# Identify the structure of compound with C₉H₁₀ formula using IR and NMR data [closed]

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

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 $$2.1.$$

• There are peaks beyond $\pu{3000 cm-1}$ in FT-IR. What's that means? – Mathew Mahindaratne Jun 13 '20 at 16:09
• 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. – Mathew Mahindaratne Jun 13 '20 at 16:14
• @Andrews: I have given the explanation for peaks in NMR for you to learn. – Mathew Mahindaratne 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).