JEE Mains Chemistry · Organic Reaction Mechanisms

Carbonyl Reactions: Enolates, Haloform and Cannizzaro

A carbonyl compound with an α-hydrogen forms an enolate that attacks another carbonyl, an ester or a C=C next to a C=O; one with a CH₃CO group loses that carbon in the haloform reaction; one with no α-hydrogen disproportionates in the Cannizzaro reaction.

All Organic Reaction Mechanisms notes JEE Mains Chemistry strategy

Why this matters

Eight PYQs, two of them asking for a number. Four are enolate chemistry: a Claisen-Schmidt yield where two molecules of benzaldehyde condense with one of acetone, a ring closed by an intramolecular Claisen, a thiol adding to acrylonitrile, and an amino alcohol closed into a cyclic carbamate. Four cut carbons or shift a hydride: the haloform reaction of a methyl ketone, an intramolecular Cannizzaro, a haloform followed by soda-lime decarboxylation, and a 1,2-diol heated with oxalic acid.

Concept 1 of 2: Aldol, Claisen-Schmidt, Claisen and Michael reactions

A hydrogen on the carbon next to a C=O (the α-carbon) is acidic, because the anion left behind spreads its charge onto the oxygen. Base removes it and gives an enolate, a carbon nucleophile. The enolate can attack the C=O of another aldehyde or ketone (aldol), the C=O of an ester (Claisen), or the far carbon of a C=C that is joined to a C=O or C≡N (Michael). On heating, an aldol product loses water and becomes an α,β-unsaturated carbonyl.

Definition

Aldol: two carbonyl compounds with α-H, dilute NaOH: a β-hydroxy aldehyde or ketone; heat removes water to give the α,β-unsaturated compound.
Claisen-Schmidt: an aromatic aldehyde (no α-H, so it only accepts) with a ketone or aldehyde that has α-H. Acetone has α-H on BOTH sides, so with excess ArCHO both sides react: $\mathrm{2\,C_6H_5CHO + CH_3COCH_3 \to C_6H_5CH{=}CHCOCH{=}CHC_6H_5 + 2H_2O}$ (dibenzalacetone).
Claisen condensation: an enolate attacks an ester and ethoxide leaves, giving a 1,3-dicarbonyl. Done inside one molecule (Dieckmann), it closes the five- or six-membered ring.
Intramolecular aldol: a diketone closes the five- or six-membered ring too; smaller rings do not form.
Michael (conjugate) addition: a nucleophile adds to the β-carbon of $\mathrm{C{=}C{-}C{=}O}$ or $\mathrm{C{=}C{-}C{\equiv}N}$ . Sulfur is a better nucleophile than oxygen, so a thiol adds through S.
Amine before alcohol: $\mathrm{-NH_2}$ attacks a carbonyl faster than $\mathrm{-OH}$ . A 1,2-amino alcohol with diethyl carbonate gives a five-membered cyclic carbamate (an oxazolidin-2-one) and two ethanol.
Percentage yield = actual mass ÷ theoretical mass × 100, with the theoretical mass from the limiting reagent and the balanced equation.

Claisen-Schmidt condensation and percentage yield

\mathrm{ArCHO + CH_3COR \xrightarrow{OH^-} ArCH{=}CHCOR + H_2O} \qquad \%\,\text{yield} = \dfrac{\text{actual mass}}{\text{theoretical mass}} \times 100

Worked example

6.0 g of 4-methylbenzaldehyde (M = 120 g/mol) is condensed with excess acetophenone in dilute NaOH, and 8.88 g of the chalcone

\mathrm{CH_3C_6H_4CH{=}CHCOC_6H_5}

is collected. Find the percentage yield.

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The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q123Moderate

Example 1 · Organic Reaction Mechanisms · Carbonyl Reactions: Enolates, Haloform and Cannizzaro

In the Claisen-Schmidt reaction to prepare dibenzalacetone from 5.3 g benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was

\_\_\_\_\_\_

Acetone condenses on both sides

With excess aromatic aldehyde, both

\mathrm{CH_3}

groups of acetone react, so two moles of aldehyde make one mole of dibenzalacetone. Taking a 1:1 ratio doubles the theoretical yield and halves the percentage.

Benzaldehyde cannot form an enolate

Benzaldehyde has no α-hydrogen. In a crossed aldol it can only be attacked; with concentrated alkali and nothing else, it undergoes the Cannizzaro reaction instead.

Michael addition goes to the β-carbon

A soft nucleophile such as a thiolate adds to the carbon at the far end of the C=C, not to the C=O or the C≡N. Acrylonitrile,

\mathrm{CH_2{=}CHCN}

, gains the new group on its

\mathrm{CH_2}

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Concept 2 of 2: Haloform, Cannizzaro and decarboxylation

Two reactions of a carbonyl in strong base change the count of carbons or the oxidation state. In the haloform reaction, the three H of a

\mathrm{CH_3CO}

group are replaced by halogen one at a time, and hydroxide then cuts off

\mathrm{CX_3^-}

: the chain loses one carbon and becomes a carboxylate. In the Cannizzaro reaction, an aldehyde with no α-H cannot form an enolate, so hydroxide adds to it and the adduct passes a hydride to a second aldehyde: one molecule is oxidised, the other reduced.

Definition

Haloform: needs $\mathrm{CH_3CO{-}}$ , or $\mathrm{CH_3CH(OH){-}}$ , which the hypohalite first oxidises to $\mathrm{CH_3CO{-}}$ . Products: $\mathrm{CHX_3}$ and a carboxylate with one carbon fewer. Iodoform is a yellow precipitate.
The reaction happens at the $\mathrm{CH_3}$ next to C=O; an aromatic ring in the molecule is not halogenated.
A tertiary alcohol, or a ketone with no $\mathrm{CH_3CO}$ group (pentan-3-one), gives no haloform.
Soda-lime decarboxylation: $\mathrm{RCOONa + NaOH \xrightarrow{CaO,\ heat} RH + Na_2CO_3}$ ; one more carbon is lost.
Cannizzaro: an aldehyde with no α-H (HCHO, $\mathrm{C_6H_5CHO}$ , $\mathrm{(CH_3)_3CCHO}$ ) in concentrated NaOH gives the alcohol and the carboxylate. In a crossed Cannizzaro, HCHO is the one oxidised, to formate.
An α-keto aldehyde does the Cannizzaro reaction inside one molecule: the CHO becomes $\mathrm{COO^-}$ and the C=O next to it becomes CH(OH).
1,2-Diol with oxalic acid, strongly heated: both OH are removed and the two carbons become a C=C (glycerol gives allyl alcohol at about 530 K).

Haloform and Cannizzaro equations

\mathrm{RCOCH_3 + 3X_2 + 4NaOH \to RCOONa + CHX_3 + 3NaX + 3H_2O} \qquad \mathrm{2\,ArCHO + NaOH \to ArCH_2OH + ArCOONa}

Worked example

Pentan-2-ol is warmed with iodine and NaOH, and the organic salt formed is then heated with soda lime. Name the precipitate and the final organic product.

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The same idea in a real exam question:

JEE Mains · 2021 · Paper 18 · Q34Moderate

Example 2 · Organic Reaction Mechanisms · Carbonyl Reactions: Enolates, Haloform and Cannizzaro

The major products formed in the following reaction sequence

A

and

B

are:

The haloform reaction removes one carbon only

Only the

\mathrm{CH_3}

of the

\mathrm{CH_3CO}

group leaves, as

\mathrm{CHX_3}

. A methyl ketone with n carbons gives a carboxylate with n − 1 carbons; soda lime then removes one more.

Cannizzaro needs an aldehyde with no α-hydrogen

Ethanal or propanal in alkali forms an enolate and gives an aldol product, not a Cannizzaro mixture. Formaldehyde, benzaldehyde and 2,2-dimethylpropanal have no α-H and disproportionate.

In a crossed Cannizzaro, formaldehyde is oxidised

Formaldehyde is the most reactive aldehyde, so hydroxide adds to it first and it gives up the hydride. It ends as formate, and the other aldehyde ends as the alcohol.

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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)

Aldol, Claisen-Schmidt, Claisen and Michael reactions
Claisen-Schmidt condensation and percentage yield
$\mathrm{ArCHO + CH_3COR \xrightarrow{OH^-} ArCH{=}CHCOR + H_2O} \qquad \%\,\text{yield} = \dfrac{\text{actual mass}}{\text{theoretical mass}} \times 100$
Haloform, Cannizzaro and decarboxylation
Haloform and Cannizzaro equations
$\mathrm{RCOCH_3 + 3X_2 + 4NaOH \to RCOONa + CHX_3 + 3NaX + 3H_2O} \qquad \mathrm{2\,ArCHO + NaOH \to ArCH_2OH + ArCOONa}$

Watch out for (6)

Acetone condenses on both sides
→ Aldol, Claisen-Schmidt, Claisen and Michael reactions
Benzaldehyde cannot form an enolate
→ Aldol, Claisen-Schmidt, Claisen and Michael reactions
Michael addition goes to the β-carbon
→ Aldol, Claisen-Schmidt, Claisen and Michael reactions
The haloform reaction removes one carbon only
→ Haloform, Cannizzaro and decarboxylation
Cannizzaro needs an aldehyde with no α-hydrogen
→ Haloform, Cannizzaro and decarboxylation
In a crossed Cannizzaro, formaldehyde is oxidised
→ Haloform, Cannizzaro and decarboxylation

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