Comprehension

Replacement of a hydrogen atom in a hydrocarbon by an alkoxy or carboxyl group yields a class of compounds known as ethers. Ethers are classified as symmetrical or unsymmetrical on the basis of groups attached to the oxygen atoms. Diethyl ether, a symmetrical ether, has been widely used as an inhalation anesthetic. Ethers can be prepared by acid catalyzed intermolecular dehydration of alcohols and Williamson's synthesis. Acid catalyzed dehydration of alcohols is not generally preferred as it gives a mixture of elimination and substitution products. In Williamson's synthesis, an alkyl halide is allowed to react with sodium alkoxide. Ethers containing substituted Alkyl groups may also be prepared by this method. The C-O bond in ether is weakly polar and is cleaved under drastic conditions with excess of hydrogen halides. In electrophilic substitution, the alkoxy group deactivates the aromatic ring and directs the incoming group to ortho and para positions. 

Question: 1

When ethanol is dehydrated in the presence of \( \text{H}_2\text{SO}_4 \) at 443K and 413K respectively, the products formed are:

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The temperature during dehydration reactions plays a key role in determining whether elimination or substitution occurs.
Updated On: Feb 16, 2026
  • Ethane and ethoxythane
  • Ethylmethyl ether and butene
  • Ethylmethyl ether and propene
  • Ethene and ethoxythane
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The Correct Option is D

Solution and Explanation

Step 1: Reaction Mechanism Elucidation.
The dehydration of ethanol using concentrated \( \text{H}_2\text{SO}_4 \) yields distinct products based on reaction temperature. Specifically, at 443 K, the reaction favors ethene production (dehydration), whereas at 413 K, the formation of ethoxyethane (an ether) is promoted.
Step 2: Product Distribution Analysis.
Elevated temperatures (443 K) promote elimination reactions leading to ethene. Conversely, lower temperatures (413 K) facilitate substitution, resulting in ethoxyethane, a byproduct of ethanol dehydration.

Step 3: Summary of Findings.
Ethene is the primary product at 443 K, while ethoxyethane is produced at 413 K.

Final Answer: \[\boxed{\text{The correct answer is (4) Ethene and ethoxythane.}}\]
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Question: 2

The major product in the reaction of anisole with bromine in ethanoic acid is:

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Electron-donating groups like methoxy (\(-OCH_3\)) direct electrophilic substitution to the ortho and para positions of the benzene ring.
Updated On: Mar 27, 2026
  • o- bromoanisole
  • p- bromoanisole
  • m- bromoanisole
  • o-bromoanisole and p-bromoanisole
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The Correct Option is D

Solution and Explanation

Step 1: Reaction Analysis.
Anisole reacts with bromine via electrophilic aromatic substitution. The methoxy group (\(-OCH_3\)) is electron-donating, activating the aromatic ring for substitution at the ortho and para positions.
Step 2: Product Identification.
Consequently, the primary products formed are o-bromoanisole (ortho-substituted) and p-bromoanisole (para-substituted).
Final Answer: \[\boxed{\text{The correct answer is (4) o-bromoanisole and p-bromoanisole.}}\]
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Question: 3

In Williamson synthesis, the alkoxide ion attacks the alkyl halide via which pathway?

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The Williamson synthesis typically follows the SN\(_2\) mechanism, which involves a backside attack by the nucleophile, leading to inversion of configuration.
Updated On: Mar 27, 2026
  • SN\(_2\)
  • SN\(_1\)
  • Depends on the nature of alkoxide ion
  • Depends on the nature of Alkyl halide
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The Correct Option is A

Solution and Explanation

Step 1: Reaction Mechanism.
The Williamson ether synthesis involves an alkoxide ion (\( R-O^- \)) acting as a nucleophile. It attacks an alkyl halide (R-X) via a bimolecular nucleophilic substitution (SN\(_2\)) mechanism. This entails a backside assault by the alkoxide on the carbon atom bearing the halogen.
Step 2: Reaction Pathway.
The reaction mechanism is identified as SN\(_2\). This pathway is a one-step process where bond formation and bond breaking occur concurrently.
Final Answer: \[\boxed{\text{The correct answer is (1) SN\(_2\).}}\]
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Question: 4

Which is the most reactive hydrogen halide for cleavage of ethers?

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The reactivity of hydrogen halides towards ethers increases with the size of the halide ion. HI is the most reactive due to the weak HI bond.
Updated On: Mar 27, 2026
  • HF
  • HCl
  • HBr
  • HI
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The Correct Option is D

Solution and Explanation

Step 1: Ether Cleavage Mechanism
Hydrogen halides cleave ethers. Halide ion size correlates positively with hydrogen halide reactivity in ether cleavage.
Step 2: Halide Reactivity Assessment
- HF: Less reactive than other hydrogen halides due to a strong H-F bond.
- HCl: More reactive than HF, but less reactive than HBr or HI.
- HBr: Exhibits moderate reactivity.
- HI: Most reactive due to a weak H-I bond, making it the most efficient for ether cleavage.
Step 3: Determination
HI is the most reactive hydrogen halide for ether cleavage.
Final Answer: \[\boxed{\text{The correct answer is (4) HI.}}\]
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Question: 5

Which type of ether is anisole?

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Anisole is a diaryl ether, which means it has two aromatic rings (phenyl groups) bonded to an oxygen atom.
Updated On: Mar 27, 2026
  • Dialkyl ether
  • Diaryl ether
  • Phenyl Alkyl ether
  • Alkoxy Alkyl ether
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The Correct Option is B

Solution and Explanation

Step 1: Anisole Structure Analysis.
Anisole is an ether characterized by an oxygen atom linked to a phenyl group (C₆H₅) and a methoxy group (C₆H₄OCH₃), classifying it as a diaryl ether.
Step 2: Classification Confirmation.
Given that anisole's structure involves two aromatic rings joined by an oxygen atom, it is definitively categorized as a diaryl ether.
Final Answer: \[\boxed{\text{The correct answer is (2) Diaryl ether.}}\]
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