Question

Complete the following equations : \[ \text{(i)} @@RAW0@@ \] \[ \text{(ii)} @@RAW1@@ \]

Hint:

The methoxy group \((-OCH_3)\) in anisole is: \[ \boxed{\text{Electron donating}} \] \[ \boxed{\text{Ring activating}} \] \[ \boxed{\text{Ortho-para directing}} \] Therefore, electrophilic substitution reactions such as alkylation, nitration, halogenation and sulphonation occur mainly at the ortho and para positions, with the para product generally predominating due to lower steric hindrance.

Answer 1

Concept: Anisole \((C_6H_5OCH_3)\) is an aromatic ether. The methoxy group \((-OCH_3)\) attached to the benzene ring exhibits a strong electron-donating resonance effect \((+M)\). Due to this electron-releasing effect, the electron density of the benzene ring increases, making the ring more reactive towards electrophilic substitution reactions. The methoxy group is an ortho- and para-directing group. Therefore, incoming electrophiles preferentially enter the ortho and para positions of the aromatic ring.

Part (i): Friedel-Crafts Alkylation

Step 1: Generation of electrophile. Methyl chloride reacts with anhydrous aluminium chloride to generate the electrophile. \[ CH_3Cl + AlCl_3 \rightarrow CH_3^{+} + AlCl_4^{-} \] The electrophile formed is \[ \boxed{CH_3^{+}} \]

Step 2: Electrophilic attack on anisole. The methoxy group activates the benzene ring and directs the incoming methyl group to the ortho and para positions. Therefore, two products are formed: \[ o\text{-Methylanisole} \] and \[ p\text{-Methylanisole} \] Because of less steric hindrance, the para product is formed in greater amount. \[ \boxed{ C_6H_5OCH_3 + CH_3Cl \xrightarrow[\text{CS}_2]{AlCl_3} o\text{-}CH_3C_6H_4OCH_3 + p\text{-}CH_3C_6H_4OCH_3 } \]

Part (ii): Nitration of Anisole

Step 3: Generation of electrophile. Concentrated nitric acid and concentrated sulphuric acid generate the nitronium ion. \[ HNO_3 + H_2SO_4 \rightarrow NO_2^{+} + HSO_4^{-} + H_2O \] The electrophile is \[ \boxed{NO_2^{+}} \]

Step 4: Electrophilic substitution. The nitronium ion attacks the ortho and para positions of anisole because the methoxy group is ortho-para directing. Hence, the products formed are: \[ o\text{-Nitroanisole} \] and \[ p\text{-Nitroanisole} \] with para-nitroanisole as the major product. \[ \boxed{ C_6H_5OCH_3 \xrightarrow{HNO_3/H_2SO_4} o\text{-}NO_2C_6H_4OCH_3 + p\text{-}NO_2C_6H_4OCH_3 } \]

Final Answer: \[ \boxed{ C_6H_5OCH_3 + CH_3Cl \xrightarrow[\text{CS}_2]{AlCl_3} o\text{-}CH_3C_6H_4OCH_3 + p\text{-}CH_3C_6H_4OCH_3 } \] \[ \boxed{ C_6H_5OCH_3 \xrightarrow{HNO_3/H_2SO_4} o\text{-}NO_2C_6H_4OCH_3 + p\text{-}NO_2C_6H_4OCH_3 } \]