Am I using these spectra to identify the compound correctly?

Question

3. What is the structure of the molecule whose data is shown below? Explain your answer [...] EI mass spectrum: The highest peak observed is $m/z$ 210 and its abundance is low. The only other significant peak is observed at 167 Th (base peak). [...]

Full question text: https://i.stack.imgur.com/MQoUD.jpg

Based on the IR value, I reasoned that it is a ketone. Nothing else in the IR spectrum really stood out to me, so I stuck to the idea that it must be a ketone. I assumed the mass of the compound is 210 g/mol so I subtracted the mass of oxygen and used the rule of 13:

Oxygen's mass = 16 g/mol

(210-16)/13 = 14 + 12/13

So the molecular formula is $\ce{C14H_{(14+12)}O = C14H26O}$

I noted that there were 7 signals recorded in the carbon NMR spectrum. I assumed that meant there were equivalent carbons in the molecule. The molecule I ended up drawing to match this is 7-tetradecanone.

Do you think this is the correct molecule?

Answer 1

Structure elucidation from spectrocopic analysis needs some experience. I have feeling that you are novice to this field by your explanation of the structure. You cannot determine the molecular formula by the data given and, hence you cannot determine the degree of unsaturation of the molecule. Yet, based on $\ce{^1H}$-$\mathrm{NMR}$, you can propose there are at least two aromatic nuclei (probably mono-substituted) in the molecule, be case there is a 10-proton multiplet in the aromatic region ($\pu{7-8 ppm}$). This assignment is supported by the 4 resonances in the region of $\pu{110-160 ppm}$ in $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum. That also suggest these two aromatic nuclei are identical. Two aromatic rings are accounted for 8 degree of unsaturation.

The given FT-IR spectrum suggest there is a carbonyl group in the molecule, which is confirmed by the resonance at about $\pu{205 ppm}$ in $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum. The peak at about $\pu{205 ppm}$ would also suggest that the carbonyl is not an aldehyde, but a ketone by its chemical shift ($\gt \pu{200 ppm}$), which would be confirmed by the lack of aldehydic peak at about $\pu{10 ppm}$ in $\ce{^1H}$-$\mathrm{NMR}$. The carbonyl group is also accounted for another 1 degree of unsaturation.

Now, let's analyze $\ce{^1H}$-$\mathrm{NMR}$ spectrum. There is a 3-proton singlet at around $\pu{2.2 ppm}$ for isolated methyl group (without a neighboring hydrogen), indicating it is a benzylic methyl or next to a carbonyl group. Also, there is another one-proton singlet at about $\pu{5.1 ppm}$ for a methine group (-$\ce{CH}$-), indicating its highly deshielding nature. That deshielding must come from two aromatic ring and carbonyl group together. Thus, based on these data, I can deduct the structure as 1,2-diphenylpropn-2-one as depicted in the diagram below (blue box):

The highest peak observed in mass spectrum of the compound at $\pu{210 m/z}$ with law abundance and the base peak ($100\%$ intensity) observed at $\pu{167 m/z}$ confirmed the structure. As depicted in the red box (vide supra), the fragmented cationic radical at $\pu{167 m/z}$ is very stable, which explained the absence of any other fragmentations (as stated by OP, it is the only other significant peak in MS).

Answer 2

My personal opinion is: screw IR, it does not give useful information – but NMR does.

The first thing to check in the proton NMR is the integrals of your signals (I used a ruler on my PC screen – which interestingly enough seems to be good enough for this case!). You have clearly aromatic protons, one singlet at around $\pu{5 ppm}$ and one larger one at around $\pu{2.2ppm}$. The signal at $\pu{2.2ppm}$ can safely be considerd a methyl group so I took its integral to be 3. That gives me an integral of 1 for the proton at around $\pu{5ppm}$ and 10 for the aromatic signals.

With that, I can assume that we have two phenyl groups, possibly chemically equivalent, one $\ce{CH}$ and one methyl group. Next, let’s take a look at the $\ce{^13C}$ spectrum. $\pu{206ppm}$ is clearly a ketone as we have no aldehyde proton. The next four carbon signals are probably the aromatic rings; since we only see one set of four but they seem rather large, we can conclude that the phenyl groups are indeed chemically equivalent. Finally, there are two more signals for the methyl and the $\ce{CH}$ group.

Thus, we have:

• 2 times $\ce{C6H5-{}}$
• $\ce{-C(=O)-{}}$
• $\ce{-CH3}$
• $\ce{R3CH}$

We also know that the aliphatic protons are all isolated as they do not couple. Adding masses for the individual fragments gives us:

$$\underset{\text{phenyl}}{2\times 77} + \underset{\text{carbonyl}}{(12+16)} + \underset{\ce{CH}}{13} + \underset{\ce{CH3}}{15} = 154 + 28 + 28 = 210$$

Thus, we have all atoms accounted for and we just need to put them together.

The solution is $\ce{Ph2-CH-CO-CH3}$ or sum formula $\ce{C15H14O}$. You were quite a bit off.

Answer 3

Do you think this is the correct molecule?

No, for three main reasons:

1. As Ezze states in the comment, there are strong indications for phenyl groups in the NMR spectra. One phenyl group corresponds to 4 double bond equivalents, which is incompatible with your molecular formula (too many hydrogen atoms).
2. The H-NMR has too few multiplets for your proposed structure. You need an isolated methyl group (singlet) and a CH or OH group that is not coupled strongly to anything else.
3. Your suggested molecule lacks the symmetry to explain why there are only 7 signals in the C-NMR spectrum. Also, the alkyl chains in it don't show up in the H-NMR (would expect multiplets at 1-3 ppm).

Finally, the molecular formula for tetradecanone is $\ce{C14H28O}$, not $\ce{C14H26O}$, so the mass is off.

Ways to solve this

1. Fix the molecular formula. If there are less hydrogen atoms than in the proposed molecular formula, there must be more carbon or oxygen atoms. One simple change is to replace 12 hydrogen atoms by one additional carbon atom, giving the same mass ($\ce{C15H14O}$ rather than $\ce{C14H26O}$). This turns out to be correct, but at this stage it is tentative.
2. Consider the number of signals in the H-NMR in the aromatic region vs the XH and CH3 signal. See Jan's answer.
3. Find a structure where the CH3 signal can't couple with anything. Also, make sure you get seven signals in the C-NMR spectrum even though there are more than seven carbon atoms in the molecular structure, as you already realized. Having two phenyl rings helps to keep the number of signals low, and also fits with the large number of double bond equivalents deducted from the tentative molecular formula. See Mathew Mahindaratne's or Jan's answer for the actual structure.