(a) (i) Identify A, B and C in the following reactions : Toluene (C₆H₅CH₃, drawn ring bearing -CH₃) + CrO₃ xri

Question

Q. (a)

(i) Identify A, B and C in the following reactions : Toluene (

C_6H_5CH_3

, drawn ring bearing

-CH_3

)

+ CrO_3 \xrightarrow[273-283\,K]{(CH_3CO)_2O}

\xrightarrow{H_3O^+}

\xrightarrow{\text{Conc. } NaOH}

+

Sodium benzoate (

C_6H_5COONa

, drawn ring bearing

-COONa

)

(ii) Give reasons for the following : (I) Carboxylic acids do not give the characteristic reactions of carbonyl group. (II) Ethanoic acid is a stronger acid than ethanol.

(OR)

(b)

(i) Write the product(s) in the following reactions : (I)

2\,CH_3COOH \xrightarrow[\Delta]{P_4O_{10}}

(II) Benzoyl chloride (

C_6H_5COCl

, drawn ring bearing

-COCl

)

+ (CH_3)_2Cd \longrightarrow

(III) Benzene-1,2-dicarboxamide (drawn: benzene ring bearing two adjacent

-CONH_2

groups)

\xrightarrow{\text{strong heating}}

(ii) Write the reaction involved in the following : (I) Wolff-Kishner reduction (II) Decarboxylation reaction

CBSECBSE Class XII Board 2025Subjective· 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 — Wolff-Kishner Reduction

Wolff-Kishner Reduction: Turning C=O into CH₂

Imagine you have a ketone or an aldehyde — a molecule with a carbonyl group (C=O) — and you want to remove that oxygen entirely, replacing it with two hydrogens. You want to go from C=O to CH₂. That's a reduction, but not the kind that gives you an alcohol. You want to strip the oxygen out completely.

The problem is that simply adding hydrogen gas (H₂) with a metal catalyst usually stops at the alcohol stage. You need a different strategy.

The Wolff-Kishner reduction solves this by first converting the carbonyl into a hydrazone, then using a strong base to break it apart, releasing nitrogen gas (N₂) and leaving behind the CH₂ group. The nitrogen gas bubbles away, driving the reaction forward.

The Intuition: Why It Works

The key insight is that nitrogen is an excellent leaving group when it can form N₂. The reaction proceeds in two main stages:

Formation of a hydrazone. Hydrazine (NH₂–NH₂) attacks the carbonyl carbon. After losing water, you get a C=N–NH₂ group — the hydrazone. This is like making an imine, but with an extra NH₂ attached.
Base-induced decomposition. A strong base (like KOH or NaOEt) in a high-boiling solvent (like ethylene glycol) deprotonates the hydrazone. The resulting anion rearranges, kicking out N₂ gas. The carbon that was double-bonded to oxygen now picks up two protons from the solvent, becoming CH₂.

The driving force is the formation of the extremely stable N≡N triple bond. That's why the reaction works so cleanly under basic conditions.

The Precise Statement

\ce{R2C=O ->[1. NH2NH2][2. KOH, heat] R2CH2 + N2 + H2O}

The carbonyl compound is treated with hydrazine (NH₂NH₂) to form a hydrazone. This intermediate is then heated with a strong base (typically KOH) in a high-boiling solvent like diethylene glycol or ethylene glycol. The product is the corresponding alkane (CH₂ group in place of C=O), along with nitrogen gas and water.

Step-by-Step Mechanism (Simplified)

Step 1: Hydrazone formation (acid-catalysed, but the overall conditions are basic)

The carbonyl oxygen is protonated (if acid is present), making the carbon more electrophilic. Hydrazine attacks, then water is eliminated to give the hydrazone.

\ce{R2C=O + NH2NH2 <=> R2C=N-NH2 + H2O}

Step 2: Deprotonation and elimination

The base abstracts a proton from the NH₂ group of the hydrazone, generating a negatively charged nitrogen. This anion undergoes a series of rearrangements, ultimately expelling N₂ and forming a carbanion intermediate.

\ce{R2C=N-NH2 + OH- -> R2C=N-NH- + H2O}

\ce{R2C=N-NH- -> R2C^- + N2 + H+}

Step 3: Protonation

The carbanion picks up a proton from the solvent (or from water present), giving the final alkane.

\ce{R2C^- + H2O -> R2CH2 + OH-}

Important Exam Points

Watch out

The Wolff-Kishner reduction requires strongly basic conditions and high temperatures (around 200°C). It is not suitable for base-sensitive compounds. If your molecule has other base-labile groups (esters, nitro groups, etc.), this reaction will destroy them.

Tip

| Feature | Wolff-Kishner | Clemmensen Reduction |

|---------|---------------|----------------------|

| Conditions | Basic, high temp | Acidic, Zn(Hg)/HCl |

| Mechanism | Via hydrazone | Via carbocation | …

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Part (a): A = benzylidene diacetate, B = benzaldehyde, C = benzyl alcohol (Cannizzaro); carboxyl C=O is resonance-stabilised so it gives no carbonyl reactions; acetic acid > ethanol because acetate is resonance-stabilised.

Part (b): $2CH_3COOH/P_4O_{10}\rightarrow$ acetic anhydride; $C_6H_5COCl + (CH_3)_2Cd \rightarrow$ acetophenone; dicarboxamide $\rightarrow$ phthalimide; Wolff–Kishner (C=O $\rightarrow$ CH $_2$ ); decarboxylation ( $CH_3COONa\rightarrow CH_4$ ).

Part (a)

(i) This is the Etard-type oxidation followed by Cannizzaro:

Toluene with chromyl-type reagent $CrO_3$ in acetic anhydride at 273–283 K gives A = benzylidene diacetate, $C_6H_5CH(OOCCH_3)_2$ (the low temperature and acetic anhydride stop oxidation at this stage, protecting the aldehyde).
Acid hydrolysis ( $H_3O^+$ ) of A gives B = benzaldehyde, $C_6H_5CHO$ .
Benzaldehyde has no $\alpha$ -hydrogen, so with concentrated NaOH it undergoes the Cannizzaro reaction (self oxidation–reduction): one molecule is reduced to C = benzyl alcohol ( $C_6H_5CH_2OH$ ) and another oxidised to sodium benzoate ( $C_6H_5COONa$ ).

(ii)(I) In a carboxylic acid, the carbonyl carbon is stabilised by resonance involving the lone pair of the –OH oxygen (the two C–O bonds become equivalent). This delocalisation lowers the electrophilic character of the carbon, so carboxylic acids do not give the nucleophilic-addition reactions characteristic of the aldehyde/ketone carbonyl group. …

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

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 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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CBSE 2018Set ANNUAL1 markMCQ
$Q. The cannizzaro's reaction is not given by -$
(a) Formaldehyde
(b) Acetaldehyde
(c) Benzaldehyde
(d) Furfural
›Reveal solutionSolution

Cannizzaro needs no α-H; acetaldehyde has α-H, so it is the exception.

Cannizzaro's reaction: 2 RCHO + conc. NaOH → RCH₂OH + RCOO⁻ (self oxidation–reduction). It occurs only for aldehydes lacking an α-hydrogen. …

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CBSE 2018Set ANNUAL1 mark
$Q. Between formaldehyde and acetaldehyde which gives Cannizzaro's reaction.$
›Reveal solutionSolution

Formaldehyde undergoes Cannizzaro's reaction; acetaldehyde does not.

The Cannizzaro reaction is a disproportionation of an aldehyde in concentrated alkali: one molecule is oxidised to a carboxylate ion and another is reduced to an alcohol. This reaction is possible only for aldehydes that lack an alpha-hydrogen (an H on the carbon next to the -CHO), because aldehydes with alpha-hydrogens undergo aldol condensation instead.

…

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Q.(a)

Part (a)

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