Animated Solution for Chemistry - Organic Compounds Containing Oxygen: Considering the above chemical reaction, identify the product 'X'.

The oxidation of alkylbenzenes is one of the most elegant and high-yield transformations in organic chemistry. In this problem, we are tasked with predicting the product of a specific oxidation reaction. Let's break down the chemistry step-by-step and understand exactly what happens when our starting material meets a powerful oxidizing agent.

Our journey begins with the reactant, $p$-methoxytoluene. If we look closely at its structure, we see a central benzene ring adorned with two distinct substituents: 1. A methyl group ($-{\text{CH}}_3$) at the top position. 2. A methoxy group ($-{\text{OCH}}_3$) at the bottom, para to the methyl group.

Understanding the nature of these two groups is crucial because they will react very differently when exposed to our reagent.

The reagent provided is alkaline potassium permanganate (${\text{KMnO}}_4/{\text{OH}}^{-}$), followed by an acidic workup (${\text{H}}^{+}$). Potassium permanganate is a classic, heavy-duty oxidizing agent.

When alkaline ${\text{KMnO}}_4$ encounters an alkyl group attached directly to a benzene ring, it acts ruthlessly. As long as the carbon atom directly attached to the ring (the benzylic carbon) has at least one hydrogen atom, the entire alkyl chain is cleaved and oxidized down to a carboxylic acid group ($-\text{COOH}$).

In our reactant, the methyl group ($-{\text{CH}}_3$) has three benzylic hydrogens. Therefore, it is a perfect target for this oxidation and will be transformed into a $-\text{COOH}$ group.

But what about the methoxy group ($-{\text{OCH}}_3$)? This is an ether linkage. Ethers are generally quite robust and chemically inert under many conditions. Specifically, they are highly resistant to oxidation by strong agents like ${\text{KMnO}}_4$.

Because the methoxy group lacks any benzylic hydrogens and is inherently stable, it will simply watch the reaction happen without participating. It remains completely unaffected throughout the process.

Putting these two pieces of logic together, we can easily predict the final product: - The $-{\text{CH}}_3$ group is oxidized to $-\text{COOH}$. - The $-{\text{OCH}}_3$ group remains unchanged.

The resulting molecule has a carboxyl group and a methoxy group para to each other on a benzene ring. This compound is $p$-methoxybenzoic acid.

Comparing our deduced structure with the given options, we find a perfect match with option (c). This problem beautifully illustrates the chemoselectivity of alkaline ${\text{KMnO}}_4$, attacking the benzylic position while leaving the stable ether linkage intact!