[Arait]

Hi,

For my organic chem lab, we were given IR, 1HNMR, 13CNMR, and mass spectra and were told to come up with the unknown compound.  Well, we've never really been properly taught how to do this, so I've been trying to use my lab textbook to piece together the answer.  I have a guess of what it is, but I wanted some outside input to see if I was even in the ballpark.  Here are the four graphs I was given:



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Based on these graphs, my guess was that the chemical formula was C8H11N because the mass should be around 122-ish g/mol and the 1HNMR flat out tells you that there are 11 H atoms.  The 8 C was kind of a guess. I knew that there should be 8 different kinds of C based on the 12CNMR, so I just put that as the total number.  The IR according to my book indicated an amide group, but I can't really get that to work with my other assumptions.  I'm thinking that the compound is some kind of benzenamine with a couple methyl groups coming off of it, but I'm not sure about the arrangement since there are other possible isomers.

Am I on the right track here at all?

[azmanam]

try starting here:

http://www.chemicalforums.com/index.php?topic=33208.0
http://www.chemicalforums.com/index.php?topic=46816.0

[nox]

It's a substituted aniline, so yes you're on the right track.

[Arait]

Thanks to both of you!

Could it be a dimethyl aniline?  The more I looked at everything, that is what I'm leaning towards.  I think I would have an easier time confirming my idea if the 1HNMR wasn't such a mess.  I've had to use 1HNMR before, but I've never had one where it was so hard to tell what the multiplicity was on a lot of the peaks. 


Also, because of the 13CNMR, it makes me think that it's not a symmetric dimethyl aniline, which would get rid of a couple isomers. I'm just not sure where to go from there to confirm or deny my prediction.

[azmanam]

here's my guess at multiplicity

either: 7.04 (d, 1H), 6.59 (d, 1H), 6.52 (t, 1H), 6.47 (t, 1H), 3.45 (s, 2H), 2.538 (q, 2H), 1.20 (t, 3H)
or: 7.04 (d, 1H), 6.59 (s, 1H), 6.52 (t, 1H), 6.47 (d, 1H), 3.45 (s, 2H), 2.538 (q, 2H), 1.20 (t, 3H)

the multiplicity of the 6-8 region is not nearly as important as the multiplicity of the 1-4 region.

What kind of puzzle pieces can you draw from the 1-4 region?

[nox]

Look at the peak at 3.45 ppm in the proton NMR. Do you see anything special about its shape/peak width?

[Arait]

To answer azmanam, I'm not sure what you're asking... 

Nox, it's just a single peak?  I don't know what else to get out of it.  It seems like that peak is in the right range for a H-N-R aromatic group according to my book...

[Arait]

Okay, so after looking more, now I think it's 2-ethylaniline.  It seems pretty consistent (depending on how you interpret the multiplicity with a couple of the bunched up peaks) with the 1HNMR and mass spectra.  This compound makes the most sense with the HNMR of any compound I've found so far.  Dimethylaniline just doesn't work for me no matter what isomer I use.

Someone please throw me a bone, tell me flat out if this prediction is right or wrong so I don't waste my time on it.

[azmanam]

You answer is logical.  How did you get to an ethyl group instead of 2 methyl groups, though?

[Arait]

A couple reasons.  One of the main reasons was the mass spectrum. If you look at it, the farthest peak is at about 122 for the molar mass of the whole compound, if you go to the next peak, it's at around 106, which would make sense for ethylbenzene (removing the amine group), and the next peak is at 77 for C6H5 (a difference of 29, which is about the mass of an ethyl group).  

Also, an ethyl group made more sense for the 1HNMR.  It's hard for me to explain because I'm basically self-taught at this stuff and I probably think about things really weird (due to the fact that my lab TA last semester was terrible at explaining this stuff).  Basically, if I used methyl groups I thought that there would need to be more "single" peaks.  If you had two methyl groups, you would end up with a ton of protons (8, I think) next to carbons with no hydrogens attached.  There's only one single peak on this spectrum, so I couldn't make the dimethyl idea work.

[azmanam]

make sure you read my explanation here - it will help you unambiguously determine whether you have 2 methyls or 1 ethyl group

http://www.chemicalforums.com/index.php?topic=33208.0

[Arait]

Will do.  I'll keep at it.  Ultimately, my report is due tomorrow, so I'll have to decide.  The good thing is we don't get a tremendous amount of points taken off for the wrong compound.  As long as I'm able to come up with a halfway logical explanation, I should be fine.

Thanks to everyone for all the *delete me*  I certainly hope I get better at interpreting spectra as time goes on!