(a) (i) Carry out the following conversions : (1) Ethanal to But-2-en-1-al (2) Propanoic acid to 2-chloropropa

Question

Q. (a)

(i) Carry out the following conversions :

(1) Ethanal to But-2-en-1-al

(2) Propanoic acid to 2-chloropropanoic acid

(ii) An alkene with molecular formula

C_5H_{10}

on ozonolysis gives a mixture of two compounds 'B' and 'C'. Compound 'B' gives positive Fehling test and also reacts with iodine and NaOH solution. Compound 'C' does not give Fehling solution test but forms iodoform. Identify the compounds 'A', 'B' and 'C'.

(OR)

(b)

(i) Distinguish with a suitable chemical test :

(1)

CH_3COCH_2CH_3

and

CH_3CH_2CH_2CHO

(2) Ethanal and Ethanoic acid

(ii) Write the structure of oxime of acetone.

(iii) Identify A to D :

CH_3COOH \xrightarrow{PCl_5} A \xrightarrow{H_2/Pd\text{-}BaSO_4} B \xrightarrow[(ii)\ H_3O^+]{(i)\ CH_3MgBr} C \xrightarrow{LiAlH_4} D

CBSECBSE Class XII Board 2023Subjective· 5mImportance★★★★★

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Part (a)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 …

Part (b)Concept understanding — Rosenmund Reduction

Rosenmund Reduction: From Intuition to Precision

Imagine you have an acyl chloride — a highly reactive molecule with a chlorine atom attached to a carbonyl carbon. You want to turn it into an aldehyde, which has a hydrogen atom in place of that chlorine. The obvious idea is to add hydrogen gas ( $H_2$ ) in the presence of a metal catalyst. But here's the problem: if you use plain palladium or platinum, the reaction doesn't stop at the aldehyde. The catalyst is too aggressive — it keeps reducing the aldehyde further, all the way to a primary alcohol.

So you need a way to tame the catalyst. You want it to be just active enough to replace the chlorine with hydrogen, but not so active that it attacks the aldehyde group you just made. That's the core intuition behind the Rosenmund reduction.

Note

The key insight: a poisoned catalyst is not a broken catalyst — it's a selective catalyst. The poison blocks the most reactive sites on the metal surface, leaving only milder sites that can still reduce the acyl chloride but cannot reduce the aldehyde.

The Precise Statement

Rosenmund reduction is the catalytic hydrogenation of an acyl chloride ( $RCOCl$ ) over palladium metal supported on barium sulfate ( $BaSO_4$ ), where the catalyst is partially deactivated (poisoned) to stop the reaction at the aldehyde stage.

The general reaction:

RCOCl + H_2 \xrightarrow{Pd/BaSO_4, \text{poison}} RCHO + HCl

The barium sulfate acts as a support, spreading the palladium into fine particles. But the real trick is the poison — typically a small amount of sulfur or a sulfur-containing compound (like thiourea or quinoline-sulfur). This poison binds to the most active sites on the palladium surface, preventing over-reduction.

RCOCl + H_2 \xrightarrow{Pd/BaSO_4, \text{quinoline-S}} RCHO + HCl

Why It Works — The Mechanism in Simple Terms

Adsorption: The acyl chloride adsorbs onto the palladium surface. The $C=O$ and $C-Cl$ bonds both interact with the metal.
Hydrogenolysis: Hydrogen gas dissociates on the metal surface into atomic hydrogen. These hydrogen atoms attack the $C-Cl$ bond, replacing chlorine with hydrogen. This step is fast and happens even on the poisoned sites.
Desorption: The aldehyde forms and desorbs from the catalyst surface. Because the most active sites are blocked, the aldehyde cannot re-adsorb and undergo further reduction to an alcohol.

Watch out

A common mistake: thinking the poison stops the reaction. It doesn't — it slows down the unwanted second step. If you use too much poison, the first step also stops. If you use too little, you get alcohol. The art is in the precise amount of poison.

Why Barium Sulfate?

You might wonder: why $BaSO_4$ specifically? It's not just any support. Barium sulfate is insoluble and chemically inert under the reaction conditions. It provides a large surface area for the palladium without interfering with the chemistry. More importantly, it doesn't adsorb the aldehyde product, so the aldehyde leaves the catalyst quickly — another factor that prevents over-reduction.

Exam-Relevant Points

Aspect	Detail
Reactant	Acyl chloride ( $RCOCl$ )
Reagent	$H_2$ gas
Catalyst	$Pd$ supported on $BaSO_4$ , poisoned
Poison	Sulfur / quinoline-sulfur / thiourea
Product	Aldehyde ( $RCHO$ )
Byproduct	$HCl$ gas
Limitation	Only works for acyl chlorides; not for other acid derivatives

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Part (a): ethanal → but‑2‑en‑1‑al by aldol condensation; propanoic acid → 2‑chloropropanoic acid by HVZ; and (matching the printed labels) A = 2‑methylbut‑2‑ene (the alkene), B = ethanal, C = propanone. Part (b): distinguishing tests (Fehling/Tollens; $\text{NaHCO}_3$ ), acetone oxime $(\text{CH}_3)_2\text{C}{=}\text{NOH}$ , and A = $\text{CH}_3\text{COCl}$ , B = $\text{CH}_3\text{CHO}$ , C = D = propan‑2‑ol.

Part (a)

(i)(1) Ethanal → but‑2‑en‑1‑al. Two ethanal molecules undergo aldol condensation: the base‑generated enolate of one adds to the carbonyl of the other giving 3‑hydroxybutanal, which dehydrates on warming to the conjugated enal:

2\,\text{CH}_3\text{CHO} \xrightarrow[\Delta]{\text{dil. NaOH}} \text{CH}_3\text{CH}{=}\text{CHCHO} \;(\text{but‑2‑en‑1‑al, crotonaldehyde}) + \text{H}_2\text{O}

(i)(2) Propanoic acid → 2‑chloropropanoic acid. Direct halogenation fails; use the Hell–Volhard–Zelinsky (HVZ) reaction — $\text{Cl}_2$ with a catalytic amount of red P (forms $\text{PCl}_3$ ), which α‑chlorinates the acid:

\text{CH}_3\text{CH}_2\text{COOH} \xrightarrow[\text{red P}]{\text{Cl}_2} \text{CH}_3\text{CHClCOOH} + \text{HCl}

(ii) Structure elucidation from ozonolysis. The alkene A ( $\text{C}_5\text{H}_{10}$ ) is cleaved to two carbonyls B and C.

B gives a positive Fehling test ⇒ it is an aldehyde; it also gives a positive iodoform test ⇒ it has a $\text{CH}_3\text{CO}$ – unit. The only aldehyde satisfying both is ethanal, $\text{CH}_3\text{CHO}$ .
C does not give Fehling ⇒ a ketone; it gives iodoform ⇒ a methyl ketone. With B being $C_2$ , C must be $C_3$ = propanone, $\text{CH}_3\text{COCH}_3$ . …

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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 27 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The catalyst used in Rosenmund's reduction is:$
(a) Zn-Hg
(b) Raney-Ni
(c) Na in ether
(d) Pd-BaSO4
›Reveal solutionSolution

Rosenmund reduction converts an acid chloride to an aldehyde using $H_2$ over a poisoned Pd-BaSO4 catalyst.

Rosenmund's reduction is the controlled catalytic hydrogenation of an acyl (acid) chloride to an aldehyde:

$R-COCl + H_2 \xrightarrow{Pd/BaSO_4} R-CHO + HCl$

The catalyst used is palladium deposited on barium sulphate, and crucially it is 'poisoned' (partially deactivated) — usually with a small amount of sulphur or quinoline (Lindlar-type deactivation). This poisoning is essential because it prevents the catalyst from being active enough to further reduce the newly-formed aldehyde all the way down to a primary alcohol; the reaction is thereby halted cleanly at the aldehyde stage.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Name the reaction: C6H5-COCl (benzoyl chloride) is converted to C6H5-CHO (benzaldehyde) using H2 / Pd-BaSO4.$
›Reveal solutionSolution

Selective reduction of an acid chloride to an aldehyde using $H_2$ over a poisoned (partially deactivated) Pd–BaSO $_4$ catalyst is called the Rosenmund reduction.

$C_6H_5COCl \xrightarrow[Pd - BaSO_4\ (\text{poisoned with S/quinoline})]{H_2} C_6H_5CHO + HCl$

…

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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 2026Set ANNUAL1 markMCQ
$Q. Acid chloride is hydrogenated over suspended palladium as catalyst on barium sulphate and form aldehydes. This reaction is called:$
(a) Stephen reaction
(b) Carbyl amine reaction
(c) Hunsdiecker reaction
(d) Rosenmund reduction
›Reveal solutionSolution

Acid chloride + H2 over Pd/BaSO4 -> aldehyde: this is the Rosenmund reduction.

In the Rosenmund reduction, an acyl chloride is reduced to an aldehyde by hydrogen gas over a palladium catalyst supported on barium sulphate:

R-COCl + H2 --(Pd/BaSO4)--> R-CHO + HCl

…

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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 markMCQ
$Q. Which of the following method is not used to prepare Ketone?$
(a) Oxidation of Secondary alcohol
(b) Catalytic dehydrogenation of secondary alcohol
(c) Dry distillation of calcium salt of fatty acids
(d) Rosenmund's reduction
›Reveal solutionSolution
Rosenmund reduction (acid chloride + H2, Pd/BaSO4 poisoned catalyst) stops at the aldehyde stage; it is a method for preparing aldehydes, not ketones.

Oxidation of a secondary alcohol (with, e.g., PCC or acidified K2Cr2O7): gives a ketone — a valid ketone-preparation method.

Catalytic dehydrogenation of a secondary alcohol (passing vapours over hot Cu): gives a ketone — a valid method.

Dry distillation of the calcium salt of a fatty acid (Ca salt of a carboxylic acid, heated): gives a ketone (e.g., calcium acetate → acetone) — a valid method. …
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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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Q.(a)

Part (a)

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