When liquid 'A' is treated with a freshly prepared ammoniacal silver nitrate solution, it gives a bright silve

NCERT Exemplar · Q35

Q. When liquid 'A' is treated with a freshly prepared ammoniacal silver nitrate solution, it gives a bright silver mirror. The liquid forms a white crystalline solid on treatment with sodium hydrogen sulphite. Liquid 'B' also forms a white crystalline solid with sodium hydrogen sulphite but it does not give a test with ammoniacal silver nitrate. Which of the two liquids is an aldehyde? Write the chemical equations of these reactions also.

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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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The key idea is that aldehydes reduce Tollens’ reagent (ammoniacal silver nitrate) to give a silver mirror, while ketones do not. Both aldehydes and ketones form bisulfite addition products. Therefore, liquid A (which gives the silver mirror) is the aldehyde, and liquid B is a ketone.

Why This Approach Works

The problem tests your understanding of two classic carbonyl reactions: the Tollens’ test and the bisulfite addition.

Tollens’ reagent (ammoniacal silver nitrate) is a mild oxidizing agent. Only aldehydes can be oxidized to carboxylic acids, reducing $Ag^+$ to metallic silver — that’s the “silver mirror”. Ketones lack the terminal $–CHO$ group and cannot be oxidized under these conditions.
Sodium hydrogen sulphite ( $NaHSO_3$ ) adds across the carbonyl group in both aldehydes and many ketones (especially methyl ketones) to form a white crystalline bisulfite addition product. So this test alone cannot distinguish between the two.

Thus, the Tollens’ test is the decisive clue: whichever liquid gives a silver mirror must be an aldehyde.

Step-by-Step Reasoning

Identify the distinguishing test

The problem gives two liquids, A and B. Both form a white crystalline solid with $NaHSO_3$ — that tells us both are carbonyl compounds (aldehyde or ketone). But only A gives a silver mirror with Tollens’ reagent. Since only aldehydes give a positive Tollens’ test, A must be an aldehyde and B must be a ketone.
Write the reaction for the Tollens’ test (liquid A)

The aldehyde is oxidized to a carboxylate ion, while $Ag^+$ is reduced to silver metal.

A general equation (using $RCHO$ for the aldehyde):

RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow RCOO^- + 2Ag \downarrow + 4NH_3 + 2H_2O

The silver deposits as a shiny mirror on the test tube walls.

Write the reaction for the bisulfite addition (both liquids) The carbonyl group adds $HSO_3^-$ to form a hydroxy sulfonate salt. For an aldehyde (A):

RCHO + NaHSO_3 \rightarrow RCH(OH)SO_3^-Na^+ \ \text{(white crystalline solid)}

For a ketone (B):

$R_2CO + NaHSO_3 \rightarrow R_2C(OH)SO_3^-Na^+ \ \text{(white crystalline solid)}$ …

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This is a classic distinction between aldehydes and ketones using chemical tests. Here is a breakdown of the common mistakes students make and how to avoid them.

🧪 The Core Concept

Ammoniacal silver nitrate (Tollens’ test) → positive only for aldehydes (and some α-hydroxy ketones, but not simple ketones).
Sodium hydrogen sulphite (NaHSO₃) → forms a white crystalline addition product with both aldehydes and many ketones (especially methyl ketones).

So:

Liquid A → gives both tests → aldehyde
Liquid B → gives only NaHSO₃ test → ketone

✗ Common Mistake #1

Thinking NaHSO₃ test is exclusive to aldehydes

Why it’s wrong:

Many students memorise “NaHSO₃ gives white crystals with aldehydes” and forget that ketones (especially methyl ketones like acetone) also give this test.

✓ How to avoid:

Remember:

NaHSO₃ test is positive for both aldehydes and many ketones.

It is not a distinguishing test — Tollens’ test is the real decider.

✗ Common Mistake #2

Confusing Tollens’ test with Fehling’s test

Why it’s wrong:

Both tests detect aldehydes, but:

Tollens’ uses ammoniacal AgNO₃ → silver mirror
Fehling’s uses Cu²⁺ in alkaline tartrate → red precipitate of Cu₂O

Students sometimes write Fehling’s equation when Tollens’ is asked.

✓ How to avoid:

Read the question carefully. If it says “ammoniacal silver nitrate”, it’s Tollens’ test only. Write the correct equation:

Tollens’ test (aldehyde):

RCHO + 2[Ag(NH_3)_2]^+ + 2OH^- \rightarrow RCOO^- + 2Ag \downarrow + 4NH_3 + 2H_2O

✗ Common Mistake #3

Writing the wrong product for NaHSO₃ addition

Why it’s wrong:

Students often write the product as a simple bisulfite salt, forgetting the addition product structure.

✓ How to avoid:

The reaction is a nucleophilic addition of HSO₃⁻ to the carbonyl carbon. The correct product is:

RCHO + NaHSO_3 \rightarrow RCH(OH)SO_3Na

For a ketone (e.g., acetone):

CH_3COCH_3 + NaHSO_3 \rightarrow (CH_3)_2C(OH)SO_3Na

Both are white crystalline solids.

✗ Common Mistake #4

Not identifying which liquid is which

Why it’s wrong: …

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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

Why This Approach Works

Step-by-Step Reasoning

Concept: Distinguishing Aldehydes from Ketones using Chemical Tests

Method: Silver Mirror Test (Tollens' Test) & Bisulfite Addition Test

Step-by-Step Reasoning

1. Tollens' Test (for aldehyde A)

2. Bisulfite Addition (for both A and B)

🧪 The Core Concept

✗ Common Mistake #1

Why it’s wrong:

✓ How to avoid:

✗ Common Mistake #2

Why it’s wrong:

✓ How to avoid:

✗ Common Mistake #3

Why it’s wrong:

✓ How to avoid:

✗ Common Mistake #4

Why it’s wrong: …

Why This Approach Works

Step-by-Step Reasoning

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