Carboxylic Acid Derivatives JEE Main previous year questions with solutions

5 solved JEE Main questions on Carboxylic Acid Derivatives, free to read — no sign-in needed. The full chapter has 75 questions; sign in to attempt the remaining 70 in the exam simulator.

1. Q1JEE Main 2026 (05 Apr, Shift 2)Preparation and Properties of Carboxylic Acids
   Complete the following reaction sequence and give the name of major product 'P'. $CH_3-CH_2-C\equiv N \xrightarrow[\text{(ii) } H_3O^+]{\text{(i) } OH^-/H_2O/\Delta} \xrightarrow[\text{(iv) } H_2O]{\text{(iii) } Cl_2/\text{Red P}} P \text{ (Major product)}$
   1. A.2-Chloropropanoic acid
   2. B.3-Chloropropanoic acid
   3. C.1-Chloropropane
   4. D.2-Chloropropane
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   Answer: (A)

   Step 1: Hydrolysis of the nitrile. Propanenitrile on basic hydrolysis followed by acidic workup gives propanoic acid: $CH_3-CH_2-C \equiv N \xrightarrow[(ii)\ H_3O^+]{(i)\ OH^-/H_2O/\Delta} CH_3-CH_2-COOH$ Step 2: Hell$-$Volhard$-$Zelinsky (HVZ) reaction. Propanoic acid has an $sp^3$ $\alpha$-carbon bearing two $\alpha$-hydrogens. In the HVZ reaction, $Cl_2$ in the presence of red phosphorus selectively replaces one $\alpha$-hydrogen by chlorine, giving the $\alpha$-chloro acid after aqueous workup: $CH_3-CH_2-COOH \xrightarrow[(iv)\ H_2O]{(iii)\ Cl_2/\text{Red P}} CH_3-CH(Cl)-COOH$ Hence, the major product $P$ is $2$-chloropropanoic acid, which corresponds to option $(1)$.
2. Q2JEE Main 2004Reactions Involving Acid Group
   Which of the following undergoes reaction with $50 \%$ sodium hydroxide solution to give the corresponding alcohol and acid?
   1. A.Phenol
   2. B.Benzoic acid
   3. C.Butanal
   4. D.Benzaldehyde
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   Answer: (D)

   Benzaldehyde does not contain $\alpha$ - hydrogen. Hence goes for cannizarro's reaction forming alcohol and acid
3. Q3JEE Main 2022 (24 Jun, Shift 1)Preparation and Properties of Carboxylic Acids
   Two statements are given belwo : Statement I : The melting point of monocarboxylic acid with even number of carbon atoms is higher than that of with odd number of carbon atoms acid immediately below and above it in the series. Statement II : The solubility of monocarboxylic acids in water decreases with increase in molar mass. Choose the most appropriate option :
   1. A.Both Statement I and Statement II are correct.
   2. B.Both Statement I and Statement II are incorrect.
   3. C.Statement I is correct but Statement II is incorrect.
   4. D.Statement I is incorrect but Statement II is correct.
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   Answer: (D)

   The melting point of carboxylic acid has no trend as we can observe from the following data of metling point. Methanoic acid $=8^{\circ}C$, Ethanoic acid $=17^{\circ}C$, Propanoic acid $=2^{\circ}C$, Butanoic acid $=-6^{\circ}C$, Pentonoic acid $=-35^{\circ}C$, Hexonic acid $=-20^{\circ}C$, Heptanoic acid $=-8^{\circ}C$, Octanoic acid $=17^{\circ}C$ acid. Water solubility of carboxylic acid decreases as we increase the molar mass & hence hydrophobic part of the chain.
4. Q4JEE Main 2021 (27 Aug, Shift 2)Preparation and Properties of Carboxylic Acids
   Which one of the following reactions will not yield propanoic acid?
   1. A.${CH}_{3}{CH}_{2}{CH}_{2}Br+Mg,{CO}_{2}$ dry ether $/{H}_{3}{O}^{+}$
   2. B.${CH}_{3}{CH}_{2}{CH}_{3}+{KMnO}_{4}(Heat),{OH}^{-}/{H}_{3}{O}^{+}$
   3. C.${CH}_{3}{CH}_{2}{CCl}_{3}+{OH}^{-}/{H}_{3}{O}^{+}$
   4. D.${CH}_{3}{CH}_{2}{COCH}_{3}+{OI}^{-}/{H}_{3}{O}^{+}$
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   Answer: (A)

   ${CH}_{3}{CH}_{2}{CH}_{2}Br⟶^{Mg}{CH}_{3}{CH}_{2}{CH}_{2}MgBr⟶^{{CO}_{2}/{H}_{3}{O}^{+}}{CH}_{3}{CH}_{2}{CH}_{2}COOH_{Butanoicacid}$
5. Q5JEE Main 2019 (09 Jan, Shift 1)Preparation and Properties of Carboxylic Acids
   Acid strength given below the correct decreasing order will be :
   1. A.${NO}_{2}{CH}_{2}COOH\gt {FCH}_{2}COOH\gt {CNCH}_{2}COOH\gt {ClCH}_{2}COOH$
   2. B.${NO}_{2}{CH}_{2}COOH\gt {NCCH}_{2}COOH\gt {FCH}_{2}COOH\gt {ClCH}_{2}COOH$
   3. C.${CNCH}_{2}COOH\gt {O}_{2}{NCH}_{2}COOH\gt {FCH}_{2}COOH\gt {ClCH}_{2}COOH$
   4. D.${FCH}_{2}COOH\gt {NCCH}_{2}COOH\gt {NO}_{2}{CH}_{2}COOH\gt {ClCH}_{2}COOH$
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   Answer: (B)

   Acidic character $\alpha$ stability of conjugate base (anion) Stability of anion $\alpha –I$ Hence greater the $–I$ effect greater will be the acidity Order of -I-effect $-{NO}_{2}\gt CN\gt -F\gt -Cl$ Hence, order of acidity will be $\begin{matrix}{O}_{2}{NCH}_{2}COOH\gt NC{CH}_{2}COOH\gt {FCH}_{2}COOH\gt Cl{CH}_{2}COOH \\ \end{matrix}$