Chemistry · Practical Organic Chemistry

JEE Main 2025 — 8 April, Evening Shift — Question 15

Match the List-I with List-II

List-I (Reagent)List-II (Functional detected)
A. Sodium bicarbonate solutionI. Double bond/unsaturation
B. Neutral ferric chlorideII. Carboxylic acid
C. Ceric ammonium nitrateIII. Phenolic - OH
D. Alkaline KMnO4\mathrm{KMnO}_{4}IV. Alcoholic - OH

Choose the correct answer from the options given below:

  1. Option A:

    A-III, B-II, C-IV, D-I

  2. Option B:

    A-II, B-IV, C-III, D-I

  3. Option C:

    A-II, B-III, C-IV, D-I

    Correct
  4. Option D:

    A-II, B-III, C-I, D-IV

Answer: C

Step-by-step solution

List-I (Reagent)List-II (Functional detected)
A. Sodium bicarbonate solutionII. Carboxylic acid
B. Neutral ferric chlorideIII. Phenolic - OH
C. Ceric ammonium nitrateIV. Alcoholic - OH
D. Alkaline KMnO4\mathrm{KMnO}_{4}I. Double bond/unsaturation

(A) Carboxylic acid gives effervescence with sodium bicarbonate

R−COOH+NaHCO3→RCOONa+CO2↑+H2O\mathrm{R}-\mathrm{COOH}+\mathrm{NaHCO}_{3} \rightarrow \mathrm{RCOONa}+\mathrm{CO}_{2} \uparrow+\mathrm{H}_{2} \mathrm{O}

(B) Phenolic-OH gives characteristic colour with neutral FeCl3\mathrm{FeCl}_{3}

(C) Alcoholic-OH gives red colour with ceric ammonium nitrate

(NH4)2[Ce(NO3)6]+2ROH→[Ce(ROH)2(NO3)4]+2NH4NO3 (Red) \left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{Ce}\left(\mathrm{NO}_{3}\right)_{6}\right]+2 \mathrm{ROH} \rightarrow \underset{\text { (Red) }}{\left[\mathrm{Ce}(\mathrm{ROH})_{2}\left(\mathrm{NO}_{3}\right)_{4}\right]+2 \mathrm{NH}_{4} \mathrm{NO}_{3}}

(D) Purple colour of alkaline KMnO4\mathrm{KMnO}_{4} is discharged by multiple bond of alkenes and alkynes

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Chemistry
Chapter
Practical Organic Chemistry
Topic
Qualitative analysis of Functional Groups