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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.

Why does Some Basic Principles of JEE 2024: Hard PYQ accept 2 or 4?

Some Basic Principles of JEE 2024: Hard PYQ has the accepted numerical answer 2 or 4 because two alkene placements satisfy every test. This is a hard JEE Advanced 2024 Paper 2 Chemistry numerical. The expected solving time is 240 seconds, based on the need to test two structures through ozonolysis, PCC oxidation and heating.

Compound P has the molecular formula:

C9H18O2

It decolourises bromine water and gives the iodoform test. Oxidative ozonolysis produces Q and R. Only Q initially gives the iodoform test. PCC oxidation followed by heating produces S and T, both of which give the test.

Question 14 asks for the total number of oxygen atoms in S and T. The complete copolymerisation statement belongs to the common paragraph, but it is not needed for this oxygen-count subquestion.

How do I convert each observation into a structural constraint?

Start by fixing one alkene and two alcohol groups. The molecular formula gives one degree of unsaturation, while bromine-water decolourisation assigns it to one carbon-carbon double bond. The positive iodoform test requires either a methyl carbinol group or a methyl ketone group.

The degree of unsaturation is:

DBE=2×9+2−182=1

One DBE, together with bromine-water decolourisation, confirms one carbon-carbon double bond.

A positive iodoform test points to either:

CH3CH(OH)−

or:

CH3CO−

Following the official route, treat P as an unsaturated diol. Oxidative ozonolysis with ozone followed by hydrogen peroxide cleaves the alkene. Each alkene carbon bearing hydrogen becomes a carboxylic acid group.

How does the first valid structure give an oxygen count of 2?

In the first arrangement, ozonolysis gives two hydroxy acids that become beta-keto acids after PCC oxidation. Heating removes carbon dioxide from both. The final products are acetone and butan-2-one, each containing one oxygen atom.

The first valid structure is:

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

Oxidative ozonolysis gives:

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

Q gives the iodoform test because it contains:

CH3CH(OH)−

R does not give the test because its alcohol-bearing carbon is not directly attached to a methyl group.

PCC oxidises the secondary alcohol in Q to a ketone:

CH3−CH(OH)−CH2−COOH→PCCCH3−CO−CH2−COOH

This is a beta-keto acid. On heating, it decarboxylates:

CH3−CO−CH2−COOH→ΔCH3−CO−CH3+CO2

Therefore, S is acetone and contains one oxygen atom.

R follows the same reaction sequence:

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

The beta-keto acid then decarboxylates:

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

T is butan-2-one and contains one oxygen atom. Acetone and butan-2-one are methyl ketones, so both give positive iodoform tests.

The first accepted total is:

1+1=2

How does the alternative valid structure give an oxygen count of 4?

In the second arrangement, ozonolysis produces lactic acid as Q. PCC converts it into pyruvic acid, an alpha-keto acid rather than a beta-keto acid. Under the official route, it remains the relevant three-oxygen product, while the other fragment forms a one-oxygen methyl ketone.

The alternative structure is:

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

Oxidative ozonolysis gives Q as lactic acid:

Q=CH3−CH(OH)−COOH

The other product is:

R=HOOC−CH2−CH(OH)−CH2−CH2−CH3

PCC oxidises Q to pyruvic acid:

CH3−CH(OH)−COOH→PCCCH3−CO−COOH

Thus:

S=CH3−CO−COOH

S is an alpha-keto acid, not a beta-keto acid. It remains the relevant product after the stated treatment in the official route. It contains three oxygen atoms and gives the iodoform test through:

CH3CO−

R undergoes PCC oxidation and beta-keto-acid decarboxylation:

HOOC−CH2−CH(OH)−CH2−CH2−CH3→PCCHOOC−CH2−CO−CH2−CH2−CH3
HOOC−CH2−CO−CH2−CH2−CH3→ΔCH3−CO−CH2−CH2−CH3+CO2

T is pentan-2-one. It is a methyl ketone, gives the iodoform test and contains one oxygen atom.

The second accepted total is:

3+1=4

The official accepted answer is:

2 or 4

Why is counting oxygens before heating incorrect?

Stopping after PCC oxidation gives an incorrect answer in the first route. At that stage, Q and R have each become beta-keto acids containing three oxygen atoms. Counting them gives 6, but the question specifies PCC oxidation followed by heating.

The incorrect calculation is:

3+3=6

Heating cannot be omitted because both beta-keto acids decarboxylate. Their carboxyl groups leave as carbon dioxide, producing ketones with one oxygen each.

Keep the two structural tests separate:

  • The presence of the methyl ketone group establishes a positive iodoform test.
  • The position of the ketone relative to the acid group decides whether standard beta-keto-acid decarboxylation occurs.

The opposite overcorrection is also wrong. Pyruvic acid contains ketone and carboxylic acid groups, but it is an alpha-keto acid. Do not decarboxylate it merely because both groups are present. Check the beta relationship explicitly.

Which related questions test the same reaction decisions?

These three problems test the exact decisions used in the PYQ: alkene cleavage, iodoform recognition and keto-acid classification. Write the carbon skeleton after every reagent. Count oxygen atoms only after applying the stated heating step.

What are the oxidative ozonolysis products of this hydroxy alkene?

Oxidative ozonolysis produces lactic acid and ethanoic acid. Only lactic acid gives the iodoform test because it contains the required methyl carbinol group.

Start with:

CH3−CH(OH)−CH=CH−CH3

The products are:

CH3−CH(OH)−COOH

and:

CH3COOH

Only the first product gives the iodoform test.

What forms after PCC oxidation and heating of this hydroxy acid?

The final product is acetone. PCC first converts the secondary alcohol into a ketone, producing a beta-keto acid. Heating then causes decarboxylation.

Start with:

CH3−CH(OH)−CH2−COOH

PCC gives:

CH3−CO−CH2−COOH

Heating gives:

CH3−CO−CH2−COOH→ΔCH3−CO−CH3+CO2

The final product is acetone.

Which keto acid undergoes the thermal decarboxylation used here?

The first compound is an alpha-keto acid. The second is a beta-keto acid and undergoes the standard thermal decarboxylation used in this question.

Compare:

CH3−CO−COOH

and:

CH3−CO−CH2−COOH

For another worked application, use Organic Compounds Containing Oxygen JEE 2025: Answer. Then search the free past-paper archive by chapter and solve similar JEE Advanced questions without looking at the answer until you have classified every keto acid as alpha or beta.

Frequently asked questions

Why is the accepted answer 2 or 4 in this JEE 2024 question?

Two placements of the alkene satisfy all the given observations. After oxidative ozonolysis, PCC oxidation and heating, one route gives two one-oxygen ketones, while the other gives a three-oxygen alpha-keto acid and a one-oxygen ketone.

Why is 6 not the correct oxygen count?

A count of 6 stops after PCC oxidation, when two beta-keto acids are present. The stated heating step decarboxylates them, so the final ketones contain one oxygen atom each.

Does pyruvic acid decarboxylate on heating in this question?

No, not by the standard beta-keto-acid decarboxylation used in the solution. Pyruvic acid is an alpha-keto acid, so it remains the three-oxygen product in the official route.

How do I identify a positive iodoform test here?

Look for either a methyl carbinol group, CH3CH(OH)–, or a methyl ketone group, CH3CO–. Apply this test to every product after writing its complete carbon skeleton.

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