PracticeHow it worksFeaturesPricingBlog Start practising free
Past Paper Solutions

Some Basic Principles of JEE 2024: Why the Answer Is 2 or 4

JEE Advanced 2024 Chemistry Some Basic Principles of Organic Chemistry Ozonolysis, iodoform test and PCC oxidation

By Founder, JEEnius - IIT Kanpur Alumni · Aug 15, 2026 · 4 min read

Hard 4 min target

PARAGRAPH I

An organic compound P with molecular formula C9H18O2 decolorizes bromine water and also shows positive iodoform test. P on ozonolysis followed by treatment with H2O2 gives Q and R. While compound Q shows positive iodoform test, compound R does not give positive iodoform test. Q and R on oxidation with pyridinium chlorochromate (PCC) followed by heating give S and T, respectively. Both S and T show positive iodoform test.
Complete copolymerization of 500 moles of Q and 500 moles of R gives one mole of a single acyclic copolymer U.
[Given, atomic mass: H = 1, C = 12, O = 16]

Q.14 Sum of number of oxygen atoms in S and T is ____.

Figure for this Chemistry question
Show answerAnswer

2 or 4

Explanation

The molecular formula C9H18O2 has one degree of unsaturation, consistent with one alkene double bond. Decolourisation of bromine water confirms the presence of a C=C bond. Positive iodoform test indicates the presence of a CH3CH(OH)− group or a CH3CO− group. Since P has two oxygens and undergoes oxidative ozonolysis to acids, P can be an unsaturated diol.

One valid structure of P is:

CH3−CH(OH)−CH2−CH=CH−CH2−CH(OH)−CH2−CH3

Oxidative ozonolysis using O3 followed by H2O2 cleaves the double bond and converts each alkene carbon into a carboxylic acid group.

Thus, Q is:

CH3−CH(OH)−CH2−COOH

This gives positive iodoform test because it contains the CH3CH(OH)− group.

R is:

HOOC−CH2−CH(OH)−CH2−CH3

This does not give positive iodoform test because the alcohol carbon is not attached to a methyl group in the required CH3CH(OH)− form.

PCC oxidizes secondary alcohols to ketones. Hence Q gives the β-keto acid:

CH3−CO−CH2−COOH

On heating, β-keto acids decarboxylate to ketones, so S is acetone:

CH3−CO−CH3

S has 1 oxygen atom.

Similarly, R gives after PCC oxidation:

HOOC−CH2−CO−CH2−CH3

On heating, this β-keto acid decarboxylates to T:

CH3−CO−CH2−CH3

T has 1 oxygen atom.

Therefore, in this case, the sum of oxygen atoms in S and T is:

1+1=2

An alternative valid structure of P also satisfies the given conditions:

CH3−CH(OH)−CH=CH−CH2−CH(OH)−CH2−CH2−CH3

Oxidative ozonolysis gives Q as lactic acid:

CH3−CH(OH)−COOH

PCC oxidation gives pyruvic acid S:

CH3−CO−COOH

S contains 3 oxygen atoms and gives positive iodoform test because it contains the CH3CO− group.

The corresponding R is:

HOOC−CH2−CH(OH)−CH2−CH2−CH3

After PCC oxidation and heating, it gives a methyl ketone T with 1 oxygen atom.

Thus, in this alternative case, the sum is:

3+1=4

Hence the accepted answer is 2 or 4.

Watch the full solution, worked step by step.

What was the JEE Advanced 2024 question on ozonolysis, PCC and iodoform test?

The some basic principles of jee 2024 problem has two legitimate answers, 2 or 4, because the evidence permits two structures of P. This was a hard JEE Advanced 2024 Paper 2 Chemistry numerical question, with an expected solving time of 240 seconds based on the need to audit both structural branches through several reactions.

P has the molecular formula:

C9H18O2

It decolourises bromine water and gives a positive iodoform test. Oxidative ozonolysis produces Q and R, of which only Q initially gives the iodoform test.

PCC oxidation of Q and R, followed by heating, produces S and T. Both final products give the iodoform test. Complete copolymerisation of 500 moles each of Q and R gives one mole of a single acyclic copolymer, but no polymer mass calculation is required.

The numerical answer asks for the total number of oxygen atoms in S and T.

How do the given tests constrain the structure of P?

Start with the molecular formula rather than guessing a carbon chain. P has one degree of unsaturation, which bromine-water decolourisation assigns to one carbon-carbon double bond. Its positive iodoform test and two oxygen atoms support an unsaturated diol containing the required secondary alcohol group.

The degree of unsaturation is:

DBE=2×9+2−182=1

Decolourisation of bromine water confirms that this single unsaturation is one carbon-carbon double bond.

A positive iodoform test points to either:

CH3CH(OH)−

or:

CH3CO−

In P, the relevant group is:

CH3CH(OH)−

Oxidative ozonolysis uses:

O3, followed by H2O2

Ozone cleaves the alkene. The hydrogen peroxide work-up oxidises each alkene carbon bearing hydrogen into a carboxylic acid group. Q and R are therefore hydroxy acids.

Their copolymerisation is consistent with molecules containing both alcohol and carboxylic acid groups. The polymer statement checks the proposed products, but it is not the quantity being calculated.

How does the first valid structure of P give the answer 2?

In the first valid branch, both ozonolysis products become β-keto acids after PCC oxidation. Heating removes carbon dioxide from each. The final products are acetone and butan-2-one, each containing one oxygen atom, so the required total is 2.

Take P as:

CH3−CH(OH)−CH2−CH=CH−CH2−CH(OH)−CH2−CH3

It has nine carbons, one alkene and two alcohol groups. Oxidative ozonolysis gives:

Q=CH3−CH(OH)−CH2−COOH
R=HOOC−CH2−CH(OH)−CH2−CH3

Q is iodoform-positive because it contains:

CH3CH(OH)−

R is iodoform-negative at this stage because its secondary alcohol carbon is not directly bonded to a methyl group.

PCC oxidises the secondary alcohol in Q:

CH3−CH(OH)−CH2−COOH⟶CH3−CO−CH2−COOH

This is a β-keto acid. It decarboxylates on heating:

CH3−CO−CH2−COOH→ΔCH3−CO−CH3+CO2

Thus, S is acetone and has one oxygen atom.

For R:

HOOC−CH2−CH(OH)−CH2−CH3→PCCHOOC−CH2−CO−CH2−CH3

Heating causes β-keto-acid decarboxylation:

HOOC−CH2−CO−CH2−CH3→ΔCH3−CO−CH2−CH3+CO2

T is butan-2-one and has one oxygen atom. Acetone and butan-2-one are methyl ketones, so both give the iodoform test.

Oxygen count=1+1=2

Why does the alternative structure of P give the answer 4?

The alkene can be placed one carbon closer to the iodoform-positive alcohol without violating any condition. This makes Q lactic acid. Its PCC product is pyruvic acid, an α-keto acid that does not undergo the β-keto-acid decarboxylation used in the first branch, so S retains three oxygen atoms.

The alternative P is:

CH3−CH(OH)−CH=CH−CH2−CH(OH)−CH2−CH2−CH3

It also has nine carbons, one alkene, two alcohol groups and the required iodoform-positive group. Oxidative ozonolysis gives:

Q=CH3−CH(OH)−COOH
R=HOOC−CH2−CH(OH)−CH2−CH2−CH3

Q is lactic acid and gives the iodoform test because it contains:

CH3CH(OH)−

R does not contain this group and is iodoform-negative at this stage.

PCC oxidises lactic acid to pyruvic acid:

CH3−CH(OH)−COOH→PCCCH3−CO−COOH

Pyruvic acid is an α-keto acid, not the β-keto acid formed in the first branch. S therefore remains pyruvic acid and contains three oxygen atoms. It remains iodoform-positive because it contains:

CH3CO−

For R:

HOOC−CH2−CH(OH)−CH2−CH2−CH3→PCCHOOC−CH2−CO−CH2−CH2−CH3

Heating causes β-keto-acid decarboxylation and gives the corresponding methyl ketone, pentan-2-one:

T=CH3−CO−CH2−CH2−CH3

T has one oxygen atom.

Oxygen count=3+1=4

The official accepted numerical answer is 2 or 4.

Why is counting oxygen before heating incorrect?

Counting immediately after PCC can give 6 in the first branch because each oxidation product contains three oxygen atoms. That calculation stops one reaction too early. S and T are defined after PCC oxidation followed by heating, not immediately after PCC oxidation.

The incorrect count is:

3+3=6

Both intermediates are β-keto acids. Heating removes carbon dioxide and converts each into a one-oxygen ketone:

1+1=2

A second method error is forcing one unique structure of P. After obtaining 2, shift the alkene while preserving every test. The alternative valid placement produces an α-keto acid and gives the accepted value 4.

Which related questions should you practise from some basic principles of jee 2024?

These three problems test the same decisions: PCC oxidation of secondary alcohols, β-keto-acid decarboxylation, oxidative ozonolysis and iodoform recognition. Write the intermediate after each reaction. That prevents oxygen counting before heating and makes the α-keto-acid versus β-keto-acid distinction explicit.

  1. Treat the following compound with PCC and then heat it:
CH3−CH(OH)−CH2−COOH

PCC gives a β-keto acid:

CH3−CO−CH2−COOH

Heating causes decarboxylation.

Checkpoint: The final product is acetone. It has one oxygen atom and gives a positive iodoform test.

  1. Apply the same sequence to:
HOOC−CH2−CH(OH)−CH2−CH3

PCC gives:

HOOC−CH2−CO−CH2−CH3

Checkpoint: Heating gives butan-2-one after β-keto-acid decarboxylation.

  1. Predict the oxidative ozonolysis products of:
CH3−CH(OH)−CH=CH−CH2OH

The products are lactic acid and glycolic acid:

CH3−CH(OH)−COOH
HOOC−CH2OH

Checkpoint: Lactic acid gives the positive iodoform test. Glycolic acid does not.

For one more closely matched problem, use Organic Compounds Containing Oxygen JEE 2025: Answer.

Frequently asked questions

What is the correct answer to this JEE Advanced 2024 question?

The officially accepted numerical answer is 2 or 4. Two structures of P satisfy every stated condition, and they produce different oxygen counts after PCC oxidation and heating.

Why is 6 not the correct oxygen count?

A count of 6 comes from counting the oxygen atoms immediately after PCC oxidation. The products must also be heated, causing the beta-keto acids to decarboxylate and form ketones.

Why does only Q initially give the iodoform test?

Q contains the CH3CH(OH)– group required for a positive iodoform test. R has a secondary alcohol, but its alcohol-bearing carbon is not directly attached to a methyl group.

Why does the alternative structure give an answer of 4?

The alternative structure produces lactic acid as Q, which PCC oxidises to pyruvic acid. Pyruvic acid retains three oxygen atoms, while R ultimately forms a one-oxygen methyl ketone, giving a total of 4.

iodoform testjee advanced 2024organic chemistryozonolysispcc oxidation

Practise this with JEEnius AI

25 years of PYQs, AI doubt solving, and the 2027 prediction paper.

Start practising free