## Step 1: **Analyze the Molecular Formula** - **Given molecular formula:** \( \mathrm{C_9H_{11}Br} \) - **Degree of Unsaturation (DU):** \[ \text{DU} = \frac{2C + 2 - H - X}{2} = \frac{2(9) + 2 - 11 - 1}{2} = \frac{18 + 2 - 11 - 1}{2} = \frac{8}{2} = 4 \] - **Explanation:** 4 units of unsaturation suggest either a benzene ring (aromatic, which is 4 DU) or a combination of rings/double bonds. --- ## Step 2: **Interpret the NMR Spectrum** - **Aromatic region (7–8 ppm):** Multiple peaks, indicating aromatic protons, consistent with a benzene ring. - **Integration:** The aromatic region likely integrates to 5 hydrogens, consistent with a monosubstituted benzene ring. - **Non-aromatic region (2–4 ppm):** - There are signals here, suggesting the presence of alkyl groups attached to the benzene ring (e.g., ethyl or propyl). - The presence of a Br atom suggests it might be attached to a benzylic position (bromomethyl or bromoethyl group). - **Aliphatic region:** The pattern suggests a benzylic side chain (like ethyl or isopropyl). - **Explanation:** The spectrum is consistent with a structure where a benzene ring is substituted with a bromine and an ethyl group (since the integration and splitting pattern match such a substitution). --- ## **Final Answer: Structure Proposal** **The most probable structure is**: **1-bromo-2-phenylethane (benzyl ethyl bromide, C₆H₅CH₂CH₂Br)** - **Structure:**  *(Benzene ring with a -CH₂CH₂Br group attached)* **Summary:** - The molecular formula and unsaturation indicate a benzene ring. - The NMR shows aromatic protons and a benzylic ethyl group attached to Br. - Therefore, the compound is likely **phenylethyl bromide (1-bromo-2-phenylethane)**.