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 12, Part 2, page 14).
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?
The acidity of the α-hydrogen in carbonyl compounds is primarily due to:
Which of the following compounds CANNOT undergo aldol condensation?
When acetaldehyde is treated with dilute NaOH at room temperature, the initial product is:
In the aldol condensation of acetaldehyde, the dehydration step produces an α,β-unsaturated aldehyde. What drives the dehydration to be thermodynamically favourable?
A crossed aldol reaction between benzaldehyde and acetaldehyde in the presence of dilute NaOH is synthetically useful because:
Among acetaldehyde, acetone, and di-tert-butyl ketone [(CH₃)₃C–CO–C(CH₃)₃], the ease of aldol reaction follows the order:
Identify the final product when propanal (CH₃CH₂CHO) undergoes aldol condensation (i.e. aldol reaction followed by heating):
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, p. 14Stronger acid than phenol due to more effective resonance over two equivalent oxygens. EWG substituents (Cl, NO2) increase acidity.
| Symbol | Quantity | SI Unit |
|---|---|---|
| pKa | -log Ka | - |
Phenols ~10⁶× more acidic than aliphatic alcohols due to resonance stabilisation of phenoxide ion. Electron-withdrawing substituents lower pKa further.
| Symbol | Quantity | SI Unit |
|---|---|---|
| pKa | -log Ka | - |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Organic Reaction Conditions
1° alcohol: PCC/PDC → aldehyde (stops). KMnO4/K2Cr2O7 in acidic → carboxylic acid (continues). 2° alcohol: any oxidiser → ketone. 3° alcohol: not oxidised by ordinary reagents.
Question gives 1° alcohol oxidation with specified reagent.
PCC, PDC, Swern, DMP: mild → stop at aldehyde. KMnO4, K2Cr2O7, CrO3, jones: strong → carboxylic acid. Reagent choice matches desired product.
Root cause: concept gap
Aldehydes more reactive: less steric hindrance, less +I from one R group. Order: HCHO > RCHO > R₂CO.
Root cause: concept gap
Iodoform test ONLY positive for: methyl ketones (CH₃-CO-R), ethanol, secondary alcohols of form CH₃-CH(OH)-R, ethanal.
20 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
Identify the suitable reagent for the following conversion.
The correct order of decreasing acidity of the following aliphatic acids is
Identify the correct reagents that would bring about the following transformation.
In which of the following equilibria, K and K are NOT equal? p c
Taking stability as the factor, which one of the following represents correct relationship?
The right option for the statement "Tyndall effect is exhibited by", is :
Which of the following reactions is the metal displacement reaction? Choose the right option. → ↑
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
inverts aldehyde ketone reactivity
Believes ketones more reactive
includes non methyl ketones
Treats all ketones as positive
ignores resonance vs induction balance
Uses inductive only without considering resonance
Test yourself on this topic with real past-paper questions:
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