[Yufinity]

I'm having a difficult time trying to figure out a specific isomer for C₁₂H₁₄O₄.

Based on the IR spectrum I was given... There is significant absorption at 1765cm^-1 which means there is a C=O bond. There's also a benzene ring since there is absorption at 1600cm^-1. Please correct me if I'm wrong anywhere.

Based on the C NMR spectrum, there are 5 peaks meaning 5 types of carbon. With only 5 types of carbon and a total of 12 carbons in the molecular formula, the benzene ring must be disubstituted with symmetry. The first C-NMR peak is at approximately 8ppm, so there must be an R-CH₃ terminal group. The next peak is at 28ppm for an R-CH₂-R fragment. There are 3 peaks within 116ppm-129ppm for benzene ring.

For the H NMR spectrum... there are 4 peaks meaning 4 types of hydrogen. The first peak is at 1.1ppm (triplet multiplicity meaning 2 neighbours or 2 coupled hydrogens... which results in a CH₃ coupled to 2 hydrogens), next peak is at 2.4 (quartet meaning 3 neighbours/coupled hydrogens... which results in a CH₃ coupled to 3 hydrogens deshielded by an O), next peak is at 6.9 (doublet meaning 1 neighbour... results in Arene-H disubstituted with para configuration for symmetry) and last peak is at 7.2 (triplet, 2 neighbours... which results in Arene-H disubstituted with para configuration for symmetry)

As for the integration for H NMR, my TA wrote down 6H (triplet), 4H (Quartet), 3H (doublet), 1H (triplet). I'm not sure if she's right.. but her reasoning was this: "The peak at 1.1ppm is about... 6cm long so that translates to 6H....etc".

I've tried MANY MANY isomers but every single time, something didn't match up. I even found a site that gave me all 291 isomers for that molecular formula... I went through all of them and none of them matched up (unless I overlooked something). But honestly, all the people in my lab section who received this same problem have had no luck... the TA spent a good 45 minutes helping us out but in the end, she couldn't even figure it out and just said to "go home and think about it". Since there is a C=O bond, there has to be an ester functional group or carboxylic acid group on both ends of the disubstituted benzene ring. This is assuming that my entire group analyzed the spectras correctly... Any help is appreciated!

[huoshiyong]

Diethyl terephthalate?

[bromidewind]

Could you possibly scan and upload your spectra on here? It would definitely make it a lot easier as we wouldn't have to (no offense) take your word for the values you've provided.

As for the integration for H NMR, my TA wrote down 6H (triplet), 4H (Quartet), 3H (doublet), 1H (triplet). I'm not sure if she's right.. but her reasoning was this: "The peak at 1.1ppm is about... 6cm long so that translates to 6H....etc".

Tell your TA that peak integration is NOT just about peak height. It's all about the peak area.

[dunno260]

Its not diethyl terephthalate.  The CH₂ resonance at 2.4 is most likely connected to the carbonyl carbon rather than the ester oxygen.  For diethyl terephthalate I would expect the resonance to be around 4ppm (similar to ethyl acetate).

Based on the information given, I would bet its 1,4-phenylene dipropionate though the splitting you described isn't consistent with p-substitution of the ring for the proton though the carbon makes sense for para substitution.  You are missing the carbonyl carbon and that could mean you are possibly missing a carbon in the aromatic region as well which would put all the para, meta, and ortho isomers all on the table.  If the aromatic region of the proton spectrum is as you described, I would bet its meta (1,3) and that your carbon-13 spectra just didn't get enough scans to show carbons with no hydrogens attached.  Also the IR if its good quality might allow us to determine the substitution of the ring by looking at some overtone stretching at 2000cm^-1.

Really going to need scans of the spectra though to get better than that.

[AC Prabakar]

I have convenience with the 1,4-phenylene dipropionate structure except the aromatic region.
If it is para substituted how triplet in aromatic region is possible?

[orgopete]

Based on the information given, I would bet its (it might be) 1,4-phenylene dipropionate though the splitting you described isn't consistent with p-substitution of the ring for the proton though the carbon makes sense for para substitution.  You are missing the carbonyl carbon and that could mean you are possibly missing a carbon in the aromatic region as well which would put all the para, meta, and ortho isomers all on the table.  If the aromatic region of the proton spectrum is as you described, I would bet its meta (1,3) and that your carbon-13 spectra just didn't get enough scans to show carbons with no hydrogens attached. 

I also think it might be the 1,3-isomer. While 5 peaks in the carbon spectrum suggests 1,4-substitution, the ¹H-NMR is consistent with something like the 1,3-substitution. I can imagine a student missing a quaternary carbon in a ¹³C-NMR spectrum.

[demoninatutu]

Tell your TA that peak integration is NOT just about peak height.

I suspect that the TA said that the integration curve was 6cm, rather than the peak.