The correct reaction/reaction sequence that would produce a dicarboxylic acid as the major product is

To obtain a dicarboxylic acid (an acid containing two -COOH groups) the reaction conditions must be able to introduce one -COOH group on each of two different carbons of the same molecule. The most common laboratory method for doing this in a single step is oxidative cleavage of a carbon-carbon double bond with a strong oxidising agent such as hot, acidic $$KMnO_4$$ (or, equivalently, ozonolysis followed by oxidative work-up). Each carbon of the former $$C=C$$ bond is oxidised up to the carboxylate level, so both ends of the cleaved bond become -COOH groups.

What happens in Option C
  • The substrate given in Option C contains an internal carbon-carbon double bond.
  • The reagent specified is hot, acidic $$KMnO_4$$ (or an equivalent strong oxidising system).
  • Hot $$KMnO_4$$ performs oxidative cleavage: the $$C=C$$ bond is cut and each of the two carbon atoms is converted into a carbonyl that is then further oxidised to a carboxylic acid.
  • Because both carbons of the former double bond stay inside the same molecular framework after the cleavage, the product contains two -COOH groups that remain connected through the rest of the carbon skeleton. Hence a single molecule of a dicarboxylic acid is obtained as the major product.

Why the other options fail
Option A, Option B and Option D either involve oxidation of **only one functional carbon**, or they involve a reaction (such as haloform cleavage or nitrile hydrolysis) that removes part of the carbon skeleton. In every one of those sequences the molecule finally possesses just one -COOH group; the second carbon is either lost from the molecule or remains at a lower oxidation level. Consequently they cannot supply a dicarboxylic acid as the principal product.

Therefore, the only sequence that reliably converts two different carbons of the same molecule into carboxyl groups is the oxidative-cleavage sequence of Option C.

Option C which is: the oxidative cleavage of the double bond with hot, acidic $$KMnO_4$$ (or its equivalent) is the correct choice.