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: Why the OCH3 group directs ortho and para:
The oxygen has a lone pair. It shares this pair with the ring through resonance (+R effect).
Resonance puts negative charge on the ortho and para carbons.
An electrophile therefore attacks these two positions and not the meta position.

Step 2: Part (i), methylation of anisole:
$\mathrm{CH_3Cl}$ with $\mathrm{AlCl_3}$ in $\mathrm{CS_2}$ is the Friedel-Crafts alkylation.
A methyl group enters the ring at the ortho or para carbon.
\[ \mathrm{C_6H_5OCH_3 \xrightarrow[CS_2]{CH_3Cl,\ AlCl_3} o\text{-}CH_3C_6H_4OCH_3 + p\text{-}CH_3C_6H_4OCH_3} \]
The para isomer (1-methoxy-4-methylbenzene) forms in larger amount because it has less crowding.

Step 3: Part (ii), nitration of anisole:
The nitrating mixture gives $\mathrm{NO_2^+}$ as the attacking species.
It enters the ortho and para carbons in the same way as the methyl group did.
\[ \mathrm{C_6H_5OCH_3 \xrightarrow{Conc.\ HNO_3 + H_2SO_4} o\text{-}O_2NC_6H_4OCH_3 + p\text{-}O_2NC_6H_4OCH_3} \]
The para product (1-methoxy-4-nitrobenzene) is major.

Final Answer:
(i) o-methylanisole and p-methylanisole. (ii) o-nitroanisole and p-nitroanisole. In both, the para isomer is the major product.