An organic compound 'A' (molecular formula $C_8H_8O$) gives 2,4-DNP test. It does not give Tollen's test, but gives a yellow precipitate 'B' with NaOH and $I_2$. On drastic oxidation, it gives a carboxylic acid 'C' with formula $C_7H_6O_2$. Identify 'A', 'B', 'C' and write the reactions involved.

An organic compound 'A' (molecular formula C₈H₈O) gives 2,4-DNP test. It does not give Tollen's test, but give

Question

C_8H_8O

CBSECBSE Class XII Board 2025Subjective· 3mImportance★★★★★

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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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Compound A is an aromatic methyl ketone (acetophenone, $\ce{C6H5COCH3}$ ) that gives a positive iodoform test and, upon vigorous oxidation, yields benzoic acid ( $\ce{C7H6O2}$ ). The yellow precipitate B is iodoform ( $\ce{CHI3}$ ).

The problem gives us a molecular formula $\ce{C8H8O}$ and a series of chemical tests. Let’s decode each clue one by one.

1. The 2,4-DNP test is positive.

This tells us the compound has a carbonyl group ( $\ce{C=O}$ ) — either an aldehyde or a ketone. The 2,4-DNP reagent reacts with any carbonyl to give an orange-red precipitate.

2. Tollen’s test is negative.

Tollen’s reagent (ammoniacal silver nitrate) oxidises only aldehydes (and $\alpha$ -hydroxy ketones) to carboxylic acids, depositing a silver mirror. A negative result means A is not an aldehyde. So A must be a ketone.

3. Iodoform test (NaOH + $\ce{I2}$ ) gives a yellow precipitate B.

This is the classic iodoform test. A positive result requires a methyl ketone ( $\ce{CH3CO-}$ ) or a secondary alcohol with a $\ce{CH3CH(OH)-}$ group (which gets oxidised to a methyl ketone under the reaction conditions). The yellow precipitate is iodoform, $\ce{CHI3}$ .

So A must contain the $\ce{CH3CO-}$ group. That means A is a methyl ketone.

4. Drastic oxidation gives a carboxylic acid C with formula $\ce{C7H6O2}$ .

The formula $\ce{C7H6O2}$ is characteristic of benzoic acid, $\ce{C6H5COOH}$ (check: $\ce{C7H6O2}$ ). This tells us that the oxidation cleaves the molecule, leaving a benzene ring with a carboxyl group.

Now, A has $\ce{C8H8O}$ . If oxidation gives benzoic acid ( $\ce{C7H6O2}$ ), then the original molecule must have a benzene ring plus a two-carbon side chain that includes the carbonyl. The only $\ce{C8H8O}$ methyl ketone that fits is acetophenone (phenyl methyl ketone), $\ce{C6H5COCH3}$ .

Tip

The degree of unsaturation (DU) for $\ce{C8H8O}$ is $8 - \frac{8}{2} + 1 = 5$ . A benzene ring accounts for 4, and the carbonyl adds 1 — perfect match. No other $\ce{C8H8O}$ methyl ketone (like a substituted benzaldehyde) would give benzoic acid on oxidation.

Let’s confirm the reactions step by step.

Step 1: Iodoform reaction of A

Acetophenone reacts with $\ce{I2}$ in the presence of $\ce{NaOH}$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set D1 markMCQ
$Q. An aldehyde on oxidation gives$
(a) an alcohol
(b) a ketone
(c) an ether
(d) an acid
›Reveal solutionSolution

Oxidation of an aldehyde gives a carboxylic acid.

Aldehydes carry an H on the carbonyl carbon and are easily oxidised. With oxidising agents (or even mild reagents such as Tollen's or Fehling's), an aldehyde is converted to the corresponding carboxylic acid:

R-CHO + [O] -> R-COOH

…

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CBSE 2024Set ANNUAL1 mark
$Q. How would you obtain the following? Benzoic acid from ethyl benzene$
›Reveal solutionSolution

Vigorous oxidation (hot alkaline $KMnO_4$ ) of any alkylbenzene side chain, regardless of its length, converts it entirely to a single $-COOH$ group attached directly to the ring.

Ethylbenzene, $C_6H_5{-}CH_2CH_3$ , has a two-carbon side chain with benzylic hydrogens. Strong oxidising agents like hot alkaline potassium permanganate attack the side chain at the benzylic position and progressively oxidise it, cleaving off the extra carbon(s) and leaving only the ring-attached carbon as a carboxyl group — the exact chain length beyond the first carbon does not matter, the product is always benzoic acid:

…

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CBSE 2023Set 56/1/11 markMCQ
$CH_3CONH_2$
›Reveal solutionSolution
This is the Hofmann bromamide degradation reaction. An amide ( $CH_3CONH_2$ ) reacts with bromine and a base to give a primary amine with one fewer carbon atom. The product here is methylamine ( $CH_3NH_2$ ), which corresponds to option (B).

The reaction you're looking at is a classic name reaction in organic chemistry — the Hofmann bromamide degradation. It's one of the most reliable ways to convert an amide into a primary amine, and it always involves a loss of one carbon from the chain. Let's understand why.

The key idea: the amide group ( $-CONH_2$ ) gets "chopped" by bromine in the presence of a strong base. The carbonyl carbon (the one attached to oxygen) is lost as carbon dioxide, and the nitrogen ends up attached to the alkyl group that was originally next to the carbonyl. So the product has one carbon fewer than the starting amide.

Now let's walk through the reaction step by step for your specific compound, acetamide ( $CH_3CONH_2$ ).

Identify the starting material.

Acetamide has the structure $CH_3-CO-NH_2$ . The alkyl group attached to the carbonyl is a methyl group ( $CH_3-$ ). The amide carbon is the carbonyl carbon.

Recall the general outcome of Hofmann degradation.

The reaction is:

$R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + 2NaBr + Na_2CO_3 + 2H_2O$
Notice that the product $R-NH_2$ has the same $R$ group as the starting amide, but the carbonyl carbon is gone (it becomes carbonate). So the amine has one less carbon than the amide.

Apply to acetamide.

Here $R = CH_3-$ . So the amine formed is $CH_3-NH_2$ , which is methylamine.

Check the options.

(A) $CH_3COONa$ — this is sodium acetate, not an amine.

(B) $CH_3NH_2$ — methylamine, matches our prediction. …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. In Benzaldehyde + [O] --(Air)--> A, A is$
(a) Benzene
(b) Benzoic acid
(c) Benzyl alcohol
(d) None of these
›Reveal solutionSolution

Aromatic aldehydes like benzaldehyde undergo slow autoxidation in air, converting -CHO to -COOH.

C6H5CHO + [O] --(air)--> C6H5COOH

On exposure to air, benzaldehyde is slowly autoxidised at the aldehydic hydrogen, converting the -CHO group into a -COOH group and giving benzoic acid. (Th …

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CBSE 2020Set 56/1/11 markMCQ
$Q. Iodoform test is not given by (A) Ethanol (B) Ethanal (C) Pentan-2-one (D) Pentan-3-one$
›Reveal solutionSolution
The iodoform test detects the presence of a methyl carbonyl group ( $\ce{CH3CO-}$ ) or a methyl carbinol group ( $\ce{CH3CH(OH)-}$ ) that can be oxidised to a methyl carbonyl. Pentan-3-one lacks this structural feature, so it does not give the test. The correct option is (D).

The iodoform test is a classic qualitative test in organic chemistry. It's not just a random reaction — it's a specific probe for a very particular structural arrangement. When you see a question about which compound gives or doesn't give this test, you're really being asked: "Which of these molecules has a methyl group directly attached to a carbonyl carbon (or to a carbon that can be easily oxidised to a carbonyl)?"

The test works because the methyl group in $\ce{CH3CO-}$ is uniquely reactive under basic, halogenating conditions. The three hydrogens on that methyl are successively replaced by iodine, forming a triiodomethyl intermediate. This intermediate is unstable and breaks apart, yielding a yellow precipitate of iodoform ( $\ce{CHI3}$ ) — that's the visible "positive" result.

Now, there's a second pathway. A primary alcohol with the structure $\ce{CH3CH(OH)-R}$ (where R can be H or any alkyl/aryl group) can be oxidised in situ by the iodine in the basic solution to give $\ce{CH3CO-R}$ , which then undergoes the same reaction. So ethanol and any secondary alcohol with a methyl group on the alcohol carbon also give a positive test.

Let's examine each option.

Ethanol ( $\ce{CH3CH2OH}$ )

This is a primary alcohol with the structure $\ce{CH3CH2OH}$ . Under the reaction conditions (basic $\ce{I2}$ ), it gets oxidised to ethanal ( $\ce{CH3CHO}$ ), which has a methyl carbonyl group. The test is positive.

Tip

Ethanol is the classic example of a compound that gives the iodoform test after oxidation. Many students forget this pathway and wrongly think only carbonyl compounds respond.

Ethanal ( $\ce{CH3CHO}$ )

This is acetaldehyde — the simplest methyl carbonyl. It has the $\ce{CH3CO-}$ group directly. The test is strongly positive. In fact, this is the reference compound for the test.

Pentan-2-one ( $\ce{CH3COCH2CH2CH3}$ ) …
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CBSE 2020Set ANNUAL1 markMCQ
$Q. Benzaldehyde + [O] --(Air)--> A. 'A' is$
(a) Benzene
(b) Benzoic acid
(c) Benzyl alcohol
(d) None of these
›Reveal solutionSolution

Aldehydes are easily oxidised even by atmospheric oxygen (autoxidation); benzaldehyde left exposed to air slowly oxidises to benzoic acid.

Benzaldehyde (C6H5CHO) has a reactive aldehydic hydrogen. On standing in air, atmospheric O2 slowly oxidises it:

C6H5CHO + [O] --(air)--> C6H5COOH

…

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