Anisole is treated separately with the reagents given in List-I. Match List-I with List-II.
| List-I (Reagent) | List-II (Product or products) |
|---|---|
| (A) \( \ce{Br2} \) in ethanoic acid | (I) 2-Methoxyacetophenone and 4-methoxyacetophenone |
| (B) \( \ce{CH3Cl} \), anhydrous \( \ce{AlCl3} \) | (II) p-Bromoanisole as the major product |
| (C) \( \ce{CH3COCl} \), anhydrous \( \ce{AlCl3} \) | (III) 2-Nitroanisole and 4-nitroanisole |
| (D) Concentrated \( \ce{H2SO4} \) and concentrated \( \ce{HNO3} \) | (IV) 2-Methoxytoluene and 4-methoxytoluene |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-II, B-IV, C-I, D-III
The methoxy group of anisole activates the benzene ring and directs the incoming group to the ortho and para positions, as the –OH group does in phenol. Each reagent brings in its own group.
(A) \( \ce{Br2} \) in ethanoic acid — This matches (II) p-Bromoanisole as the major product. The ring is activated enough to be brominated without an iron(III) bromide catalyst; the para isomer is the main product and o-bromoanisole is the minor one.
\( \displaystyle \ce{C6H5OCH3 + Br2} \xrightarrow{\text{ethanoic acid}} \text{p-}\ce{BrC6H4OCH3 + HBr} \)
(B) \( \ce{CH3Cl} \), anhydrous \( \ce{AlCl3} \) — This matches (IV) 2-Methoxytoluene and 4-methoxytoluene. This is Friedel-Crafts alkylation: a methyl group enters the ring.
\( \displaystyle \ce{C6H5OCH3 + CH3Cl} \xrightarrow{\text{anhyd. }\ce{AlCl3}} \text{o- and p-}\ce{CH3C6H4OCH3 + HCl} \)
(C) \( \ce{CH3COCl} \), anhydrous \( \ce{AlCl3} \) — This matches (I) 2-Methoxyacetophenone and 4-methoxyacetophenone. This is Friedel-Crafts acylation: the acetyl group, CH3CO–, enters the ring.
\( \displaystyle \ce{C6H5OCH3 + CH3COCl} \xrightarrow{\text{anhyd. }\ce{AlCl3}} \text{o- and p-}\ce{CH3COC6H4OCH3 + HCl} \)
(D) Concentrated \( \ce{H2SO4} \) and concentrated \( \ce{HNO3} \) — This matches (III) 2-Nitroanisole and 4-nitroanisole. The nitrating mixture puts a nitro group at the ortho and para positions.
\( \displaystyle \ce{C6H5OCH3 + HNO3} \xrightarrow{\text{conc. }\ce{H2SO4}} \text{o- and p-}\ce{O2NC6H4OCH3 + H2O} \)
NCERT (2026-27 edition):
7.6.3 Chemical Reactions – “2. Electrophilic substitution The alkoxy group (-OR) is ortho, para directing and activates the aromatic ring towards electrophilic substitution in the same way as in phenol.”
“(i) Halogenation: Phenylalkyl ethers undergo usual halogenation in the benzene ring, e.g., anisole undergoes bromination with bromine in ethanoic acid even in the absence of iron (III) bromide catalyst. It is due to the activation of benzene ring by the methoxy group. Para isomer is obtained in 90% yield.”
“(ii) Friedel-Crafts reaction: Anisole undergoes Friedel-Crafts reaction, i.e., the alkyl and acyl groups are introduced at ortho and para positions by reaction with alkyl halide and acyl halide in the presence of anhydrous aluminium chloride (a Lewis acid) as catalyst.”
“(iii) Nitration: Anisole reacts with a mixture of concentrated sulphuric and nitric acids to yield a mixture of ortho and para nitroanisole.”