IIT-JAM Carbonyl Chemistry — Addition, Condensation and the Named Reactions
The carbonyl group is the busiest functional group in the organic syllabus, and it earns that place honestly: almost every named reaction in the unit is one of two things happening at the same carbon. Either a nucleophile attacks the electron-poor carbonyl carbon, or a base removes an α-hydrogen to make an enolate that then attacks somebody else's carbonyl carbon. Sort every reaction you meet into those two families and the list of names stops being a list. This guide does that sorting, then works through the product-prediction and structure problems that carry the marks.
The two reactive sites, and the one reason for each
Site 2 — the α-carbon. A hydrogen on the carbon next to the carbonyl is unusually acidic (pKa of the order of 20 for a simple ketone, far below an ordinary C–H) because the resulting carbanion is delocalised onto oxygen. That anion is the enolate, and it is the nucleophile in every condensation reaction below.
Everything else is detail. Notice that the two sites are in competition: a reagent that is a good nucleophile adds at site 1, and a reagent that is a strong base takes the proton from site 2.
Reactivity towards nucleophilic addition
Two effects run in the same direction, so the order is easy to justify in a written answer. Electronic: each alkyl group pushes electron density towards the carbonyl carbon by +I, reducing the δ+ and making it less attractive to a nucleophile; an aryl group does the same more strongly, because the ring conjugates with the C=O and delocalises the positive character. Steric: each substituent obstructs the approach of the nucleophile and crowds the tetrahedral product. Aldehydes therefore beat ketones on both counts, and formaldehyde — with two hydrogens and nothing else — is the most reactive of all. Conversely, electron-withdrawing groups such as –CCl3 increase reactivity, which is why chloral forms a stable hydrate.
Worked example 1 — finding the compound from percentage composition
A neutral carbonyl compound contains 62.04 % carbon and 10.41 % hydrogen by mass, the remainder being oxygen, and its molar mass is 58.08 g mol−1. It gives a positive iodoform test but no silver mirror with Tollens' reagent. Identify it.
Step 1 — oxygen by difference.
% O = 100 − 62.04 − 10.41 = 27.55 %
Step 2 — moles in 100 g, using Ar(C) = 12.011, Ar(H) = 1.008,
Ar(O) = 15.999:
C: 62.04 ÷ 12.011 = 5.1653 mol
H: 10.41 ÷ 1.008 = 10.327 mol
O: 27.55 ÷ 15.999 = 1.7220 mol
Step 3 — divide by the smallest, 1.7220.
C: 5.1653 ÷ 1.7220 = 3.000 · H: 10.327 ÷ 1.7220 = 5.997 ·
O: 1.7220 ÷ 1.7220 = 1.000
Empirical formula = C3H6O
Step 4 — empirical to molecular.
Empirical mass = (3 × 12.011) + (6 × 1.008) + 15.999 = 36.033 + 6.048 + 15.999 =
58.080 g mol−1
n = 58.08 ÷ 58.080 = 1, so the molecular formula is also
C3H6O.
Step 5 — which isomer? DBE = 3 − 6/2 + 1 = 1, consistent with one C=O. The two carbonyl isomers are propanal (CH3CH2CHO) and propanone (CH3COCH3). Tollens' reagent oxidises aldehydes and gives a silver mirror; there is none, so it is not propanal. The iodoform test is positive for a CH3CO– group, which propanone has. The compound is propanone (acetone).
Check backwards: for C3H6O, % C = 36.033 ÷ 58.080 = 62.04 %, % H = 6.048 ÷ 58.080 = 10.41 %, % O = 15.999 ÷ 58.080 = 27.55 % — the data reproduce exactly. Always run this reverse check; it catches an arithmetic slip in seconds.
Family 1 — nucleophilic addition and addition–elimination
| Reagent | Product | Point that gets examined |
|---|---|---|
| HCN (with a trace of base) | Cyanohydrin, RCH(OH)CN | CN− is the actual nucleophile, so base speeds it up and acid stops it. Adds one carbon — a classic chain-extension step. |
| NaHSO3 | Crystalline bisulphite adduct | Works only for aldehydes, methyl ketones and small cyclic ketones. Reversible with acid or base, so it is a purification trick. |
| 2 ROH, dry HCl | Acetal (or ketal) | Stable to base and to nucleophiles, hydrolysed by aqueous acid — the standard protecting group for a carbonyl. |
| RNH2 (primary amine) | Imine (Schiff base), C=NR | Addition then elimination of water. Rate is maximum near pH 4–5. |
| NH2OH / NH2NH2 / 2,4-DNP / semicarbazide | Oxime, hydrazone, 2,4-dinitrophenylhydrazone, semicarbazone | All the same addition–elimination. 2,4-DNP gives an orange to red solid used to identify the carbonyl. |
| RMgX then H3O+ | Alcohol (1° from HCHO, 2° from other aldehydes, 3° from ketones) | The carbanion adds irreversibly; water or any acidic proton destroys the reagent first. |
| NaBH4 or LiAlH4 | Alcohol | NaBH4 is mild and leaves esters and acids alone; LiAlH4 reduces almost everything and reacts violently with water. |
| Ph3P=CHR (Wittig) | Alkene | The C=O becomes C=C at exactly the carbon you choose — no rearrangement. Driven by formation of the very strong P=O bond. |
The pH 4–5 optimum for imine formation is worth being able to explain rather than recite. A little acid is needed, because the –OH of the tetrahedral intermediate must be protonated before water can leave. But too much acid protonates the amine to RNH3+, which has no lone pair and cannot attack at all. The rate therefore rises and then falls, giving a bell-shaped curve with a maximum in the weakly acidic region.
Family 2 — enolate chemistry and the condensations
- Aldol reaction. Dilute base removes an α-H; the enolate adds to a second molecule's carbonyl carbon, giving a β-hydroxy aldehyde or ketone. Warming, or stronger base, dehydrates it to an α,β-unsaturated carbonyl compound — that second step is what makes it an aldol condensation. It requires an α-hydrogen.
- Cannizzaro reaction. An aldehyde with no α-hydrogen, treated with concentrated alkali, disproportionates: one molecule is reduced to the alcohol and the other oxidised to the carboxylate salt. In the crossed version, formaldehyde is the hydride donor because it is the most reactive towards hydroxide, so it is oxidised to formate and the other aldehyde is reduced cleanly to its alcohol.
- Haloform reaction. A methyl ketone (or an alcohol oxidisable to one, such as ethanol or propan-2-ol) with X2/OH− gives CHX3 plus a carboxylate with one carbon fewer. With iodine this is the yellow-precipitate iodoform test.
- Claisen ester condensation. Two ester molecules with alkoxide give a β-keto ester. It needs a full equivalent of base, not a catalytic amount, because the reaction is driven to completion by deprotonation of the unusually acidic product.
- Perkin, Knoevenagel, Reformatsky, Mannich. Perkin: aromatic aldehyde + acid anhydride + the sodium salt of that acid → an α,β-unsaturated acid such as cinnamic acid. Knoevenagel: an active-methylene compound (malonic ester, ethyl acetoacetate) plus an amine base. Reformatsky: an α-halo ester with zinc gives an organozinc mild enough not to attack the ester, then adds to a carbonyl to give a β-hydroxy ester. Mannich: aldehyde + amine + enolisable ketone → β-amino ketone.
- Benzoin condensation. Two molecules of benzaldehyde with cyanide give benzoin. Cyanide is uniquely suited because it is a good nucleophile, it stabilises the carbanion it creates, and it is a good leaving group at the end — all three are needed.
Worked example 2 — counting the products of a crossed aldol
Ethanal and propanal are mixed with dilute NaOH. How many aldol products are possible, and why is this a poor synthetic reaction?
Both aldehydes have α-hydrogens, so both can form an enolate, and both can act as the electrophile. That gives 2 × 2 = 4 combinations:
- ethanal enolate + ethanal (self-aldol)
- ethanal enolate + propanal (crossed)
- propanal enolate + propanal (self-aldol)
- propanal enolate + ethanal (crossed)
Each can then dehydrate, so the mixture is worse still. This is why a crossed aldol is only useful when one partner has no α-hydrogen — benzaldehyde or formaldehyde, for instance — so it can only ever be the electrophile. Benzaldehyde with ethanal in dilute base gives essentially one condensation product, cinnamaldehyde, and that is the version that appears in synthesis questions.
Worked example 3 — aldol or Cannizzaro?
State what each of these gives with concentrated NaOH: methanal (HCHO), benzaldehyde (C6H5CHO), 2,2-dimethylpropanal ((CH3)3CCHO) and ethanal (CH3CHO).
The single test is: is there a hydrogen on the carbon next to the C=O?
- HCHO — the carbonyl carbon carries only hydrogens; there is no α-carbon at all. Cannizzaro → methanol + sodium formate.
- C6H5CHO — the α-carbon is part of the aromatic ring and has no hydrogen available for enolisation. Cannizzaro → benzyl alcohol + sodium benzoate.
- (CH3)3CCHO — the α-carbon is quaternary; its three methyls are on the β carbon, not the α. Cannizzaro → neopentyl alcohol + its carboxylate. This is the one students get wrong, because the molecule is full of hydrogens; count the position, not the total.
- CH3CHO — three α-hydrogens. Aldol, giving 3-hydroxybutanal and then but-2-enal on dehydration.
Extension: mix HCHO with 2,2-dimethylpropanal and you get a clean crossed Cannizzaro — formaldehyde is oxidised to formate and the other aldehyde is reduced to neopentyl alcohol.
Oxidation, reduction and rearrangement — choosing the right reagent
| Transformation | Reagent | Condition that decides the choice |
|---|---|---|
| C=O → CH2 | Clemmensen, Zn(Hg)/conc. HCl | Acidic — use when the substrate tolerates acid but not base |
| C=O → CH2 | Wolff–Kishner, H2NNH2/KOH, high boiling solvent | Basic — use for acid-sensitive substrates. Same product, opposite conditions |
| C=O → CH(OH), keeping C=C | Meerwein–Ponndorf–Verley, Al(OiPr)3/propan-2-ol | Selective; the reverse reaction, Oppenauer, oxidises an alcohol to a ketone |
| Aldehyde → carboxylic acid | Tollens', Fehling's, KMnO4, chromic acid | Fehling's works for aliphatic aldehydes but not aromatic ones — a standard distinguishing test |
| Ketone → ester (one O inserted) | Baeyer–Villiger, peroxyacid such as mCPBA | The group that migrates is set by migratory aptitude, not by size alone |
Worked example 4 — Baeyer–Villiger regiochemistry
Predict the product when (a) acetophenone and (b) 3,3-dimethylbutan-2-one are treated with a peroxyacid.
An oxygen is inserted between the carbonyl carbon and one of its two neighbours. Which one migrates follows the migratory aptitude order, which is essentially the order of ability to stabilise developing positive charge on the migrating carbon:
tertiary alkyl > cyclohexyl ≈ secondary alkyl ≈ benzyl > phenyl > primary alkyl > methyl
(a) Acetophenone, C6H5COCH3. The choice is phenyl against methyl. Phenyl is far higher in the series, so phenyl migrates to oxygen and the product is phenyl acetate, CH3COOC6H5 — not methyl benzoate.
(b) 3,3-Dimethylbutan-2-one, (CH3)3CCOCH3. The choice is tert-butyl against methyl. Tert-butyl is at the top of the series, so the product is tert-butyl acetate, CH3COOC(CH3)3.
Methyl is bottom of the order in both cases, which is the general rule worth carrying: in a methyl ketone, the other group migrates and you get an acetate ester. The migration is also stereospecific with retention at the migrating carbon, a detail that turns up in stereochemistry-flavoured questions.
Common mistakes that cost marks
- Looking for α-hydrogens in the wrong place. The α-carbon is the one bonded directly to the carbonyl carbon. 2,2-Dimethylpropanal has plenty of hydrogens and none of them is α.
- Stopping the aldol at the β-hydroxy stage, or jumping past it. Mild conditions give the aldol; heat or stronger base gives the α,β-unsaturated condensation product. Read the conditions in the question.
- Choosing Clemmensen for an acid-sensitive substrate. Clemmensen is strongly acidic and Wolff–Kishner is strongly basic; they give the same product, so the substrate decides.
- Assuming the bigger group always migrates in a Baeyer–Villiger. Use the aptitude order — phenyl beats methyl, and tert-butyl beats phenyl.
- Treating the iodoform test as a ketone test. It detects a CH3CO– group or a CH3CH(OH)– group. Ethanol and propan-2-ol are positive; methanol and benzaldehyde are negative.
- Using a Grignard reagent in the presence of –OH, –NH or –COOH. The reagent is destroyed by the acidic proton before it ever reaches the carbonyl. Protect first.
- Writing a Claisen condensation with a catalytic amount of base. A full equivalent is required, and the reason — deprotonation of the β-keto ester product — is often the actual question.
- Forgetting that Fehling's solution fails with aromatic aldehydes. Benzaldehyde gives a silver mirror with Tollens' but no red precipitate with Fehling's.
How to prepare this unit
| Theme | What you must be able to do without hesitation |
|---|---|
| Mechanism of addition | Draw nucleophile → tetrahedral alkoxide → protonation, with arrows |
| Reactivity order | Rank any set of carbonyl compounds and justify with both electronic and steric reasons |
| Addition–elimination | Produce imine, oxime, hydrazone, 2,4-DNP derivative and explain the pH 4–5 optimum |
| α-Hydrogen test | Decide aldol versus Cannizzaro for any aldehyde in one glance |
| Crossed reactions | Explain when a crossed aldol or Cannizzaro is clean and when it gives a mixture |
| Named reactions | Wittig, Claisen, Perkin, Knoevenagel, Reformatsky, Mannich, benzoin — reagent, product, driving force |
| Reduction choice | Pick between Clemmensen, Wolff–Kishner, NaBH4, LiAlH4 and MPV on substrate grounds |
| Distinguishing tests | Tollens', Fehling's, iodoform, 2,4-DNP — know exactly what each detects and what it misses |
Treat that as a revision checklist, not as a prediction of the paper. For the syllabus and the current pattern, read the official IIT-JAM notification for your year.
Speed up the formula step. Worked example 1 is the shape of a very common question — percentage composition, then molar mass, then a chemical test to choose between isomers — and the slow part is the empirical-formula arithmetic. The molar mass and composition tool computes a formula's molar mass and its percentage composition, so you can verify the reverse check in seconds and spend your time on the chemistry instead.
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