Compound (A) ($C_6H_{12}O_2$) on reduction with $LiAlH_4$ gives two compounds (B) and (C). The compound (B) on oxidation with PCC gives compound (D) which upon treatment with dilute NaOH and subsequent heating gives compound (E). Compound (E) on catalytic hydrogenation gives compound (C). The compound (D) is oxidized further to give compound (F) which is found to be a monobasic acid (Molecular weight = 60). Identify the compounds (A), (B), (C), (D), (E) and (F). (½×6=3)

Compound (A) (C₆H₁₂O₂) on reduction with LiAlH₄ gives two compounds (B) and (C). The compound (B) on oxidation

Question

C_6H_{12}O_2

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★

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Concept understanding — Aldol Condensation

Aldol Condensation – From Intuition to Precision

Imagine you have two identical aldehyde molecules. Each has a carbon–oxygen double bond (the carbonyl) that is electron-hungry — the oxygen pulls electron density toward itself, leaving the carbonyl carbon slightly positive. Now look at the carbon next to the carbonyl (the $\alpha$ -carbon). The hydrogens attached to it are unusually acidic, because if you remove one, the negative charge that forms can be stabilised by resonance with the carbonyl group.

What if you could make one molecule act as an "electrophile" (electron-poor, at its carbonyl carbon) and the other as a "nucleophile" (electron-rich, at its $\alpha$ -carbon)? That is exactly what aldol condensation does. The two molecules join together, forming a new carbon–carbon bond.

Note

The name "aldol" comes from aldehyde + alcohol — the initial product has both functional groups.

The Mechanism in Two Stages

Stage 1: The Aldol Addition (the "aldol" part)

Under base catalysis (typically dilute NaOH or KOH), the base abstracts an $\alpha$ -hydrogen from one molecule of the aldehyde or ketone. This generates an enolate ion — a carbanion that is resonance-stabilised.

The enolate then attacks the carbonyl carbon of a second, unreacted molecule. The result is a $\beta$ -hydroxy carbonyl compound — an "aldol" (if starting from an aldehyde) or a "ketol" (if starting from a ketone).

2 \text{CH}_3\text{CHO} \xrightarrow{\text{OH}^-} \text{CH}_3\text{CH(OH)CH}_2\text{CHO}

Stage 2: Dehydration (the "condensation" part)

The $\beta$ -hydroxy carbonyl compound now has an $\alpha$ -hydrogen and a $\beta$ -hydroxyl group. Under the reaction conditions (often mild heat or slightly stronger base), a molecule of water is eliminated. This creates a conjugated $\alpha,\beta$ -unsaturated carbonyl compound — a much more stable product because the double bond is in conjugation with the carbonyl.

\text{CH}_3\text{CH(OH)CH}_2\text{CHO} \xrightarrow{\Delta} \text{CH}_3\text{CH}=\text{CHCHO} + \text{H}_2\text{O}

The overall process — addition followed by dehydration — is called aldol condensation.

The Precise Statement

Aldol condensation is a base-catalysed reaction in which two molecules of an aldehyde or ketone, each possessing at least one $\alpha$ -hydrogen, combine to form a $\beta$ -hydroxy carbonyl compound (the aldol addition product), which then undergoes dehydration to yield an $\alpha,\beta$ -unsaturated carbonyl compound.

Watch out

A common mistake: students think "condensation" means the reaction stops at the $\beta$ -hydroxy stage. In fact, the term "condensation" here refers to the loss of a small molecule (water) — the dehydration step is essential to the full condensation. If no dehydration occurs, the reaction is simply an aldol addition.

Key Conditions and Limitations …

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This problem involves deducing organic compounds through a series of reactions, starting from a known molecular weight and working through reduction, oxidation, and condensation reactions. The key is to identify the monobasic acid (F) first, then sequentially determine (D), (B), (E), (C), and finally (A) as an ester. The compounds are: (A) Butyl acetate, (B) Ethanol, (C) Butan-1-ol, (D) Acetaldehyde, (E) But-2-enal, and (F) Acetic acid.

Organic chemistry problems like this are puzzles where each reaction provides a clue about the structure of the compounds involved. The most effective strategy is to start with the most specific piece of information, which often involves a molecular formula or molecular weight, and then work backward or forward through the reaction sequence. Here, the molecular weight of compound (F) is our anchor. We will use our knowledge of common organic reactions like $LiAlH_4$ reduction, PCC oxidation, Aldol condensation, and catalytic hydrogenation to unravel the structures.

Identify Compound (F):

The problem states that compound (F) is a monobasic acid with a molecular weight of $60$ . A monobasic acid has the general formula $RCOOH$ .
- If $R = H$ , the acid is formic acid ( $HCOOH$ ). Its molecular weight is $1 (H) + 12 (C) + 16 (O) + 16 (O) + 1 (H) = 46$ . This does not match.
- If $R = CH_3$ , the acid is acetic acid ( $CH_3COOH$ ). Its molecular weight is $12 (C) + 3 (H) + 12 (C) + 16 (O) + 16 (O) + 1 (H) = 60$ . This matches the given molecular weight. Therefore, (F) is Acetic acid ( $CH_3COOH$ ).
Identify Compound (D):

Compound (D) is oxidized further to give compound (F) ( $CH_3COOH$ ). Carboxylic acids are typically formed by the oxidation of aldehydes or primary alcohols. Since the problem states "oxidized further", it implies (D) is an intermediate oxidation state before the carboxylic acid. Aldehydes are oxidized to carboxylic acids.

Thus, (D) must be the aldehyde corresponding to acetic acid.

CH_3CHO \xrightarrow{[O]} CH_3COOH

Therefore, **(D) is Acetaldehyde ($CH_3CHO$)**.

3. Identify Compound (B):

Compound (B) on oxidation with PCC gives compound (D) ( $CH_3CHO$ ). PCC (Pyridinium chlorochromate) is a mild oxidizing agent used to convert primary alcohols to aldehydes and secondary alcohols to ketones, without further oxidizing the aldehyde to a carboxylic acid.

Since (D) is an aldehyde, (B) must be the primary alcohol that oxidizes to acetaldehyde.

CH_3CH_2OH \xrightarrow{PCC} CH_3CHO

Therefore, **(B) is Ethanol ($CH_3CH_2OH$)**.

4. Identify Compound (E):

Compound (D) ( $CH_3CHO$ ) upon treatment with dilute $NaOH$ and subsequent heating gives compound (E). This reaction sequence is characteristic of an Aldol condensation. Acetaldehyde has alpha-hydrogens, making it capable of undergoing Aldol condensation.

In the presence of dilute base, two molecules of acetaldehyde condense to form 3-hydroxybutanal (an aldol). Upon heating, this aldol undergoes dehydration to form an $\alpha, \beta$ -unsaturated aldehyde.

2CH_3CHO \xrightarrow{dil. NaOH} CH_3CH(OH)CH_2CHO \xrightarrow{\Delta} CH_3CH=CHCHO + H_2O

Therefore, **(E) is But-2-enal ($CH_3CH=CHCHO$)**.

5. Identify Compound (C):

Compound (E) ( $CH_3CH=CHCHO$ ) on catalytic hydrogenation gives compound (C). Catalytic hydrogenation (e.g., using $H_2$ with a catalyst like $Ni$ , $Pd$ , or $Pt$ ) reduces both carbon-carbon double bonds and carbonyl groups (aldehydes/ketones) to single bonds.

CH_3CH=CHCHO \xrightarrow{H_2/catalyst} CH_3CH_2CH_2CH_2OH

Therefore, **(C) is Butan-1-ol ($CH_3CH_2CH_2CH_2OH$)**.

6. Identify Compound (A):

Compound (A) ( $C_6H_{12}O_2$ ) on reduction with $LiAlH_4$ gives two compounds (B) ( $CH_3CH_2OH$ ) and (C) ( $CH_3CH_2CH_2CH_2OH$ ).

* First, let's analyze the molecular formula of (A), $C_6H_{12}O_2$ . The degree of unsaturation (DoU) is calculated as $DoU = C + 1 - H/2 - X/2 + N/2$ . For $C_6H_{12}O_2$ :

DoU = 6 + 1 - \frac{12}{2} = 7 - 6 = 1

    A DoU of 1 for a compound with two oxygen atoms suggests it is likely an ester ($RCOOR'$), which contains one carbonyl ($C=O$) group. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 16 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Aldol condensation does not occur between$
(a) two different aldehydes
(b) two different ketones
(c) an aldehyde and a ketone
(d) an aldehyde and an ester
›Reveal solutionSolution
Aldol condensation is defined between aldehydes/ketones bearing $\alpha$ -hydrogens; esters instead undergo the distinctly-named Claisen condensation, so the aldehyde–ester combination falls outside 'aldol condensation'.

(a) Two different aldehydes: a crossed aldol condensation is possible — e.g. base generates the enolate of one aldehyde, which attacks the carbonyl carbon of the other, followed by dehydration to the $\alpha,\beta$ -unsaturated carbonyl product. This does occur (though it can give a mixture of products).

(b) Two different ketones: a crossed aldol condensation between ketones can also occur, though it is often slower and lower-yielding (ketones are less electrophilic and more sterically hindered than aldehydes), but it is still classified as an aldol-type reaction.

(c) An aldehyde and a ketone: a crossed aldol (Claisen–Schmidt-type when aromatic) reaction readily occurs — typically the aldehyde (lacking $\alpha$ -H if aromatic, or simply more electrophilic) acts as the electrophile and the ketone's enolate as the nucleophile. …
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CBSE 2025Set D1 markMCQ
$Q. Which of the following would give Aldol condensation reaction?$
(a) CCl3CHO
(b) CH3-C(CH3)(CH3)-CHO
(c) CH3CHO
(d) HCHO
›Reveal solutionSolution
Aldol condensation needs an alpha-H; only CH3CHO has alpha-hydrogens.

Aldol condensation occurs only with aldehydes/ketones that possess at least one alpha-hydrogen (a hydrogen on the carbon next to the carbonyl group).

CCl3CHO: alpha-carbon carries three Cl, no alpha-H.

(CH3)3C-CHO (2,2-dimethylpropanal): the alpha-carbon is quaternary, no alpha-H. …
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CBSE 2025Set ANNUAL1 mark
$Q. Name the reagent which is used to convert aldehydes or ketones having alpha-hydrogen into beta-hydroxy aldehydes or beta-hydroxy ketones.$
›Reveal solutionSolution

Aldehydes/ketones with an alpha-hydrogen undergo base-catalysed aldol addition with dilute NaOH, giving a beta-hydroxy carbonyl product.

When an aldehyde or ketone possessing at least one alpha-hydrogen is treated with dilute aqueous sodium hydroxide (or another dilute alkali), the base removes an alpha-hydrogen to form an enolate, which then attacks the carbonyl carbon of a second carbonyl molecule. The result is a beta-hydroxy aldehyde (aldol) or beta-hydroxy ketone (ketol) - t …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Assertion (A): Acetaldehyde undergoes aldol-condensation with NaOH (dil). Reason (R): Aldehydes which do not contain α-hydrogen undergo aldol-condensation.$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
›Reveal solutionSolution

Acetaldehyde does undergo aldol condensation with dilute NaOH (true), but the stated reason is backwards — aldol condensation requires aldehydes WITH $\alpha$ -hydrogens, not without them.

Assertion: Acetaldehyde ( $CH_3CHO$ ) has $\alpha$ -hydrogens (on the methyl carbon adjacent to the carbonyl) and readily undergoes base-catalyzed (dil. NaOH) aldol condensation to give 3-hydroxybutanal, which on further heating dehydrates to crotonaldehyde. This is TRUE.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The product formed in Aldol condensation is ........................ .$
(a) an α, β unsaturated ester
(b) an α-hydroxy aldehyde or ketone
(c) a β-hydroxy acid
(d) a β-hydroxy aldehyde or ketone.
›Reveal solutionSolution

Base-catalysed self-addition of an aldehyde/ketone with an α-hydrogen first gives a β-hydroxy carbonyl compound (the aldol); loss of water on heating gives the α,β-unsaturated product.

In Aldol condensation, a base (e.g. dilute NaOH) removes an α-hydrogen from one molecule of aldehyde/ketone to form a carbanion (enolate), which then attacks the carbonyl carbon of a second molecule. The direct product of this nucleophilic addition — before any dehydration — is a β-hydroxy aldehyde (an 'aldol') or β-hydroxy ketone (a 'ketol'), because the newly formed –OH group is on the carbon β to the original c …

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CBSE 2024Set B1 markMCQ
$Q. The nature of α-hydrogen in aldehydes is$
(a) Acidic
(b) Basic
(c) Neutral
(d) Amphoteric
›Reveal solutionSolution

The hydrogen atoms attached to the carbon adjacent to the carbonyl group (the alpha-carbon) are acidic because the carbanion left behind after their removal is resonance-stabilised by the adjacent C=O group.

In aldehydes and ketones, the carbonyl carbon is electron-withdrawing (due to the electronegative oxygen). This polarises the C-H bonds on the alpha-carbon, and if a base removes an alpha-hydrogen as H+, the resulting carbanion is stabilised by delocalisation into the carbonyl pi-system, forming an enolate ion:

R-CO-CH2-R' (base removes H+) -> R-C(O-)=CH-R' (enolate, resonance-stabilised)

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following does not give aldol condensation reaction?$
(a) CH3CHO (acetaldehyde)
(b) C6H5-CHO (benzaldehyde, an aromatic ring with -CHO group)
(c) cyclohexanecarbaldehyde (a non-aromatic six-membered ring with -CHO group)
(d) CH3COCH3 (acetone)
›Reveal solutionSolution
Aldol condensation requires the carbonyl compound to have at least one alpha-hydrogen, which is removed to form the nucleophilic enolate that attacks a second carbonyl molecule.

Checking each for alpha-hydrogens (hydrogens on the carbon directly attached to the C=O carbon):

CH3CHO (acetaldehyde): the CH3 carbon (alpha carbon) has 3 H's -> has alpha-H -> undergoes aldol condensation. …
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CBSE 2023Set 56/2/11 markMCQ
$CH_3-CHO$
›Reveal solutionSolution
Aldol condensation requires at least one $\alpha$ -hydrogen next to the carbonyl group. Benzaldehyde ( $C_6H_5-CHO$ ) has no $\alpha$ -hydrogen, so it cannot undergo aldol condensation. The correct answer is (C).

The Cannizzaro reaction and aldol condensation are two classic carbonyl reactions that often trip students up because they seem to compete. The key difference is simple: aldol condensation needs an $\alpha$ -hydrogen (a hydrogen on the carbon atom directly adjacent to the carbonyl group), while the Cannizzaro reaction happens when there is no $\alpha$ -hydrogen at all.

Let’s see why this matters for each option.

Identify the requirement for aldol condensation

For a carbonyl compound to undergo aldol condensation, it must have at least one hydrogen atom on the $\alpha$ -carbon (the carbon next to the $C=O$ group). This hydrogen is acidic enough to be removed by a base, forming an enolate ion. That enolate then attacks another carbonyl molecule, leading to the $\beta$ -hydroxy aldehyde or ketone, which can dehydrate to give an $\alpha,\beta$ -unsaturated product.

Examine each compound

(A) $CH_3-CHO$ (acetaldehyde): The $\alpha$ -carbon is the $CH_3$ group. It has three $\alpha$ -hydrogens. So it readily undergoes aldol condensation.

(B) $C_6H_{11}-CHO$ (cyclohexanecarbaldehyde): The aldehyde group is attached to a saturated cyclohexane ring. The carbon directly attached to the $-CHO$ is a ring carbon — that carbon has at least one hydrogen (since it’s $sp^3$ hybridized and not fully substituted). So there is at least one $\alpha$ -hydrogen. This compound can undergo aldol condensation. …
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CBSE 2023Set ANNUAL1 mark
$Q. Write chemical reaction of Aldol Condensation.$
›Reveal solutionSolution

Aldol condensation: two aldehyde/ketone molecules combine (base-catalysed) to give a β-hydroxy carbonyl compound, which loses water on heating to give an α,β-unsaturated carbonyl compound.

Taking acetaldehyde as an example, in the presence of dilute NaOH, two molecules of acetaldehyde undergo a nucleophilic addition (the α-hydrogen of one molecule attacks the carbonyl carbon of another) to give 3-hydroxybutanal (an aldol):

$2CH_3CHO \xrightarrow{\text{dil. NaOH}} CH_3\text{-}CH(OH)\text{-}CH_2\text{-}CHO$ …

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CBSE 2022Set E1 markMCQ
$Q. Which of the following does not give Aldol condensation ?$
(a) Methanal
(b) Ethanal
(c) Propanone
(d) Propanal
›Reveal solutionSolution
Aldol condensation requires at least one alpha-hydrogen; methanal (HCHO) has none, so it does not give aldol.

Aldol condensation needs an alpha-hydrogen (a H on the carbon next to the carbonyl) which a base removes to form the enolate nucleophile.

Methanal, HCHO: carbonyl carbon has only H's, NO alpha-carbon and no alpha-H -> cannot form an aldol.

Ethanal (CH3CHO): has alpha-H -> gives aldol. …
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CBSE 2022Set ANNUAL1 mark
$Q. Write the reagent required (denoted '?') for the following reaction: benzaldehyde (ring-CHO) --?--> cinnamic-acid-type product (ring-CH=CHCOOH).$
›Reveal solutionSolution

Benzaldehyde condenses with acetic anhydride (catalysed by sodium acetate, on heating) to give cinnamic acid — the classic Perkin reaction.

Identifying the reaction: An aromatic aldehyde converting to an alpha,beta-unsaturated aromatic acid (ring-CH=CH-COOH) is the Perkin condensation.

Reagent and conditions: Acetic anhydride, (CH3CO)2O, in the presence of its corresponding sodium salt, sodium acetate (CH3COONa), heated together.

…

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CBSE 2022Set ANNUAL1 mark
$Q. Write the reagent required (denoted '?') for the following reaction: CH3COCH3 --?--> (CH3)2CHOH-CH2COCH3.$
›Reveal solutionSolution

Two molecules of acetone undergo a base-catalysed aldol ADDITION (not condensation, since no dehydration occurs here) to give diacetone alcohol.

Identifying the reaction: Acetone dimerising, with retention of the -OH (no C=C formed), to (CH3)2C(OH)-CH2-CO-CH3 is the aldol addition of acetone with itself.

Reagent and conditions: A dilute base such as barium hydroxide, Ba(OH)2 (or dilute NaOH), at low/room temperature — mild conditions that stop the reaction at the beta-hydroxy ketone stage without dehydrating it.

…

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