The Carbonyl World: Aldehydes, Ketones & Carboxylic Acids
Carbonyl compounds are the backbone of organic chemistry in Class 12. Once you understand the carbonyl group (C=O), everything else — nucleophilic addition, oxidation, condensation reactions — starts making sense as variations of the same theme.
The carbonyl carbon is electron-deficient. That one insight explains 80% of the reactions in this chapter. A nucleophile (electron-rich species) attacks the carbonyl carbon, and the reaction proceeds from there. Keep this mental model, and this chapter stops feeling like a list of random reactions.
This chapter carries heavy weightage: roughly 3-5 marks in CBSE boards and 1-2 questions in JEE Main almost every year. NEET typically tests 1 question, usually on distinguishing tests or reaction products. The reactions appear complex at first, but most follow patterns we’ll build together.
Key Terms & Definitions
Aldehyde — A carbonyl compound where the C=O group is at the terminal carbon, always bonded to at least one hydrogen. General formula: RCHO. Example: ethanal (CH₃CHO), benzaldehyde (C₆H₅CHO).
Ketone — A carbonyl compound where C=O is flanked by two carbon groups (never a hydrogen). General formula: RCOR’. Example: propanone (CH₃COCH₃), acetophenone (C₆H₅COCH₃).
Carboxylic Acid — Contains the carboxyl group (–COOH), which is a carbonyl bonded to a hydroxyl. General formula: RCOOH. Example: ethanoic acid (CH₃COOH), benzoic acid (C₆H₅COOH).
Nucleophilic Addition — The hallmark reaction of aldehydes and ketones. A nucleophile adds across the C=O bond. This happens because the π bond of C=O is weaker than a σ bond and the carbon is electrophilic.
Tollen’s Reagent — Ammoniacal silver nitrate solution [Ag(NH₃)₂]⁺ OH⁻. Oxidises aldehydes to carboxylate ions, depositing a silver mirror. Ketones don’t react — this is the key distinguishing test.
Fehling’s Solution — Deep blue alkaline solution of copper(II) tartrate complex. Gives brick-red precipitate (Cu₂O) with aliphatic aldehydes only. Benzaldehyde and ketones don’t react.
Cannizzaro Reaction — A disproportionation reaction shown by aldehydes without α-hydrogen. The aldehyde simultaneously acts as oxidising agent (gets reduced to alcohol) and reducing agent (gets oxidised to carboxylate salt).
Core Concepts and Reactions
Reactivity Order: Why Aldehydes > Ketones
Nucleophilic addition is easier when:
- The carbonyl carbon is more electrophilic (less electron density)
- The transition state is less sterically hindered
In ketones, two alkyl groups donate electrons to the carbonyl carbon (inductive effect) — making it less electrophilic than in aldehydes (one alkyl, one H). Also, two bulky groups in ketones create more steric hindrance. This is why aldehydes are more reactive than ketones in nucleophilic addition.
Remember this order for JEE: HCHO (methanal) > RCHO (other aldehydes) > RCOR’ (ketones). Methanal is most reactive — no alkyl group at all, maximum electrophilicity, minimum steric hindrance.
Nucleophilic Addition Reactions
Reaction with HCN (Hydrogen Cyanide)
The product is a cyanohydrin. The CN⁻ ion (nucleophile) attacks the carbonyl carbon first, then the H⁺ protonates oxygen. Cyanohydrins are useful intermediates — on hydrolysis, they give α-hydroxy acids.
Reaction with NaHSO₃ (Sodium Bisulphite)
Only methyl ketones (and cyclic ketones up to 8 carbons) react along with all aldehydes. This reaction is used to purify aldehydes — form the bisulphite adduct, filter it, then decompose with acid or base to regenerate the pure aldehyde.
Reaction with Grignard Reagent (RMgX)
This is a powerful C–C bond-forming reaction. The carbanion part (R⁻) of RMgX acts as nucleophile. On hydrolysis:
- HCHO + RMgX → primary alcohol (1°)
- RCHO + R’MgX → secondary alcohol (2°)
- RCOR’ + R”MgX → tertiary alcohol (3°)
Oxidation Reactions
Why aldehydes oxidise easily, ketones don’t:
The aldehyde has a C–H bond adjacent to carbonyl. This H can be abstracted, allowing oxidation. Ketones have no such H at the carbonyl carbon, so they resist mild oxidants.
Mild oxidants that oxidise aldehydes but not ketones:
- Tollen’s reagent (silver mirror test)
- Fehling’s solution (brick-red precipitate — aliphatic aldehydes only)
- Benedict’s solution (similar to Fehling’s)
Strong oxidants (K₂Cr₂O₇/H₂SO₄, KMnO₄) can oxidise ketones, but they break C–C bonds, giving a mixture of carboxylic acids.
Condensation Reactions
Aldol Condensation
Requires an aldehyde or ketone with α-hydrogen (H on the carbon adjacent to C=O).
The product is a β-hydroxy aldehyde (aldol). On heating, it loses water to give an α,β-unsaturated carbonyl compound (aldol condensation product).
The mechanism: the base removes an α-H, generating an enolate ion (the nucleophile), which attacks the carbonyl carbon of another molecule.
Cannizzaro Reaction
For aldehydes with no α-hydrogen (HCHO, C₆H₅CHO, (CH₃)₃CCHO):
One HCHO gets oxidised to formate (HCOO⁻), the other gets reduced to methanol.
Students often apply Cannizzaro to acetaldehyde (CH₃CHO). Wrong — acetaldehyde has α-hydrogens, so it undergoes aldol condensation, not Cannizzaro. Check for α-H first.
Carboxylic Acid Reactions
Acidity of Carboxylic Acids
Carboxylic acids are more acidic than phenols (pKa ~4.7 vs ~10) because the carboxylate anion (RCOO⁻) has better resonance stabilisation — negative charge delocalised equally over both oxygens.
Electron-withdrawing groups on R increase acidity; electron-donating groups decrease it.
Reactions of Carboxylic Acids
| Reaction | Reagent | Product |
|---|---|---|
| Esterification | ROH / H⁺ | Ester (RCOOR’) |
| Acid chloride formation | PCl₅ or SOCl₂ | RCOCl |
| Reduction | LiAlH₄ | Primary alcohol |
| Decarboxylation | NaOH + CaO (soda lime), heat | Alkane (RH) |
| Hell-Volhard-Zelinsky | Br₂ / red P | α-bromo acid |
Decarboxylation with soda lime gives an alkane with one less carbon:
Solved Examples
Example 1 (CBSE Level)
Q: How will you distinguish between propanal and propanone using a chemical test?
Solution:
Take both compounds separately in test tubes. Add Tollen’s reagent (ammoniacal silver nitrate) to each.
- Propanal (an aldehyde) → silver mirror forms on the test tube wall ✓
- Propanone (a ketone) → no reaction ✗
Why it works: Tollen’s reagent is a mild oxidant. It oxidises the aldehyde group (–CHO) to carboxylate (–COO⁻), while Ag⁺ gets reduced to Ag metal (silver mirror). Ketones lack the reducible C–H at carbonyl, so they don’t react.
Alternatively: Fehling’s test works for aliphatic aldehydes but remember — benzaldehyde gives silver mirror with Tollen’s but not a red precipitate with Fehling’s. This distinction has appeared in CBSE marking schemes.
Example 2 (JEE Main Level)
Q: Identify the product when acetaldehyde reacts with dilute NaOH, and then the product is heated.
Solution:
Step 1 — Aldol condensation (dilute NaOH):
Product: 3-hydroxybutanal (aldol)
Step 2 — Dehydration on heating:
Final product: but-2-enal (crotonaldehyde)
JEE Main 2023 had a question asking the final product of aldol condensation of acetaldehyde. The trap: students stop at the aldol (β-hydroxy aldehyde) and forget that heating gives the dehydrated α,β-unsaturated product. The question usually asks for the “condensation product” which implies dehydration has occurred.
Example 3 (JEE Advanced Level)
Q: Arrange the following in decreasing order of acidity: CH₃COOH, CF₃COOH, CCl₃COOH, CBr₃COOH. Justify.
Solution:
Acidity depends on how well the carboxylate anion is stabilised. Electron-withdrawing halogens on the α-carbon stabilise the negative charge on –COO⁻ through inductive effect, increasing acidity.
The key comparison is CF₃ vs CCl₃ vs CBr₃:
- F is most electronegative but smallest, so inductive withdrawal: F > Cl > Br
- Acidity order follows the same trend
The subtlety for JEE Advanced: Don’t confuse electronegativity with inductive effect efficiency. F has the highest electronegativity, so CF₃COOH is the strongest acid here. This was the exact basis for a JEE Advanced 2019 question on haloacetic acids.
Exam-Specific Tips
CBSE Boards (3-5 marks): CBSE loves distinguishing tests. Know all four: Tollen’s, Fehling’s, iodoform test (for CH₃COR and CH₃CHOH), and 2,4-DNP test (forms yellow/orange precipitate with both aldehydes and ketones — not distinguishing between them, but confirms carbonyl group). The iodoform test is highly scoring — it confirms both a methyl ketone and methyl carbinol structure.
JEE Main (1-2 questions): Focus on: nucleophilic addition mechanisms, reactivity order, Cannizzaro vs aldol (α-H check), and product identification in multi-step reactions. Named reactions (Clemmensen, Wolff-Kishner, Rosenmund, Stephen) appear regularly.
NEET (1 question): NEET typically tests one reaction product or one distinguishing test. Prioritise: Tollen’s test, Fehling’s test, iodoform test, and the acidity order of carboxylic acids. Don’t spend too much time on mechanisms for NEET — focus on what the reagent does and what comes out.
Named Reactions You Must Know
Clemmensen Reduction — Reduces C=O to CH₂ using Zn-Hg amalgam / conc. HCl. Used when the molecule is acid-sensitive.
Wolff-Kishner Reduction — Same conversion (C=O → CH₂) but using N₂H₄ / KOH / ethylene glycol. Used when the molecule is base-sensitive.
Rosenmund Reduction — Reduces acid chloride (RCOCl) to aldehyde (RCHO) using H₂ / Pd-BaSO₄.
Stephen Reaction — Reduces nitrile (RCN) to aldehyde (RCHO) using SnCl₂ / HCl, followed by hydrolysis.
Rosenmund and Stephen both give aldehydes from non-aldehyde precursors — this is a common source for “how to prepare aldehyde from X” questions. Rosenmund starts from acid chloride; Stephen starts from nitrile.
Common Mistakes to Avoid
Mistake 1: Applying Fehling’s test to benzaldehyde
Benzaldehyde gives a positive Tollen’s test (silver mirror) but does NOT give Fehling’s test (no red precipitate). Fehling’s works only for aliphatic aldehydes. Benzaldehyde is aromatic — its resonance stabilisation makes it less reactive toward Fehling’s Cu²⁺ complex. This distinction has caused marks loss in many CBSE papers.
Mistake 2: Forgetting the α-hydrogen check before deciding Aldol vs Cannizzaro
Before writing any reaction with NaOH, ask: does this aldehyde have α-hydrogens?
- Yes → Aldol condensation
- No → Cannizzaro reaction
HCHO (no H at carbon at all), C₆H₅CHO, and (CH₃)₃CCHO have no α-H. Everything else (CH₃CHO, C₂H₅CHO…) has α-H and does aldol.
Mistake 3: Writing the wrong product for LiAlH₄ reduction of carboxylic acids
RCOOH + LiAlH₄ → RCH₂OH (primary alcohol). Students sometimes write RCHO (aldehyde). LiAlH₄ is a powerful reductant — it goes all the way to alcohol, never stopping at aldehyde stage. If you want an aldehyde from acid, use DIBAL-H (diisobutylaluminium hydride) at low temperature — but this is beyond CBSE scope.
Mistake 4: Confusing iodoform test results
Iodoform test (I₂ / NaOH) is positive for:
- Methyl ketones: CH₃COR
- Acetaldehyde: CH₃CHO (the only aldehyde)
- Alcohols with CH₃CHOH– group (e.g., ethanol, isopropanol)
Acetone (CH₃COCH₃) gives iodoform — yes. Diethyl ketone (CH₃CH₂COCH₂CH₃) does NOT — no methyl group attached to carbonyl. Draw the structure and check.
Mistake 5: Wrong acidity comparison between carboxylic acid and carbonic acid
Many students think CO₂ + H₂O (carbonic acid, pKa ~6.4) is stronger than acetic acid (pKa ~4.7). Wrong — acetic acid is stronger. This matters when you’re asked whether carboxylate salts can be regenerated by CO₂. Yes, they can — CO₂ + H₂O is a weaker acid, so it cannot liberate carboxylic acid from its sodium salt. Wait — that’s backwards. CO₂ + H₂O is weaker than RCOOH, so CO₂ + H₂O cannot displace RCOOH. Only stronger acids displace weaker ones.
Practice Questions
Q1. Give the IUPAC name of CH₃CHO and CH₃COCH₂CH₃.
CH₃CHO = Ethanal
CH₃COCH₂CH₃ = Butane-2-one (or 2-butanone, ethyl methyl ketone)
Remember: for ketones, number the chain from the end closest to C=O.
Q2. Which of the following gives a silver mirror with Tollen’s reagent? (a) Acetone (b) Benzaldehyde (c) Propanone (d) Cyclohexanone
Answer: (b) Benzaldehyde
Tollen’s reagent reacts with all aldehydes (both aliphatic and aromatic). Acetone, propanone, and cyclohexanone are all ketones — they don’t react with Tollen’s. Benzaldehyde is an aldehyde (–CHO attached to benzene ring), so it gives the silver mirror.
Q3. Write the product of the reaction: CH₃CHO + HCN →
Product: 2-hydroxypropanenitrile (acetaldehyde cyanohydrin)
The CN⁻ (nucleophile) attacks the electrophilic carbonyl carbon. H⁺ then protonates the alkoxide oxygen. This is nucleophilic addition.
Q4. How will you convert ethanoic acid to methane?
Heat the sodium salt of ethanoic acid with soda lime (NaOH + CaO):
First convert CH₃COOH to CH₃COONa (sodium ethanoate) by reaction with NaOH. Then dry distillation with soda lime gives methane (decarboxylation — loss of CO₂ equivalent).
Q5. Arrange in increasing order of boiling point: pentane, pentan-1-ol, pentanal, pentanone.
Pentane < Pentanal ≈ Pentanone < Pentan-1-ol
Reasoning:
- Pentane: only weak van der Waals forces, lowest BP
- Pentanal and pentanone: polar C=O allows dipole-dipole interactions, higher BP than pentane
- Pentan-1-ol: can form intermolecular H-bonds (O–H···O), highest BP
Between pentanal and pentanone: pentanone has slightly higher BP due to larger dipole moment from two alkyl groups.
Q6. What happens when formaldehyde is treated with conc. NaOH?
Cannizzaro reaction (HCHO has no α-hydrogen):
One HCHO is oxidised to sodium formate (HCOONa), the other is reduced to methanol (CH₃OH). This disproportionation occurs because there’s no α-H for aldol condensation.
Q7. Predict the product: C₆H₅CHO + CH₃CHO → (dil. NaOH)
This is a cross aldol condensation (Claisen-Schmidt reaction):
Product: Cinnamaldehyde (3-phenylprop-2-enal)
C₆H₅CHO has no α-H, so it acts as the electrophile (carbonyl acceptor). CH₃CHO has α-H, so its enolate acts as the nucleophile. The aldol product immediately dehydrates (resonance stabilisation from conjugation with benzene ring drives this). This reaction appeared in JEE Main 2022.
Q8. Which acid is stronger: chloroacetic acid or fluoroacetic acid? Explain.
Fluoroacetic acid (FCH₂COOH) is stronger than chloroacetic acid (ClCH₂COOH).
F is more electronegative than Cl, so it exerts a stronger electron-withdrawing inductive effect on the carboxylate group. Greater electron withdrawal = better stabilisation of the carboxylate anion (FCH₂COO⁻) = lower pKa = stronger acid.
pKa values: FCH₂COOH ≈ 2.66, ClCH₂COOH ≈ 2.86 (lower pKa = stronger acid, confirming F > Cl in inductive withdrawal).
FAQs
Why can’t ketones be oxidised by Tollen’s or Fehling’s reagent?
Oxidation of a carbonyl compound requires a C–H bond at the carbonyl carbon. Aldehydes have that (–CHO), ketones don’t (–COR’). Tollen’s and Fehling’s are mild oxidants — they can only oxidise the aldehyde C–H. Ketones are unreactive toward these mild oxidants because breaking a C–C bond (which would be needed to oxidise the ketone carbon) requires harsh conditions.
What is the difference between aldol condensation and Cannizzaro reaction?
Both are reactions of aldehydes with NaOH, but:
- Aldol: Requires α-hydrogen. Two molecules of the same (or different) aldehyde/ketone combine. One becomes a nucleophile (enolate), one acts as electrophile. Product has more carbons than starting material.
- Cannizzaro: Requires NO α-hydrogen. Disproportionation — one molecule gets oxidised, one gets reduced. No new C–C bonds formed.
The α-H check is the deciding factor.
Why does acidity of carboxylic acids increase with electron-withdrawing substituents?
Acidity means ease of losing H⁺. After losing H⁺, you have the carboxylate anion (RCOO⁻). Electron-withdrawing groups (like Cl, F, NO₂) pull electron density away from the –COO⁻, dispersing the negative charge and stabilising the anion. More stable anion → equilibrium favours ionisation → stronger acid.
What is the iodoform test used for?
It detects the CH₃CO– group (methyl ketone) or the CH₃CHOH– group (secondary alcohol where one group is methyl). With I₂/NaOH, these get converted to CHI₃ (iodoform) — a yellow precipitate with a distinctive antiseptic smell. Acetaldehyde (CH₃CHO) is the only aldehyde that gives a positive iodoform test.
How is acetic acid converted to acetone in one step?
Heat calcium acetate (dry distillation):
This is a classic preparation of ketones from carboxylate salts. Using calcium formate instead gives formaldehyde; using one calcium acetate + one calcium formate gives acetaldehyde.
Why is formic acid (HCOOH) different from other carboxylic acids?
Formic acid has a unique structure: the “R” group is actually a hydrogen, so it has a –CHO-like portion built into the –COOH. This means it can act as both a reducing agent (like an aldehyde) AND an acid. Formic acid gives a positive Tollen’s test — something no other carboxylic acid does. It also gets oxidised to CO₂ + H₂O by mild oxidants.
What is the role of soda lime in decarboxylation?
Soda lime is a mixture of NaOH (80%) and CaO (20%). NaOH provides the base for the reaction; CaO serves two purposes — it absorbs the moisture (preventing NaOH from becoming syrupy and blocking the tube) and it raises the melting point of the mixture so the reaction can proceed at higher temperatures. Pure NaOH alone at high temperature would melt, foam, and cause a mess in the reaction tube.