Animated Solution for Chemistry - Hydrocarbons: For above chemical reactions, identify the correct statement from the following

When studying organic chemistry, you quickly realize that reagents are like tools in a toolbox. Some are delicate scalpels, while others are heavy sledgehammers. Potassium permanganate ($KMn{O}_4$) is a fascinating reagent because it can be both, depending entirely on the conditions under which it is used.

In this problem, we are presented with a single starting material: cyclohexene. It is a simple six-membered ring containing one carbon-carbon double bond. The question asks us to predict the major products when this alkene is subjected to two very different sets of conditions using the same primary reagent, $KMn{O}_4$.

The Sledgehammer

Let's look at the first reaction pathway to form Product 'A'. The reagents are $KMn{O}_4$ in an acidic medium ($H_{2}S{O}_4$), accompanied by heat ($\Delta$). This combination creates an incredibly harsh and vigorous oxidizing environment.

Under these conditions, $KMn{O}_4$ acts like a sledgehammer. It doesn't just add atoms to the double bond; it completely shatters it. This process is known as oxidative cleavage. Both the $\pi$ bond and the underlying $\sigma$ bond of the alkene are broken. Because cyclohexene is a cyclic molecule, breaking this bond causes the ring to pop open, forming a straight chain.

The two vinylic carbons (the ones that were part of the double bond) are oxidized to their highest possible oxidation state without breaking further carbon-carbon bonds. Since they each have one hydrogen attached initially, they are oxidized all the way to carboxylic acid groups ($-COOH$). The resulting molecule is a six-carbon chain with a carboxylic acid group at each end. This is hexanedioic acid, commonly known as adipic acid. Therefore, Compound 'A' is a dicarboxylic acid.

The Scalpel

Now, let's examine the second reaction pathway to form Product 'B'. Here, the reagents are $KMn{O}_4$ in water ($H_{2}O$) at a chilly $273K$ ($0^{\circ }C$). This cold, dilute, and neutral (or slightly alkaline) solution is famously known as Baeyer's reagent.

Without the heat and the acid, $KMn{O}_4$ behaves much more gently—like a scalpel. It is a mild oxidizing agent. It lacks the energy to break the strong carbon-carbon $\sigma$ bond. Instead, it only breaks the weaker $\pi$ bond.

As the $\pi$ bond breaks, the reagent adds two hydroxyl groups ($-OH$) across the adjacent carbons. This reaction is called syn-hydroxylation because both $-OH$ groups are added to the same face of the molecule. The carbon ring remains perfectly intact. The resulting molecule is cyclohexane-1,2-diol. Therefore, Compound 'B' is a diol.

By carefully analyzing the reaction conditions, we have deduced the nature of both products. Compound 'A' is a dicarboxylic acid resulting from oxidative cleavage, while Compound 'B' is a diol resulting from mild hydroxylation.

Matching our findings with the given options, we can confidently conclude that Compound 'A' is dicarboxylic acid and compound 'B' is diol, which corresponds to option (d). This problem beautifully illustrates how controlling temperature and pH can completely alter the course of an organic reaction!