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JEE Main 20265 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
5 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Carboxylic Acid Derivatives

JEE Main 2026 · 5 April · Evening Shift · Chemistry · Carboxylic Acid Derivatives

Complete the following reaction sequence and give the name of major...

Complete the following reaction sequence and give the name of major product 'P'.
CH3−CH2−C≡N→(ii) H3O+(i) OH−/H2O/Δ→(iv) H2O(iii) Cl2/Red PP (Major product)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)}

Options

  1. A2-Chloropropanoic acid
  2. B3-Chloropropanoic acid
  3. C1-Chloropropane
  4. D2-Chloropropane

Correct answer

A. 2-Chloropropanoic acid

Step-by-step solution

Step 1: Hydrolysis of the nitrile.
Propanenitrile on basic hydrolysis followed by acidic workup gives propanoic acid:
CH3−CH2−C≡N→(ii) H3O+(i) OH−/H2O/ΔCH3−CH2−COOHCH_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 sp3sp^3 α\alpha-carbon bearing two α\alpha-hydrogens. In the HVZ reaction, Cl2Cl_2 in the presence of red phosphorus selectively replaces one α\alpha-hydrogen by chlorine, giving the α\alpha-chloro acid after aqueous workup:
CH3−CH2−COOH→(iv) H2O(iii) Cl2/Red PCH3−CH(Cl)−COOHCH_3-CH_2-COOH \xrightarrow[(iv)\ H_2O]{(iii)\ Cl_2/\text{Red P}} CH_3-CH(Cl)-COOH

Hence, the major product PP is 22-chloropropanoic acid, which corresponds to option (1)(1).
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