Comprehension
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 \( S_N2 \) attack of an alkoxide ion on a primary alkyl halide. If tertiary alkyl halide is used, elimination reaction occurs and alkene is formed instead of ether.
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 the smaller alkyl group. If one of the alkyl groups is tertiary, the alkyl halide is formed from the tertiary alkyl group because tertiary carbocation is more stable than primary carbocation. Cleavage of alkyl aryl ethers takes place at the alkyl–oxygen bond due to more stable alkyl–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 and directs the incoming group to ortho and para positions.
Question: 1

Complete the following equation: \[ \text{Anisole} + CH_3Cl \xrightarrow{\text{Anhyd. } AlCl_3} \; ? \]

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–OCH\(_3\) group is ortho–para directing and strongly activating.
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Solution and Explanation

Concept: This is a Friedel–Crafts alkylation reaction. The –OCH\(_3\) group activates the benzene ring and directs substitution to ortho and para positions.
Products: \[ \text{o-Methylanisole and p-Methylanisole (para major)} \]
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Question: 2

Complete the following reaction: \[ \text{Anisole} \xrightarrow{\text{Conc. } HNO_3 + H_2SO_4} \; ? \]

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Electrophilic substitution in anisole occurs mainly at ortho and para positions.
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Solution and Explanation

Concept: Nitration is an electrophilic substitution reaction. The methoxy group directs nitration to ortho and para positions.
Products: \[ \text{o-Nitroanisole and p-Nitroanisole (para major)} \]
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Question: 3

Write the names of alkyl halide and sodium alkoxide used to prepare tert-butyl ethyl ether by Williamson synthesis.

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In Williamson synthesis use primary alkyl halide to avoid elimination.
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Solution and Explanation

Concept: In Williamson synthesis, primary alkyl halide should be used to avoid elimination. To prepare tert-butyl ethyl ether: \[ (C_2H_5)Cl + (CH_3)_3CONa \]
Answer:
  • Alkyl halide: Ethyl chloride
  • Sodium alkoxide: Sodium tert-butoxide
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Question: 4

Anisole on reaction with HI gives phenol and CH\(_3\)I and not methanol and iodobenzene. Justify the statement.

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Alkyl aryl ether cleavage occurs at alkyl–O bond, not aryl–O bond.
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Solution and Explanation

Concept: Cleavage of alkyl aryl ether occurs at the alkyl–oxygen bond because:
  • The aryl–O bond has partial double bond character.
  • It is stronger and does not break easily.
\[ C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I \] Therefore:
  • Phenol is formed.
  • Methyl iodide is formed.
Methanol and iodobenzene are not formed because cleavage does not occur at aryl–oxygen bond.
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Question: 5

Why is C–O–C bond angle in ethers slightly greater than tetrahedral angle?

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Greater steric repulsion → Slightly larger bond angle than tetrahedral.
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Solution and Explanation

Concept: The tetrahedral bond angle is 109.5\(^\circ\). In ethers:
  • Two bulky alkyl groups attached to oxygen repel each other.
  • This increases bond angle slightly.
Therefore: \[ \text{C–O–C bond angle} \approx 111^\circ \]
Conclusion: Due to repulsion between bulky alkyl groups, bond angle is slightly greater than 109.5\(^\circ\).
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