Match List-I with List-II.
| List-I (Reaction) | List-II (Type of reaction) |
|---|---|
| (A) \( \ce{CH3CH2Br + KOH(aq) -> CH3CH2OH + KBr} \) | (I) Elimination |
| (B) \( \ce{CH3CH2Br + KOH} \xrightarrow{\text{alc.}} \ce{CH2=CH2 + KBr + H2O} \) | (II) Electrophilic substitution |
| (C) \( \ce{C6H5Cl + HNO3} \xrightarrow{\text{conc. }\ce{H2SO4}} \ce{ClC6H4NO2 + H2O} \) | (III) Addition |
| (D) \( \ce{CH3CH=CH2 + HBr -> CH3CHBrCH3} \) | (IV) Nucleophilic substitution |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-I, C-II, D-III
(A) \( \ce{CH3CH2Br + KOH(aq) -> CH3CH2OH + KBr} \) — This matches (IV) Nucleophilic substitution. The electron rich OH− attacks the carbon that carries the bromine and replaces Br, which leaves as Br−. One group is swapped for another.
(B) \( \ce{CH3CH2Br + KOH} \xrightarrow{\text{alc.}} \ce{CH2=CH2 + KBr + H2O} \) — This matches (I) Elimination. In alcoholic KOH, H is removed from the β-carbon and Br from the α-carbon, so a double bond forms (β-elimination).
(C) \( \ce{C6H5Cl + HNO3} \xrightarrow{\text{conc. }\ce{H2SO4}} \ce{ClC6H4NO2 + H2O} \) — This matches (II) Electrophilic substitution. The electrophile, the nitronium ion NO2+ formed from the nitrating mixture, replaces a hydrogen of the ring; the chlorine already present directs it to the ortho- and para-positions.
(D) \( \ce{CH3CH=CH2 + HBr -> CH3CHBrCH3} \) — This matches (III) Addition. H and Br add across the double bond, and nothing is lost. By Markovnikov’s rule, 2-bromopropane is the main product.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “…In this type of reaction, a nucleophile reacts with haloalkane (the substrate) having a partial positive charge on the carbon atom bonded to halogen. A substitution reaction takes place and halogen atom, called leaving group departs as halide ion. Since the substitution reaction is initiated by a nucleophile, it is called nucleophilic substitution reaction.”
“When a haloalkane with β-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from β-carbon and a halogen atom from the α-carbon atom. As a result, an alkene is formed as a product.”
6.7.2 Reactions of Haloarenes – “Haloarenes undergo the usual electrophilic reactions of the benzene ring such as halogenation, nitration, sulphonation and Friedel-Crafts reactions.”
6.4.2 From Hydrocarbons – “(i) Addition of hydrogen halides: An alkene is converted to corresponding alkyl halide by reaction with hydrogen chloride, hydrogen bromide or hydrogen iodide.”