ETHERS General methods of preparation, properties, reactions, and uses. Ethers – Structure: NEET MCQs with Answers

9 free questions on ETHERS General methods of preparation, properties, reactions, and uses. Ethers – Structure for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Structure.

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Question 1

Previous-year question
Anisole on cleavage with HI gives [NEET 2020]
  1. iodobenzene + CH₃OH
  2. phenol + C₂H₅I
  3. iodobenzene + C₂H₅OH
  4. phenol + CH₃I
Show answer
Answer: (D) phenol + CH₃I
Explanation: Iodide attacks the methyl carbon because the aryl–oxygen bond has partial double-bond character and resists attack. The products are therefore phenol and methyl iodide.

Question 2

Previous-year question
The compound that is most difficult to protonate is [NEET 2019]
  1. H–O–H
  2. H₃C–O–H
  3. H₃C–O–CH₃
  4. Ph–O–H
Show answer
Answer: (D) Ph–O–H
Explanation: Protonation requires an available lone pair on oxygen. In phenol one lone pair is delocalised into the aromatic ring, so the oxygen is much less electron rich and hardest to protonate. The two methyl groups of dimethyl ether, by contrast, make its oxygen the most basic.

Question 3

Previous-year question
The compound A on treatment with Na gives B, and with PCl₅ gives C. B and C react together to give diethyl ether. A, B and C are in the order [NEET 2018]
  1. C₂H₅OH, C₂H₆, C₂H₅Cl
  2. C₂H₅OH, C₂H₅Cl, C₂H₅ONa
  3. C₂H₅OH, C₂H₅ONa, C₂H₅Cl
  4. C₂H₅Cl, C₂H₆, C₂H₅OH
Show answer
Answer: (C) C₂H₅OH, C₂H₅ONa, C₂H₅Cl
Explanation: Ethanol with sodium gives sodium ethoxide (B) and with PCl₅ gives chloroethane (C). The ethoxide then displaces chloride from chloroethane in a Williamson synthesis, giving diethyl ether.

Question 4

Previous-year question
The heating of phenyl methyl ethers with HI produces [NEET 2017]
  1. iodobenzene
  2. phenol
  3. benzene
  4. ethyl chlorides
Show answer
Answer: (B) phenol
Explanation: Anisole is cleaved at the alkyl–oxygen bond because the aryl–oxygen bond has partial double-bond character and cannot be attacked. Iodide therefore takes the methyl group as CH₃I, leaving phenol.

Question 5

Previous-year question
The reaction cyclopentanol →(NaOH) cyclopentyl oxide →(MeI) methyl cyclopentyl ether can be classified as [NEET 2016]
Figure for question 5
  1. Williamson ether synthesis
  2. alcohol formation reaction
  3. dehydration reaction
  4. Williamson alcohol synthesis
Show answer
Answer: (A) Williamson ether synthesis
Explanation: An alkoxide formed from the alcohol attacks the primary carbon of methyl iodide and displaces iodide, which is precisely Williamson's ether synthesis.

Question 6

Previous-year question
The reaction (CH₃)₃C–ONa + CH₃CH₂Cl → (CH₃)₃C–O–CH₂CH₃ + NaCl is called [AIPMT 2015]
  1. Williamson continuous etherification process
  2. Etard reaction
  3. Gatterman–Koch reaction
  4. Williamson synthesis
Show answer
Answer: (D) Williamson synthesis
Explanation: An alkoxide displacing a halide from a primary alkyl halide is the classical Williamson ether synthesis. Note that the bulky group is on the alkoxide and the primary carbon carries the leaving group, which is exactly the right arrangement.

Question 7

Previous-year question
Identify Z in the sequence CH₃CH₂CH=CH₂ →(HBr/H₂O₂) Y →(C₂H₅ONa) Z. [AIPMT 2014]
  1. CH₃(CH₂)₃–O–CH₂CH₃
  2. (CH₃)₂CH–CH₂–O–CH₂CH₃
  3. CH₃(CH₂)₄–O–CH₃
  4. CH₃CH₂–CH(CH₃)–O–CH₂CH₃
Show answer
Answer: (A) CH₃(CH₂)₃–O–CH₂CH₃
Explanation: With peroxide, HBr adds anti-Markownikoff, so bromine goes to the terminal carbon giving 1-bromobutane. Sodium ethoxide then displaces bromide by Williamson synthesis to give butyl ethyl ether.
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Question 8

Previous-year question
Among the following sets of reactants, which one produces anisole? [AIPMT 2014]
  1. CH₃CHO and RMgX
  2. C₆H₅OH, NaOH and CH₃I
  3. C₆H₅OH and neutral FeCl₃
  4. C₆H₅CH₃ and CH₃COCl/AlCl₃
Show answer
Answer: (B) C₆H₅OH, NaOH and CH₃I
Explanation: Sodium hydroxide converts phenol into sodium phenoxide, whose oxygen then attacks methyl iodide in a Williamson synthesis. The product C₆H₅OCH₃ is anisole.

Question 9

Ethoxyethane, butan-1-ol and n-pentane have nearly the same molar mass. Which statement correctly compares them?
  1. Butan-1-ol is miscible with water in all proportions while ethoxyethane is essentially immiscible, since ether oxygen cannot form a hydrogen bond with water
  2. Ethoxyethane and butan-1-ol dissolve in water to a similar extent while pentane is essentially immiscible, yet butan-1-ol boils far above ethoxyethane
  3. All three dissolve in water to a similar extent, because solubility in water is governed mainly by molar mass
  4. Ethoxyethane dissolves in water better than butan-1-ol and also boils higher, because it has two alkyl groups
Show answer
Answer: (B) Ethoxyethane and butan-1-ol dissolve in water to a similar extent while pentane is essentially immiscible, yet butan-1-ol boils far above ethoxyethane
Explanation: The oxygen of an ether can form hydrogen bonds with water just as the oxygen of an alcohol can, so ethoxyethane and butan-1-ol are miscible with water to almost the same extent. Pentane has no such oxygen and is essentially immiscible. Ether molecules cannot hydrogen bond with one another, however, so ethoxyethane boils close to pentane and much lower than butan-1-ol.

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