α-Hydrogen acidity and aldol condensation is one of the most productive reaction pathways in carbonyl chemistry — and a topic where NEET questions test whether you truly understand why the α-hydrogen is acidic, not just that it is.
Why is the α-hydrogen acidic?
The hydrogen on the carbon adjacent to the carbonyl group (the α-carbon) is acidic because the conjugate base — the enolate ion — is stabilised by resonance. The negative charge delocalises across both the α-carbon and the carbonyl oxygen, forming a resonance-stabilised anion. This is the key fact: ordinary C–H bonds (pKa ~50) are not acidic, but α-C–H bonds in aldehydes and ketones have pKa values around 19–20, making them removable by strong bases like dilute NaOH or LDA (NCERT Class 12 Chemistry, Chapter 8, Part 2, page 241).
The aldol reaction
When an aldehyde possessing an α-hydrogen is treated with dilute NaOH at low temperature, the base abstracts the α-hydrogen to form the enolate. This enolate acts as a nucleophile and attacks the carbonyl carbon of a second molecule, yielding a β-hydroxy aldehyde — the "aldol" (aldehyde + alcohol). On heating, the aldol readily undergoes dehydration to give an α,β-unsaturated aldehyde (the aldol condensation product).
Acetaldehyde (ethanal) is the textbook example: two molecules of CH₃CHO with dilute NaOH give 3-hydroxybutanal (the aldol), which on heating gives crotonaldehyde (but-2-enal).
Ketones undergo the same reaction, though less favourably due to steric hindrance at the carbonyl carbon. Acetone gives diacetone alcohol, then mesityl oxide on dehydration.
Watch-out: Formaldehyde (HCHO) has no α-hydrogen — it cannot undergo the aldol reaction. Similarly, benzaldehyde has no α-hydrogen. When NEET asks "which compound cannot undergo aldol condensation?", check for the α-hydrogen first.
Crossed aldol reactions (between two different aldehydes) give mixtures unless one partner has no α-hydrogen (e.g. HCHO or PhCHO as the electrophilic partner).
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following statements correctly describes the α-hydrogen in a carbonyl compound?
Answer: A. The α-carbon is the carbon directly adjacent to the carbonyl carbon, and the hydrogens attached to it are called α-hydrogens (NCERT Class 12 Chemistry, Chapter 8, Part 2).
Why B is wrong: B describes the hydrogen on the carbonyl carbon itself — that would be the aldehydic hydrogen in RCHO, not the α-hydrogen.
Why C is wrong: C describes a β-hydrogen (two carbons away from the carbonyl), not the α-hydrogen.
Why D is wrong: D is incorrect — the carbonyl oxygen in aldehydes and ketones does not carry a hydrogen.
The acidity of the α-hydrogen in carbonyl compounds is primarily due to:
Answer: C. Removal of the α-hydrogen gives an enolate ion where the negative charge is delocalised over both the α-carbon and the carbonyl oxygen through resonance, significantly stabilising the conjugate base (NCERT Class 12 Chemistry, Chapter 8, Part 2, page 241).
Why A is wrong: A is incomplete — the carbonyl carbon is electrophilic, but α-hydrogen acidity is explained by the stability of the product (enolate), not by the electronegativity of the carbonyl carbon alone.
Why B is wrong: B is a minor contributing factor at best; hyperconjugation does not account for the ~30 pKa unit difference between α-C–H and ordinary C–H bonds.
Why D is wrong: D describes the inductive effect, which contributes weakly. The dominant factor is resonance stabilisation of the enolate, not σ-bond induction.
Which of the following compounds CANNOT undergo aldol condensation?
Answer: A. Benzaldehyde has no α-hydrogen (the carbon adjacent to the carbonyl is part of the aromatic ring with no C–H bond available for enolisation), so it cannot form an enolate and cannot undergo the aldol reaction by itself (NCERT Class 12 Chemistry, Chapter 8, Part 2).
Why B is wrong: B is wrong — propanal has two α-hydrogens on the CH₂ group adjacent to the carbonyl and readily undergoes aldol condensation.
Why C is wrong: C is wrong — acetaldehyde has three α-hydrogens on the methyl group and is the textbook example of the aldol reaction.
Why D is wrong: D is wrong — acetone has six α-hydrogens (three on each methyl group) and undergoes the aldol reaction, though less readily than aldehydes due to steric factors.
When acetaldehyde is treated with dilute NaOH at room temperature, the initial product is:
Answer: C. The aldol reaction of acetaldehyde first produces 3-hydroxybutanal (the β-hydroxy aldehyde or "aldol"). Crotonaldehyde forms only on subsequent heating (dehydration step). The question specifies room temperature, so the aldol is the initial product (NCERT Class 12 Chemistry, Chapter 8, Part 2).
Why A is wrong: A describes the dehydration product (α,β-unsaturated aldehyde) formed on heating — not the initial room-temperature product. Confusing the aldol with the condensation product is a common trap.
Why B is wrong: B would require oxidation, not base-catalysed aldol reaction. Dilute NaOH is a base, not an oxidising agent.
Why D is wrong: D would require reduction (e.g., NaBH₄ or LiAlH₄), not treatment with NaOH.
In the aldol condensation of acetaldehyde, the dehydration step produces an α,β-unsaturated aldehyde. What drives the dehydration to be thermodynamically favourable?
Answer: D. The dehydration product (crotonaldehyde) has a conjugated system where the newly formed C=C double bond is in conjugation with the C=O bond. This extended conjugation provides additional thermodynamic stability, driving the elimination of water (NCERT Class 12 Chemistry, Chapter 8, Part 2).
Why A is wrong: A is incorrect — no new C–O bond forms during dehydration; a C–O bond (the β-hydroxy group) is actually broken as water is lost.
Why B is wrong: B is incorrect — the α-carbon goes from sp³ to sp², not sp. sp hybridisation would correspond to a triple bond or allene system, which is not formed here.
Why C is wrong: C is incorrect — dehydration eliminates water (H₂O), not hydrogen gas (H₂). No gaseous H₂ is released.
A crossed aldol reaction between benzaldehyde and acetaldehyde in the presence of dilute NaOH is synthetically useful because:
Answer: D. Benzaldehyde lacks α-hydrogens (the adjacent carbon is part of the aromatic ring), so it cannot form an enolate. It can only serve as the electrophilic carbonyl partner. Acetaldehyde provides the enolate. This limits the product mixture, making the crossed aldol synthetically useful (NCERT Class 12 Chemistry, Chapter 8, Part 2).
Why A is wrong: A is incorrect on two counts: benzaldehyde has no α-hydrogen, and when both partners have α-hydrogens, crossed aldol gives a mixture of products, not a single product.
Why B is wrong: B is incorrect — the aldol reaction proceeds through an ionic (enolate nucleophilic addition) mechanism, not a free-radical pathway.
Why C is wrong: C is incorrect — benzaldehyde acts as the electrophile (accepts the nucleophilic enolate), not as a nucleophile. The carbonyl carbon of benzaldehyde is the electrophilic centre.
Among acetaldehyde, acetone, and di-tert-butyl ketone [(CH₃)₃C–CO–C(CH₃)₃], the ease of aldol reaction follows the order:
Answer: B. Aldol reaction ease depends on both the acidity of the α-hydrogen (enolate formation) and the electrophilicity of the carbonyl carbon (nucleophilic attack by the enolate). Acetaldehyde has less steric hindrance at the carbonyl and only one alkyl group (less +I effect), so it reacts most readily. Acetone is intermediate. Di-tert-butyl ketone has extreme steric bulk around both the α-carbon and the carbonyl, making enolisation and nucleophilic attack very difficult (NCERT Class 12 Chemistry, Chapter 8).
Why A is wrong: A reverses the correct order — di-tert-butyl ketone is the least reactive due to severe steric hindrance at both the α-position and the carbonyl carbon.
Why C is wrong: C incorrectly places acetone above acetaldehyde. Aldehydes are more reactive than ketones in the aldol reaction because the carbonyl carbon in aldehydes is less sterically hindered and less stabilised by alkyl +I effects.
Why D is wrong: D is incorrect — steric and electronic effects cause significant differences in reactivity. These are not equivalent substrates.
Identify the final product when propanal (CH₃CH₂CHO) undergoes aldol condensation (i.e. aldol reaction followed by heating):
Answer: B. Step 1: The enolate of propanal attacks the carbonyl of another propanal molecule at the α-carbon, giving 3-hydroxy-2-methylpentanal (the aldol, a β-hydroxy aldehyde). Step 2: Heating causes dehydration (loss of H₂O from the β-hydroxy group and the α-hydrogen), yielding 2-methylpent-2-enal, an α,β-unsaturated aldehyde. The methyl branch at C-2 arises because the α-carbon of the nucleophilic propanal retains its ethyl substituent and gains the new C–C bond (NCERT Class 12 Chemistry, Chapter 8, Part 2).
Why A is wrong: A is the aldol (β-hydroxy aldehyde) before dehydration. The question asks for the condensation product, which requires the heating/dehydration step — so the unsaturated aldehyde is the answer, not the hydroxy intermediate.
Why C is wrong: C (pentanal) is a straight-chain aldehyde that would result from a simple carbon-chain extension without branching. The aldol mechanism introduces a branch at the α-position because the new C–C bond forms at the α-carbon, not at the terminal carbon.
Why D is wrong: D has the right six carbons but the wrong skeleton: the new C–C bond joins the α-carbon of one propanal (which carries a methyl) to the carbonyl carbon of the other, giving a methyl branch on a pent-2-enal chain, not an ethyl branch on a but-2-enal chain.
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
Given
• Substrate: Acetaldehyde (CH₃CHO), two molecules• Base: dilute NaOH (catalytic)• Conditions: first at low temperature (aldol step), then heating (dehydration step)
Required
Identify the aldol intermediate and the final condensation product. Name both.
Concept
The aldol reaction occurs in two stages:
1. Base abstracts an α-hydrogen from one molecule of CH₃CHO, forming the resonance-stabilised enolate ion.
2. The enolate (nucleophile) attacks the carbonyl carbon of a second CH₃CHO molecule (electrophile), forming a new C–C bond. Protonation gives the β-hydroxy aldehyde (the aldol).
3. On heating, the aldol loses water (dehydration) to form an α,β-unsaturated aldehyde.
Formula / Reaction scheme
CH₃CHO + CH₃CHO → (dilute NaOH, low temp) → CH₃CH(OH)CH₂CHO → (heat, –H₂O) → CH₃CH=CHCHO
Substitution / Mechanism outline
1. NaOH abstracts one α-hydrogen from CH₃CHO:
CH₃CHO + OH⁻ → ⁻CH₂CHO (enolate) + H₂O
2. Enolate attacks the carbonyl C of the second CH₃CHO:
⁻CH₂CHO + CH₃CHO → CH₃CH(O⁻)CH₂CHO
3. Protonation by water:
CH₃CH(O⁻)CH₂CHO + H₂O → CH₃CH(OH)CH₂CHO + OH⁻
(Aldol intermediate: 3-hydroxybutanal)
4. Heating causes E1cb-type elimination of water:
CH₃CH(OH)CH₂CHO → CH₃CH=CHCHO + H₂O
(Final product: but-2-enal, commonly called crotonaldehyde)
Calculation
This is a reaction-prediction problem, not a numerical calculation. The key reasoning steps are:• Identify the α-hydrogen (on the –CH₃ group of acetaldehyde) — three equivalent α-hydrogens available.• Recognise that the enolate is the nucleophile and the intact carbonyl is the electrophile.• Count carbons: 2 (from first CH₃CHO) + 2 (from second CH₃CHO) = 4-carbon product.
Final answer
• Aldol intermediate: 3-hydroxybutanal (β-hydroxy aldehyde)• Condensation product: but-2-enal (crotonaldehyde, an α,β-unsaturated aldehyde)
Common trap
Confusing the aldol (3-hydroxybutanal) with the condensation product (crotonaldehyde). NEET questions sometimes ask for the "aldol product" (meaning the intermediate before dehydration) versus the "aldol condensation product" (meaning after dehydration). Read the question carefully — if it says "aldol condensation," the answer includes the dehydration step.
Similar NEET-style question
"When propanal is treated with dilute NaOH followed by heating, the major product is:" — apply the same two-step approach (aldol formation → dehydration) to a three-carbon aldehyde.
---
Two molecules of aldehyde or ketone with α-H, in presence of dilute base, combine: forms β-hydroxy carbonyl (aldol), which dehydrates on heating to α,β-unsaturated carbonyl. Acetaldehyde → 3-hydroxybutanal → but-2-enal.
-- NCERT Class 12 Chemistry, Ch. 8, p. 241More in Alcohols, Phenols, Ethers and Carbonyl Compounds: 3 exam traps and mistakes · 2 formulas · 3 question patterns from its other lessons.
2 questions from NEET 2020, 2026. Answers verified against NTA official keys.
Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as :
All 23 past-paper questions from Alcohols, Phenols, Ethers and Carbonyl Compounds →
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →