See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Cold, dilute, alkaline KMnO4 (Baeyer's reagent) is a mild oxidizing agent that reacts with alkenes via syn-dihydroxylation, adding two OH groups to the double bond carbons in a cis (syn) fashion without breaking the carbon-carbon bond. Step 1: Identify the substrate. 2-Butene is CH3-CH=CH-CH3 (the double bond is between C2 and C3). Step 2: Apply the reaction conditions. Cold alkaline KMnO4 performs syn-dihydroxylation across the C2=C3 double bond, adding an -OH group to each of the two carbons of the double bond. Step 3: Determine the product. Adding -OH to C2 and -OH to C3 of 2-butene gives CH3-CH(OH)-CH(OH)-CH3, which is 2,3-butanediol (a vicinal diol / glycol). This corresponds to option (d). Why other options fail: - (a) Butanal (an aldehyde) would result from ozonolysis (O3 followed by reductive workup) or from oxidative cleavage conditions, not cold alkaline KMnO4. - (b) Butan-2-one (a ketone) would also result from oxidative cleavage or Wacker oxidation, not from cold alkaline KMnO4 dihydroxylation. - (c) An epoxide is formed by reaction with a peracid (e.g., mCPBA), not with cold alkaline KMnO4. Cold alkaline KMnO4 is specifically known as Baeyer's reagent and its hallmark reaction is the formation of vicinal diols from alkenes. Therefore, the correct answer is D.