Short Answer 
To analyze the proton NMR spectrum of the compound C14H14O2, it is important to understand its structure, which consists of an aromatic system formed from 1-bromo-4-methoxybenzene and m-cresol. The compound exhibits 7 distinct proton NMR signals: 4 from aromatic positions, 1 from the methoxy group, 1 from the methyl group, and 1 from the hydroxyl group.
Step 1: Understand the Compound Structure
To analyze the proton NMR spectrum of the organic compound C14H14O2, we first need to comprehend its structure. This compound is synthesized from 1-bromo-4-methoxybenzene and m-cresol. In the coupling reaction, we can assume that the resulting compound will form a single connected aromatic structure, which is key to determining unique hydrogen environments.
Step 2: Identify Distinct Hydrogen Environments
Next, we will identify the different hydrogen environments present in the compound. The aromatic rings of both 1-bromo-4-methoxybenzene and m-cresol contribute various unique signals due to their substitution patterns. We should consider:
- Aromatic hydrogens from the aromatic rings create distinct environments.
- Effects of substituents, such as electron-donating methoxy and electron-withdrawing bromine, impact signal formation.
Step 3: Count and Summarize Unique Signals
Finally, we count the number of distinct NMR signals based on our earlier analysis. The expected unique signals come from:
- 4 distinct signals from different aromatic positions.
- 1 signal from the methoxy group (‚à ÃOCH3).
- 1 signal from the methyl group (‚à ÃCH3).
- 1 signal from the hydroxyl group (‚à ÃOH).
Thus, the total expected number of unique proton NMR signals in the compound is 7.