See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The iodoform reaction. When a compound reacts with iodine (I2) and dilute NaOH to give a yellow precipitate, the yellow precipitate is iodoform (CHI3). This is the iodoform test, which is positive for: (1) methyl ketones (compounds with CH3CO- group), (2) secondary alcohols of the type CH3CH(OH)- (i.e., compounds with CH3CHOH- group), and (3) ethanol itself. Reasoning: The molecular formula C8H10O with one degree of unsaturation from the benzene ring (4 degrees) plus the oxygen-containing functional group. We need a compound that gives a positive iodoform test. Option (a): Acetophenone (C6H5COCH3) — this is a methyl ketone and would give a positive iodoform test. Molecular formula: C6H5COCH3 = C8H8O. This does NOT match C8H10O (two fewer hydrogens), so option (a) is ruled out on the basis of molecular formula. Option (b): C6H5CHOHCH3 (1-phenylethan-1-ol) — this has the structure CH3CH(OH)C6H5, which contains the CH3CHOH- group. Molecular formula: C6H5 + CHOH + CH3 = C8H10O. This matches C8H10O. This compound has a secondary alcohol with a methyl group on the carbinol carbon (CH3-CH(OH)-), which is exactly the structural requirement for a positive iodoform test. NaOH/I2 oxidizes the -CHOH- to -CO-, giving CH3CO-C6H5 in situ, and then the methyl ketone undergoes iodoform reaction to give CHI3 (yellow precipitate). Option (c): 4-methylbenzyl alcohol (CH3-C6H4-CH2OH) — molecular formula C8H10O (matches). However, this is a primary benzylic alcohol with no alpha methyl group on the carbinol carbon, so it does NOT give a positive iodoform test. Option (d): The structure shown has a benzene ring with CH3 (para), OH, and CH3 substituents — this appears to be a phenol derivative (cresol with extra methyl). Phenols do not give the iodoform test under these mild conditions (dilute NaOH/I2 would give tri-iodination of the ring for phenols, not iodoform precipitate in the same sense). Also the -OH is directly on the ring (phenol), not a secondary alcohol, so no iodoform reaction. Conclusion: Only option (b), C6H5CHOHCH3, satisfies both the correct molecular formula C8H10O and gives a positive iodoform test (yellow CHI3 precipitate) with I2/dilute NaOH. Therefore, the correct answer is B.