Step 1: Understanding the Question:
The problem asks to determine which starting alkene yields a six-carbon dicarboxylic acid (adipic acid) upon vigorous oxidative cleavage with hot, acidified potassium permanganate ($\text{KMnO}_4$).
Step 2: Key Formula or Approach:
Vigorous oxidation of an alkene using acidified $\text{KMnO}_4$ completely cleaves the carbon-carbon double bond (C=C).
Open-chain alkenes break apart into separate fragment molecules (ketones or carboxylic acids).
Cyclic alkenes undergo ring-opening oxidative cleavage at the double bond site to yield a single straight-chain dicarboxylic acid molecule containing the same total number of carbon atoms.
Step 3: Detailed Explanation:
Let's consider the structure of
cyclohexene. It is a six-carbon cyclic alkene.
When cyclohexene is treated with $\text{KMnO}_4$ in dilute $\text{H}_2\text{SO}_4$, the double bond between carbon-1 and carbon-2 is entirely broken. Both alkene carbons are completely oxidized into carboxylic acid functional groups ($-\text{COOH}$):
$$\text{Cyclohexene} \xrightarrow{\text{KMnO}_4 / \text{dil. }\text{H}_2\text{SO}_4} \text{HOOC-CH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-COOH}$$
The resulting molecule has six carbons arranged linearly with a carboxylic acid group at each terminal end. This compound is systematically named hexane-1,6-dioic acid, commonly known as
adipic acid.
Open-chain isomers like hex-1-ene, hex-2-ene, or hex-3-ene would yield smaller fragmented short-chain acids (e.g., pentanoic acid, butanoic acid, or propanoic acid) instead of a single continuous dicarboxylic chain.
Step 4: Final Answer:
The alkene that oxidizes to form adipic acid is cyclohexene, matching option (D).