Assertion (A): Aromatic aldehydes and formaldehyde undergo Cannizzaro's reaction. Reason (R): Aromatic aldehyd

NCERT Exemplar · Q49

Q. Assertion (A): Aromatic aldehydes and formaldehyde undergo Cannizzaro's reaction. Reason (R): Aromatic aldehydes are almost as reactive as formaldehyde.

(i) Assertion and Reason both are correct and Reason is the correct explanation of Assertion.

(ii) Assertion and Reason both are wrong.

(iii) Assertion is correct but Reason is wrong.

(iv) Assertion is wrong but Reason is correct.

Sikkim CbseMCQ· 1mImportance★★★★★

94% · 82/87 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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.

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

Start your 14-day free trial to unlock the full solution →

The key idea: Cannizzaro reaction requires an aldehyde with no α-hydrogen. Both aromatic aldehydes and formaldehyde satisfy this, so the Assertion is correct. But the Reason claims they are "almost as reactive" — this is false because formaldehyde is far more reactive due to its small size and lack of steric hindrance. Hence, Assertion is correct, Reason is wrong → option (iii).

What is the Cannizzaro reaction?

It is a disproportionation reaction where two molecules of an aldehyde (with no α-hydrogen) react in the presence of a strong base to give one molecule of alcohol and one molecule of carboxylic acid (as its salt). The key requirement: the aldehyde must not have a hydrogen atom on the carbon adjacent to the –CHO group (i.e., no α-hydrogen). Why? Because if α-hydrogens were present, the base would instead deprotonate them, leading to aldol condensation — not Cannizzaro.
Why do aromatic aldehydes and formaldehyde undergo this reaction?
- Formaldehyde ( $HCHO$ ) has no α-carbon at all — the carbon of the –CHO group is bonded only to two hydrogens. So no α-hydrogen exists.
- Aromatic aldehydes like benzaldehyde ( $C_6H_5CHO$ ) have the –CHO group directly attached to the benzene ring. The α-carbon would be the ring carbon, which has no hydrogen (it's part of the aromatic system). So again, no α-hydrogen. Both satisfy the condition, so the Assertion is correct.
Now examine the Reason: "Aromatic aldehydes are almost as reactive as formaldehyde."

This is about reactivity toward nucleophilic addition (the first step of Cannizzaro involves hydroxide attacking the carbonyl carbon). Reactivity depends on:
- Electrophilicity of the carbonyl carbon: Formaldehyde has no alkyl or aryl group to donate electron density — its carbonyl carbon is highly electron-deficient, making it extremely reactive. …

Start your 14-day free trial to unlock the full solution →

Here’s a breakdown of the common mistakes students make on this question and how to avoid each.

Mistake 1: Thinking all aldehydes undergo Cannizzaro’s reaction

The error: Students assume that because formaldehyde and aromatic aldehydes undergo Cannizzaro’s reaction, all aldehydes do. They forget that aliphatic aldehydes (except formaldehyde) have alpha-hydrogens and undergo aldol condensation instead.
How to avoid: Remember the key condition for Cannizzaro’s reaction: the aldehyde must have no alpha-hydrogen atom. Only then can it undergo disproportionation in the presence of a strong base.
- Formaldehyde ( $HCHO$ ) — no alpha-H ✓
- Aromatic aldehydes (e.g., benzaldehyde $C_6H_5CHO$ ) — no alpha-H ✓
- Aliphatic aldehydes like acetaldehyde ( $CH_3CHO$ ) — have alpha-H ✗

Mistake 2: Confusing reactivity order of aldehydes

The error: The Reason (R) states that aromatic aldehydes are “almost as reactive as formaldehyde.” Many students accept this without checking. In reality, formaldehyde is far more reactive than aromatic aldehydes toward nucleophilic addition.
How to avoid: Know the reactivity trend for nucleophilic addition:
- Formaldehyde > Aliphatic aldehydes > Aromatic aldehydes
- Why? Aromatic aldehydes have the carbonyl group attached to an electron-rich benzene ring. The ring donates electron density via resonance, making the carbonyl carbon less electrophilic (less positive). Formaldehyde has no such electron-donating group, so its carbonyl carbon is much more reactive.
- So, the Reason is wrong — aromatic aldehydes are less reactive than formaldehyde.

Mistake 3: Misinterpreting the Assertion-Reason logic

The error: Students pick option (A) because both statements seem true, without checking if the Reason actually explains the Assertion. Even if both were true, the Reason here is false, so (A) is automatically invalid.
How to avoid: Always test the causal link:
- Does the Reason correctly explain why the Assertion happens? …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ48 — Assertion (A): The α-hydrogen atom in carbonyl compounds is less acidi…→Next questionQ50 — Assertion (A): Aldehydes and ketones both react with Tollen's reagent …

Q.Assertion (A): Aromatic aldehydes and formaldehyde undergo Cannizzaro's reaction.
Reason (R): Aromatic aldehydes are almost as reactive as formaldehyde.

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

Concept: Cannizzaro Reaction

Method: Factual Verification + Logical Check

Mistake 1: Thinking all aldehydes undergo Cannizzaro’s reaction

Mistake 2: Confusing reactivity order of aldehydes

Mistake 3: Misinterpreting the Assertion-Reason logic

Q.Assertion (A): Aromatic aldehydes and formaldehyde undergo Cannizzaro's reaction. Reason (R): Aromatic aldehydes are almost as reactive as formaldehyde.

🎓Unlock everything free for 14 days

Q.Assertion (A): Aromatic aldehydes and formaldehyde undergo Cannizzaro's reaction.
Reason (R): Aromatic aldehydes are almost as reactive as formaldehyde.

Unlock everything free for 14 days