Match List-I with List-II List-I Reaction List-II Type of reaction (A…

The correct answer is Option (1) → (A)-(I), (B)-(IV), (C)-(II), (D)-(III)

List-I Reaction	List-II Type of reaction
(A) Hydrolysis of Ethyl Chloride	(I) Substitution reaction
(B) 2-Propanol is treated with Conc.Sulphuric acid	(IV) Elimination reaction
(C) Reaction of HCN with Acetone	(II) Addition reaction
(D) Treatment of acetamide with NaOH and $Br_2$	(III) Degradation reaction

Hydrolysis of Ethyl Chloride

C₂H₅Cl + OH⁻ → C₂H₅OH + Cl⁻

Here, the Cl atom is replaced by OH⁻, so this is a substitution reaction.

Thus,

Hydrolysis of Ethyl Chloride → Substitution reaction (I)

2-Propanol with Concentrated H₂SO₄

When 2-propanol is heated with concentrated sulphuric acid, dehydration occurs forming propene.

CH₃–CHOH–CH₃ → CH₃–CH=CH₂ + H₂O

Since a molecule of water is removed, this is an elimination reaction.

Thus,

2-Propanol + Conc. H₂SO₄ → Elimination reaction (IV)

Reaction of HCN with Acetone

Acetone reacts with HCN to form cyanohydrin.

(CH₃)₂CO + HCN → (CH₃)₂C(OH)CN

Here, HCN adds across the C=O double bond, so it is an addition reaction.

Thus,

HCN + Acetone → Addition reaction (II)

Acetamide with NaOH and Br₂

Acetamide reacts with bromine and NaOH in Hofmann bromamide reaction, producing a primary amine with one carbon less.

CH₃CONH₂ → CH₃NH₂

Since the carbon chain length decreases, this is a degradation reaction.

Thus,

Acetamide + NaOH + Br₂ → Degradation reaction (III)