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

Step 1: Activating effect of methoxy group.
In anisole ($C_6H_5OCH_3$), the $-OCH_3$ group donates electrons to the ring by resonance ($+M$ effect), activating ortho and para positions toward electrophilic substitution.
Step 2: Friedel-Crafts alkylation (i).
$CH_3Cl + AlCl_3$ generates $CH_3^+$. Attack at ortho/para positions gives: \[ C_6H_5OCH_3 + CH_3Cl \xrightarrow{AlCl_3} o\text{-}CH_3C_6H_4OCH_3 + p\text{-}CH_3C_6H_4OCH_3 \] Para product is major (less steric hindrance).
Step 3: Nitration of anisole (ii).
$HNO_3 + H_2SO_4$ generates $NO_2^+$. Attack at ortho/para positions gives: \[ C_6H_5OCH_3 \xrightarrow{HNO_3/H_2SO_4} o\text{-}NO_2C_6H_4OCH_3 + p\text{-}NO_2C_6H_4OCH_3 \] \[ \boxed{\text{(i) o- and p-methylanisole; (ii) o- and p-nitroanisole (para major in both)}} \]