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Aldehydes, Ketones and Carboxylic Acids · Question 4

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

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.

Answer

(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.

Solution

Kolbe electrolysis: two propyl radicals (from 2 RCOO⁻, –2CO₂) couple: [figure] electrolysis [figure]

Aromatisation (cyclisation + dehydrogenation): [figure] [ 10-20 atm]V2O5, 773 K [figure] = Q

Friedel–Crafts acylation (anhydride opens, one C=O acylates benzene): [figure] + [figure] anhyd. AlCl3 [figure] = R

–COOH → –COCl, then Rosenmund reduction –COCl → –CHO: [figure] PCl5 [figure] H2-Pd/BaSO4 [figure] = S

Both C=O groups (1,4-related) condense with the two –NH₂ of hydrazine: [figure] [-2H2O]NH2NH2, heat [figure] = T (1-phenylphthalazine)

(A) S has a –CHO group, so it gives a silver mirror with Tollens' reagent. Correct.

(B) C6H6 + 3Cl2 UV C6H6Cl6 (BHC, gammaxane). Correct.

(C) T has an N–N containing ring, so it is heterocyclic. Correct.

(D) [figure] [-H2O] H+ [figure] Zn-Hg/HCl [figure]

Clemmensen reduction removes the C=O groups (gives 9,10-dihydroanthracene), not 9,10-dihydroxyanthracene. Wrong.

Answer: (A), (B), (C)

JEE Advanced 2026 Paper 1

Keep practising — Aldehydes, Ketones and Carboxylic Acids

2

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
3

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
4

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
5

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
6

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
7

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
8

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
9

List-I contains various reaction sequences and List-II contains the possible products. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

  1. P-3, Q-5, R-4, S-1
  2. P-3, Q-2, R-4, S-1
  3. P-3, Q-5, R-1, S-4
  4. P-5, Q-2, R-4, S-1
JEE Advanced 2024 Paper 1
10

In the following reaction sequence, the major product P is formed. [figure] Glycerol reacts completely with excess P in the presence of an acid catalyst to form Q. Reaction of Q with excess NaOH followed by the treatment with CaCl2 yields Ca-soap R, quantitatively. Starting with one mole of Q, the amount of R produced in gram is . [Given, atomic weight: H = 1, C = 12, N = 14, O = 16, Na = 23, Cl = 35, Ca = 40]

JEE Advanced 2024 Paper 1
11

The option(s) with correct sequence of reagents for the conversion of P to Q is(are) [figure]

  1. i) Lindlar’s catalyst, H2; ii) SnCl2/HCl; iii) NaBH4; iv) H3O+
  2. i) Lindlar’s catalyst, H2; ii) H3O+; iii) SnCl2/HCl; iv) NaBH4
  3. i) NaBH4; ii) SnCl2/HCl; iii) H3O+; iv) Lindlar’s catalyst, H2
  4. i) Lindlar’s catalyst, H2; ii) NaBH4; iii) SnCl2/HCl; iv) H3O+
JEE Advanced 2024 Paper 2

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