Chemistry - Halogen Derivative Question with Solution | TestHub
Match the reactions given in Column I with the types of reactions given in Column II.
Column I | Column II
(i) Chlorobenzene + Cl₂/Fe → two products shown in source figure (second printed product is 1,2,4-trichlorobenzene) | (a) Nucleophilic aromatic substitution
(ii) CH₃CH=CH₂ + HBr → CH₃CH(Br)CH₃ | (b) Electrophilic aromatic substitution
(iii) 1-Iodoethylbenzene + OH⁻ → 1-phenylethanol | (c) Saytzeff elimination
(iv) p-nitrochlorobenzene + NaOH → p-nitrophenol | (d) Electrophilic addition
(v) CH₃CH₂CH(Br)CH₃ + alc. KOH → CH₃CH=CHCH₃ | (f) Nucleophilic substitution (Sₙ1)
Options:
Answer:
Solution:
Ring chlorination is electrophilic aromatic substitution; alkene HBr is electrophilic addition; benzylic substitution is Sₙ1; nitroactivated aromatic substitution is nucleophilic; alcoholic KOH promotes elimination. Source drawing shows the second product of (i) as 1,2,4-trichlorobenzene, although ordinary ring chlorination would give the para dichloro product. Preserve its drawing. Printed NCERT key says (iii)–(e), but Column II has no (e) and labels Sₙ1 as (f); corrected to (f) on Rahul’s instruction.
