Phenylmethanal is reacted with concentrated NaOH to give two products X and Y. X reacts with metallic sodium t

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Q. Phenylmethanal is reacted with concentrated NaOH to give two products X and Y. X reacts with metallic sodium to liberate hydrogen. X and Y are a) sodium benzoate and phenol b) sodium benzoate and phenylmethanol c) phenylmethanol and sodium benzoate d) none of these

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Concept understanding — Cannizzaro Reaction

Clemmensen Reduction & Cannizzaro Reaction: Two Completely Different Reactions

These two reactions are often grouped together in textbooks because they both involve carbonyl compounds (C=O), but they do entirely different things. Let's take them one at a time.

Clemmensen Reduction

Intuition first. Imagine you have a ketone or aldehyde — a molecule with a C=O group. You want to remove that oxygen entirely and replace the C=O with two hydrogen atoms, turning it into a simple hydrocarbon chain. That's a reduction (adding hydrogen, removing oxygen). The Clemmensen reduction is a brute-force way to do this using a strongly acidic, reducing environment.

The precise reaction:

A ketone or aldehyde is heated with zinc amalgam (Zn-Hg) and concentrated hydrochloric acid (HCl). The C=O group is reduced to a CH₂ group.

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

\text{Aldehyde or Ketone} \xrightarrow{\text{Zn(Hg), conc. HCl, } \Delta} \text{Hydrocarbon}

Key points for exams:

Works only for ketones and aldehydes that are stable in strong acid.
Does not work for compounds that get destroyed by conc. HCl (e.g., acid-sensitive groups like esters, nitriles).
The mechanism is complex and not usually tested in detail — just know it's a reductive removal of C=O.
The product is always a saturated hydrocarbon (alkane).

Watch out

Clemmensen reduction cannot reduce carboxylic acids, esters, or amides. Only aldehydes and ketones.

Example:

Acetophenone ( $\ce{C6H5-CO-CH3}$ ) → Ethylbenzene ( $\ce{C6H5-CH2-CH3}$ )

Cannizzaro Reaction

Intuition first. This is a disproportionation reaction — one molecule of aldehyde gets oxidised (to a carboxylic acid) while another gets reduced (to an alcohol). It happens only with aldehydes that have no alpha-hydrogen atoms (i.e., the carbon next to the C=O has no H). Why? Because if there were alpha-hydrogens, the aldehyde would undergo aldol condensation instead.

The precise reaction:

An aldehyde without $\alpha$ -hydrogen is treated with concentrated aqueous or alcoholic base (NaOH/KOH). Two molecules of aldehyde react: one becomes a carboxylate salt, the other becomes a primary alcohol.

\ce{2 R-CHO + OH- -> R-COO- + R-CH2OH}

After acidification, the carboxylate salt gives the carboxylic acid.

2\,\text{HCHO} \xrightarrow{\text{conc. NaOH}} \text{HCOONa} + \text{CH}_3\text{OH}

Key points for exams:

Only works for aldehydes with no $\alpha$ -hydrogen: formaldehyde, benzaldehyde, trimethylacetaldehyde, etc.
The base must be concentrated (dilute base won't work).
Formaldehyde is the most common example — it gives formic acid (as formate) and methanol.
Crossed Cannizzaro: When formaldehyde is mixed with another aldehyde (like benzaldehyde), formaldehyde is always the one that gets oxidised (to formate), and the other aldehyde gets reduced (to alcohol). This is because formaldehyde is the strongest reducing agent among aldehydes.

Tip

In a crossed Cannizzaro, formaldehyde always becomes the carboxylate. The other aldehyde becomes the alcohol. This is a common exam question.

Example: …

Why this formula?

Cannizzaro Reaction: Why the Key Formulas Hold

The Cannizzaro reaction is a disproportionation reaction of aldehydes (without α-hydrogens) in the presence of a strong base. Let's build the understanding from the ground up.

1. What Happens in the Reaction?

An aldehyde (like formaldehyde or benzaldehyde) reacts with concentrated base to give:

One molecule is oxidized to a carboxylic acid (or its salt)
Another molecule is reduced to a primary alcohol

General equation (for two identical aldehydes):

2 \text{RCHO} + \text{OH}^- \rightarrow \text{RCOO}^- + \text{RCH}_2\text{OH}

2. Why Does Disproportionation Occur?

The Key Insight: No α-Hydrogen

Aldehydes with α-hydrogens undergo aldol condensation instead.
Without α-hydrogens, the only available reaction path is hydride transfer.

The Mechanism (Step-by-Step Reasoning)

Nucleophilic attack: $\text{OH}^-$ attacks the carbonyl carbon of one aldehyde molecule.
- Forms a tetrahedral intermediate (a gem-diolate).
Hydride shift: The intermediate acts as a hydride donor ( $\text{H}^-$ ) to a second aldehyde molecule.
- This is the rate-determining step.
- The hydride comes from the C–H bond of the intermediate (not from the OH).
Products:
- The donor aldehyde becomes a carboxylate ion (oxidized).
- The acceptor aldehyde becomes an alkoxide ion (reduced).
Protonation: In workup, the carboxylate gives the acid, and the alkoxide gives the alcohol.

3. The Key Formula(e) and Their Derivation

Formula 1: Stoichiometry

2 \text{RCHO} + \text{OH}^- \rightarrow \text{RCOO}^- + \text{RCH}_2\text{OH}

Why this holds:

One aldehyde loses a hydride ( $\text{H}^-$ ) → gains an oxygen → oxidation state increases by 2.
The other aldehyde gains a hydride → oxidation state decreases by 2.
The base ( $\text{OH}^-$ ) is consumed stoichiometrically (one per two aldehydes).

Formula 2: Oxidation State Change

For an aldehyde carbon (carbonyl carbon):

In $\text{RCHO}$ : oxidation state = +1
In $\text{RCOO}^-$ : oxidation state = +3 (gain of +2)
In $\text{RCH}_2\text{OH}$ : oxidation state = -1 (loss of -2)

Net change: $+2$ (oxidation) + $(-2)$ (reduction) = 0 — consistent with disproportionation.

Formula 3: Rate Law (for the hydride transfer step)

\text{Rate} = k[\text{aldehyde}]^2[\text{OH}^-]

Why:

First aldehyde reacts with $\text{OH}^-$ to form the hydride donor (first order in each).
Second aldehyde accepts the hydride (first order in aldehyde).
Overall: second order in aldehyde, first order in base.

4. Why Only Certain Aldehydes Work?

Condition: Aldehyde must have no α-hydrogen atoms.

Examples: $\text{HCHO}$ (formaldehyde), $\text{C}_6\text{H}_5\text{CHO}$ (benzaldehyde), $\text{(CH}_3\text{)}_3\text{CCHO}$ (pivalaldehyde). …

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Step 1. Phenylmethanal (benzaldehyde), treated with concentrated NaOH, undergoes the Cannizzaro reaction (Section 12.5.F.2), giving two products: benzyl alcohol (phenylmethanol, C6H5CH2OH) and sodium benzoate (C6H5COONa).

Step 2. The question states that X specifically reacts with metallic sodium to liberate hydrogen gas -- this is the classic reaction of an ALCOHOL's -OH with active sodium metal (Unit 11: 2 R-OH + 2 Na gives 2 R-ONa + H2), a reaction sodium benzoate (already a fully-formed sodium SALT, with no free O-H left to react with more sodium) cannot undergo. …

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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 A1 markMCQ
$Q. With which of the following does Cannizzaro's reaction take place ?$
(a) CH3CHO
(b) HCHO
(c) HCOOH
(d) CH3COCH3
›Reveal solutionSolution

The Cannizzaro reaction occurs only with aldehydes that have NO alpha-hydrogen; formaldehyde (HCHO) qualifies.

In the Cannizzaro reaction, two molecules of an aldehyde lacking an alpha-hydrogen undergo self oxidation-reduction (disproportionation) with concentrated alkali to give an alcohol and a salt of a carboxylic acid:

2 HCHO + NaOH --> CH3OH + HCOONa

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following compounds gives Cannizaro reaction?$
(a) cyclohexanecarbaldehyde (cyclohexane ring with a -CHO substituent, structure drawn)
(b) CH3CHO
(c) CH3-CH(CH3)-CHO (2-methylpropanal, structure drawn)
(d) benzaldehyde (benzene ring with a -CHO substituent, structure drawn)
›Reveal solutionSolution
The Cannizzaro reaction (base-induced disproportionation into an alcohol and a carboxylate) only happens for aldehydes with NO alpha-hydrogen — because an alpha-H aldehyde instead undergoes the much faster aldol condensation.

Check each option for an alpha-hydrogen (an H on the carbon directly attached to -CHO):

cyclohexanecarbaldehyde — the ring carbon attached to CHO carries an alpha-H → undergoes aldol, not Cannizzaro.

CH3CHO (acetaldehyde) — the CH3 carbon has alpha-H's → aldol.

2-methylpropanal, (CH3)2CH-CHO — the CH carbon has an alpha-H → aldol. …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. Cannizaro's reaction is not given by :$
(a) Structure (A): a cyclohexane ring bearing a –CHO group and an adjacent –CH3 group on the ring (2-methylcyclohexanecarbaldehyde-type structure, an aldehyde with an alpha-hydrogen)
(b) Structure (B): a benzene ring (drawn as a hexagon with an inscribed circle) bearing a –CHO group (benzaldehyde)
(c) HCHO
(d) CH3CHO
›Reveal solutionSolution
The Cannizzaro reaction is given only by aldehydes with NO $\alpha$ -hydrogen; CH3CHO has $\alpha$ -hydrogens (on its methyl group), so it is oxidised/reduced by the aldol route instead and does not undergo Cannizzaro.

The Cannizzaro reaction is a base-mediated disproportionation (self oxidation–reduction) of an aldehyde lacking $\alpha$ -hydrogens: two molecules of the aldehyde react with concentrated $NaOH$ , one being oxidised to the carboxylate and the other reduced to the alcohol. If $\alpha$ -hydrogens ARE present, the base instead deprotonates the $\alpha$ -carbon and the aldehyde undergoes base-catalysed aldol condensation, which is much faster.

(B) benzaldehyde ( $C_6H_5CHO$ ): the carbon bonded to CHO is an aromatic ring carbon with no removable $\alpha$ -H for enolisation — undergoes Cannizzaro (classic example, gives benzyl alcohol + sodium benzoate).

(C) $HCHO$ (formaldehyde): has no $\alpha$ -carbon at all — undergoes Cannizzaro (gives methanol + sodium formate). …
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CBSE 2023Set F1 markMCQ
$Q. Which of the following gives Cannizzaro's reaction?$
(a) CH3CHO
(b) HCHO
(c) HCOOH
(d) CH3COCH3
›Reveal solutionSolution
The Cannizzaro reaction is a self oxidation-reduction of aldehydes that have NO alpha-hydrogen; formaldehyde (HCHO) qualifies.

In the Cannizzaro reaction, two molecules of an aldehyde lacking alpha-hydrogen react in concentrated alkali — one is oxidised to the acid (salt) and the other reduced to the alcohol.

HCHO has no alpha-H → gives Cannizzaro reaction: …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. Which of the following compounds does not participate in Cannizzaro reaction?$
(a) structure
(a)
(b) CH3CHO
(c) HCHO
(d) structure (d)
›Reveal solutionSolution
Cannizzaro reaction (base-catalysed disproportionation of an aldehyde to an alcohol and a carboxylate) only happens for aldehydes that have NO alpha-hydrogen.

Cannizzaro's reaction requires an aldehyde lacking alpha-hydrogens, because if alpha-H were present, concentrated base would instead deprotonate it and trigger aldol condensation, which is much faster.

(a) o-methylbenzaldehyde: the carbonyl carbon is attached directly to the aromatic ring — no alpha-carbon with H exists on that side, so it undergoes Cannizzaro.

(c) HCHO (formaldehyde): the carbonyl carbon has only H atoms on it, no alpha-carbon at all — undergoes Cannizzaro (indeed reacts fastest of all). …
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CBSE 2021Set A1 markMCQ
$Q. Formaldehyde on heating with NaOH solution gives$
(a) Formic acid
(b) Acetone
(c) Methyl alcohol
(d) Ethyl formate
›Reveal solutionSolution

Formaldehyde has no α-H → Cannizzaro disproportionation with NaOH giving methanol + sodium formate.

Formaldehyde (HCHO) has no alpha-hydrogen atom, so it cannot undergo aldol condensation. Instead, with concentrated NaOH it undergoes the Cannizzaro reaction — a self oxidation-reduction (disproportionation):

2 HCHO + NaOH → CH3OH + HCOONa

One molecule of formaldehyde is reduced to methyl alcohol (CH3OH) and the other is oxidised to sodium formate (the sodium salt of formic acid).

…

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Q.Phenylmethanal is reacted with concentrated NaOH to give two products X and Y. X reacts with metallic sodium to liberate hydrogen. X and Y are a) sodium benzoate and phenol b) sodium benzoate and phenylmethanol c) phenylmethanol and sodium benzoate d) none of these

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