Animated Solution for Chemistry - Carbonyl Compounds: On vigorous oxidation by permangnate solution $(C{H}_3{)}_{2}C=CHC{H}_{2}CHO$ gives

Imagine you are a molecular surgeon, and your scalpel is potassium permanganate ($KMn{O}_4$). When we talk about "vigorous oxidation" using $KMn{O}_4$, we are not talking about a gentle nudge. We are talking about a powerful chemical transformation that aggressively attacks specific vulnerabilities in a molecule.

In our reactant, $(C{H}_3{)}_{2}C=CH-C{H}_2-CHO$, there are two primary targets for this aggressive oxidizing agent: the carbon-carbon double bond ($C=C$) and the aldehyde group ($-CHO$).

The first major event is the oxidative cleavage of the double bond. $KMn{O}_4$ acts like a pair of molecular scissors, completely snapping the $C=C$ bond in half.

Let's look at the left fragment after the snap: $(C{H}_3{)}_{2}C=$. Because this carbon atom is bonded to two methyl groups and has no hydrogen atoms attached to it, it gets capped with an oxygen atom to form a stable ketone. Specifically, it forms acetone, $(C{H}_3{)}_{2}C=O$. Ketones are tough; they resist further oxidation under these conditions, so this part of the molecule is done reacting.

Now, let's turn our attention to the right fragment: $=CH-C{H}_2-CHO$. This part is in for a double whammy.

First, the carbon that was part of the double bond ($=CH-$) gets oxidized. Because it has a hydrogen atom attached, it doesn't stop at an aldehyde; it gets fully oxidized into a carboxylic acid group ($-COOH$).

Second, the molecule already had an aldehyde group ($-CHO$) at the other end. $KMn{O}_4$ makes quick work of this as well, oxidizing it straight into another carboxylic acid group.

The $-C{H}_2-$ group in the middle is relatively unreactive and stays intact. So, the entire right fragment transforms into a dicarboxylic acid: $HOOC-C{H}_2-COOH$, which is commonly known as malonic acid.

By carefully analyzing the action of vigorous $KMn{O}_4$ on both reactive sites, we've successfully predicted the outcome. The molecule splits and oxidizes to give us two distinct products: acetone and malonic acid.

Therefore, the final products are $(C{H}_3{)}_{2}CO$ and $C{H}_2(COOH{)}_2$. This perfectly matches option (d).

Always remember: strong oxidizing agents don't just break bonds; they push every reactive carbon to its highest possible oxidation state!