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JEE Advanced Chemistry Carboxylic Acid Derivatives 2023 JEE Advanced 2023 (Paper 1)

JEE Advanced Chemistry Question (2023) — Solution

Question

Match the reactions (in the given stoichiometry of the reactants) in List-I with one of their products given in List-II and choose the correct option.   List-I   List-II (P) P 2 O 3 + 3 H 2 O → (1) P ( O ) OCH 3 Cl 2 (Q) P 4 + 3 NaOH + 3 H 2 O → (2) H 3 PO 3 (R) PCl 5 + CH 3 COOH → (3) PH 3 (S) H 3 PO 2 + 2 H 2 O + 4 AgNO 3 → (4) POCl 3     (5) H 3 PO 4  

Options

  1. A. P → 2 ;   Q → 3 ;   R → 1 ;   S → 5
  2. B. P → 3 ;   Q → 5 ;   R → 4 ;   S → 2
  3. C. P → 5 ;   Q → 2 ;   R → 1 ;   S → 3
  4. D. P → 2 ;   Q → 3 ;   R → 4 ;   S → 5

Answer

D. P → 2 ;   Q → 3 ;   R → 4 ;   S → 5

Step-by-step solution

Phosphorus trioxide reacts with water to form phosphorous acid, P 2 O 3 + 3 H 2 O → 2 H 3 PO 3 When acetic acid is heated with phosphorus pentachloride, acetyl chloride is obtained. In this reaction, − COOH group is converted to − COCl group. The side products are phosphorus oxychloride and HCl . PCl 5 + CH 3 COOH → CH 3 COCl + POCl 3 + HCl P 4  undergo disproportionation reaction with sodium hydroxide. P 4 + 3 NaOH + 3 H 2 O   → PH 3 + 3 NaH 2 PO 2 Due to the presence two  P - H  bonds,  H 3 PO 2  acts as reducing agent. It reduces  Ag +  to  Ag . H 3 PO 2 + H 2 O + AgNO 3 → H 3 PO 4 + HNO 3 + Ag ↓

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Related: Chemistry — Carboxylic Acid Derivatives · All PYQ Banks