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Aldehydes, Ketones and Carboxylic Acids · Question 12

JEE Advanced 2024 Paper 1MatchingHard📈 Topic in 17% papers+4−1
1

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

Answer

(1) P-3, Q-5, R-4, S-1

Solution

Ethylene glycol/PTSA protects a C=O as a cyclic acetal/ketal (so it survives the next step); H3O+ removes it; NaBH4 then reduces the freed C=O to CH–OH. Hydroboration–oxidation is anti-Markovnikov; oxymercuration–demercuration is Markovnikov.

P: ozonolysis, then intramolecular aldol condensation (5-membered ring):

[figure] O3, Zn [figure] aq. NaOH, [figure]

ethylene glycol, PTSA [figure] BH3; H2O2, NaOH [figure]

H3O+ [figure] NaBH4 [figure] → (3)

Q: the ketone α-CH2 (enolate) attacks the CHO, giving 1-acetylcyclopentene:

[figure] O3, Zn [figure] aq. NaOH, [figure]

ethylene glycol, PTSA [figure] BH3; H2O2, NaOH [figure]

H3O+ [figure] NaBH4 [figure] → (5)

R: Markovnikov hydration puts OH on the CH3-bearing carbon:

[figure] ethylene glycol, PTSA [figure] Hg(OAc)2, H2O; NaBH4 [figure]

H3O+ [figure] NaBH4 [figure] → (4)

S: anti-Markovnikov hydration puts OH on C-3 (the less substituted alkene carbon):

[figure] ethylene glycol, PTSA [figure] BH3; H2O2, NaOH [figure]

H3O+ [figure] NaBH4 [figure] → (1)

P-3, Q-5, R-4, S-1

Answer: (A)

JEE Advanced 2024 Paper 1

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2

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
3

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
4

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

JEE Advanced 2024 Paper 2

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