For alkene and alkyne reactions, pay close attention to the reagents to determine the mechanism and regiochemistry:
- H\(_2\)O/H\(^+\) on alkene: Markovnikov hydration.
- H\(_2\)O/Hg\(^{2+}\) on alkyne: Markovnikov hydration (gives ketone).
- Dilute cold KMnO\(_4\) on alkene: Syn-dihydroxylation.
- Kolbe's electrolysis of RCOONa gives R-R.
Let's analyze each reaction in List-1 and identify its major product from List-2. A. Hydration of Propyne: \( \text{CH}_3\text{-C}\equiv\text{CH} \) reacts with water in the presence of Hg\(^{2+}\)/H\(^+\). This is the Kucherov reaction. It follows Markovnikov's rule, adding -OH to the more substituted carbon, forming an unstable enol which tautomerizes to a ketone.
\( \text{CH}_3\text{C(OH)=CH}_2 \rightarrow \text{CH}_3\text{C(=O)CH}_3 \) (Propanone).
So, A matches with II. B. Kolbe's Electrolysis: Electrolysis of an aqueous solution of sodium acetate (\(\text{CH}_3\text{COONa}\)). The acetate ion loses CO\(_2\) and the resulting methyl radicals combine to form ethane.
At anode: \( 2\text{CH}_3\text{COO}^- \rightarrow 2\text{CH}_3\cdot + 2\text{CO}_2 + 2e^- \), then \( \text{CH}_3\cdot + \cdot\text{CH}_3 \rightarrow \text{CH}_3\text{-CH}_3 \) (Ethane).
So, B matches with IV. C. Hydration of Propene: \( \text{CH}_3\text{-CH=CH}_2 \) reacts with water in the presence of an acid catalyst (H\(^+\)). This is acid-catalyzed hydration of an alkene. It follows Markovnikov's rule, where the H\(^+\) adds to the carbon with more hydrogens (C1), and the -OH group adds to the more substituted carbon (C2).
The product is \( \text{CH}_3\text{-CH(OH)-CH}_3 \) (Propan-2-ol).
So, C matches with I. D. Dihydroxylation of Propene: \( \text{CH}_3\text{-CH=CH}_2 \) reacts with dilute, cold, alkaline KMnO\(_4\) (Baeyer's reagent). This is a syn-dihydroxylation reaction, where two -OH groups are added across the double bond.
The product is \( \text{CH}_3\text{-CH(OH)-CH}_2\text{(OH)} \) (Propane-1,2-diol).
So, D matches with III.
The correct set of matches is: A-II, B-IV, C-I, D-III.
