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Aldehydes, Ketones and Carboxylic Acids · Question 7

JEE Advanced 2025 Paper 1NumericalHard📈 Topic in 17% papers+4
1

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

Answer

175

Solution

Step 1 (Fittig coupling, then acetal hydrolysis): two aryl bromides join; H3O+ frees both CHO groups.

[figure] [(ii) H3O+](i) Na, dry ether [figure] (P)

P = C14H10O2, M = 210. 16 mol X gives 8 mol P (100 %).

Step 2 (intramolecular Cannizzaro): one CHO is reduced, the other oxidised.

P [(ii) H3O+](i) NaOH, [figure] (Q), 50 %: 4 mol

Step 3 (decarboxylation with soda lime): COOH is replaced by H.

Q [(ii) ](i) NaOH, CaO [figure] (R), 50 %: 2 mol

Step 4: OH is replaced by Br.

R PBr3 [figure] (T), 50 %: 1 mol

Step 5 (Williamson synthesis): NaH makes the alkoxide of R, which displaces Br⁻ from T.

T + R NaH [figure] (S), 50 %: 0.5 mol

S = C26H22O, M = 312 + 22 + 16 = 350

Mass of S = 0.5 × 350 = 175 g

Answer: 175

JEE Advanced 2025 Paper 1

Keep practising — Aldehydes, Ketones and Carboxylic Acids

2

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
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3

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
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4

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
5

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
6

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
7

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
8

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+
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9

Sum of number of oxygen atoms in S and T is .

JEE Advanced 2024 Paper 2

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