Pent-$2$-ene is treated with hot, acidic KMnO$_4$. What are the products?

**Reagent behaviour.** *Cold, dilute* KMnO₄ (Baeyer's reagent) gives the diol — that's option D. *Hot, acidic* KMnO₄ is much more aggressive: it oxidatively cleaves the double bond, giving carboxylic acids (or CO₂ if the carbon was terminal). **Apply to pent-2-ene.** $\mathrm{CH_3-CH=CH-CH_2-CH_3}$. Cleaving the C2=C3 bond, each side becomes its own acid: - Left ($\mathrm{CH_3-CH}=$) → acetic acid $\mathrm{CH_3COOH}$. - Right ($=\mathrm{CH-CH_2-CH_3}$) → propanoic acid $\mathrm{CH_3CH_2COOH}$. **Why option D is tempting.** It's the *cold* dilute KMnO₄ answer (syn-diol). Read the reagent conditions carefully — temperature and acidity decide which product surfaces.