Carboxylic Acid Derivatives JEE Advanced previous year questions with solutions

1 solved JEE Advanced questions on Carboxylic Acid Derivatives, free to read — no sign-in needed. The full chapter has 18 questions; sign in to attempt the remaining 17 in the exam simulator.

1. Q1JEE Advanced Adv 2023 (Paper 1)
   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) ${P}_{2}{O}_{3}+3{H}_{2}O\rightarrow$ (1) $P(O)\left({OCH}_{3}\right){Cl}_{2}$ (Q) ${P}_{4}+3NaOH+3{H}_{2}O\rightarrow$ (2) ${H}_{3}{PO}_{3}$ (R) ${PCl}_{5}+{CH}_{3}COOH\rightarrow$ (3) ${PH}_{3}$ (S) ${H}_{3}{PO}_{2}+2{H}_{2}O+4{AgNO}_{3}\rightarrow$ (4) ${POCl}_{3}$ (5) ${H}_{3}{PO}_{4}$
   1. A.$P\rightarrow 2;Q\rightarrow 3;R\rightarrow 1;S\rightarrow 5$
   2. B.$P\rightarrow 3;Q\rightarrow 5;R\rightarrow 4;S\rightarrow 2$
   3. C.$P\rightarrow 5;Q\rightarrow 2;R\rightarrow 1;S\rightarrow 3$
   4. D.$P\rightarrow 2;Q\rightarrow 3;R\rightarrow 4;S\rightarrow 5$
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   Answer: (D)

   Phosphorus trioxide reacts with water to form phosphorous acid, ${P}_{2}{O}_{3}+3{H}_{2}O\rightarrow 2{H}_{3}{PO}_{3}$ 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 $HCl$. ${PCl}_{5}+{CH}_{3}COOH\rightarrow {CH}_{3}COCl+{POCl}_{3}+HCl$ ${P}_{4}$ undergo disproportionation reaction with sodium hydroxide. ${P}_{4}+3NaOH+3{H}_{2}O\rightarrow ^{}{PH}_{3}+3{NaH}_{2}{PO}_{2}$ Due to the presence two $P-H$ bonds, ${H}_{3}{PO}_{2}$ acts as reducing agent. It reduces ${Ag}^{+}$ to $Ag$. ${H}_{3}{PO}_{2}+{H}_{2}O+{AgNO}_{3}\rightarrow {H}_{3}{PO}_{4}+{HNO}_{3}+Ag↓$