Match the alkenes in List-I with their respective major products...

Vigorous oxidation of alkenes with hot acidic $\text{KMnO}_4$ leads to the cleavage of the $\text{C}=\text{C}$ double bond.

A. 2-Methylbut-2-ene ($\text{CH}_3-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3$) has one vinylic carbon with no hydrogen and one with a hydrogen. It cleaves to form a ketone (propanone) and a carboxylic acid (ethanoic acid). (A $\rightarrow$ III)

B. Pent-2-ene ($\text{CH}_3-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_3$) is an internal alkene with hydrogens on both vinylic carbons. It cleaves to form propanoic acid and ethanoic acid. (B $\rightarrow$ II)

C. 3,3-Dimethylbut-1-ene ($(\text{CH}_3)_3\text{C}-\text{CH}=\text{CH}_2$) is a terminal alkene. The terminal $=\text{CH}_2$ group is oxidized to $\text{CO}_2$ and $\text{H}_2\text{O}$, while the other part becomes 2,2-dimethylpropanoic acid. (C $\rightarrow$ IV)

D. Cyclobutene is a cyclic alkene. Oxidative cleavage opens the four-membered ring to form a dicarboxylic acid with four carbons, which is butanedioic acid (succinic acid). (D $\rightarrow$ I)

Therefore, the correct match is A-III, B-II, C-IV, D-I.

Answer: A-III, B-II, C-IV, D-I