Q19 / 75 · JEE Main 2026 (23 January Shift 1) · Chemistry

Match List - I with List - II.
$\begin{array}{l} \text{List - I} & \text{List - II} \\ \text{Functional group (detection)} & \text{Change observed during detection} \\ \text{A. Unsaturation (Baeyer's test)} & \text{I. Red colour appears} \\ \text{B. Alcoholic group} \\ \text{(Ceric ammonium nitrate test)} & \text{II. Silver mirror appears} \\ \text{C. Aldehyde group (Tollen's reagent)} & \text{III. Violet colour appears} \\ \text{D. Phenolic group } \left(\mathrm{FeCl}_{3}\right. \text{ test)} & \text{IV. Discharge of pink colour} \end{array}$
Choose the correct answer from the options given below :

  • A
    A-III, B-IV, C-II, D-I
  • Correct answer
    B
    A-IV, B-I, C-II, D-III
  • C
    A-IV, B-III, C-II, D-I
  • D
    A-III, B-IV, C-I, D-II

Solution

Unsaturation (Baeyer's test) involves decolorization of the reagent, represented by discharge of color (IV).
Alcoholic groups (Ceric ammonium nitrate test) produce a red color (I).
Aldehyde groups (Tollen's reagent) produce a characteristic silver mirror effect (II).
Phenolic groups (FeCl₃ test) produce a violet or pink color (III).
The correct matching is A-IV, B-I, C-II, D-III.

Hence, the correct answer is option: 'B'.

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