MHT-CET Chemistry · Aldehydes, Ketones and Carboxylic Acids

Carboxylic Acids: Properties, Reactions and Derivatives

Carboxylic acids boil highest of all comparable compounds because they hydrogen-bond as dimers; they are made from Grignard reagents and CO₂ or by oxidative cleavage, converted to acid chlorides by PCl₅ or SOCl₂, and their derivatives — chlorides, anhydrides, esters — hydrolyse back to the acid.

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Why this matters

21 PYQs, none HARD. Nine are properties and preparation — the boiling-point orders (acid > alcohol > amine; formic lowest, valeric highest of the C1–C5 acids), Grignard + dry ice giving the acid with one extra carbon, cyclohexene to adipic acid; twelve are the derivatives — acid chloride or anhydride plus water, PCl₅ by its by-products, ester hydrolysis then acidification, aspirin from salicylic acid, paracetamol's amide link, diborane reducing COOH. Two cards.

Concept 1 of 2

Boiling Points of Acids, and Making Them

Intuition

Two acid molecules pair up through two hydrogen bonds into a dimer, so acids boil above alcohols of the same size, which boil above amines. Within the acids boiling point rises with molar mass. To MAKE an acid: a Grignard reagent attacks dry ice and the work-up gives an acid one carbon longer than R; hot acidic KMnO₄ cleaves a C=C, so cyclohexene opens to adipic acid; nitriles hydrolyse; aldehydes and primary alcohols oxidise.

Definition

Boiling point: carboxylic acid > alcohol > amine (ethanoic acid > propan-1-ol > propanone); formic < acetic < butyric < valeric.
$\text{RMgX} + \text{CO}_2 \to \text{RCOOMgX} \xrightarrow{\text{H}_3\text{O}^+} \text{RCOOH}$ : CH₃MgBr → ethanoic acid; C₂H₅MgBr → propanoic acid; $(\text{CH}_3)_2\text{CHMgBr}$ → 2-methylpropanoic acid.
Cyclohexene $\xrightarrow{\text{KMnO}_4/\text{H}_2\text{SO}_4}$ adipic acid (the ring opens to HOOC(CH₂)₄COOH); hex-3-ene would give two propanoic acids.
R–CN + H₃O⁺, Δ → RCOOH; RCHO + [O] → RCOOH; toluene + KMnO₄ → benzoic acid.

Grignard to acid

\text{R-MgX} \xrightarrow{\text{(i) CO}_2\text{ (dry ice); (ii) H}_3\text{O}^+} \text{R-COOH} \quad (\text{one carbon more than R})

Worked example

Which Grignard reagent gives butanoic acid with dry ice, and what does hex-3-ene give with hot acidic KMnO₄?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Aldehydes, Ketones and Carboxylic AcidsMODERATE

Which of the following alkenes on oxidation by KMnO

_4

in dil. H

_2

_4

forms adipic acid?

[Q87 · May Shift 1 · 2021]

Forgetting the carbon that CO₂ brings

Ethyl-MgBr does not give ethanoic acid. The CO₂ carbon becomes the COOH: ethyl → propanoic acid, isopropyl → 2-methylpropanoic acid.

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Concept 2 of 2

Acid Chlorides, Anhydrides, Esters and Amides

Intuition

Every derivative is the acid with OH replaced — by Cl, by another acyl-O, by OR, by NH₂ — and water puts the OH back. PCl₅ makes the chloride and announces itself by its by-products POCl₃ + HCl (SOCl₂ gives SO₂ + HCl). Ester hydrolysis in base gives the acid's SALT; acid then frees the acid. Acetic anhydride acetylates salicylic acid to aspirin; paracetamol is the amide on the list.

Definition

$\text{RCOOH} + \text{PCl}_5 \to \text{RCOCl} + \text{POCl}_3 + \text{HCl}$ (by-products name the reagent); $+ \text{SOCl}_2 \to \text{RCOCl} + \text{SO}_2 + \text{HCl}$ .
Hydrolysis: $\text{RCOCl} + \text{H}_2\text{O} \to \text{RCOOH}$ (ethanoyl chloride → ethanoic acid; benzoyl chloride → benzoic acid); $(\text{CH}_3\text{CO})_2\text{O} + \text{H}_2\text{O} \to 2\text{CH}_3\text{COOH}$ . Ethanoic acid + water alone: no change.
Ester: methyl propanoate $\xrightarrow{\text{dil. NaOH}}$ sodium propanoate (A) $\xrightarrow{\text{conc. HCl}}$ propanoic acid $\text{C}_2\text{H}_5\text{COOH}$ (B).
Salicylic acid + acetic anhydride (H⁺) → aspirin + acetic acid (ester link). Paracetamol carries the amide link; aspirin, methyl salicylate are esters; curcumin has neither.
Diborane $\text{B}_2\text{H}_6$ reduces COOH (to CH₂OH) but not esters, nitro groups or halides.

Derivative ↔ acid

\text{RCOOH} \xrightarrow{\text{PCl}_5} \text{RCOCl} \xrightarrow{\text{H}_2\text{O}} \text{RCOOH};\qquad \text{RCOOR'} \xrightarrow{\text{NaOH}} \text{RCOONa} \xrightarrow{\text{HCl}} \text{RCOOH}

Worked example

Identify A and B: ethyl ethanoate → (dil. NaOH) A → (conc. HCl) B, and name the reagent whose by-products with benzoic acid are SO₂ and HCl.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Aldehydes, Ketones and Carboxylic AcidsEASY

Identify the reagent used in the following reaction. Benzoic acid

\xrightarrow{\text{Reagent},\,\Delta}

Benzoyl chloride + Phosphorous oxychloride + Hydrogen chloride

[Q52 · 14th May Shift 1 · 2024]

Stopping at the sodium salt

Base hydrolysis of an ester gives the carboxylate; the question then adds concentrated HCl, which protonates it. The answer is the ACID, not sodium propanoate — offered as option (a).

Drill 11 more on acid chlorides, anhydrides, esters and amides

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Boiling Points of Acids, and Making Them
Grignard to acid
$\text{R-MgX} \xrightarrow{\text{(i) CO}_2\text{ (dry ice); (ii) H}_3\text{O}^+} \text{R-COOH} \quad (\text{one carbon more than R})$
Acid Chlorides, Anhydrides, Esters and Amides
Derivative ↔ acid
$\text{RCOOH} \xrightarrow{\text{PCl}_5} \text{RCOCl} \xrightarrow{\text{H}_2\text{O}} \text{RCOOH};\qquad \text{RCOOR'} \xrightarrow{\text{NaOH}} \text{RCOONa} \xrightarrow{\text{HCl}} \text{RCOOH}$

Watch out for (2)

Forgetting the carbon that CO₂ brings
→ Boiling Points of Acids, and Making Them
Stopping at the sodium salt
→ Acid Chlorides, Anhydrides, Esters and Amides

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