In Clemmensen reduction the carbonyl compound is treated with ______. (i) zinc amalgam + HCl (ii) sodium amalg

NCERT Exemplar · Q12

Q. In Clemmensen reduction the carbonyl compound is treated with ____________.

(i) zinc amalgam + HCl

(ii) sodium amalgam + HCl

(iii) zinc amalgam + nitric acid

(iv) sodium amalgam +

HNO_3

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Concept understanding — Clemmensen Reduction 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?

Okay, let's break down these two very different reactions. They are often studied together because they both involve carbonyl compounds ( $\ce{C=O}$ ), but their mechanisms and purposes are completely opposite.

The Core Idea: Two Paths from a Carbonyl

Think of a carbonyl group ( $\ce{C=O}$ ) as a reactive hub. The carbon is electrophilic (electron-loving) because the oxygen pulls electron density away. The reactions it undergoes depend entirely on the conditions (acidic, basic, reducing) and the structure of the molecule (does it have an $\alpha$ -hydrogen?).

Clemmensen Reduction is about removing the oxygen entirely.
Cannizzaro Reaction is about disproportionating the molecule (one gets reduced, one gets oxidized).

1. Clemmensen Reduction: Why it Removes Oxygen

What it does: Converts a carbonyl group ( $\ce{C=O}$ ) in an aldehyde or ketone into a methylene group ( $\ce{CH2}$ ).

\ce{R2C=O ->[Zn(Hg)/HCl, heat] R2CH2}

Why this formula holds (The Mechanism):

The key is the reducing power of zinc amalgam in a strongly acidic environment.

Protonation: The carbonyl oxygen is basic. In the strong $\ce{HCl}$ , it gets protonated first.

\ce{R2C=O + H+ -> R2C=OH+}

This makes the carbon *even more* electrophilic.

2. Electron Transfer from Zinc: Zinc metal ( $\ce{Zn}$ ) is a good reducing agent. It donates electrons to the electron-deficient carbon. This is a single electron transfer (SET) process, not a simple hydride transfer.

- The zinc inserts itself, forming an organozinc intermediate (a carbenoid species).

- This intermediate is highly reactive.

Protonation and Elimination: The acidic medium provides plenty of $\ce{H+}$ ions. The intermediate gets protonated, and the oxygen (now as $\ce{H2O}$ ) is eliminated. The zinc is oxidized to $\ce{Zn^{2+}}$ .

The "Why" in a nutshell: The strong acid activates the carbonyl, and the zinc metal provides the electrons needed to break the $\ce{C=O}$ bond and replace it with two $\ce{C-H}$ bonds. The reaction does not work under basic conditions because you need the acid to protonate the oxygen first.

2. Cannizzaro Reaction: Why it Disproportionates

What it does: An aldehyde without an $\alpha$ -hydrogen (like formaldehyde $\ce{HCHO}$ or benzaldehyde $\ce{C6H5CHO}$ ) reacts with a strong base to give a carboxylic acid and an alcohol.

\ce{2 HCHO ->[conc. NaOH] HCOONa + CH3OH}

Why this formula holds (The Mechanism):

The key is the absence of $\alpha$ -hydrogens. If there were an $\alpha$ -hydrogen, the base would deprotonate that instead, leading to an aldol reaction. Here, the base has no choice but to attack the carbonyl itself.

Nucleophilic Attack: The strong base ( $\ce{OH-}$ ) attacks the electrophilic carbonyl carbon.

\ce{RCHO + OH- -> R-CH(OH)O-}

This forms a **tetrahedral intermediate** (an alkoxide).

2. The Crucial Hydride Transfer: This is the unique step. The tetrahedral intermediate is unstable. It can't lose $\ce{OH-}$ (that would just give back the aldehyde). Instead, it acts as a hydride donor ( $\ce{H-}$ ).

- The carbon bearing the negative charge (from the $\ce{OH-}$ attack) is very electron-rich. It kicks out a hydride ion ( $\ce{H-}$ ) to a second molecule of aldehyde.

- This is a hydride shift.

$\ce{R-CH(OH)O- + RCHO -> RCOOH + RCH2O-}$ …

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Clemmensen reduction uses zinc amalgam (Zn-Hg) and concentrated HCl to reduce a carbonyl group to a methylene group. The correct option is (i).

Zinc amalgam is the reducing agent of choice because zinc is a good electron donor and the mercury coating keeps it from reacting too violently with the acid. Concentrated HCl supplies the protons needed to protonate the carbonyl oxygen.

Eliminating the wrong options:

(ii) sodium amalgam + HCl: sodium is too reactive with HCl and is not the Clemmensen reagent.
(iii) zinc amalgam + nitric acid: nitric acid is an oxidising agent, incompatible with a reduction.
(iv) sodium amalgam + HNO3: wrong metal and wrong (oxidising) acid. …

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Here are the common mistakes students make with this specific question and the broader concepts of Clemmensen Reduction and Cannizzaro Reaction, along with how to avoid each.

1. Confusing the Reagent (Zinc vs. Sodium Amalgam)

The Mistake:

Choosing option (B) or (D) — thinking sodium amalgam is used. Students often mix up Clemmensen (uses Zn) with other reductions (e.g., sodium amalgam is used in some other reactions like the reduction of nitriles or in the Bouveault–Blanc reduction of esters).

Why it happens:

Both are “amalgams” (metal + mercury), so the names sound similar. The key difference is the metal: zinc for Clemmensen, sodium for other processes.

How to avoid:

Memorise the exact phrase: “Zinc amalgam (Zn–Hg) + conc. HCl”.
Link the Z in Zinc to Clemmensen (both have a “Z”/“C” connection in your mind: Zn → Clemmensen).
Remember: Clemmensen is for acidic conditions (HCl). Sodium amalgam is usually used in neutral or basic conditions.

✓ Correct answer: (A) zinc amalgam + HCl

2. Confusing the Acid (HCl vs. HNO₃)

The Mistake:

Choosing (C) or (D) — thinking nitric acid ( $HNO_3$ ) is used. This is a dangerous error because nitric acid is a strong oxidising agent, while Clemmensen reduction is a reduction reaction.

Why it happens:

Students see “acid” and pick any strong acid without checking its chemical nature.

How to avoid:

Rule of thumb: In Clemmensen, the acid must be non-oxidising. HCl is perfect. $HNO_3$ would oxidise the zinc and destroy the reaction.
Write this on your formula sheet: Clemmensen = Zn–Hg + conc. HCl (never $HNO_3$ or $H_2SO_4$ conc.)

3. Confusing Clemmensen with Wolff–Kishner Reduction

The Mistake:

Thinking Clemmensen uses hydrazine ( $NH_2NH_2$ ) + base (which is actually Wolff–Kishner).

Why it happens:

Both reactions reduce a carbonyl ( $C=O$ ) to a methylene ( $CH_2$ ) group. Students memorise the purpose but swap the conditions.

How to avoid:

Clemmensen = acidic (Zn–Hg / HCl) → for acid-stable compounds.
Wolff–Kishner = basic ( $NH_2NH_2$ / KOH) → for base-stable compounds.
Use a mnemonic:
- Clemmensen → Conc. HCl
- Wolff–Kishner → With hydrazine (the “h” in Wolff hints at hydrazine)

4. Confusing Clemmensen with Cannizzaro Reaction

The Mistake:

Thinking Clemmensen is used for aldehydes without $\alpha$ -hydrogen (that’s Cannizzaro). Or thinking Cannizzaro uses Zn–Hg.

Why it happens:

Both involve carbonyl compounds, but the outcomes are completely different:

Clemmensen: $C=O$ → $CH_2$ (complete reduction)
Cannizzaro: $C=O$ → alcohol + carboxylic acid (disproportionation)

How to avoid:

Clemmensen = removes oxygen entirely (reduction). …

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

GUJCET 2024Set 131 markMCQ
$C_6H_5COCl$
›Reveal solutionSolution

Acid chloride → aldehyde with $H_2$ /Pd–BaSO $_4$ (poisoned catalyst) is Rosenmund reduction.

Concept. Rosenmund reduction converts an acyl chloride to an aldehyde using $H_2$ over palladium supported on barium sulphate (partially poisoned to stop over-reduction to the alcohol) …
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GUJCET 2023Set 091 markMCQ
$C_6H_5-CHO$
›Reveal solutionSolution
Cannizzaro requires no α-hydrogen; acetaldehyde $CH_3CHO$ has α-H, so it fails the test.

Concept — Cannizzaro reaction. Aldehydes lacking an α-hydrogen undergo self-oxidation–reduction (disproportionation) with concentrated alkali to give an alcohol + a carboxylate salt. Aldehydes with α-hydrogens instead undergo aldol condensation and do not give Cannizzaro.

Checking each:

(A) Benzaldehyde $C_6H_5CHO$ — no α-H → gives Cannizzaro. …
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GUJCET 2021Set 151 markMCQ
$CH_3CHO$
›Reveal solutionSolution
Cannizzaro = disproportionation of an aldehyde that has no α-hydrogen.

Concept: In presence of strong base, aldehydes lacking an α-H undergo self oxidation–reduction (Cannizzaro). An α-H would instead permit aldol.

(A) $CH_3CHO$ — 3 α-H.

(B) $CH_2ClCHO$ — 2 α-H. …
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GUJCET 2020Set 071 markMCQ
$LiAlH_4$
›Reveal solutionSolution

Amide → amine with loss of one carbon is Hofmann bromamide degradation: $\text{NaOH}+\text{Br}_2$ . …

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GUJCET 2014Set A1 markMCQ
$Q. Which of the following compound does not react with concentrated alkali to give corresponding alcohol and salt of carboxylic acid? (A) Benzaldehyde (B) Trimethyl acetaldehyde (C) Dimethyl acetaldehyde (D) Formaldehyde$
›Reveal solutionSolution
[!TLDR]

The Cannizzaro reaction requires an aldehyde lacking alpha-H. Dimethyl acetaldehyde has an alpha-H, so it is the odd one out.

Concept

In the Cannizzaro reaction, aldehydes without an alpha-hydrogen undergo base-induced disproportionation to give an alcohol and the salt of a carboxylic acid (NCERT Aldehydes/Ketones). Aldehydes possessing alpha-hydrogen instead prefer aldol-type reactions.

Solution

Check for alpha-hydrogen (H on the carbon next to $-CHO$ ):

Benzaldehyde $C_6H_5CHO$ : no alpha-H (ring carbon) - undergoes Cannizzaro.

Trimethyl acetaldehyde $(CH_3)_3C\text{-}CHO$ : alpha-carbon is quaternary, no alpha-H - undergoes Cannizzaro. …
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Q.In Clemmensen reduction the carbonyl compound is treated with ____________.

Concept understanding — Clemmensen Reduction Cannizzaro Reaction

Clemmensen Reduction & Cannizzaro Reaction: Two Completely Different Reactions

Clemmensen Reduction

Cannizzaro Reaction

Why this formula?

The Core Idea: Two Paths from a Carbonyl

1. Clemmensen Reduction: Why it Removes Oxygen

2. Cannizzaro Reaction: Why it Disproportionates

Clemmensen Reduction — Method & Steps

Steps of the method:

Key result:

Answer to the fill-in-the-blank

Why the other options are wrong:

Quick comparison with Cannizzaro Reaction …

1. Confusing the Reagent (Zinc vs. Sodium Amalgam)

2. Confusing the Acid (HCl vs. HNO₃)

3. Confusing Clemmensen with Wolff–Kishner Reduction

4. Confusing Clemmensen with Cannizzaro Reaction

Concept

Solution

Q.In Clemmensen reduction the carbonyl compound is treated with ____________.

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