(a) (i) Write the product(s) when : (I) One mol of ethanal is treated with 1 mol of CH₃OH in the presence of d

Question

Q. (a)

(i) Write the product(s) when : (I) One mol of ethanal is treated with 1 mol of

CH_3OH

in the presence of dry HCl gas. (II) Benzaldehyde is treated with conc. NaOH. (III) Ethanoic acid is heated in the presence of

P_2O_5

(ii) Write a simple chemical test to distinguish between Ethanal and Propanal.

(iii) Write the name of the reagent to transform Allyl alcohol to Propenal.

(OR)

(b)

(i) Draw the structure of the semicarbazone of acetone.

(ii) Why are

\alpha

-hydrogen atoms of aldehydes and ketones acidic in nature ?

(iii) Arrange the following compounds in increasing order of their reactivity towards HCN :

CH_3COCH_3

CH_3CHO

(CH_3)_3C-\overset{O}{\underset{\|}{C}}-CH_3

(iv) Write the reaction involved in Etard reaction.

(v) Write the product when Cyclohexanecarbaldehyde (

C_6H_{11}CHO

, a cyclohexane ring bearing a

-CHO

group) reacts with

Zn(Hg)/\text{conc. } HCl

CBSECBSE Class XII Board 2026Subjective· 5mImportance★★★★★

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Part (a)Concept understanding — Cannizzaro Reaction

The Cannizzaro Reaction: An Intuition

Imagine two identical molecules of an aldehyde meeting in a strongly basic solution. Normally, aldehydes with a hydrogen on the carbon next to the carbonyl (the $\alpha$ -carbon) undergo aldol condensation. But what if that $\alpha$ -carbon has no hydrogen at all? The molecule cannot do the usual reaction. Instead, something remarkable happens: one aldehyde molecule gets reduced to an alcohol, while the other gets oxidized to a carboxylic acid (which, in base, exists as its salt). One molecule gives away electrons; the other accepts them. This is a disproportionation — a single species (the aldehyde) acts as both the oxidising and the reducing agent.

Note

The reaction requires concentrated alkali (typically NaOH or KOH). Dilute base will not work.

The Precise Statement

The Cannizzaro reaction is the base-catalysed disproportionation of an aldehyde that lacks an $\alpha$ -hydrogen atom (i.e., the carbon adjacent to the $-\text{CHO}$ group has no hydrogen attached). In the presence of concentrated aqueous or alcoholic alkali, two molecules of such an aldehyde yield one molecule of a primary alcohol and one molecule of the salt of a carboxylic acid.

The general equation (using benzaldehyde as the classic example):

2\,\text{C}_6\text{H}_5\text{CHO} + \text{NaOH} \xrightarrow{\text{conc.}} \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{C}_6\text{H}_5\text{COONa}

Benzaldehyde gives benzyl alcohol and sodium benzoate.

2\,\text{RCHO} \xrightarrow[\text{conc. alkali}]{\text{no }\alpha\text{-H}} \text{RCH}_2\text{OH} + \text{RCOO}^- \text{M}^+

Why "No $\alpha$ -Hydrogen" Matters

The key is the mechanism. The first step is the attack of hydroxide ion ( $\text{OH}^-$ ) on the carbonyl carbon. This forms a tetrahedral intermediate. If an $\alpha$ -hydrogen were present, this intermediate would lose water and form an enolate — leading to aldol condensation. Without that hydrogen, the intermediate cannot do that. Instead, it transfers a hydride ion ( $\text{H}^-$ ) to a second molecule of aldehyde. That second molecule gets reduced to the alkoxide (which later picks up a proton to become the alcohol), while the first molecule becomes the carboxylate.

Watch out

A common mistake: thinking the reaction works for any aldehyde. It does not. If the aldehyde has even one $\alpha$ -hydrogen, the aldol pathway dominates. Only aldehydes like formaldehyde ( $\text{HCHO}$ ), benzaldehyde ( $\text{C}_6\text{H}_5\text{CHO}$ ), and trimethylacetaldehyde ( $(\text{CH}_3)_3\text{CCHO}$ ) undergo the Cannizzaro reaction.

Crossed Cannizzaro Reaction …

Part (b)Concept understanding — Clemmensen Reduction

Clemmensen Reduction: From Intuition to Precision

Imagine you have a ketone or an aldehyde — a molecule with a C=O group. You want to rip that oxygen out entirely and replace it with two hydrogens, turning the carbonyl carbon into a plain CH₂ group. That is exactly what the Clemmensen reduction does.

Why would you want that? Because sometimes you need a hydrocarbon chain where a carbonyl used to be. For example, if you have a ketone attached to a benzene ring, the Clemmensen reduction gives you an alkylbenzene — a common starting material in organic synthesis.

The reaction uses zinc amalgam (Zn-Hg) and concentrated hydrochloric acid (HCl). The zinc amalgam is simply zinc metal that has been treated with mercury to form an alloy — this makes the zinc surface more reactive and prevents side reactions.

\ce{R-C(=O)-R' + 4[H] ->[Zn-Hg, conc. HCl] R-CH2-R' + H2O}

The "4[H]" is a shorthand for the reducing equivalent supplied by the zinc in acid. The carbonyl oxygen leaves as water, and two hydrogens attach to the carbon.

The Mechanism (What Actually Happens)

The mechanism is not fully settled, but the most widely accepted path involves a carbocation intermediate. Here is the key insight: the reaction works because concentrated HCl protonates the carbonyl oxygen, making the carbon highly electrophilic. Zinc then donates electrons, and the oxygen leaves as water, leaving behind a positively charged carbon (a carbocation). That carbocation then picks up two hydride-like hydrogens from the zinc surface.

Watch out

The Clemmensen reduction only works for aldehydes and ketones. It does not reduce carboxylic acids, esters, or amides — those require different conditions (like LiAlH₄). Also, the reaction requires strong acid, so any functional group that is acid-sensitive (like an alcohol or an alkene) will be destroyed.

When to Use It — and When Not To

The Clemmensen reduction is a classic method, but it has a major limitation: the strongly acidic conditions. If your molecule contains an acid-sensitive group (a tertiary alcohol, an acetal, a double bond that might rearrange), this reaction will ruin it.

For those cases, you use the Wolff-Kishner reduction instead — that uses hydrazine and strong base, so it works under basic conditions. The two reactions are complementary: Clemmensen for acid-stable substrates, Wolff-Kishner for base-stable ones. …

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ethanal + 1 CH₃OH → hemiacetal; benzaldehyde + NaOH → benzyl alcohol + sodium benzoate (Cannizzaro); acetic acid + $P_2O_5$ → acetic anhydride; iodoform distinguishes ethanal from propanal; allyl alcohol → propenal with PCC.
acetone semicarbazone is $(CH_3)_2C{=}N{-}NH{-}CO{-}NH_2$ ; $\alpha$ -H acidity is due to enolate resonance; HCN reactivity $(CH_3)_3CCOCH_3 < CH_3COCH_3 < CH_3CHO$ ; Etard gives benzaldehyde; Clemmensen turns cyclohexanecarbaldehyde into methylcyclohexane.

Part (a)

(i) Products.

(I) With only one equivalent of methanol and dry HCl, ethanal adds a single alcohol molecule across the C=O to give the hemiacetal $CH_3CH(OH)(OCH_3)$ (1-methoxyethanol). A second equivalent (with removal of water) would be needed for the full acetal, which is not the case here.
(II) Benzaldehyde has no $\alpha$ -hydrogen, so instead of aldol it undergoes the Cannizzaro reaction (a disproportionation) in concentrated base: one molecule is oxidised, another reduced.

2\,C_6H_5CHO + \text{conc. NaOH} \rightarrow C_6H_5CH_2OH + C_6H_5COONa

(III) $P_2O_5$ is a strong dehydrating agent; heating acetic acid with it removes water from two molecules to give acetic anhydride: $2CH_3COOH \xrightarrow{P_2O_5} (CH_3CO)_2O + H_2O$ .

(ii) Distinguishing ethanal from propanal. Both are aldehydes (both give Tollens'/Fehling's), so use the iodoform test: only a compound with a $CH_3CO-$ (methyl ketone/acetaldehyde-type) group responds. Ethanal ( $CH_3CHO$ ) gives a yellow precipitate of iodoform ( $CHI_3$ ) with $I_2$ /NaOH; propanal ( $CH_3CH_2CHO$ ) does not. …

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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 18 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. In Clemmensen reduction, carbonyl compound is reduced in the presence of$
(a) Zn-Hg + HCl
(b) Na + C2H5OH
(c) Zn-Hg + HNO3
(d) Na + dry ether
›Reveal solutionSolution

Clemmensen reduction converts an aldehyde or ketone's C=O group completely to CH2 using zinc amalgam and concentrated hydrochloric acid.

In the Clemmensen reduction, the carbonyl compound (aldehyde or ketone) is refluxed with zinc amalgam (Zn-Hg) and concentrated HCl:

R2C=O --(Zn-Hg / conc. HCl)--> R2CH2

…

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CBSE 2026Set ANNUAL1 mark
$Q. Name the reaction in which aldehydes with no alpha-hydrogen atom undergo self-oxidation and reduction reaction in presence of conc. alkali.$
›Reveal solutionSolution

The Cannizzaro reaction is a base-mediated disproportionation unique to aldehydes with no alpha-hydrogen, where one molecule is reduced to an alcohol and another is oxidised to a carboxylate salt.

Aldehydes that have no alpha-hydrogen (i.e., no hydrogen atom on the carbon adjacent to the -CHO group, such as benzaldehyde, formaldehyde, or trimethylacetaldehyde) cannot undergo the usual base-catalysed aldol condensation, since that requires removing an alpha-hydrogen to form an enolate.

…

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CBSE 2025Set 56/5/11 markMCQ
$CH_3{-}CH(CH_3){-}CHO$
›Reveal solutionSolution

The Cannizzaro reaction requires an aldehyde with no alpha-hydrogen atoms (i.e., not enolizable). Among the given options, only $(CH_3)_3C{-}CHO$ (pivalaldehyde) has no $\alpha$ -H, so it alone undergoes the reaction. The correct option is (B).

Why the Cannizzaro reaction happens — and when it doesn’t

The Cannizzaro reaction is a disproportionation of an aldehyde in concentrated base: one molecule is reduced to a primary alcohol, the other is oxidised to a carboxylate salt. But this reaction only works if the aldehyde cannot form an enolate. Why? Because if there is even one hydrogen on the carbon next to the carbonyl (the $\alpha$ -carbon), the base will preferentially pull that hydrogen off, leading to aldol condensation instead. So the key condition is: no $\alpha$ -hydrogen atoms.
$R{-}CHO$
Now let’s examine each option.

1. Option (A): $CH_3{-}CH(CH_3){-}CHO$

Draw the structure: the carbonyl carbon is at the end. The $\alpha$ -carbon is the one directly attached to the CHO group — that’s the carbon bearing the $CH_3$ and $CH(CH_3)$ groups. Count its hydrogens: it has one hydrogen (since it’s a tertiary carbon with one H). That one $\alpha$ -H makes this aldehyde enolizable. In strong base, that H will be abstracted, and the resulting enolate will undergo aldol condensation — not Cannizzaro.

Watch out

A common mistake: thinking that a branched aldehyde automatically has no $\alpha$ -H. Check the $\alpha$ -carbon carefully — even one H is enough to block the Cannizzaro path.

2. Option (B): $(CH_3)_3C{-}CHO$ …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following aldehydes undergo Cannizzaro reaction -$
(a) CH3-CHO
(b) CH3-CH2-CHO
(c) (CH3)2-CH-CHO
(d) C6H5-CHO
›Reveal solutionSolution
The Cannizzaro (base-catalysed self oxidation-reduction/disproportionation) reaction occurs only with aldehydes that lack an alpha-hydrogen atom.

An aldehyde with alpha-hydrogens instead undergoes base-catalysed aldol condensation (since the alpha-H is acidic enough to be removed by strong base, forming an enolate that attacks another aldehyde molecule), NOT the Cannizzaro reaction.

Checking each option:

CH3-CHO (acetaldehyde): has alpha-H (on the CH3 carbon) - undergoes aldol, not Cannizzaro.

CH3-CH2-CHO (propanal): has alpha-H - undergoes aldol.

(CH3)2-CH-CHO (isobutyraldehyde): has alpha-H (on the CH carbon) - undergoes aldol. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Cannizzaro's reaction is not given by -$
(a) 1-methylcyclohexane-1-carbaldehyde (a cyclohexane ring drawn with both a -CHO group and a -CH3 group attached to the same ring carbon, i.e. no alpha-hydrogen on the carbonyl carbon)
(b) Benzaldehyde (a benzene ring drawn with a -CHO group attached, no alpha-hydrogen)
(c) HCHO (formaldehyde, printed as a plain chemical formula, no ring drawn)
(d) $CH_3CHO$ (acetaldehyde, printed as a plain chemical formula, no ring drawn)
›Reveal solutionSolution
Cannizzaro's reaction needs an aldehyde with NO alpha-hydrogen; acetaldehyde has alpha-hydrogens and undergoes a different reaction (aldol condensation) instead.

Cannizzaro's reaction (base-mediated disproportionation into one molecule of alcohol and one of the carboxylate salt) occurs only for aldehydes that have no $\alpha$ -hydrogen — because if $\alpha$ -hydrogens were present, the far faster aldol condensation pathway would dominate instead.

Checking each option for $\alpha$ -hydrogens:

(a) 1-methylcyclohexane-1-carbaldehyde: the carbon bearing $-CHO$ has no H on it (fully substituted, quaternary-type carbon holding both $-CHO$ and $-CH_3$ plus two ring bonds) → no $\alpha$ -H → gives Cannizzaro.

(b) Benzaldehyde: the carbonyl carbon is attached directly to the aromatic ring, with no adjacent sp³ carbon bearing H → no $\alpha$ -H → gives Cannizzaro (the classic textbook example). …
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CBSE 2024Set ANNUAL1 markMCQ
$\alpha$
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is incorrect.
(d) Both A and R are incorrect.
›Reveal solutionSolution

Formaldehyde and benzaldehyde undergo Cannizzaro reaction precisely because they LACK an alpha-hydrogen, not because they have one.

Assertion (A) is correct: aldehydes that have no $\alpha$ -hydrogen atom (such as HCHO and $C_6H_5CHO$ ) cannot undergo aldol condensation; instead, when treated with concentrated alkali, two molecules of such an aldehyde undergo a self oxidation-reduction (disproportionation) reaction called the Cannizzaro reaction, giving one molecule of the corresponding alcohol and one molecule of the corresponding carboxylate salt.

$2HCHO \xrightarrow{conc.\,NaOH} CH_3OH + HCOONa$ …

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CBSE 2024Set ANNUAL1 mark
$Q. Aldehydes which do not contain ______ give the Cannizzaro reaction.$
›Reveal solutionSolution

Only aldehydes without any alpha-hydrogen atom (so they cannot undergo aldol condensation) undergo the Cannizzaro reaction on treatment with concentrated alkali.

The Cannizzaro reaction is a self-oxidation-reduction (disproportionation) reaction in which two molecules of an aldehyde lacking alpha-hydrogens, when treated with concentrated (50%) NaOH or KOH, give one molecule of an alcohol (by reduction) and one molecule of a carboxylate salt (by oxidation).

Example: 2HCHO + NaOH -> CH3OH + HCOONa (formaldehyde disproportionates to methanol and sodium formate).

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. In Clemmensen reduction, carbonyl compound is reacted with$
(a) Zinc amalgam + HCl
(b) Sodium amalgam + HCl
(c) Zinc amalgam + HNO3
(d) Sodium amalgam + HNO3
›Reveal solutionSolution

Clemmensen reduction is the reagent combination Zn-Hg (zinc amalgam) with concentrated HCl, used to reduce an aldehyde or ketone's C=O group all the way to CH2 (a strongly acidic reduction, unsuitable for acid-sensitive substrates).

R2C=O --(Zn-Hg / conc. HCl)--> R2CH2 + H2O

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. In Acetaldehyde + 4[H] --(Zn-Hg/HCl)--> A, A is$
(a) Methane
(b) Ethane
(c) Propane
(d) None of these
›Reveal solutionSolution

Zn-Hg amalgam with concentrated HCl is the Clemmensen reduction, which converts an aldehyde/ketone carbonyl (C=O) all the way to CH2, i.e. a hydrocarbon.

CH3CHO + 4[H] --(Zn-Hg/HCl)--> CH3-CH3 + H2O …

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CBSE 2022Set ANNUAL1 mark
$Q. Give the structure of the product expected from the following reaction: 2-butanone is treated with Zn/Hg and conc. HCl.$
›Reveal solutionSolution

Zinc amalgam with concentrated HCl is the classic Clemmensen reduction, which reduces a ketone's carbonyl group all the way down to a methylene ( $-CH_2-$ ) group.

2-Butanone (methyl ethyl ketone, $CH_3-CO-CH_2-CH_3$ ) treated with zinc amalgam ( $Zn/Hg$ ) and concentrated hydrochloric acid undergoes the Clemmensen reduction, which completely deoxygenates the carbonyl carbon, converting $C=O$ directly into $-CH_2-$ :

$CH_3-CO-CH_2-CH_3 \xrightarrow{Zn(Hg)/conc.\,HCl} CH_3-CH_2-CH_2-CH_3$

…

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CBSE 2022Set ANNUAL1 mark
$Q. Give the structure of the product expected from the following reaction: Two molecules of benzaldehyde are treated with conc. NaOH.$
›Reveal solutionSolution

Benzaldehyde has no $\alpha$ -hydrogen, so it cannot undergo aldol condensation; instead, concentrated alkali makes it undergo the Cannizzaro reaction — a self-oxidation-reduction (disproportionation) between two molecules of the aldehyde.

Benzaldehyde ( $C_6H_5CHO$ ) has no hydrogen on the carbon adjacent to the carbonyl (the ring carbon takes that position), so it cannot enolise and cannot undergo aldol condensation. When treated with a concentrated (strong) base such as NaOH, it instead undergoes the Cannizzaro reaction: hydroxide ion adds to the carbonyl of one molecule, and the resulting alkoxide intermediate transfers a hydride ion to the carbonyl carbon of a second benzaldehyde molecule. This is a dispropor …

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CBSE 2020Set ANNUAL1 mark
$Q. Write the name of the reagent that reacts with formaldehyde to give sodium formate and methyl alcohol.$
›Reveal solutionSolution

Formaldehyde has no alpha-hydrogen, so concentrated NaOH disproportionates it via the Cannizzaro reaction into sodium formate and methanol.

Aldehydes lacking an alpha-hydrogen (such as formaldehyde, HCHO, and benzaldehyde) cannot undergo aldol condensation. Instead, when treated with concentrated alkali (typically 50% NaOH), they undergo a self-oxidation-reduction (disproportionation) called the Cannizzaro reaction: one molecule is oxidised to the carboxylate salt while another is reduced to the alcohol.

…

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Q.(a)

Part (a)

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