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JEE Mains Chemistry · Aldehydes, Ketones and Carboxylic Acids

Enols and Aldol Condensation

A hydrogen on the carbon next to a C=O is acidic; its enolate adds to a second carbonyl group (the aldol reaction), and the product loses water on heating, between two molecules or within one.

All Aldehydes, Ketones and Carboxylic Acids notesJEE Mains Chemistry strategy

Why this matters

Seventeen PYQs, two numerical, two from 2026. Six rank α-hydrogen acidity or enol content; six predict the product of a self-aldol condensation; five close a ring by an intramolecular aldol condensation.

Concept 1 of 3: Acidity of α-hydrogens and enol content

An α-hydrogen is acidic because the anion left behind, the enolate, spreads its negative charge onto the carbonyl oxygen. A second C=O on the same carbon spreads it further, so the hydrogen between two C=O groups is by far the most acidic.

Definition

  • Approximate pKa of the α-H: β-diketone RCOCH2COR\mathrm{RCOCH_2COR} about 9; β-keto ester about 11; malonic ester about 13; simple ketone about 19–20.
  • An ester C=O acidifies less than a ketone C=O, because the OR oxygen already feeds electrons into it.
  • Enol content follows the same features. Propanone contains only a trace of enol. Pentane-2,4-dione is largely enolised, because its enol is conjugated and held by an internal O–H···O hydrogen bond.
  • Cyclohexane-1,3,5-trione exists almost entirely as its enol, benzene-1,3,5-triol (phloroglucinol), because that enol is aromatic.
  • The enolate is a carbon nucleophile: with an alkyl halide it is alkylated on the α-carbon.

Acidity of the α-hydrogen

RCOCH2COR>RCOCH2COOR′>R′OOCCH2COOR′>RCOCH3\mathrm{RCOCH_2COR > RCOCH_2COOR' > R'OOCCH_2COOR' > RCOCH_3}

Worked example

Arrange in order of acidity of the most acidic hydrogen: propanone, ethyl 3-oxobutanoate, pentane-2,4-dione, diethyl propanedioate.
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The same idea in a real exam question:

JEE Mains · 2024 · 27 Jan 2024 · Q36Moderate

Example 1 · Aldehydes, Ketones and Carboxylic Acids · Enols and Aldol Condensation

Which of the following has highly acidic hydrogen?

The most acidic H sits between two C=O groups

In a 1,3-dicarbonyl compound the CH2\mathrm{CH_2} between the carbonyls is far more acidic than a terminal CH3\mathrm{CH_3} next to only one. In a 1,4- or 1,5-diketone no carbon has two C=O neighbours.

An ester group helps less than a ketone group

Replacing one ketone of a β-diketone by an ester lowers the acidity by about two pKa units, and replacing both lowers it further.
Drill 5 more on acidity of α-hydrogens and enol content

Concept 2 of 3: Self-aldol condensation: predicting the product

Dilute base removes an α-hydrogen. The enolate carbon attacks the carbonyl carbon of a second molecule of the same compound, giving a β-hydroxy aldehyde or ketone (the aldol). On heating, the new OH and an α-hydrogen leave as water, and the C=C that forms is conjugated with the C=O.

Definition

  • The new C–C bond joins the α-carbon of one molecule to the carbonyl carbon of the other.
  • Aldol addition gives the β-hydroxy carbonyl compound; heating removes water to give the α,β-unsaturated compound. Addition plus dehydration is the aldol condensation.
  • Ethanal gives 3-hydroxybutanal, then but-2-enal. Propanone with Ba(OH)2\mathrm{Ba(OH)_2} gives 4-hydroxy-4-methylpentan-2-one (diacetone alcohol), then 4-methylpent-3-en-2-one (mesityl oxide).
  • A cyclic ketone gives a C=C outside the ring it came from: cyclohexanone gives 2-cyclohexylidenecyclohexan-1-one.
  • Dehydration needs a hydrogen on the α-carbon of the aldol. If that carbon ends up with no H, the product stays as the β-hydroxy compound.

Aldol addition, then dehydration

2 RCH2CHO→dil. OH−RCH2CH(OH)CH(R)CHO→Δ, −H2ORCH2CH=C(R)CHO\mathrm{2\,RCH_2CHO \xrightarrow{dil.\ OH^-} RCH_2CH(OH)CH(R)CHO \xrightarrow{\Delta,\ -H_2O} RCH_2CH{=}C(R)CHO}

Worked example

Butanal is warmed with dilute NaOH. Give the aldol and the condensation product.
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The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q39Moderate

Example 2 · Aldehydes, Ketones and Carboxylic Acids · Enols and Aldol Condensation

' x ' is the product which is obtained by the hydrolysis of prop-1-yne in the presence of mercuric sulphate under dilute acidic medium at 333 K . ' y ' is the product which is obtained by the reaction of ethane nitrile with methyl magnesium bromide in dry ether followed by hydrolysis. IUPAC name of product obtained from ' x ' and ' y ' in the presence of barium hydroxide followed by heating is:

Number the product from the new chain

Join the α-carbon of one molecule to the carbonyl carbon of the other, draw the whole chain, then number from the C=O that survives. Naming each half separately gives wrong locants.

No α-hydrogen left, no dehydration

Water leaves from the OH and an α-hydrogen. If the α-carbon of the aldol carries two alkyl groups and the C=O, the aldol cannot condense further.
Drill 5 more on self-aldol condensation: predicting the product

Concept 3 of 3: Intramolecular aldol: which ring closes

When one molecule carries two carbonyl groups, its own enolate can attack its other C=O. Several enolates are possible, but the ring that forms is the one with five or six atoms; three-, four- and seven-membered rings lose.

Definition

  • Count the ring from the enolate carbon to the attacked carbonyl carbon, both included.
  • A 1,4-diketone closes a five-membered ring (a cyclopentenone). A 1,5-diketone closes a six-membered ring (a cyclohexenone).
  • A 1,6-dicarbonyl compound also closes a five-membered ring, with the other carbonyl group left outside the ring: hexanedial gives cyclopent-1-ene-1-carbaldehyde.
  • Both carbons of the new C=C (the enolate carbon and the old carbonyl carbon) are ring atoms. So the C=C of an intramolecular aldol product is always in the ring; an exocyclic =CH2\mathrm{{=}CH_2} next to the C=O cannot come from this reaction.

Ring size

ring atoms=enolate carbon to attacked carbonyl carbon, both counted;5 or 6 wins\text{ring atoms} = \text{enolate carbon to attacked carbonyl carbon, both counted}; \quad 5 \text{ or } 6 \text{ wins}

Worked example

Heptane-2,6-dione is heated with dilute NaOH. Give the product.
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The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q29Moderate

Example 3 · Aldehydes, Ketones and Carboxylic Acids · Enols and Aldol Condensation

Aldol condensation is a popular and classical method to prepare α,β\alpha,\beta- unsaturated carbonyl compounds. This reaction can be both intermolecular and intramolecular. Predict which one of the following is not a product of intramolecular aldol condensation?

Count atoms in the ring, not bonds

Include both the enolate carbon and the carbonyl carbon. Miscounting by one turns a five-membered ring into a 'four' or a 'six' and sends you to the wrong option.

Pick the enolate that makes a five- or six-membered ring

A diketone usually has two or more α-carbons. Try each, count the ring, and keep only the one that gives five or six atoms.
Drill 4 more on intramolecular aldol: which ring closes

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 (3)

  • Acidity of α-hydrogens and enol content

    Acidity of the α-hydrogen

    RCOCH2COR>RCOCH2COOR′>R′OOCCH2COOR′>RCOCH3\mathrm{RCOCH_2COR > RCOCH_2COOR' > R'OOCCH_2COOR' > RCOCH_3}
  • Self-aldol condensation: predicting the product

    Aldol addition, then dehydration

    2 RCH2CHO→dil. OH−RCH2CH(OH)CH(R)CHO→Δ, −H2ORCH2CH=C(R)CHO\mathrm{2\,RCH_2CHO \xrightarrow{dil.\ OH^-} RCH_2CH(OH)CH(R)CHO \xrightarrow{\Delta,\ -H_2O} RCH_2CH{=}C(R)CHO}
  • Intramolecular aldol: which ring closes

    Ring size

    ring atoms=enolate carbon to attacked carbonyl carbon, both counted;5 or 6 wins\text{ring atoms} = \text{enolate carbon to attacked carbonyl carbon, both counted}; \quad 5 \text{ or } 6 \text{ wins}

Watch out for (6)

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