Q.How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.

Concept: Oxidation Reactions (side-chain oxidation, nitration, esterification)

Reasoning steps:

1. Methyl benzoate – Benzene undergoes Friedel–Crafts acylation with acetyl chloride ($\ce{CH3COCl}$, $\ce{AlCl3}$) to give acetophenone. Then haloform reaction ($\ce{I2/NaOH}$) oxidises the methyl ketone to benzoic acid. Finally, esterify with $\ce{CH3OH/H+}$.
2. m-Nitrobenzoic acid – Nitrate benzoic acid (already prepared) with $\ce{HNO3/H2SO4}$; the $-\ce{COOH}$ group is meta-directing, giving exclusively m-nitrobenzoic acid.
3. p-Nitrobenzoic acid – First nitrate benzene to nitrobenzene ($\ce{HNO3/H2SO4}$). Then oxidise the side chain? No — nitrobenzene has no alkyl group. Instead, use a different route: convert benzene to toluene (Friedel–Crafts alkylation with $\ce{CH3Cl/AlCl3}$), nitrate toluene ($\ce{HNO3/H2SO4}$) to get a mixture of o- and p-nitrotoluene, separate p-nitrotoluene, then oxidise the methyl group with $\ce{KMnO4/H+}$ to get p-nitrobenzoic acid.
4. Phenylacetic acid – Benzene + $\ce{CH3COCl/AlCl3}$ → acetophenone. Reduce the carbonyl to ethylbenzene? No — use Arndt–Eistert homologation? Simpler: prepare benzyl chloride (chloromethylation: $\ce{HCHO/HCl/ZnCl2}$), then react with $\ce{KCN}$ to give benzyl cyanide, hydrolyse ($\ce{H3O+}$) to phenylacetic acid. …

Concept and Intuition

Benzene is an aromatic ring that undergoes electrophilic substitution -- not addition -- because its pi-system is too stable to break. To attach a side chain, we use reactions like Friedel-Crafts alkylation/acylation or nitration. Once a group is attached, we can oxidize or reduce it to get the desired product. The constraint 'reagents with <=1 carbon' means we cannot use Grignard reagents with longer chains or complex organometallics; we rely on simple one-carbon units like CH3Cl, HCHO, etc.

A point that decides several of these routes: which existing group is a meta-director and which is an ortho/para-director. -COOH, -CHO and -COCH3 are all electron-withdrawing and meta-directing; -CH3 (and -NH2, after protection) are electron-donating and ortho/para-directing. Whenever the target needs a nitro group PARA to a meta-director like -COOH or -CHO, nitrating that meta-director directly will not work (it gives the meta product) -- the practical route is to build the target via a -CH3 group instead (which sends the new nitro group ortho/para), separate the desired para isomer, and only then oxidize the methyl group into the -COOH or -CHO that was actually wanted.

Step-by-Step Solutions

(i) Methyl benzoate

1. Introduce a carbonyl via Friedel-Crafts acylation. React benzene with acetyl chloride (CH3COCl) and anhydrous AlCl3 to form acetophenone (C6H5COCH3).
2. Oxidize the methyl ketone to a carboxylic acid. Treat acetophenone with I2/NaOH (haloform-type oxidative cleavage) or alkaline KMnO4 to oxidize -COCH3 to -COOH, giving benzoic acid (C6H5COOH).
3. Esterify with methanol. React benzoic acid with CH3OH in the presence of conc. H2SO4 (Fischer esterification) to form methyl benzoate (C6H5COOCH3).

(ii) m-Nitrobenzoic acid

1. First make benzoic acid (as in part i, steps 1-2).
2. Nitrate benzoic acid. Treat benzoic acid with a nitrating mixture (conc. HNO3 + conc. H2SO4). The -COOH group is meta-directing, so the nitro group goes predominantly to the meta position, giving m-nitrobenzoic acid directly.

(iii) p-Nitrobenzoic acid

Nitrating benzoic acid directly (as in part ii) gives overwhelmingly the META isomer, because -COOH is meta-directing -- there is no useful para fraction to isolate that way. To reach the PARA isomer, the nitro group must be introduced while an ortho/para-director is on the ring, and the -COOH built afterward.

1. Alkylate benzene to toluene. Benzene + CH3Cl/anhydrous AlCl3 (Friedel-Crafts alkylation) gives toluene (C6H5CH3).
2. Nitrate toluene and separate the para isomer. Toluene + HNO3/H2SO4 gives a mixture of o- and p-nitrotoluene (the -CH3 group is ortho/para-directing); separate p-nitrotoluene by fractional distillation/crystallization.
3. Oxidize the methyl group. Treat p-nitrotoluene with hot KMnO4/H+ (or alkaline KMnO4 then acidify) to oxidize -CH3 to -COOH, giving p-nitrobenzoic acid.

(iv) Phenylacetic acid

1. Introduce a -CH2Cl group via chloromethylation. React benzene with formaldehyde (HCHO) and HCl in the presence of ZnCl2 (Blanc chloromethylation) to form benzyl chloride (C6H5CH2Cl).
2. Convert to benzyl cyanide. Treat benzyl chloride with KCN (or NaCN) in aqueous ethanol to get benzyl cyanide (C6H5CH2CN). …

Here is the clear, concept-first solution for each conversion from benzene.

Core Concept: Benzene is electron-rich and undergoes Electrophilic Aromatic Substitution (EAS). To introduce a carboxylic acid ($-COOH$) or aldehyde ($-CHO$) directly, we use oxidation of a pre-attached alkyl side chain. The directing effects of substituents (activating/deactivating, ortho/para vs meta) determine the sequence of steps.

(i) Methyl benzoate from Benzene

Method: Friedel-Crafts Acylation followed by Oxidation and Esterification.

Steps:

1. Friedel-Crafts Acylation: Treat benzene with acetyl chloride ($CH_3COCl$) and anhydrous $AlCl_3$ to form acetophenone ($C_6H_5COCH_3$).
2. Oxidation: Oxidize the methyl ketone ($-COCH_3$) to a carboxylic acid ($-COOH$) using a strong oxidizing agent like $KMnO_4$ / $H^+$ (alkaline $KMnO_4$ followed by acidification). This gives benzoic acid ($C_6H_5COOH$).
3. Esterification: React benzoic acid with methanol ($CH_3OH$) in the presence of a few drops of concentrated $H_2SO_4$ (acid catalyst) to form methyl benzoate ($C_6H_5COOCH_3$).

Key Equation:

(ii) m-Nitrobenzoic acid from Benzene

Method: Nitration followed by Oxidation.

Steps:

1. Nitration: Treat benzene with a nitrating mixture (conc. $HNO_3$ + conc. $H_2SO_4$) to form nitrobenzene ($C_6H_5NO_2$). The $-NO_2$ group is a meta-directing deactivator.
2. Oxidation: This is tricky because $-NO_2$ is not an alkyl group. We cannot oxidize it directly. The correct approach is to first introduce a methyl group, then nitrate, then oxidize.
   • Friedel-Crafts Alkylation: React nitrobenzene with methyl chloride ($CH_3Cl$) and anhydrous $AlCl_3$. Because $-NO_2$ is meta-directing, the methyl group goes to the meta position, giving m-nitrotoluene.
   • Oxidation: Oxidize the methyl group ($-CH_3$) on m-nitrotoluene to a carboxylic acid ($-COOH$) using $KMnO_4$ / $H^+$ (alkaline $KMnO_4$ then acid). This yields m-nitrobenzoic acid.

Key Equation:

(iii) p-Nitrobenzoic acid from Benzene

Method: Friedel-Crafts Alkylation, then Nitration, then Oxidation.

Steps:

1. Friedel–Crafts methylation: Treat benzene with $CH_3Cl$ and anhydrous $AlCl_3$ to obtain toluene ($C_6H_5CH_3$).
2. Side-chain chlorination: Chlorinate toluene at the side chain ($Cl_2$, sunlight/hv) to obtain benzyl chloride ($C_6H_5CH_2Cl$).
3. Cyanide substitution: Treat benzyl chloride with KCN to obtain benzyl cyanide ($C_6H_5CH_2CN$) — this adds the one extra carbon.
4. Hydrolysis: Hydrolyse the nitrile ($H_3O^+$, heat) to obtain phenylacetic acid ($C_6H_5CH_2COOH$).

Key Equation:

(v) p-Nitrobenzaldehyde from Benzene

Method: Friedel-Crafts Acylation, then Nitration, then Reduction (or use a different aldehyde synthesis).

Steps:

1. Friedel-Crafts Acylation: Treat benzene with acetyl chloride ($CH_3COCl$) and anhydrous $AlCl_3$ to form acetophenone ($C_6H_5COCH_3$).
2. Nitration: Treat acetophenone with a nitrating mixture (conc. $HNO_3$ + conc. $H_2SO_4$). The $-COCH_3$ group is a meta-directing deactivator. The major product is m-nitroacetophenone, not the para isomer.
   • Correct approach: We need the $-NO_2$ group para to the aldehyde. So, we must first introduce the aldehyde group, then nitrate.
   • Aldehyde introduction: Use Gattermann-Koch formylation: Treat benzene with $CO$ + $HCl$ in the presence of $AlCl_3$ and $CuCl$ to form benzaldehyde ($C_6H_5CHO$). …

Here are the common mistakes students make when solving this specific type of benzene conversion problem, along with how to avoid each.

General Mistake: Ignoring the “One Carbon” Rule

The Mistake: Using a reagent with more than one carbon atom (e.g., using ethanol, $CH_3CH_2OH$, or acetyl chloride, $CH_3COCl$).

Why it’s wrong: The problem explicitly restricts you to organic reagents having not more than one carbon atom. This means you can only use:

• $CH_3X$ (methyl halides)
• $HCHO$ (formaldehyde)
• $HCOOH$ (formic acid)
• $CH_3OH$ (methanol)
• $CO_2$ (inorganic, but often forgotten)
• $COCl_2$ (phosgene — rarely needed here)

How to avoid: Before writing any step, check the carbon count of every organic reagent you plan to use. If it has 2 or more carbons, stop and find a one-carbon alternative.

(i) Methyl benzoate — Mistake: Direct Esterification

The Mistake: Trying to make benzoic acid first, then reacting with methanol and $H_2SO_4$.

Why it’s wrong: This is actually correct in principle, but students often forget that benzene does not directly give benzoic acid in one step. They write:

• Benzene $\rightarrow$ Benzoic acid (wrong — no direct oxidation of benzene to benzoic acid)

How to avoid: Remember the correct sequence:

1. Friedel-Crafts acylation with $CO$ + $HCl$ + $AlCl_3$ (Gattermann-Koch) to get benzaldehyde.
2. Oxidize benzaldehyde to benzoic acid (using $KMnO_4$ or $K_2Cr_2O_7$).
3. Esterify with $CH_3OH$ / $H_2SO_4$ to get methyl benzoate.

Key point: Benzene must first be converted to a mono-carbon side-chain compound (like benzaldehyde or benzoic acid) before esterification.

(ii) m-Nitrobenzoic acid — Mistake: Wrong Order of Substitution

The Mistake: Nitrating benzoic acid directly (expecting meta product) but forgetting that benzoic acid is meta-directing — so this is actually correct. The real mistake is trying to oxidize m-nitrotoluene but using the wrong oxidising agent or conditions.

Why it’s wrong: Students often write:

• Benzene $\rightarrow$ Nitrobenzene $\rightarrow$ m-Nitrobenzoic acid (impossible — nitrobenzene cannot be directly oxidised to benzoic acid)

How to avoid: Use the correct route:

1. Friedel-Crafts alkylation with $CH_3Cl$ / $AlCl_3$ to get toluene.
2. Nitrate toluene (mixed acid) to get a mixture of o- and p-nitrotoluene. Separate m-nitrotoluene? No — nitration of toluene gives mostly o- and p-, not meta. So this route fails.

Correct approach:

1. Nitrate benzene to nitrobenzene.
2. Reduce nitrobenzene to aniline ($Sn/HCl$).
3. Protect amine (acetylation with $CH_3COCl$ — but that’s 2 carbons! So use formylation? No — better: use diazotisation followed by Sandmeyer to get chlorobenzene? That’s too long.)

Simplest correct method:

• Benzene $\rightarrow$ Benzoic acid (via Gattermann-Koch + oxidation)
• Nitrate benzoic acid with mixed acid → m-Nitrobenzoic acid (since -COOH is meta-directing)

Common error: Forgetting that nitration of benzoic acid gives meta product directly.

(iii) p-Nitrobenzoic acid — Mistake: Assuming Direct Nitration of Benzoic Acid

The Mistake: Trying to nitrate benzoic acid and expecting para product.

Why it’s wrong: Benzoic acid is meta-directing. Nitration gives only m-nitrobenzoic acid, not para.

How to avoid: Use a para-directing group first:

1. Nitrate benzene to nitrobenzene.
2. Reduce to aniline.
3. Protect amine (acetylation — but again, 2-carbon problem). Instead, use diazotisation and replace with -CN (Sandmeyer) to get benzonitrile.
4. Hydrolyse benzonitrile to benzoic acid.
5. Nitrate? No — now the -COOH is meta-directing again.

Better route:

• Benzene $\rightarrow$ Toluene (Friedel-Crafts alkylation with $CH_3Cl$)
• Nitrate toluene → mixture of o- and p-nitrotoluene
• Separate p-nitrotoluene (by distillation or crystallisation)
• Oxidise the methyl group to -COOH using $KMnO_4$ → p-Nitrobenzoic acid

Key mistake: Forgetting that separation of isomers is required and that oxidation of p-nitrotoluene works.

(iv) Phenylacetic acid — Mistake: Trying to Make it from Benzene Directly

The Mistake: Writing benzene $\rightarrow$ benzyl chloride $\rightarrow$ phenylacetic acid (via $NaCN$ + hydrolysis). This is correct but students forget the one-carbon rule for the cyanide source.

Why it’s wrong: $NaCN$ is inorganic — fine. But the step to get benzyl chloride from benzene requires chloromethylation:

• Benzene + $HCHO$ + $HCl$ + $ZnCl_2$ → Benzyl chloride

Common error: Using $CH_3Cl$ + $AlCl_3$ to get toluene, then free radical chlorination to get benzyl chloride — this works but gives mixtures.

How to avoid: Use chloromethylation (Blanc reaction) — it’s a one-carbon addition:

• $C_6H_6 + HCHO + HCl \xrightarrow{ZnCl_2} C_6H_5CH_2Cl$
• Then $C_6H_5CH_2Cl + NaCN \rightarrow C_6H_5CH_2CN$
• Hydrolysis → $C_6H_5CH_2COOH$

Key point: Remember that formaldehyde ($HCHO$) is a one-carbon reagent.

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