Preparation of Carboxylic Acids

Ask a synthesis question in the "Aldehydes, Ketones and Carboxylic Acids" unit and it usually ends the same way: build a –COOH\text{–COOH} group onto some starting material. There is no single reaction for this — five genuinely different routes converge on the same functional group, and CBSE Class 12 expects you to write balanced equations for each one and to know which route to reach for. Oxidise a primary alcohol or an aldehyde far enough and you land on a carboxylic acid. Hydrolyse a nitrile or an amide and you also land on one. Carbonate a Grignard reagent and you land on one too — but with an extra carbon. Oxidise the side chain of an alkylbenzene and, no matter how long that chain was, you land on benzoic acid. This article organises the five routes around one idea worth internalising: carboxylic acid preparation is really a story about oxidation state and carbon-count bookkeeping, not five unrelated reactions to memorise separately.

The core insight

Carboxylic acids, R–COOH\text{R–COOH}, form by pushing an alcohol/aldehyde carbon to its most oxidised state, by hydrolysing a nitrile or amide back to that same oxidation state, by adding one new carbon via Grignard-plus-CO2\text{CO}_2 carbonation, or by oxidatively stripping any alkylbenzene side chain down to a ring-bound –COOH\text{–COOH}.

The core idea

Five routes, one destination. Group them by what changes: the first three keep the carbon skeleton fixed and change only the oxidation state at one carbon; the fourth adds a carbon; the fifth destroys everything except the ring carbon the group sits on.

1. Oxidation of primary alcohols and aldehydes. Acidic or alkaline KMnO4\text{KMnO}_4, or acidic K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7, oxidises a primary alcohol all the way to the acid, passing through the aldehyde on the way; the same reagents take an aldehyde straight to the acid.

R–CH2OH→ΔKMnO4/H+R–CHO→[O]R–COOH\text{R–CH}_2\text{OH} \xrightarrow[\Delta]{\text{KMnO}_4/\text{H}^+} \text{R–CHO} \xrightarrow{[\text{O}]} \text{R–COOH}

where: R\text{R} is the carbon skeleton attached to the oxidised carbon, held fixed throughout; [O][\text{O}] denotes the oxidising agent supplying the second oxygen. A mild oxidant such as PCC stops at the aldehyde stage — it cannot make this second jump.

2. Hydrolysis of nitriles and amides. A nitrile, R–CN\text{R–CN}, hydrolysed under acidic or basic aqueous conditions with heat, passes through the amide, R–CONH2\text{R–CONH}_2, and lands on the acid (or its salt, under base). The nitrogen leaves as NH4+\text{NH}_4^+ or NH3\text{NH}_3; the carbon count of R\text{R} never changes.

3. Carbonation of Grignard reagents. A Grignard reagent, R–MgX\text{R–MgX}, attacks the electrophilic carbon of CO2\text{CO}_2 (usually solid CO2\text{CO}_2, to keep the reaction anhydrous), giving a magnesium carboxylate that is acidified to the free acid. This is the one route on the list that adds a carbon — the acid has one more carbon than the halide it started from.

R–MgX+CO2⟶R–COOMgX→H3O+R–COOH\text{R–MgX} + \text{CO}_2 \longrightarrow \text{R–COOMgX} \xrightarrow{\text{H}_3\text{O}^+} \text{R–COOH}

where: X\text{X} is the halogen of the original alkyl halide; the new –COOH\text{–COOH} carbon comes entirely from CO2\text{CO}_2, so if R\text{R} has nn carbons the product has n+1n+1.

4. Side-chain oxidation of alkylbenzenes. Hot acidic or alkaline KMnO4\text{KMnO}_4 attacks the benzylic C–H of any monoalkylbenzene and oxidises the whole side chain away, leaving only –COOH\text{–COOH} bonded to the ring — regardless of how long or branched that chain was.

C6H5CH2R′→KMnO4/H+, ΔC6H5COOH\text{C}_6\text{H}_5\text{CH}_2\text{R}' \xrightarrow{\text{KMnO}_4/\text{H}^+,\ \Delta} \text{C}_6\text{H}_5\text{COOH}

A side chain with no benzylic hydrogen at all — such as the tert-butyl group in C6H5C(CH3)3\text{C}_6\text{H}_5\text{C(CH}_3)_3 — resists this oxidation, since there is no benzylic C–H for the oxidant to remove first.

Figure
Five arrows converging on the carboxylic acid functional group from a primary alcohol, an aldehyde, a nitrile, an amide, a Grignard reagent and an alkylbenzene
Five routes converge on R-COOH R-COOH carboxylic acid R-CH2OH R-CHO R-CN R-CONH2 R-MgX + CO2 Ar-CH2R' side-chain oxidation oxidation, hydrolysis and carbonation routes (blue), all skeleton-preserving except Grignard (+1 carbon)
Derivation ACarbon-count bookkeeping for the Grignard route

Carbon counting is the one place synthesis questions on this topic most often go wrong, so it deserves a standalone rule rather than being folded into the reagent list above. Given a starting alkyl halide with a known number of carbons, predict how many carbons the resulting Grignard-carbonation acid will have, and why every other route on this page leaves that count unchanged.

Step 1. Fix the reference: count the carbons in the starting material R\text{R} before any reaction. For a route that keeps the same carbon skeleton — alcohol/aldehyde oxidation, or nitrile/amide hydrolysis — the acid's carbon count equals R\text{R}'s carbon count, because oxidation only adds oxygen and removes hydrogen; it inserts no new carbon–carbon bond anywhere in the chain.

Step 2. Isolate the one route that differs: Grignard carbonation. The Grignard carbon, R–MgX\text{R–MgX}, attacks CO2\text{CO}_2 and bonds to its carbon, so the new –COOH\text{–COOH} carbon is not part of the original R\text{R} group at all — it is imported whole from CO2\text{CO}_2. If R\text{R} contributes nn carbons, the product R–COOH\text{R–COOH} has n+1n+1 carbons.

Step 3. Apply it. Starting from 1-bromopropane (n=3n=3 carbons: CH3CH2CH2Br\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}), the Grignard route gives n+1=4n+1 = 4 carbons — butanoic acid, CH3CH2CH2COOH\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} — not propanoic acid. To reach a target acid with mm carbons by the Grignard route, the halide must be chosen with m−1m-1 carbons, one short of the target.

Step 4. Contrast with the skeleton-preserving routes as a check. Oxidising propan-1-ol (n=3n=3) with hot acidic KMnO4\text{KMnO}_4 gives propanoic acid, also 33 carbons — confirming that only the Grignard route shifts the count, and it always shifts it by exactly +1+1. Rule: skeleton-preserving routes (oxidation, hydrolysis) give carbons(product) = carbons(R); Grignard carbonation gives carbons(product) = carbons(R) + 1.

Worked example

Prepare propanoic acid from 1-bromopropane using the Grignard route, and separately from propan-1-ol by direct oxidation; compare the carbon counts.

Step 1. Form the Grignard reagent from 1-bromopropane in dry ether: CH3CH2CH2Br+Mg→dry etherCH3CH2CH2MgBr\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{Mg} \xrightarrow{\text{dry ether}} \text{CH}_3\text{CH}_2\text{CH}_2\text{MgBr}.

Step 2. Carbonate with dry CO2\text{CO}_2, then acidify: CH3CH2CH2MgBr+CO2→CH3CH2CH2COOMgBr→H3O+CH3CH2CH2COOH\text{CH}_3\text{CH}_2\text{CH}_2\text{MgBr} + \text{CO}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{COOMgBr} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}. Applying the bookkeeping rule, n=3n = 3 for the halide gives n+1=4n+1 = 4 carbons in the product — this is butanoic acid, not propanoic acid.

Step 3. Try the direct oxidation route instead: CH3CH2CH2OH→ΔKMnO4/H+CH3CH2COOH\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow[\Delta]{\text{KMnO}_4/\text{H}^+} \text{CH}_3\text{CH}_2\text{COOH}. This skeleton-preserving route keeps the carbon count at 33, giving propanoic acid directly.

Step 4. State the fix for the Grignard route: to reach a 3-carbon acid by carbonation, start one carbon shorter — from 1-bromoethane (n=2n=2), so that n+1=3n+1=3. Answer: oxidation of propan-1-ol gives propanoic acid directly (3 carbons preserved); the Grignard route from 1-bromopropane gives butanoic acid (4 carbons) because carbonation always adds one carbon.

Competitive stretch

Toluene oxidised by hot alkaline KMnO4\text{KMnO}_4, then acidified, gives potassium benzoate and then benzoic acid, C6H5COOH\text{C}_6\text{H}_5\text{COOH}. Ethylbenzene under the same conditions gives the same benzoic acid, not phenylacetic acid — the oxidant attacks the benzylic C–H first and cleaves the entire side chain down to the ring carbon, so chain length is destroyed information; this route only makes the aromatic acid, never a chain-length-specific product.

Common traps

  • Forgetting that the Grignard-plus-CO2\text{CO}_2 route always adds one carbon relative to the starting halide — the single most common numerical slip in synthesis questions on this topic.
  • Reaching for a mild oxidant like PCC to make an acid from a primary alcohol — PCC stops at the aldehyde; only strong oxidants (KMnO4\text{KMnO}_4/K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7) or a second oxidation step reach the acid.
  • Assuming side-chain oxidation of an alkylbenzene preserves chain length — it does not; any monoalkylbenzene with a benzylic hydrogen converges on the same ring-bound acid.
  • Leaving the answer as a carboxylate salt after basic hydrolysis of a nitrile or amide, without the separate acidification step needed to isolate the free acid.

Ask the tutor: want help picking the right starting halide for a target acid via the Grignard route, or checking a multi-step synthesis carbon count? Ask the AI tutor to walk through your plan.

Why does the Grignard route add a carbon while alcohol oxidation does not?

Oxidation of an alcohol or aldehyde only changes what is attached to an existing carbon — it removes hydrogens and adds an oxygen at the same carbon that was already there, so the skeleton is untouched. The Grignard route is different in kind: the organometallic carbon forms a brand-new carbon–carbon bond to the carbon of CO2\text{CO}_2, so that carbon is imported from the reagent, not oxidised in place. That is why only the carbonation route shifts the carbon count, always by exactly one.

Can I use the alkylbenzene oxidation route to make a specific chain-length acid, like phenylacetic acid from a two-carbon side chain?

No. Hot strong oxidants attack the benzylic C–H first and keep oxidising until the entire side chain is gone except the ring-bound carbon, so ethylbenzene, propylbenzene and toluene itself all converge on the same benzoic acid. This route is only useful for making the aromatic acid itself; if you need to preserve a specific side-chain length, you need a different, milder strategy that this syllabus does not require you to name.

Both acidic and basic hydrolysis of a nitrile give a carboxylic acid — is there any real difference?

Yes, in the form of the product you isolate directly. Acidic hydrolysis (aqueous H3O+\text{H}_3\text{O}^+, heat) delivers the free acid straight away. Basic hydrolysis delivers the carboxylate salt first, because the base deprotonates the acid as fast as it forms; you then need a separate acidification step to liberate the free R–COOH\text{R–COOH}. Both routes pass through the amide intermediate on the way, and both need heating.

Before this

Where this leads