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Match the reactions in List-I with the correct products in List-II. List-I: (A) Aniline treated with (i) NaNO2 + HCl at 0-5 deg C, then (ii) warm water (B) Phenol treated with Na2Cr2O7 / H2SO4 (C) Phenol treated with (i) CHCl3 + aq. NaOH, then (ii) H+ (D) Phenol treated with (i) NaOH, (ii) CO2 (pressure), then (iii) H+ List-II: (I) o-Hydroxybenzaldehyde (II) Phenol (III) o-Hydroxybenzoic acid (IV) p-Benzoquinone

1. (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
2. (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
3. (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
4. (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Correct answer: (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Solution

(A) Aniline + NaNO2/HCl forms diazonium salt; warming with water hydrolyzes it to phenol (II). (B) Phenol + Na2Cr2O7/H2SO4 (oxidizing agent) gives p-benzoquinone (IV). (C) Phenol + CHCl3/NaOH, then H+ is the Reimer-Tiemann reaction giving o-hydroxybenzaldehyde (I). (D) Phenol + NaOH, CO2, H+ is Kolbe's reaction giving o-hydroxybenzoic acid = salicylic acid (III).

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