Structural Elucidation via NMR Spectroscopy & ESI-MS

Firstly, molecule (D) cannot be the correct molecule, as there are nine carbon resonances observed in the 13C NMR spectrum, and molecule (D) only has 7 carbons.

Molecule (B) cannot be the correct molecule, as it has two methoxy moieties, which would give rise to two closely spaced singlets in the 1H spectrum. This is not observed, thus molecule (B) cannot be the correct molecule.

Molecule (A) cannot be the correct molecule, as it has only 2 distinct proton environments in the aromatic ring (which would appear as two 1:1 doublets), yet the provided 1H spectrum clearly exhibits 4 distinct resonances in the aromatic region.

The molecule that gives rise to the provided NMR spectra is molecule (C): ethyl 2-hydroxybenzoate (ethyl salicylate). The molecular structure of ethyl salicylate is shown below.

The terminal methyl group is adjacent to a methylene group, therefore (n + 1) = 3. A 1:2:1 triplet is observed at ca. 1.4 ppm, with an relative integral of 3. Thus, this is assigned as the methyl peak.

The methylene group is adjacent to a terminal methyl, and an ester oxygen. Therefore (n + 1) = 4. A 1:2:2:1 quartet is oberved at ca. 4.4 ppm, with a relative integral of 2. Thus, this is assigned as the methylene peak.

The aromatic ring possesses 4 distinct proton environments, two of which will appear as doublets, as they are only adjacent to 1 proton bearing carbon atom. Four proton resonances are observed in the 1H spectrum in the aromatic region, two of which are doublets. (Note: the aromatic proton resonances are slightly split further via long-range coupling).

The hydroxyl proton will naturally appear as singlet, which is observed in the 1H spectrum at ca. 10.87 ppm.

To assign the spectrum of ethyl salicylate from first principles, one would need to acquire a set of different NMR experiments, including: 1H,1H-COSY, DEPT(45, 90 and 135), HMBC, HSQC, and possibly SELTOCSY spectra. But as the other three molecules could be excluded by using simple chemical logic, these were not required to answer the question.

My working is like this (Quick way)

The mass from Mass Spectroscopy eliminates option $D$ .

Then C13 shows 9 peaks while in $A$ and $B$ we have 2 pairs of symmetrical carbon ( in A and B , only 7 peaks would be there )

Hence $C$ is the answer .

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Extra help and detailed solution (Confirming our selection)

Remember Given mass is of Sodium adduct , hence actual mass is 189.15 - 23 = 166.15 .

Alcoholic group is confirmed by the braod peak in IR Spectroscopy (due to Hydrogen bonding).

That broad peak in 50 to 100 region is of solvent (not of our use)

Also , C13 show

• 1 peak in 0 to 50 implies one methyl group (attached to C ) (end CH3 group in ethyl substituent in our compound)
• 1 peak in 50 to 100 implies one saturated C attached to O. (Our ethyl carbon in ester)
• 5 peaks in 100 to 150 are that of 5 unsaturated carbons (except those attached to oxygen)(in our compound , the 5 C of aromatic ring , disincluding the phenolic one .
• 2 peaks in 150 to 200 is that of 2 unsaturated carbons attached to Oxygen .(one to ester and one to phenol type in our compound)

The IR spectra was useful. The broad peak at 3000 ${cm}^{-1}$ indicates hydrogen bonding and hence B is eliminated.

The broad peaks in the 1600-1500 ${cm}^{-1}$ eliminates the possibility of an aldehyde, so D is eliminated.

The transmittance of the 3000 ${cm}^{-1}$ peak is characteristic of a hydroxyl group not a carboxylic acid, this eliminates A .

This indicates that the required compound is C and both NMR spectra are consistent with the deduction.