Q.Show how each of the following compounds can be converted to benzoic acid.

The key idea is oxidation of the side chain to a carboxyl group, regardless of its length or complexity, provided the benzylic carbon has at least one hydrogen.

Step 1: Ethylbenzene — Vigorous oxidation (e.g., alkaline $\ce{KMnO4}$ or $\ce{K2Cr2O7/H+}$) cleaves the ethyl group at the benzylic carbon, converting it directly to benzoic acid.

Step 2: Acetophenone — The carbonyl is already at the benzylic position. Haloform reaction (with $\ce{NaOX}$) or strong oxidation cleaves the $\ce{-COCH3}$ group to $\ce{-COOH}$.

Step 3: Bromobenzene — This requires a different approach: first convert to a Grignard reagent ($\ce{Mg/ether}$), then react with dry ice ($\ce{CO2}$) followed by acid hydrolysis. …

The Concept: Oxidation of Side Chains on Benzene

Benzoic acid is benzene with a carboxylic acid group ($-\text{COOH}$) attached. The problem asks how to get there from four different starting compounds. The core principle is that any carbon side chain directly attached to the benzene ring can be oxidised to a carboxyl group, provided the carbon attached to the ring has at least one hydrogen atom (i.e., it is not a tertiary carbon). The benzene ring itself is resistant to oxidation under normal conditions — it’s the side chain that gets chopped down.

For bromobenzene, oxidation won’t work because there’s no carbon side chain to oxidise. Instead, we replace the bromine with a group that can be turned into $-\text{COOH}$.

Let’s go through each one.

1. Ethylbenzene to Benzoic Acid

Ethylbenzene has a $- \text{CH}_2\text{CH}_3$ group. The carbon directly attached to the ring (the benzylic carbon) has two hydrogens. Strong oxidising agents like alkaline potassium permanganate ($\text{KMnO}_4$) or acidic potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4$) will oxidise the entire side chain to $-\text{COOH}$, regardless of its length. The reaction is:

The mechanism involves initial formation of a benzylic radical or carbocation, then stepwise oxidation through an alcohol and aldehyde to the acid. The key point: the entire ethyl group is converted to a single carboxyl carbon. The extra carbon is lost as $\text{CO}_2$ or carbonate.

2. Acetophenone to Benzoic Acid

Acetophenone is $\text{C}_6\text{H}_5\text{COCH}_3$. The carbon attached to the ring is already partially oxidised (it’s a carbonyl carbon). But the methyl group ($-\text{CH}_3$) still has hydrogens. Strong oxidising agents will cleave the bond between the carbonyl carbon and the methyl group, oxidising the methyl to $\text{CO}_2$ and leaving the ring carbon as $-\text{COOH}$.

The reaction is:

Notice that the cleaved methyl carbon is fully oxidised — it leaves as carbonate ($\text{K}_2\text{CO}_3$, i.e. $\text{CO}_2$ on acidification). The benzene ring remains intact.

3. Bromobenzene to Benzoic Acid

Bromobenzene has no carbon side chain — just a bromine atom. Oxidation won’t give you a carboxyl group. Instead, we need to introduce a carboxyl group by replacing the bromine.

The classic method is the Grignard reaction followed by carbonation:

1. Form the Grignard reagent:

2. React with carbon dioxide ($\text{CO}_2$):

3. Hydrolyse with acid:

Alternatively, you can do a nucleophilic aromatic substitution with $\text{CuCN}$ (the Rosenmund–von Braun reaction) to get benzonitrile, then hydrolyse:

Both routes work. The Grignard route is more common in introductory organic chemistry. …

Method: Side-Chain Oxidation (or Oxidation of Alkyl/Acyl/Haloarenes)

This method uses strong oxidizing agents (like alkaline $\ce{KMnO4}$ or acidic $\ce{K2Cr2O7}$) to convert an alkyl/alkenyl/acyl group attached to a benzene ring into a carboxylic acid group ($-\ce{COOH}$). The benzene ring itself remains intact.

General Principle

Any carbon side chain directly attached to the benzene ring, regardless of length or unsaturation, is oxidized to a single benzoic acid group — provided the $\alpha$-carbon (the carbon attached to the ring) has at least one hydrogen atom.

Step-by-Step for Each Compound

(i) Ethylbenzene → Benzoic acid

Reaction:

$\ce{C6H5-CH2CH3 ->[KMnO4/OH-, \Delta] C6H5-COOH}$

Steps:

1. Take ethylbenzene in an alkaline solution of $\ce{KMnO4}$.
2. Heat under reflux (strong oxidation).
3. The ethyl side chain ($-\ce{CH2CH3}$) is completely oxidized to $-\ce{COOH}$.
4. Acidify the reaction mixture (e.g., with dilute $\ce{HCl}$) to get benzoic acid as a precipitate.

Key point: The entire two-carbon chain is converted to one carboxyl group — no carbon is left.

(ii) Acetophenone → Benzoic acid

Reaction:

$\ce{C6H5-COCH3 ->[KMnO4/OH-, \Delta] C6H5-COOH}$

Steps:

1. Treat acetophenone with alkaline $\ce{KMnO4}$ and heat.
2. The acyl group ($-\ce{COCH3}$) is oxidized — the carbonyl carbon becomes part of the carboxyl group, and the methyl group is oxidized away.
3. Acidify to obtain benzoic acid.

Key point: Even though there is a carbonyl group, the $\alpha$-carbon (methyl) has hydrogens, so oxidation proceeds.

(iii) Bromobenzene → Benzoic acid

Reaction:

$\ce{C6H5-Br ->[1. Mg/ether][2. CO2][3. H3O+] C6H5-COOH}$

Steps:

1. Grignard reaction: React bromobenzene with magnesium in dry ether to form phenylmagnesium bromide ($\ce{C6H5MgBr}$).
2. Carboxylation: Pass dry $\ce{CO2}$ gas through the Grignard reagent — forms benzoic acid magnesium salt.
3. Hydrolysis: Add dilute acid ($\ce{H3O+}$) to liberate benzoic acid. …

Here are the common mistakes students make when converting these compounds to benzoic acid, along with how to avoid each.

(i) Ethylbenzene → Benzoic acid

Common mistake: Using mild/cold conditions and stopping at an intermediate (benzaldehyde or acetophenone), or forgetting the acidification step after the oxidation — the printed reagent is hot alkaline $KMnO_4$, then $H_3O^+$.

Why it happens: Students forget that the side chain must be fully oxidised to a carboxylic acid group. Ethylbenzene has a two-carbon side chain; partial oxidation gives a ketone or aldehyde, not the acid.

How to avoid:

• Use strong oxidising conditions: alkaline $KMnO_4$ with heat, or $K_2Cr_2O_7/H_2SO_4$ with heat.
• Remember: any alkyl side chain attached to benzene, regardless of length, is oxidised to $-COOH$ under vigorous conditions.
• Write the full reaction:

(ii) Acetophenone → Benzoic acid

Common mistake: Using a reducing agent (like $NaBH_4$ or $LiAlH_4$) instead of an oxidising agent.

Why it happens: Students confuse the carbonyl group ($C=O$) with a functional group that needs reduction. Acetophenone already has a carbonyl; to get benzoic acid, you must cleave the C–C bond between the carbonyl and the methyl group.

How to avoid:

• Use the printed route: hot alkaline $KMnO_4$, then acidification — the C–C bond to the methyl is cleaved and the methyl carbon leaves as carbonate.
• Write:

• (The haloform reaction — $I_2/NaOH$, then $H^+$ — is a valid alternative that cleaves the same bond via iodoform.)
• Never use $NaBH_4$ or $LiAlH_4$ — those reduce the ketone to an alcohol, not oxidise it.

(iii) Bromobenzene → Benzoic acid

Common mistake: Attempting direct oxidation of bromobenzene (which fails because the C–Br bond is not easily oxidised).

Why it happens: Students think “oxidation” means directly attacking the benzene ring. But bromobenzene is resistant to oxidation; you must first replace Br with a functional group that can be oxidised.

How to avoid:

• First, convert bromobenzene to benzonitrile via nucleophilic aromatic substitution (using $CuCN$ at high temperature), then hydrolyse.
• Or, convert to Grignard reagent ($C_6H_5MgBr$) and react with $CO_2$, then acidify.
• Write the Grignard route:

• Key point: You cannot oxidise bromobenzene directly — you must introduce a carbon atom (via $CO_2$ or $CN^-$) that becomes the carboxyl carbon.

(iv) Phenylethene (Styrene) → Benzoic acid

Common mistake: Using mild oxidation (e.g., $KMnO_4$ cold, neutral) which gives a diol, or using ozonolysis which gives benzaldehyde. …