Propan-2-ol as a secondary alcohol

Steps:

• Propan-2-ol (CH3CH(OH)CH3) is a secondary alcohol.
• Oxidation with Cr2O72- yields propanone (CH3COCH3), a ketone with boiling point 56°C (lower than propan-2-ol's 82°C), so no higher-boiling product.
• Esterification with ethanoic acid and H2SO4 forms isopropyl ethanoate, a sweet-smelling ester.
• These reactions match option B.

Why B is correct:

• Secondary alcohols oxidize to ketones (formula R2CHOH → R2C=O), which resist further oxidation and have lower boiling points due to reduced hydrogen bonding; they also esterify via Fischer reaction to fruity esters.

Why the others are wrong:

• A: Higher-boiling oxidation product indicates primary alcohol (to carboxylic acid), but alcohols always esterify.
• C: Higher-boiling product fits primary alcohol, but no esterification contradicts alcohol reactivity.
• D: Higher-boiling product and esterification describe a primary alcohol like propan-1-ol.

Final answer: B