The major product of the following reaction is: Ethylbenzene + (i) Alkaline KMnO4 (ii) H3O+ - ?

Q: Why doesn't the benzene ring get oxidized by $KMnO_4$?

The benzene ring is exceptionally stable due to its aromaticity (delocalized pi electrons). $KMnO_4$ is strong, but not strong enough to break the aromatic stability under these conditions.

Q: What happens to the extra carbon atom from the ethyl group?

The terminal methyl group ($-CH_3$) is cleaved off during the vigorous oxidation process and is ultimately oxidized to carbon dioxide ($CO_2$) and water.

Q: Would isopropylbenzene give the same product?

Yes! Isopropylbenzene (cumene) has one benzylic hydrogen. Therefore, the entire isopropyl group will be oxidized down to a single $-COOH$ group, yielding benzoic acid.

Q: What if the reactant was tert-butylbenzene?

Tert-butylbenzene has no benzylic hydrogens (the carbon attached to the ring is bonded to three other carbons). Because it lacks this crucial hydrogen, it resists oxidation by $KMnO_4$, and no reaction occurs.

Q: Why is the second step ($H_3O^+$) necessary?

The first step uses *alkaline* $KMnO_4$. In a basic medium, the carboxylic acid formed immediately loses a proton to become a water-soluble carboxylate salt (like potassium benzoate). The acidic hydrolysis ($H_3O^+$) provides the proton back to form the neutral, insoluble benzoic acid.

Question

The major product of the following reaction is

Sigma Percentile · Accepted Answer

The oxidation of alkylbenzenes is one of the most powerful and visually satisfying reactions in organic chemistry. It demonstrates how a seemingly complex side chain can be completely dismantled by a strong oxidizing agent, leaving behind a stable aromatic core. Let's dive into the mechanics of this fascinating transformation!

### The Setup: Meet the Reactants

In this problem, our starting material is **ethylbenzene** ($C_6H_5CH_2CH_3$). It consists of a highly stable benzene ring attached to a two-carbon ethyl chain.

We are treating this molecule with **alkaline potassium permanganate** ($KMnO_4$), followed by an acidic workup using hydronium ions ($H_3O^+$). Potassium permanganate is a notoriously strong oxidizing agent, famous for its deep purple color and its ability to ruthlessly add oxygen to organic molecules.

### The Golden Rule: The Benzylic Hydrogen

When alkaline $KMnO_4$ encounters an alkylbenzene, it doesn't just attack randomly. It looks for a specific vulnerability: the **benzylic carbon**. This is the carbon atom directly attached to the benzene ring.

For oxidation to occur, this benzylic carbon *must* have at least one hydrogen atom attached to it (a benzylic hydrogen). If we look at ethylbenzene, the benzylic carbon is the $-CH_2-$ group. It has two hydrogens! This means the molecule is primed and ready for oxidation.

### The Mechanism: Cleavage and Protonation

Because the benzylic hydrogens are present, $KMnO_4$ initiates a vigorous oxidation process. The exact mechanism involves complex radical and anion intermediates, but the macroscopic result is beautifully simple: **the entire alkyl chain is cleaved**.

It doesn't matter if the chain has two carbons (like our ethyl group), three carbons, or ten carbons. The strong oxidizing agent will chew through the carbon-carbon bonds until only the benzylic carbon remains attached to the ring.

In the alkaline medium, this remaining carbon is fully oxidized to a carboxylate ion ($-COO^-$).

$$C_6H_5CH_2CH_3 \xrightarrow{KMnO_4, \ OH^-} C_6H_5COO^-$$

To get our final, neutral product, we perform an acidic workup. The $H_3O^+$ provides a proton ($H^+$) to the carboxylate ion, converting it into a stable carboxylic acid group ($-COOH$).

$$C_6H_5COO^- + H_3O^+ \rightarrow C_6H_5COOH + H_2O$$

### The Final Verdict

The major product of this reaction is **benzoic acid** ($C_6H_5COOH$).

This reaction is a fantastic synthetic tool. It reminds us that the aromatic ring is incredibly stable and acts as an anchor, while the aliphatic side chain is vulnerable to severe oxidation—provided it has that crucial benzylic hydrogen. If we had used tert-butylbenzene, which lacks a benzylic hydrogen, the $KMnO_4$ would have simply stared at it, and no reaction would have occurred!

Animated Solution for Chemistry - Organic Chemistry: The major product of the following reaction is

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Visualized Solution

$R e a c t i o n S e t u p$

$R o l e o f K M n O_{4}$

$T h e B e n z y l i c H y d r o g e n C o n d i t i o n$

$O x i d a t i o n in A l k a l in e M e d i u m$

$A c i d i c H y d r o l y s i s$

$F ina l P r o d u c t$

$E x ce pt i o n : t er t - B u t y l b e n z e n e$

The Sigma Insight: Carbonyl Compounds

The Setup

The Golden Rule

The Mechanism

The Final Verdict

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