Use IR and NMR data to define the structure of the molecule, the formula of the given structure $\ce{C9H10}.$

¹H NMR spectra

IR spectrum

On what basis do we conclude that its structure is that (the answer) only? On what points can we identify the structure of this molecule?

I have found out that since its frequency is just less than $\pu{3000 cm-1}$ it is an alkane. Also, the approximate value of the chemical shift is near $7.2,$ hence it is aromatic.

But what I did not get was the meaning of two peaks at $7.2$ and the shifts at $2.9$ and $2.1.$

  • $\begingroup$ There are peaks beyond $\pu{3000 cm-1}$ in FT-IR. What's that means? $\endgroup$ Jun 13 '20 at 16:09
  • $\begingroup$ Good. Actually they are $\mathrm{sp^2}$-$\ce{C-H}$ stretching. You have an aromatic ring so that's it for IR. Now count how many unsaturation in your compound. You can count out 4 for aromatic ring. $\endgroup$ Jun 13 '20 at 16:14
  • $\begingroup$ @Andrews: I have given the explanation for peaks in NMR for you to learn. $\endgroup$ Jun 13 '20 at 17:09

The molecular formula of the compound is $\ce{C9H10}$. The number of unsaturations in the compound of $\ce{C_nH_m}$ is $\frac{2n+1-n}{2}$. Thus, the number of unsaturations in $\ce{C9H10}$ is $\frac{2\times 9+1-10}{2}=5$. You already found out that compound is aromatic by FT-IR, and hence $\ce{C6}$ of the formula counted out with 4 unsaturaations (3 double bond and a ring for aromatic nucleus).The remaining number of unsaturarion is $5-4=1$, and remaining number of $\ce{C}$ in the formula is 3. The remaining unsaturation could be a double bond or a ring.

When look at $\ce{^1H}$-$\mathrm{NMR}$ spectrum, chemical shifts of all but aromatic peaks are less than $\pu{3.00 ppm}$. Therefore, it is safe to conclude that the compound is a aromatic bicyclic rather than aromatic alkene. Therefore, the compound is indane:


The protons on 1-$\ce{C}$ and 3-$\ce{C}$ are chemically and magnetically equivalent, and benzylic, hence the their relevant signal would be down-field to that of protons on 2-$\ce{C}$, which is usual acyclic aliphatic carbon. The benzylic protons have two vicinal protons, thus the corresponding peak is a triplet. Meanwhile, protons on 2-$\ce{C}$ have four vicinal protons (benzylic protons), thus the corresponding peak is a quintet (5 peaks).

Note: The integration given is a ratio, not correct number of protons.


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