Match the reactions given in Column I with the suitable reagents given in Column II. Column I: (i) Benzophenon

NCERT Exemplar · Q44

Q. Match the reactions given in Column I with the suitable reagents given in Column II. Column I:

(i) Benzophenone

\rightarrow

Diphenylmethane

(ii) Benzaldehyde

\rightarrow

1-Phenylethanol

(iii) Cyclohexanone

\rightarrow

Cyclohexanol

(iv) Phenyl benzoate

\rightarrow

Benzaldehyde
Column II:

(a)

LiAlH_4

(b) DIBAL-H

(c)

Zn(Hg)

/ Conc. HCl

(d)

CH_3MgBr

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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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This problem tests your ability to match carbonyl reduction reactions with the correct reagents. The key is recognising that Clemmensen reduction (Zn(Hg)/HCl) removes the carbonyl oxygen entirely, while hydride reagents (LiAlH₄, DIBAL-H) add hydrogen across the C=O bond, and Grignard reagents add an alkyl group. The correct matches are: (i)→(c), (ii)→(d), (iii)→(a), (iv)→(b).

Let’s build the intuition first. Every reaction here involves a carbonyl compound (C=O) being transformed into something else. The reagent determines whether the carbonyl is:

Completely removed (replaced by two hydrogens) — that’s Clemmensen reduction.
Reduced to an alcohol (C=O becomes CH–OH) — that’s a hydride reduction.
Extended by one carbon (a Grignard addition) — that’s an organometallic addition.
Selectively reduced to an aldehyde from an ester — that’s DIBAL-H’s specialty.

Now let’s match each one step by step.

1. Benzophenone → Diphenylmethane

Benzophenone is $(C_6H_5)_2C=O$ . Diphenylmethane is $(C_6H_5)_2CH_2$ . Notice the carbonyl oxygen is gone, and the carbon now has two hydrogens instead of the double-bonded oxygen. This is a complete deoxygenation — the C=O is replaced by CH₂.

The classic reagent for this is Clemmensen reduction: zinc amalgam ( $Zn(Hg)$ ) in concentrated HCl. It works under strongly acidic conditions and is specific for converting aryl alkyl ketones (and diaryl ketones) to the corresponding hydrocarbon.

Watch out

A common mistake is to think LiAlH₄ does this. It doesn’t — LiAlH₄ reduces C=O to CH–OH, not to CH₂. Clemmensen reduction is the only one here that removes oxygen entirely.

So (i) matches with (c).

2. Benzaldehyde → 1-Phenylethanol

Benzaldehyde is $C_6H_5CHO$ . 1-Phenylethanol is $C_6H_5CH(OH)CH_3$ . Compare: the aldehyde carbon has gained a methyl group and the oxygen has become an OH. This is a Grignard addition — the methyl group comes from a Grignard reagent, and the carbonyl is reduced to an alcohol in the process.

The reagent $CH_3MgBr$ (methylmagnesium bromide) adds its methyl group to the carbonyl carbon. After aqueous workup, you get the secondary alcohol.

Tip

Grignard reagents always add an alkyl (or aryl) group to the carbonyl, making a new C–C bond. They are not simple reducing agents — they are alkylating agents.

So (ii) matches with (d).

3. Cyclohexanone → Cyclohexanol

Cyclohexanone is a cyclic ketone ( $C_6H_{10}O$ ). Cyclohexanol is the corresponding secondary alcohol ( $C_6H_{11}OH$ ). The carbonyl is simply reduced to a CH–OH group — no carbon added, no oxygen removed. …

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Here’s a clear, concept-first solution using the Reagent-Mapping Method.

Method: Reagent-Mapping by Functional Group Transformation

Why this works:

Each reaction in Column I is a specific functional group conversion. By identifying the type of change (e.g., carbonyl → alkane, carbonyl → alcohol, ester → aldehyde), you can match it to the reagent in Column II that performs exactly that transformation.

Step 1: Identify the transformation in each row of Column I

Row	Starting Material → Product	Type of Change
(i)	Benzophenone → Diphenylmethane	Ketone → Alkane (removal of C=O)
(ii)	Benzaldehyde → 1-Phenylethanol	Aldehyde → Secondary alcohol (adds one carbon)
(iii)	Cyclohexanone → Cyclohexanol	Ketone → Secondary alcohol (no carbon added)
(iv)	Phenyl benzoate → Benzaldehyde	Ester → Aldehyde (partial reduction)

Step 2: Match each transformation to the correct reagent from Column II

Clemmensen Reduction uses Zn(Hg) / Conc. HCl to convert a carbonyl (ketone/aldehyde) directly to an alkane.

→ (i) matches with (c).
Grignard reaction (here with CH₃MgBr) adds a methyl group to an aldehyde, giving a secondary alcohol.

→ (ii) matches with (d).
LiAlH₄ reduces ketones to secondary alcohols (no carbon added).

→ (iii) matches with (a).
DIBAL-H at low temperature reduces an ester to an aldehyde (stops before full reduction to alcohol).

→ (iv) matches with (b).

Step 3: Write the final matched pairs …

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Here is a breakdown of the common mistakes students make when matching these reactions, along with the correct reasoning to avoid them.

Mistake 1: Confusing Clemmensen Reduction with Wolff-Kishner Reduction

The Mistake: Students often see a carbonyl group ( $C=O$ ) being converted to a methylene group ( $CH_2$ ) and immediately think of the Wolff-Kishner reduction ( $NH_2NH_2$ , KOH). They might incorrectly match (i) with a hydrazine-based reagent.
Why it’s wrong: The question specifies Benzophenone $\rightarrow$ Diphenylmethane. This is a classic Clemmensen Reduction.
- Clemmensen: Uses $Zn(Hg)$ / Conc. HCl. It works best for aromatic ketones (like benzophenone) because the acidic conditions don't harm the benzene rings.
- Wolff-Kishner: Uses strong base ( $KOH$ ) and high heat. It is preferred for acid-sensitive compounds (like those with ester groups).
How to Avoid: Memorize the condition that dictates the choice.
- Acidic medium (Conc. HCl) → Clemmensen ( $Zn(Hg)$ ).
- Basic medium (KOH, heat) → Wolff-Kishner ( $NH_2NH_2$ ).
- Result: Match (i) with (c).

Mistake 2: Forgetting that Grignard Reagents Attack Esters Twice

The Mistake: Students see "Phenyl benzoate $\rightarrow$ Benzaldehyde" and think a Grignard reagent ( $CH_3MgBr$ ) will simply add once to the ester to give an aldehyde.
Why it’s wrong: Grignard reagents are strong nucleophiles. They do not stop at the aldehyde stage. They will attack the newly formed aldehyde to give a tertiary alcohol.
- $PhCOOPh + CH_3MgBr \rightarrow$ (after workup) Triphenylmethanol (a tertiary alcohol), not benzaldehyde.
How to Avoid: Remember the reactivity hierarchy.
- $LiAlH_4$ reduces esters all the way to primary alcohols.
- DIBAL-H (Diisobutylaluminium hydride) is a bulky, milder reducing agent. At low temperatures (-78°C) , it stops at the aldehyde stage.
- Result: Match (iv) with (b).

Mistake 3: Assuming $LiAlH_4$ Reduces Everything to Alcohols (Blindly)

The Mistake: Students know $LiAlH_4$ is a powerful reducing agent and assume it is the correct answer for (ii) Benzaldehyde $\rightarrow$ 1-Phenylethanol.
Why it’s wrong: $LiAlH_4$ reduces benzaldehyde to benzyl alcohol ( $PhCH_2OH$ ), not 1-phenylethanol ( $PhCH(OH)CH_3$ ). To get 1-phenylethanol, you need to add a carbon atom (the methyl group).
How to Avoid: Distinguish between reduction (removing oxygen/adding hydrogen) and carbon-carbon bond formation.
- $LiAlH_4$ is for reduction (no new C-C bonds).
- $CH_3MgBr$ (Grignard) is for adding a methyl group to a carbonyl.
- Reaction: $PhCHO + CH_3MgBr \rightarrow$ (after hydrolysis) $PhCH(OH)CH_3$ .
- Result: Match (ii) with (d). …

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Q.Match the reactions given in Column I with the suitable reagents given in Column II. Column I:

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Q.Match the reactions given in Column I with the suitable reagents given in Column II.
Column I:

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