[tearingmyhairout]

Hi guys ,

I was hoping someone could help me with this NMR sample of an unknown compound . I 've been given a mass spectrum , 13C and 1H nmr to analyse it . I dont know how to analyse the mass spectrum but have included it anyway.

Mass Spectrum:


13C NMR:


1H NMR:


Here is my analysis so far for the carbon NMR :

PPM        DEPT                Assignment
26.1        CH3                 CH3
113.7       CH                   CH
127.5       Quarternary
130.8       CH                   CH
151.5       Quarternary
196.7       Quarternary

As 4 of these signals are in the aromatic region for 13C (from a carbon shifts table) I reckon this is a benzene ring with 3 substituents attached due to the 3 quarternary carbons . CH3 is one of the substituents and the signal at 196.7 may indicate a CHO aldehyde group as a substituent ( from a carbon shifts table). Only two CH signals are shown as two of the carbons with one hydrogen group are equivalent . ( Please correct me if any of these assumptions are wrong)

Here is my 1H NMR analysis so far :

Chemical shift(ppm)   No. of Hydrogens   Multiplicity                     Assignment
2.49                                3                singlet                            CH3
4.3                                  2                singlet                            OH , NH , NH2 ?  due to D2O shake
6.64 - 6.61                       2                doublet of triplets
7.80 - 7.77                       2                doublet of triplets               

This shows that the CH3 at 2.49 will have a group with no hydrogens attached as its neighbour.
Also the signal at 4.3 will have a group with no hydrogens attached as its neighbour.

Now here is where i need help:

Will the signal at 4.3 be an NH2 ? (Due to the integral showing 2 hydrogens) . I wasn't sure if the D2O shake replaces both hydrogens of the NH2.

If the answer to my question above is yes then I think the three substituents attached to the benzene may be a CH3 , NH2 and an aldehyde ( although wouldn't the aldehyde proton show up on the NMR?) - what could this 3rd substituent be instead of the aldehyde?

A doublet of triplets arises when a atom with a H attached is between 2 atoms containing 1 and 2 H's respectively ? right?Which one of my substituents would give rise to these (  doublet of triplet )  signals and where would they be located on the benzene ? According to the NMR the two substituents will require two H's to be attached to them . NH2 will have two H's so this could be one substituent. What would be the other?


So basically I want to confirm the substituents attached to the benzene and need help with the positions of the substituents to get the desired signal splitting.

Thank you kindly for any help you give .




                             

[fledarmus]

This is where you need to start using your mass spec. It will (with some work) give you the molecular formula of your molecule. That will narrow down your options.

For example, does your compound have any nitrogens? How many carbons does it have? What about sulfur or halogens? Answers to these questions can almost be read directly from the mass spec with practice.

[tearingmyhairout]

OK

So i think i roughly understand how to analyze my mass spectrum and think I've got the compound but I still need help.

I think the compound is a benzene with a CH3 group attached and an aldehyde (CHO) group and and an NH2 group attached on both carbons meta to the CH3. My mass spectrum sort of backs this up I think ( please correct me if my results are wrong) .
The mass of the whole molecule would be 135 and this corresponds to the 135 on the mass spectrum . This would be the molecular ion.
The mass at 120 on the mass spectrum corresponds to the whole molecule minus the CH3 group which gives a mass of 120 . This is also the base peak as it is the largest.
The mass at 106 on the mass spectrum corresponds to the whole molecule minus the NH2 and CH3 groups which gives a mass of 106.
92 on the mass spectrum corresponds to the whole molecule minus the NH2 and aldehyde COH which gives a mass of 92.
77 on the mass spectrum is just the benzene with a mass of 77.
about 29 on the mass spectrum is just the CHO aldehyde  with a mass of 29
15 on the mass spectrum is the CH3 . I'm not sure about the rest of the signals.

OK , now are my questions.

On my carbon NMR i get a signal at 196.7 which corresponds to an aldehyde but why doesn't the aldehyde H show up on my H NMR ? what else could give rise to this signal?

Also are the doublets of triplet signals from the carbons attached to the NH2 and CHO aldehyde ( i.e the carbons in the meta position to the CH3)?

Any help will be very appreciated.

[tearingmyhairout]

I think I know how to handle the aldehyde problem. It isn't an aldehyde at all and could be a ketone . So now I think my compound is a benzene with a ketone C=OCH3 attached to it with an NH2 in the position para to it . This agrees with my mass spectrum and the readings on my 13C NMR but I still don't know how to get the doublet of triplets in the 1H NMR . Any help?

[fledarmus]

That's a much better answer 

It's a little hard to predict aromatic splitting patterns. In aromatic systems, you can get splitting from protons ortho, meta, and para to the one you are looking at. Ortho splittings are usually much larger than meta splittings, and para splittings are usually very small.

What you are seeing is typical of para substitution - two doublets caused by each proton having one adjacent proton, and fine structure from splitting by the protons on the other side of the ring.

Once you have seen a lot of 1,2-, 1,3-, and 1,4- substituted aromatic systems, you will start to recognize the splitting patterns.