Primary alcohol oxidation to carboxylic acid under excess conditions

Steps:

• Identify J as branched C4H9OH: options A, C, D share CH3CH(CH3)CH2OH (2-methylpropan-1-ol, primary); B is straight-chain secondary (butan-2-ol).
• Excess Cr2O7^2-/H+ reflux oxidizes primary alcohols fully to carboxylic acids, stopping at ketones for secondary.
• K's IR (implied) shows carboxylic acid features (broad O-H ~3000 cm⁻¹, C=O ~1710 cm⁻¹), not aldehyde, ketone, or ester.
• Match: J oxidizes to CH3CH(CH3)COOH, fitting branched primary alcohol and IR.

Why C is correct:

• Primary alcohols with excess dichromate oxidize to RCOOH (formula: 3RCH2OH + Cr2O7^2- + 8H+ → 3RCOOH + 2Cr^3+ + 7H2O), confirmed by IR's broad O-H and lowered C=O stretch.

Why the others are wrong:

• A: Product is aldehyde (stops early without excess/distillation; IR lacks broad O-H).
• B: Alcohol is unbranched; product is ketone (IR shows sharp C=O ~1715 cm⁻¹, no O-H).
• D: Product is ester (requires acid + alcohol, not oxidation; IR shows C=O ~1735 cm⁻¹, C-O ~1100 cm⁻¹).

Final answer: C