See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The iodoform test involves reaction with I2 and dilute NaOH (or NaI/NaOH) to give a yellow precipitate of iodoform (CHI3). This test is positive for compounds containing the CH3CH(OH)- group (secondary alcohols where the OH is on the carbon adjacent to CH3) or the CH3CO- group (methyl ketones), as well as ethanol and acetaldehyde. Step 1: Identify the structural requirement for a positive iodoform test. A compound gives a positive iodoform test if it has the structural unit CH3-CH(OH)- or CH3-CO- (i.e., a methyl carbinol or methyl ketone). The key feature is a methyl group attached to the carbon bearing the OH (secondary alcohol with CH3CHOH- unit). Step 2: Check molecular formula C8H10O. Degree of unsaturation = (2×8 + 2 - 10)/2 = (16+2-10)/2 = 8/2 = 4. This corresponds to a benzene ring (4 degrees of unsaturation), so the compound is aromatic with no additional rings or double bonds. Step 3: Evaluate option (d): Phenyl-CHOH-CH3, i.e., 1-phenylethan-1-ol (C6H5CHOHCH3). Molecular formula: C6H5 (C6H5) + CHOH + CH3 = C8H10O. This matches C8H10O. This compound has the CH3-CH(OH)- unit attached to a phenyl group, which is exactly the structural unit that gives a positive iodoform test. Step 4: Evaluate why other options fail: - Option (a): C6H5CH2CH2OH is 2-phenylethanol. The OH is on a -CH2- group (not adjacent to CH3), so no CH3CHOH unit. Iodoform test negative. - Option (b): CH3-C6H4-CH2OH is 4-methylbenzyl alcohol. The OH is on a -CH2- group with no adjacent CH3 on the same carbon. Iodoform test negative. - Option (c): This is a dimethylphenol (xylenol), a phenol. Phenols do not give positive iodoform test under these conditions. Step 5: Option (d) is C6H5-CH(OH)-CH3 (1-phenylethanol), which contains the CH3CH(OH)- unit and gives a positive iodoform test (yellow precipitate of CHI3). Therefore, the correct answer is D.