Which of the following conversions can be carried out by Clemmensen reduction? (Two or more than two options m

NCERT Exemplar · Q38

Q. Which of the following conversions can be carried out by Clemmensen reduction? (Two or more than two options may be correct.)

(i) Benzaldehyde into benzyl alcohol

(ii) Cyclohexanone into cyclohexane

(iii) Benzoyl chloride into benzaldehyde

(iv) Benzophenone into diphenylmethane

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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 Zn(Hg)/HCl to convert a carbonyl group directly into a methylene (CH2) group. It works only on ketones and aldehydes, not on acid chlorides or for partial reduction to alcohols. The correct options are (ii) and (iv).

Why (i) fails: benzaldehyde into benzyl alcohol needs a reduction that stops at the alcohol stage (e.g. NaBH4). Clemmensen removes the oxygen entirely, giving toluene instead.

Why (ii) works: cyclohexanone's ring C=O is reduced straight to CH2, giving cyclohexane -- exactly what Clemmensen does.

Why (iii) fails: converting an acid chloride to an aldehyde is the job of Rosenmund reduction (H2/Pd-BaSO4), not Clemmensen; acid chlorides are not suitable Clemmensen substrates. …

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Clemmensen Reduction — Concept & Application

Method: Clemmensen Reduction

What it does:

Clemmensen reduction converts a carbonyl group ( $\ce{C=O}$ ) in an aldehyde or ketone into a methylene group ( $\ce{CH2}$ ) using zinc amalgam (Zn-Hg) and concentrated HCl.

General reaction:

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

Step-by-step logic for the question

Identify the functional group in the starting compound — only aldehydes and ketones undergo Clemmensen reduction.
Check the product — the carbonyl oxygen is completely removed, replaced by two hydrogens.
Eliminate options where the starting compound is not a carbonyl compound (e.g., acid chlorides, alcohols).

Applying to the options

Option	Starting compound	Product	Possible by Clemmensen?	Reason
(A)	Benzaldehyde (aldehyde)	Benzyl alcohol (alcohol)	✗ No	Clemmensen gives hydrocarbon, not alcohol. Alcohol formation requires reduction with NaBH₄ or LiAlH₄.

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Common Mistakes in Clemmensen Reduction & How to Avoid Them

Mistake 1: Confusing Clemmensen with other carbonyl reductions

The error: Students often think Clemmensen reduction converts aldehydes to alcohols (like option A — Benzaldehyde → Benzyl alcohol).

Why it’s wrong:

Clemmensen reduction uses Zn(Hg) / conc. HCl — it completely removes the carbonyl oxygen, replacing it with two hydrogens.

Aldehyde ( $\ce{RCHO}$ ) → Alkane ( $\ce{RCH3}$ )
Ketone ( $\ce{RCOR'}$ ) → Alkane ( $\ce{RCH2R'}$ )

How to avoid:

Remember the “all the way down” rule: Clemmensen goes from carbonyl to hydrocarbon, not to alcohol.

For alcohol formation, think NaBH₄ or LiAlH₄ (reduction stops at alcohol).
For complete removal of oxygen, think Clemmensen (acidic) or Wolff-Kishner (basic).

Mistake 2: Forgetting that Clemmensen fails on acid-sensitive groups

The error: Students apply Clemmensen to compounds with acid-sensitive groups (e.g., esters, acid chlorides) without checking.

Why it’s wrong:

Clemmensen uses hot, concentrated HCl. This will:

Hydrolyse esters and acid chlorides
Cause side reactions with other functional groups

How to avoid:

Check the substrate:

✓ Works on simple ketones/aldehydes (stable in acid)
✗ Fails on esters, acid chlorides, amides (they hydrolyse first)

For option (C) — Benzoyl chloride → Benzaldehyde — this is not Clemmensen; it’s Rosenmund reduction (H₂ / Pd-BaSO₄).

Mistake 3: Missing that Clemmensen works on both aldehydes and ketones

The error: Students think Clemmensen works only on ketones.

Why it’s wrong:

Both aldehydes and ketones undergo Clemmensen reduction.

Cyclohexanone (ketone) → Cyclohexane ✓
Benzophenone (ketone) → Diphenylmethane ✓

How to avoid:

Memorise: Any carbonyl (C=O) that can tolerate hot acid can be reduced to the corresponding alkane.

Mistake 4: Not recognising the product structure correctly

The error: Students misidentify the product when the carbonyl is attached to an aromatic ring.

Why it’s wrong:

Benzophenone ( $\ce{Ph-CO-Ph}$ ) → Diphenylmethane ( $\ce{Ph-CH2-Ph}$ ) — correct
But some think it gives benzene + toluene (wrong — no C–C bond cleavage)

How to avoid:

Draw the structure:

The carbonyl carbon becomes a $\ce{CH2}$ group
The two phenyl rings remain attached to that carbon

Mistake 5: Selecting options that require reduction to alcohol

The error: Choosing (A) Benzaldehyde → Benzyl alcohol as a valid Clemmensen product.

Why it’s wrong:

Clemmensen gives toluene ( $\ce{C6H5CH3}$ ), not benzyl alcohol ( $\ce{C6H5CH2OH}$ ).

Benzaldehyde ( $\ce{C6H5CHO}$ ) → Toluene ( $\ce{C6H5CH3}$ )
Benzyl alcohol requires a milder reducing agent

How to avoid: …

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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.Which of the following conversions can be carried out by Clemmensen reduction? (Two or more than two options may be correct.)

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