See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: Na2Cr2O7 (sodium dichromate) is an oxidizing agent. The reaction converts an alcohol to a carbonyl compound. The molecular formula change from C9H12O to C9H10O represents a loss of 2 hydrogen atoms (H2), which is consistent with oxidation of an alcohol to a ketone or aldehyde. Step 1: Analyze the degree of unsaturation change. C9H12O has DBE = (2×9 + 2 - 12)/2 = (18+2-12)/2 = 8/2 = 4. C9H10O has DBE = (18+2-10)/2 = 10/2 = 5. The DBE increases by 1, confirming formation of a new C=O bond (oxidation of alcohol to carbonyl). Step 2: Analyze option (a): phenyl-CH2-CH2-CH2-OH. This is a primary alcohol (3-phenyl-1-propanol). Oxidation of a primary alcohol with Na2Cr2O7 gives an aldehyde (phenyl-CH2-CH2-CHO) or further to carboxylic acid. The product formula would be C9H10O for the aldehyde. However, primary alcohols with Na2Cr2O7 under typical conditions tend to oxidize further to carboxylic acids (C9H10O2), not stop cleanly at the aldehyde. Also, the product C9H10O from option (a) would be an aldehyde. Step 3: Analyze option (b): phenyl ring - CH(OH) - CH2 - CH3. This is 1-phenyl-1-propanol, a secondary alcohol. Secondary alcohols are oxidized by Na2Cr2O7 to ketones. The product would be phenyl-CO-CH2-CH3 (1-phenyl-1-propanone / propiophenone), formula C9H10O. Secondary alcohols oxidize cleanly to ketones and cannot be further oxidized under normal conditions. Step 4: Why option (b) is correct. Option (b) is a secondary alcohol. Oxidation with Na2Cr2O7 cleanly gives the ketone (propiophenone, C9H10O), with a net loss of 2H. This matches the given product formula exactly. The oxidation stops at the ketone stage. Step 5: Why option (a) is less suitable. A primary alcohol would oxidize to an aldehyde (C9H10O) but Na2Cr2O7 in aqueous acidic conditions typically over-oxidizes primary alcohols to carboxylic acids (C9H10O2), so option (a) would not reliably give C9H10O. Therefore, the correct answer is B.