Questions & explanations
1. Which of the following statements about the physical properties of ethers is correct?
- Ethers have higher boiling points than corresponding alcohols due to hydrogen bonding.
- Ethers are moderately soluble in water because they can form hydrogen bonds with water molecules.
- Ethers are completely non-polar compounds.
- Diethyl ether is denser than water.
Answer: Ethers are moderately soluble in water because they can form hydrogen bonds with water molecules.
Ethers are polar compounds with a bent structure and can accept hydrogen bonds from water, making them moderately soluble. Alcohols have higher boiling points than ethers because alcohols can both donate and accept hydrogen bonds, while ethers only accept them. Ethers have lower boiling points than alcohols of similar molecular weight. Diethyl ether is less dense than water (density ≈ 0.71 g/mL).
2. What is the IUPAC name of CH₃–O–C₂H₅?
- Methoxyethane
- Ethoxymethane
- Ethyl methyl ether
- Dimethyl ether
Answer: Methoxyethane
In IUPAC naming, ethers are named as alkoxyalkanes. The smaller alkyl group (methyl) is treated as the alkoxy substituent, and the larger alkyl group (ethyl) forms the parent alkane. Therefore, CH₃–O–C₂H₅ is methoxyethane. 'Ethoxymethane' would imply the ethoxy group on methane, which reverses the order. 'Ethyl methyl ether' is the common name, not IUPAC. 'Dimethyl ether' would be CH₃–O–CH₃.
3. Which of the following reactions correctly represents Williamson ether synthesis?
- C₂H₅ONa + CH₃I → C₂H₅OCH₃ + NaI
- C₂H₅OH + H₂SO₄ → C₂H₅OC₂H₅ + H₂O
- CH₃OH + HI → CH₃I + H₂O
- C₂H₅Cl + NaOH → C₂H₅OH + NaCl
Answer: C₂H₅ONa + CH₃I → C₂H₅OCH₃ + NaI
Williamson ether synthesis involves the reaction of an alkoxide ion (sodium ethoxide) with a primary alkyl halide (methyl iodide) to form an ether (methoxyethane). The second option is acid-catalyzed dehydration of alcohol. The third is nucleophilic substitution of alcohol to form alkyl halide. The fourth is hydrolysis of alkyl halide to form alcohol.
4. What are the products formed when methoxybenzene (anisole) is treated with excess HI and heated?
- Phenol and methyl iodide
- Benzene and methanol
- Iodobenzene and methanol
- Aniline and methyl iodide
Answer: Phenol and methyl iodide
In anisole, the C–O bond to the methyl group is weaker and undergoes cleavage by HI via an SN2 mechanism (methyl is primary), giving methyl iodide (iodomethane) and phenol. The aromatic C–O bond is not cleaved under these conditions because the phenoxide leaving group is unfavourable. Benzene, methanol, iodobenzene, or aniline are not formed.
5. Anisole undergoes electrophilic substitution reactions preferentially at which positions?
- Ortho and para
- Meta
- Only ortho
- Only para
Answer: Ortho and para
The methoxy group (–OCH₃) is an activating group that donates electrons by resonance, increasing electron density at the ortho and para positions of the benzene ring. Therefore, electrophilic substitution (e.g., nitration, bromination) in anisole occurs predominantly at the ortho and para positions.
6. When diethyl ether is treated with excess concentrated HI and heated, the final organic products are:
- Ethanol and ethyl iodide
- Two moles of ethyl iodide
- Ethene and water
- Ethane and iodine
Answer: Two moles of ethyl iodide
Cleavage of the C–O bond in diethyl ether by excess HI proceeds in two steps: first, one mole of HI gives ethanol and ethyl iodide; second, ethanol reacts with another mole of HI to give a second mole of ethyl iodide. Thus, the overall reaction yields two moles of ethyl iodide (iodoethane).
7. Which alcohol, upon dehydration with concentrated H₂SO₄ at 140°C, gives diethyl ether as the major product?
- Ethanol
- Methanol
- Propan-1-ol
- Butan-1-ol
Answer: Ethanol
Diethyl ether (C₂H₅–O–C₂H₅) is formed by the intermolecular dehydration of two molecules of ethanol at 140°C in the presence of concentrated H₂SO₄. Methanol would give dimethyl ether. Propanol and butanol would yield their respective symmetrical ethers or alkenes depending on conditions.