Assertion (A): Aldehydes and ketones both react with Tollen's reagent to form a silver mirror. Reason (R): Bot

NCERT Exemplar · Q50

Q. Assertion (A): Aldehydes and ketones both react with Tollen's reagent to form a silver mirror. Reason (R): Both aldehydes and ketones contain a carbonyl group.

(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.

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Concept understanding — Oxidation Reactions

Alcohol Oxidation: The Intuition First

Imagine you have a molecule of ethanol — the alcohol in your hand sanitizer or a drink. It has a carbon atom bonded to an –OH group. Now picture that –OH group as a "handle" that can be transformed. Oxidation, in organic chemistry, doesn't always mean adding oxygen — it often means removing hydrogen from a carbon that already has a bond to oxygen. For alcohols, oxidation is like "stripping away" hydrogen atoms from the carbon that holds the –OH, turning the alcohol into a more oxidized functional group.

Think of it this way: a primary alcohol (R–CH₂–OH) has two hydrogens on the carbon with the –OH. If you remove one hydrogen and the hydrogen from the –OH, you get an aldehyde (R–CHO). Remove both hydrogens (and the –OH hydrogen), and you get a carboxylic acid (R–COOH). A secondary alcohol (R–CHOH–R') has only one hydrogen on that carbon — remove it, and you get a ketone (R–CO–R'). A tertiary alcohol has no hydrogen on that carbon — so it cannot be oxidized without breaking the carbon skeleton.

That's the core intuition: oxidation of an alcohol is about removing hydrogens from the carbon bearing the –OH group. The more hydrogens you can remove, the more oxidized the product.

The Precise Statement

Alcohol oxidation is the process in which an alcohol loses hydrogen atoms (dehydrogenation) from the carbon bonded to the -OH group, increasing the number of C-O bonds (or decreasing C-H bonds). The outcome depends on the class of the alcohol:

Alcohol Class	Structure	Product after oxidation	Reagent example
Primary (1°)	R–CH₂–OH	Aldehyde (R–CHO) then Carboxylic acid (R–COOH)	PCC (stops at aldehyde); K₂Cr₂O₇/H⁺ (goes to acid)
Secondary (2°)	R–CHOH–R'	Ketone (R–CO–R')	K₂Cr₂O₇/H⁺, CrO₃, etc.
Tertiary (3°)	R₃C–OH	No reaction (under normal conditions)	—

Watch out

A common mistake: students think "oxidation" always adds oxygen. For alcohols, it's removal of hydrogen from the carbon with the –OH. The oxygen from the –OH stays — it's the hydrogens that leave.

Why Does Tertiary Alcohol Not Oxidize?

Look at the carbon with the –OH in a tertiary alcohol: it has three carbon groups attached and no hydrogen. To form a C=O bond, you'd need to remove a hydrogen from that carbon — but there is none. The only way to oxidize a tertiary alcohol is to break a C–C bond (strong and difficult), which is not typical oxidation. So in standard organic chemistry, tertiary alcohols are inert to mild oxidizing agents.

A Real-World Analogy

Think of the alcohol carbon as a "parking spot" with a certain number of hydrogen "cars." Primary alcohol has two cars parked. Oxidation is like towing away one car (→ aldehyde) or both cars (→ carboxylic acid). Secondary alcohol has one car — tow it away, and you get a ketone. Tertiary alcohol has zero cars — nothing to tow, so no reaction.

Key Reagents to Remember (for exams)

PCC (pyridinium chlorochromate): oxidizes 1° alcohols to aldehydes only — stops there.
K₂Cr₂O₇ / H₂SO₄ (Jones reagent): oxidizes 1° alcohols all the way to carboxylic acids; 2° alcohols to ketones.
KMnO₄: similar to dichromate, but stronger — can over-oxidize. …

Why this formula?

Oxidation Reactions: Why the Key Principles Hold

Oxidation reactions are fundamental to chemistry, and understanding why they work the way they do is essential for mastering Indian board exams (Class 11, 12, JEE, NEET). Let's break down the core ideas from first principles.

1. The Core Definition: What Does "Oxidation" Really Mean?

Historically, oxidation meant "adding oxygen." But that's too narrow. The modern, exam-correct definition is:

Oxidation is the loss of electrons by a species.

This is the electronic concept (given by the ionic theory). The why behind this definition comes from the behavior of atoms during chemical reactions.

Why do atoms lose electrons?

Atoms seek stability. They achieve this by having a full outer electron shell (octet, duplet, or pseudo-inert gas configuration).

Metals (like Na, Mg, Fe) have few valence electrons (1, 2, or 3). It's energetically easier for them to lose these electrons than to gain 5, 6, or 7.
Non-metals (like O, Cl, F) have many valence electrons (5, 6, or 7). It's energetically easier for them to gain electrons.

So, when a metal reacts with a non-metal, the metal loses electrons (gets oxidized), and the non-metal gains electrons (gets reduced).

Example:

2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}

Na loses 1 electron: $\text{Na} \rightarrow \text{Na}^+ + e^-$ (Oxidation)
Cl gains 1 electron: $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$ (Reduction)

Key takeaway: Oxidation and reduction always happen together (Redox reactions). You cannot have one without the other.

2. The Key Formula(e): Oxidation Number Rules

The oxidation number (O.N.) is a bookkeeping tool. It's not a real charge (except in ionic compounds), but it helps track electron flow.

Why do we assign oxidation numbers?

Because in covalent compounds (like $\text{CH}_4$ or $\text{H}_2\text{O}$ ), electrons are shared, not transferred. We need a way to pretend they are transferred to see which atom "owns" the electrons more.

The Rules (and why they exist)

Rule	Statement	Why this rule?
1	O.N. of an element in its free state = 0	No electron transfer has occurred.
2	O.N. of a monatomic ion = its charge	The atom has actually lost/gained that many electrons.
3	O.N. of H = +1 (except in metal hydrides where it's -1)	H is less electronegative than O, F, Cl, but more electronegative than metals.
4	O.N. of O = -2 (except in peroxides where it's -1, superoxides -1/2, and with F where it's +2)	O is highly electronegative (3.44 on Pauling scale). It "pulls" electrons toward itself.
5	Sum of O.N. in a neutral compound = 0	The compound has no net charge.
6	Sum of O.N. in a polyatomic ion = charge of the ion	The ion's overall charge must be accounted for.

The Derivation of a Key Formula: Finding O.N. of an Unknown Element

Suppose you need to find the O.N. of S in $\text{H}_2\text{SO}_4$ .

Step 1: Write known O.N.s:

H: +1 (rule 3)
O: -2 (rule 4)
S: let it be $x$ (unknown)

Step 2: Apply rule 5 (neutral compound sum = 0):

2(+1) + x + 4(-2) = 0

Step 3: Solve:

2 + x - 8 = 0

x - 6 = 0

x = +6

Why this works: The oxidation number is a mathematical consequence of the electronegativity hierarchy. Oxygen is more electronegative than sulfur, so it "takes" the electrons. Hydrogen is less electronegative than sulfur, so it "gives" electrons to sulfur. The net result is that sulfur appears to have lost 6 electrons.

3. The Key Formula(e): Balancing Redox Equations

Two methods are exam-critical: Oxidation Number Method and Ion-Electron Method (Half-Reaction Method).

Why do we need these methods?

Because in a redox reaction, the total number of electrons lost (oxidation) must equal the total number of electrons gained (reduction). This is the Law of Conservation of Charge.

The Ion-Electron Method (for acidic medium) — Step-by-step why

Example: Balance $\text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+}$ (acidic)

Step 1: Write half-reactions.

Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$

Why? Fe loses 1 electron (O.N. goes from +2 to +3).
Reduction: $\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$

Why? Mn gains electrons (O.N. goes from +7 to +2).

Step 2: Balance atoms other than H and O.

Mn is already balanced (1 on each side).

Step 3: Balance O by adding $\text{H}_2\text{O}$ .

Left: 4 O atoms. Right: 0 O atoms.
Add 4 $\text{H}_2\text{O}$ to the right:

\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 4: Balance H by adding $\text{H}^+$ (because acidic medium).

Right: 8 H atoms (from 4 $\text{H}_2\text{O}$ ). Left: 0 H atoms.
Add 8 $\text{H}^+$ to the left:

8\text{H}^+ + \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 5: Balance charge by adding electrons.

Left: $8(+1) + (-1) = +7$ charge.
Right: $+2$ charge.
To make left = right, add 5 electrons to the left:

8\text{H}^+ + \text{MnO}_4^- + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 6: Multiply half-reactions to equalize electrons.

Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (×5)
Reduction: $8\text{H}^+ + \text{MnO}_4^- + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ (×1)

Step 7: Add them:

5\text{Fe}^{2+} + 8\text{H}^+ + \text{MnO}_4^- \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}

Why this works: Every step is driven by conservation laws:

Mass balance: Same number of each atom on both sides.
Charge balance: Net charge on left = net charge on right.
Electron balance: Electrons lost = electrons gained.

4. The Key Formula(e): Electrochemical Series and Cell Potential

For a galvanic cell (voltaic cell), the cell potential $E^\circ_{\text{cell}}$ is:

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

Why this formula? …

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The assertion is false because only aldehydes (not ketones) give a silver mirror with Tollen's reagent; the reason is true but does not explain the assertion. The correct option is (iv).

Understand the chemistry of Tollen's reagent.

Tollen's reagent is an ammoniacal solution of silver oxide ( $\text{Ag}_2\text{O}$ in $\text{NH}_3$ ). It acts as a mild oxidizing agent. The key reaction is the oxidation of an aldehyde to a carboxylate ion, while silver ions are reduced to metallic silver, which deposits as a mirror on the test tube.
Why only aldehydes work.

The aldehyde group has a hydrogen atom directly attached to the carbonyl carbon ( $\text{R–CHO}$ ). This hydrogen is easily removed during oxidation, converting the aldehyde into a carboxylic acid (or its salt). Ketones ( $\text{R–CO–R'}$ ) lack this hydrogen — both substituents are carbon groups. Without that hydrogen, ketones cannot be oxidized by mild agents like Tollen's reagent.
$\text{R–CHO} + 2[\text{Ag(NH}_3\text{)}_2]^+ + 2\text{OH}^- \rightarrow \text{R–COO}^- + 2\text{Ag} \downarrow + 4\text{NH}_3 + 2\text{H}_2\text{O}$
Evaluate the Assertion (A).

"Aldehydes and ketones both react with Tollen's reagent to form a silver mirror."

This is wrong — only aldehydes give the silver mirror. Ketones (except α-hydroxy ketones, which are a special case) do not react. So the assertion is false.
Evaluate the Reason (R).

"Both aldehydes and ketones contain a carbonyl group."

This is correct — every aldehyde and ketone has a $\text{C=O}$ group. But the presence of a carbonyl group alone does not guarantee reaction with Tollen's reagent; the specific structure (the aldehyde hydrogen) is what matters. So the reason is true, but it does not explain the (false) assertion. …

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Here are the common mistakes students make with this Assertion-Reason question, and how to avoid each.

Mistake 1: Assuming both statements are true because both involve a carbonyl group

Why it happens:

Students see that both aldehydes and ketones have a $\ce{C=O}$ group (Reason R is correct), and then incorrectly assume that both must react with Tollen’s reagent (making Assertion A seem correct).

How to avoid:

Know the functional group difference: Aldehydes have a $\ce{-CHO}$ group (carbonyl at the end of a chain), while ketones have $\ce{-CO-}$ (carbonyl in the middle).
Recall the specific reaction: Tollen’s reagent ( $\ce{[Ag(NH3)2]+}$ ) oxidises only aldehydes to carboxylic acids, reducing $\ce{Ag+}$ to metallic silver (the mirror). Ketones do not undergo this oxidation because they lack a hydrogen atom on the carbonyl carbon.
Key fact: The presence of a carbonyl group alone does not guarantee reaction with Tollen’s reagent — the structure matters.

Mistake 2: Thinking the Reason explains the Assertion (choosing option A)

Why it happens:

Students match “both contain carbonyl” with “both react” and assume a cause-effect link.

How to avoid:

Check the logical link: Even if both statements were true, the Reason (carbonyl group) does not explain why both would react — because ketones also have a carbonyl but do not react.
Use the “because” test: Ask yourself: “Does the Reason directly cause the Assertion to be true?” Here, the answer is no — the carbonyl group is necessary but not sufficient for the reaction.

Mistake 3: Forgetting that ketones do NOT give a silver mirror

Why it happens:

Students memorise “Tollen’s test = silver mirror for aldehydes” but forget that ketones are negative in this test.

How to avoid:

Create a clear contrast table:

| Compound | Tollen’s test result |

|----------|----------------------| …

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Q.Assertion (A): Aldehydes and ketones both react with Tollen's reagent to form a silver mirror.
Reason (R): Both aldehydes and ketones contain a carbonyl group.

Concept understanding — Oxidation Reactions

Alcohol Oxidation: The Intuition First

The Precise Statement

Why Does Tertiary Alcohol Not Oxidize?

A Real-World Analogy

Key Reagents to Remember (for exams)

Why this formula?

Oxidation Reactions: Why the Key Principles Hold

1. The Core Definition: What Does "Oxidation" Really Mean?

Why do atoms lose electrons?

2. The Key Formula(e): Oxidation Number Rules

Why do we assign oxidation numbers?

The Rules (and why they exist)

The Derivation of a Key Formula: Finding O.N. of an Unknown Element

3. The Key Formula(e): Balancing Redox Equations

Why do we need these methods?

The Ion-Electron Method (for acidic medium) — Step-by-step why

4. The Key Formula(e): Electrochemical Series and Cell Potential

Why this formula? …

Method: Conceptual Verification with Fact-Checking

Concept: Oxidation of Carbonyl Compounds by Tollen’s Reagent

Steps:

Mistake 1: Assuming both statements are true because both involve a carbonyl group

Mistake 2: Thinking the Reason explains the Assertion (choosing option A)

Mistake 3: Forgetting that ketones do NOT give a silver mirror

Q.Assertion (A): Aldehydes and ketones both react with Tollen's reagent to form a silver mirror. Reason (R): Both aldehydes and ketones contain a carbonyl group.

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Q.Assertion (A): Aldehydes and ketones both react with Tollen's reagent to form a silver mirror.
Reason (R): Both aldehydes and ketones contain a carbonyl group.

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