Question

Ethers are prepared by the dehydration of alcohols in presence of protic acids at 413 K. Symmetrical and unsymmetrical ethers can also be prepared by Williamson synthesis. This reaction involves \( \mathrm{S_N2} \) attack of an alkoxide ion on primary alkyl halide. If a tertiary alkyl halide is used, elimination reaction occurs and an alkene is formed, no ether is formed. C - O bond in ethers are cleaved under drastic conditions with excess of HI. When unsymmetrical ethers react with HI, the alkyl halide is formed from smaller alkyl group. If one of the alkyl group is tertiary, the alkyl halide is formed from the tertiary alkyl group because tertiary alkyl carbocation is more stable than the primary carbocation. Cleavage of alkyl aryl ethers takes place at the alkyl-oxygen bond due to more stable aryl-oxygen bond. The order of reactivity of hydrogen halides is HI > HBr > HCl. Aromatic ethers undergo electrophilic substitution reactions. The alkoxy group attached to the aromatic ring activates the ring towards electrophilic substitution and directs the incoming group to ortho- and para-positions.

Answer the following questions :

(a) Complete the following equations :

(i) \( \mathrm{C_6H_5{-}OCH_3 + CH_3Cl \xrightarrow[CS_2]{Anhyd.\ AlCl_3}} \)

(ii) \( \mathrm{C_6H_5{-}OCH_3 \xrightarrow{Conc.\ (HNO_3 + H_2SO_4)}} \)

Hint:

The OCH3 group is an activating, ortho and para directing group. Write both the ortho and para products for each reaction.

Answer 1

Step 1: Use the passage:
The passage says the alkoxy group activates the ring and directs the new group to the ortho and para positions.
Anisole is methoxybenzene, \( \mathrm{C_6H_5OCH_3} \). So both reactions give ortho and para products.

Step 2: Part (i), Friedel-Crafts alkylation:
\( \mathrm{AlCl_3} \) pulls Cl from \( \mathrm{CH_3Cl} \) and makes the electrophile \( \mathrm{CH_3^+} \) (as a polarised complex).
This electrophile attacks the electron-rich ortho and para positions of anisole.
The para product is major because the ortho position is more hindered.
\[ \mathrm{C_6H_5OCH_3 + CH_3Cl \xrightarrow[CS_2]{Anhyd.\ AlCl_3} 2\text{-}CH_3C_6H_4OCH_3 + 4\text{-}CH_3C_6H_4OCH_3 + HCl} \]
The products are 2-methylanisole (minor) and 4-methylanisole (major).

Step 3: Part (ii), nitration:
Conc. \( \mathrm{HNO_3} \) and conc. \( \mathrm{H_2SO_4} \) form the nitronium ion \( \mathrm{NO_2^+} \).
\( \mathrm{NO_2^+} \) attacks the ortho and para positions, which have the highest electron density from the +R effect of OCH3.
\[ \mathrm{C_6H_5OCH_3 + HNO_3 \xrightarrow{H_2SO_4} 2\text{-}O_2NC_6H_4OCH_3 + 4\text{-}O_2NC_6H_4OCH_3 + H_2O} \]
The products are 2-nitroanisole (minor) and 4-nitroanisole (major).

Final Answer:
(i) A mixture of 2-methylanisole and 4-methylanisole (para major). (ii) A mixture of 2-nitroanisole and 4-nitroanisole (para major).