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Aldehydes, Ketones and Carboxylic Acids · Question 1

JEE Advanced 2026 Paper 1MatchingHard📈 Topic in 17% papers+4−1
1

The List-II contains products obtained from the reaction of compounds in List-I with O3/Zn-H2O followed by cyclization (via more stable enolate) in the presence of aqueous NaOH. Match each entry in List-I with appropriate entry in List-II and choose the correct option.

  1. P ⟶ 2; Q ⟶ 4; R ⟶ 1; S ⟶ 3
  2. P ⟶ 3; Q ⟶ 4; R ⟶ 5; S ⟶ 2
  3. P ⟶ 2; Q ⟶ 1; R ⟶ 5; S ⟶ 3
  4. P ⟶ 3; Q ⟶ 5; R ⟶ 4; S ⟶ 2

Answer

(3) P ⟶ 2; Q ⟶ 1; R ⟶ 5; S ⟶ 3

Solution

Ozonolysis cleaves the ring-fusion C=C, giving a ten-membered 1,6-dione with a CH₃ on a carbon α to a C=O. NaOH forms the more stable (more substituted) enolate at that α-CH(CH₃), which attacks the other C=O; the ring that closes as a 5- or 6-membered ring decides the product (C=O next to the CH₃-bearing fusion carbon, OH on the other fusion carbon).

(P) [figure] O3/Zn-H2O [figure] aq. NaOH [figure]

A five-membered ring closes, giving a 5 + 7 fused ketol with the second CH₃ in the 7-ring next to C–OH → (2)

(Q) [figure] O3/Zn-H2O [figure] aq. NaOH [figure]

A six-membered ring closes, giving a 6 + 6 ketol with the second CH₃ next to C–OH in the C=O ring → (1)

(R) [figure] O3/Zn-H2O [figure] aq. NaOH [figure]

A six-membered ring closes, giving a 6 + 6 ketol with the second CH₃ next to C–OH in the ring without C=O → (5)

(S) [figure] O3/Zn-H2O [figure] aq. NaOH [figure]

A five-membered ring closes, giving a 5 + 7 ketol with C=O in the 7-ring and the second CH₃ on the 5-ring next to C–OH → (3)

P → 2; Q → 1; R → 5; S → 3

Answer: (C)

JEE Advanced 2026 Paper 1

Keep practising — Aldehydes, Ketones and Carboxylic Acids

2

Match the major products obtained in the reactions given in List-I with the corresponding structures in List-II and choose the correct option.

  1. P → 2; Q ⟶ 1; R → 5; S ⟶ 4
  2. P → 1; Q ⟶ 2; R → 4; S ⟶ 5
  3. P → 1; Q ⟶ 2; R → 3; S ⟶ 4
  4. P → 2; Q ⟶ 1; R → 3; S ⟶ 5
JEE Advanced 2026 Paper 1
3

Considering LiBH4 reduces an ester group to the corresponding alcohol and does not reduce a carboxylic acid group, the correct statement about the major products P, Q, R and S is [figure]

  1. P & Q are identical, and R & S are diastereomers.
  2. P & Q are diastereomers, and R & S are identical.
  3. P & Q are diastereomers, and R & S are diastereomers.
  4. P & Q are identical, and R & S are identical.
JEE Advanced 2026 Paper 1
4

In the following reaction sequence, Q, R, S and T are the major products. [figure] The correct statement(s) about Q, R, S and T is(are)

  1. S on warming with ammoniacal AgNO3 results in the formation of silver mirror.
  2. Q on treatment with Cl2(excess)/UV gives gammaxane.
  3. T is a heterocyclic compound.
  4. R on acid catalyzed intramolecular cyclization followed by treatment with Zn-Hg/HCl gives 9,10-dihydroxyanthracene.
JEE Advanced 2026 Paper 1
5

The sum of total number of carbonyl groups ( >C=O ) present in the major products X and Y in the following reactions is . [figure]

JEE Advanced 2026 Paper 1
6

Natural rubber on complete ozonolysis ( O3/Zn-H2O) gives compound X as the major product. X gives positive iodoform and Tollen's tests. X on heating with aqueous NaOH gives Y as the major product. Y is

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Advanced 2026 Paper 2
7

The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is. [figure] Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80

JEE Advanced 2025 Paper 1
8

The correct reaction/reaction sequence that would produce a dicarboxylic acid as the major product is

  1. [figure] [ (iii) H3O+] (i) NaCN; (ii) HO-, H2O
  2. arraycCHO\ |\ (CHOH)4\ |\ CH2OH array Br2, H2O
  3. [figure] [(ii) KMnO4, H2SO4, ](i) KOH, EtOH
  4. [figure] H2CrO4
JEE Advanced 2025 Paper 2
9

Monocyclic compounds P, Q, R and S are the major products formed in the reaction sequences given below. [figure] The product having the highest number of unsaturated carbon atom(s) is

  1. P
  2. Q
  3. R
  4. S
JEE Advanced 2025 Paper 2
10

Reaction of iso-propylbenzene with O2 followed by the treatment with H3O+ forms phenol and a by-product P. Reaction of P with 3 equivalents of Cl2 gives compound Q. Treatment of Q with Ca(OH)2 produces compound R and calcium salt S. The correct statement(s) regarding P, Q, R and S is(are)

  1. Reaction of P with R in the presence of KOH followed by acidification gives [figure]
  2. Reaction of R with O2 in the presence of light gives phosgene gas
  3. Q reacts with aqueous NaOH to produce Cl3CCH2OH and Cl3CCOONa
  4. S on heating gives P
JEE Advanced 2024 Paper 1
11

Complete reaction of acetaldehyde with excess formaldehyde, upon heating with conc. NaOH solution, gives P and Q. Compound P does not give Tollens’ test, whereas Q on acidification gives positive Tollens’ test. Treatment of P with excess cyclohexanone in the presence of catalytic amount of p-toluenesulfonic acid (PTSA) gives product R. Sum of the number of methylene groups (-CH2-) and oxygen atoms in R is .

JEE Advanced 2024 Paper 1

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