Isomeric compounds \(\mathbf{A}\) and \(\mathbf{B}\) have the composition \(\mathrm{C}_{11} \mathrm{H}_{12} \mathrm{O}_{4}\). Spectral data are summarized below. Deduce structures for \(\mathbf{A}\) and \(\mathbf{B}\) and explain your reasoning. Compound A IR (KBr): 1720 (s), 1240 (s) \(\mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3}\right): \delta 2.46(\mathrm{~s}, 3 \mathrm{H})\) $$ \begin{aligned} &3.94(\mathrm{~s}, 6 \mathrm{H}) \\ &8.05(\mathrm{~d}, J=2 \mathrm{~Hz}, 2 \mathrm{H}) \\ &8.49(\mathrm{t}, J=2 \mathrm{~Hz}, 1 \mathrm{H}) \end{aligned} $$ Compound B IR (Nujol): 1720 (s), 1245 (s) \(\mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3}\right): \delta 2.63(\mathrm{~s}, 3 \mathrm{H})\) $$ \begin{aligned} &3.91(\mathrm{~s}, 6 \mathrm{H}) \\ &7.28(\mathrm{~d}, J=8 \mathrm{~Hz}, 1 \mathrm{H}) \\ &8.00(\mathrm{dd}, J=8 \mathrm{~Hz}, 2 \mathrm{~Hz} ; 1 \mathrm{H}) \\ &\mathrm{dd}=\text { doublet of doublets } \\ &8.52(\mathrm{~d}, J=2 \mathrm{~Hz}, 1 \mathrm{H}) \end{aligned} $$

Molecular formula analysis is the first step in understanding the structure of a compound. The molecular formula gives us the number of each type of atom in a molecule. For instance, with the formula \(\text{C}_{11}\text{H}_{12}\text{O}_{4}\), we know there are 11 carbon atoms, 12 hydrogen atoms, and 4 oxygen atoms.
This molecular formula can provide clues about the possible functional groups and the degree of unsaturation without knowing the structure. The degree of unsaturation helps us understand the number of rings and/or double bonds in the molecule.
We calculate the degree of unsaturation using the formula: \[ \text{Degree of Unsaturation} = \frac{2C + 2 - H}{2} \] Here, \(C\) is the number of carbons, and \(H\) is the number of hydrogens.
By plugging in our values: \[ \text{Degree of Unsaturation} = \frac{2(11) + 2 - 12}{2} = 5 \] So, the compound has 5 degrees of unsaturation, indicating multiple double bonds or rings might be present. This analysis is pivotal when determining if the molecule contains aromatic rings or multiple bonds.

IR spectroscopy helps identify functional groups in a compound by measuring the absorption of infrared light, which causes molecular vibrations. Each type of bond vibrates at a characteristic wavenumber.
In the given problem, compounds A and B show strong absorption at 1720 cm^-1 and around 1240 cm^-1. The absorption at 1720 cm^-1 is indicative of a carbonyl (C=O) group. This is because carbonyl stretches typically occur around this frequency.
The absorption around 1240 cm^-1 corresponds to the C-O stretching of an ester or an ether group. This helps us deduce that the compounds likely contain ester functionalities.

Proton NMR spectroscopy provides detailed information about the hydrogen atoms in a molecule. It reveals the number, environment, and interactions of hydrogen atoms. It does this by showing signals at different chemical shifts (\textbackslash{}\textbackslash\textbackslash(\textdelta\textbackslash\textbackslash) values).
For Compound A, the proton NMR spectrum shows signals at:
- \( \textbackslash\textdelta\textbackslash\textdelta 2.46 (s, 3 H)\): Likely a methyl group (CH₃) attached to an electronegative element.
- \( \textbackslash\textdelta\textbackslash\textdelta 3.94 (s, 6 H)\): Suggestive of two methoxy groups (OCH₃).
- \( \textbackslash\textdelta\textbackslash\textdelta 8.05 (d, J=2 Hz, 2 H)\): Likely aromatic protons in a 1,2,4-substituted benzene ring.
- \( \textbackslash\textdelta\textbackslash\textdelta 8.49 (t, J=2 Hz, 1 H)\): Likely the central proton in a 1,2,4-substituted benzene ring.
For Compound B, we find:
- \( \textbackslash\textdelta\textbackslash\textdelta 2.63 (s, 3 H)\): Likely a methyl group (CH₃) connected to an electronegative element.
- \( \textbackslash\textdelta\textbackslash\textdelta 3.91 (s, 6 H)\): Indicates two methoxy groups (OCH₃).
- \( \textbackslash\textdelta\textbackslash\textdelta 7.28 (d, J=8 Hz, 1 H)\): Likely an aromatic proton ortho to a substituent.
- \( \textbackslash\textdelta\textbackslash\textdelta 8.00 (dd, J=8 Hz, 2 Hz, 1 H)\): Suggestive of an aromatic proton meta-coupled.
- \( \textbackslash\textdelta\textbackslash\textdelta 8.52 (d, J=2 Hz, 1 H)\): Likely an aromatic proton ortho-coupled.
This breakdown helps in determining the precise structure of the compounds.

Aromatic substitution patterns in NMR spectroscopy help us understand the positioning of substituents on a benzene ring. Benzene has six carbon atoms forming an aromatic ring, and substituents can attach at different positions, affecting the chemical shift and splitting of aromatic protons.
For a monosubstituted benzene ring, NMR typically shows a characteristic pattern. A disubstituted benzene ring can be 1,2- (ortho), 1,3- (meta), or 1,4- (para) substituted, leading to different splitting patterns:

• Ortho-disubstitution: Close protons generate higher coupling constants (e.g., \( J_{\text{ortho}} \approx 8 Hz\)).
• Meta-disubstitution: Produces medium-range couplings (e.g., \( J_{\text{meta}} \approx 2 Hz\)).
• Para-disubstitution: Usually shows smaller to zero coupling constants due to symmetry.

The patterns in this problem indicate:
- Compound A shows signals consistent with a 1,2,4-substitution pattern: \( \textbackslash\textdelta\textbackslash\textdelta 8.05 (d, 2 H); \textbackslash\textdelta\textbackslash\textdelta 8.49 (t, 1 H)\).
- Compound B shows signals aligning with a different arrangement: \( \textbackslash\textdelta\textbackslash\textdelta 7.28 (d, 1 H); \textbackslash\textdelta\textdelta 8.00 (dd, 1 H); \textbackslash\textdelta\textdelta 8.52 (d, 1 H)\), pointing to a more complex substitution pattern involving both ortho and meta positions.
Understanding these patterns helps to confirm the molecular structures of the two isomeric compounds.