# What is the identity of the unknown compound using GCMS and NMR?

I've been asked to identify an unknown compound by performing several analytical techniques on it.

• The compound was a white powder and had a boiling point of 65 degrees celsius.
• I have the GC mass spectrum as well as both Carbon-13 and Proton NMR spectra (note that the compound was dissolved in chloroform for NMR analysis).
• So far I think it contains an aromatic ring because of the chemical shifts for the NMR spectra but that's all I have so far.

These are the images of the spectra (Fig 1. GC and Mass spectrum, Fig 2. Proton NMR spectrum, Fig 3. Zoomed in Proton NMR spectrum, Fig 4. Carbon NMR spectrum, Fig. 5 Zoomed in Carbon NMR spectrum):

Hints:

1. If you counted carbon signals in $\ce{^{13}C}\mathrm{~NMR}$, you'd find the compound has at least 10 $\ce{C}$ atoms. Thus, it's indicated boiling point is incorrect, unless it's determined at reduced pressure. If so, you need to give that pressure.

2. You are correct about saying your compound should be aromatic. Did you count how many aromatic hydrogens are there? At least, by doing that, you may able to speculate some framework for your compound, as a beginning point.

3. Now, if you are careful, you see no aliphatic carbon signals in your $\ce{^{13}C}\mathrm{~NMR}$. Then, what is the signal at $\mathrm{\delta~1.6}$ in your $\ce{^{1}H}\mathrm{~NMR}$ doing there?

4. Now, go to $\mathrm{GC/MS}$. In there, your parent peak is almost looks like a doublet. Do you know why? Can you recall what element have the two stable isotopes about 50:50 natural abundance with $2~m/z$ units? If you don't remember, then subtract each of your parent peaks from your base peak in your $\mathrm{MS}$. Now, you almost have the empirical formula of your compound. Now it's matter of analyzing peak patterns in $\ce{^{1}H}\mathrm{~NMR}$ and fragmentation pattern in $\mathrm{MS}$.

• Thanks for your response! I meant the melting point was 65. Apologies for any confusion. So far I think there are 10 carbons but I’m struggling to count the hydrogen atoms and determine whether the peaks at 7.733 and 7.717 are singlets or one doublet… As for the signal at 1.6 in the H NMR I have a feeling it's water in the chloroform solvent? I believe the element you're referring to is bromine? Does that then mean the compound contains two bromine atoms? If so a possibility could be C10H12Br2. But I still need to have a better look at the H NMR to try to count the hydrogen atoms – Emma Apr 19 '18 at 11:04
• Emma, you are correct about water in the chloroform and bromine. So, I felt you are on right track. Now, you have confirmed there is no aliphatic portion in your molecule. Thus, it's a substituted aromatic compound. – Mathew Mahindaratne Apr 19 '18 at 20:15
• PS: Also, you are wrong about having two bromine atom. It is a characteristic feature that if you have one bromine in the molecule, you's see equally intense two peaks ($\mathrm{2~m/z}$ apart for ebvery bromine containing fragment in $\mathrm{MS}$, each representing $\ce{^{79}Br}$ and $\ce{^{81}Br}$. Also, if you do close inspection on $\mathrm{^{1}H~NMR}$, you'd see only 7 proton signals. Now tell me what is the compound? :-) – Mathew Mahindaratne Apr 19 '18 at 20:15
• I have a feeling it might be the solid form of bromonaphthalene as that would be consistent with the formula C10H7Br and the carbon and proton NMR peaks. However, I'm not sure the proton NMR spectrum above reflects the proton coupling in bromonapthalene (I think the spectrum should show more triplets). Thanks very much for your help so far though, you've been very helpful! – Emma Apr 20 '18 at 3:42
• Bromonaphthalene has two positional isomers: 1-bromonaphthalene and 2-bromonaphthalene. Which is the solid close to mp of $65~\mathrm{^\circ C}$? And remember, each isomer has different $\ce{^1H}~\mathrm {NMR}$ patterns. – Mathew Mahindaratne Apr 20 '18 at 4:12