Which compound with molecular formula $C_9{H}_{10}{O}_2$ would show two doublets and two singlets in its $H^1$ NMR spectrum?

• A. Ethyl 3-phenylpropanoate
• B. Methyl 4-methoxybenzoate
• C. p-Methoxyacetophenone
• D. p-Xylene-1,4-dicarboxylic acid

Answer: (C)

Verified step by step guidance

1. First, identify the molecular formula given: \(\mathrm{C_9H_{10}O_2}\). This will help confirm which compounds fit this formula before analyzing their NMR spectra.
2. Next, consider the structure of each compound and the symmetry or unique environments of the hydrogen atoms. The number and type of signals in the \(^1\mathrm{H}\) NMR depend on how many distinct proton environments exist.
3. For \(^1\mathrm{H}\) NMR splitting patterns, recall that a doublet arises when a proton has one neighboring proton (n+1 rule), and a singlet arises when there are no neighboring protons causing splitting.
4. Analyze the aromatic ring protons in each compound. Para-substituted benzenes often show two doublets due to the symmetry and coupling between ortho protons, while substituents like methoxy or acetyl groups influence chemical shifts and splitting.
5. Finally, match the expected splitting pattern of two doublets and two singlets to the compound's proton environments. For example, p-methoxyacetophenone has a para-substituted benzene ring giving two doublets, and the methoxy and methyl protons appear as singlets, fitting the described NMR pattern.