenrgy]

    As for the H NMR , the compound should have something likes (CH2 CH CH2) group, because of the pattern splitting , I draw the tree diagram and confirm that.
    Up until this morning, I think that I intepret the IR peak 1200 as 3 C- N , it may be wrong because the possible structure is not fitted with MW = 121 .
   The exist of Benzyl ring is not possible , I believe so b/c of the NMR at 6.0

[Dolphinsiu]

I have not learnt 13C NMR, so I cannot help you!

[gibbeenrgy]

Ha ha. I just figurout it this morning. It is actually Allyl bromide (C3H5Br )
An interesting problem indeed.
BTW, you can check it by ChemExper website . The link is here: http://www.chemexper.com/

[Custos]

Looks right. A mass spectrum, which would show the 1:1 isotope ratio for bromine, would have made the problem much easier. It just shows that mass spec can be useful and nmr doesn't always tell all. 

[movies]

Heh, that's what I thought it was.  The NMR splitting just doesn't look right in the expansions though.

[Custos]

It's not a real spectrum - probably calculated or even just stylized for the problem. The methylene would be a doublet (with fine structure from allylic coupling), the terminal protons would both be doublets of doublets (again possibly with some fine structure) and the internal vinyl proton will be a doublet doublet triplet - but of course you would expect overlap, not cleanly separated as shown in the problem. If you do the simulated pmr in ChemDraw that proton looks like this.