Question

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

• A. Ethane and ethoxythane
• B. Ethylmethyl ether and butene
• C. Ethylmethyl ether and propene
• D. Ethene and ethoxythane
• A. o- bromoanisole
• B. p- bromoanisole
• C. m- bromoanisole
• D. o-bromoanisole and p-bromoanisole
• A. SN\(_2\)
• B. SN\(_1\)
• C. Depends on the nature of alkoxide ion
• D. Depends on the nature of Alkyl halide
• A. HF
• B. HCl
• C. HBr
• D. HI
• A. Dialkyl ether
• B. Diaryl ether
• C. Phenyl Alkyl ether
• D. Alkoxy Alkyl ether

Answer 1

The Correct Option is D

The question asks about the products formed when ethanol is dehydrated in the presence of \( \text{H}_2\text{SO}_4 \) at two different temperatures: 443K and 413K. Let's understand the dehydration reactions at these specific conditions:

1. Dehydration of Ethanol at 443K: At this higher temperature, ethanol undergoes intramolecular dehydration in the presence of concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)). This reaction leads to the formation of ethene (an alkene) and water. The reaction is as follows: \(( \text{C}_2\text{H}_5\text{OH} \rightarrow \text{C}_2\text{H}_4 + \text{H}_2\text{O} )\). Here, \(\text{C}_2\text{H}_4\) represents ethene.
2. Dehydration of Ethanol at 413K: At this lower temperature, ethanol undergoes intermolecular dehydration. This leads to the formation of an ether, specifically diethyl ether (ethoxythane), instead of an alkene. The balanced chemical reaction is: \((2\text{C}_2\text{H}_5\text{OH} \rightarrow \text{C}_2\text{H}_5\text{OC}_2\text{H}_5 + \text{H}_2\text{O})\). Here, \(\text{C}_2\text{H}_5\text{OC}_2\text{H}_5\) represents ethoxythane.

Considering these dehydration reactions, the correct combination of products for the given temperatures (443K and 413K) is ethene and ethoxythane. Therefore, the correct answer is the option:

Ethene and ethoxythane

This option is justified because:

• At 443K, ethanol produces ethene by intramolecular dehydration.
• At 413K, ethanol gives ethoxythane (diethyl ether) by intermolecular dehydration.

Other options can be ruled out because they either state incorrect products or do not align with the known mechanistic path of ethanol dehydration under given conditions.

Answer 2

The reaction of anisole with bromine in ethanoic acid is an example of an electrophilic aromatic substitution reaction. Anisole, being an ether (methoxybenzene), has a methoxy group (-OCH_3) attached to the benzene ring. This methoxy group is an electron-donating group through resonance, which increases the electron density on the benzene ring, particularly at the ortho and para positions relative to the methoxy group.

1. Understanding the role of the methoxy group: The methoxy group is an ortho-para directing activator. This means it increases the likelihood that electrophilic substitution will occur at the ortho or para positions.
2. Reaction Mechanism:
   • Step 1: Formation of the electrophile. In ethanoic acid, bromine dissociates to form bromonium ions (Br^+), which act as the electrophile.
   • Step 2: Electrophilic attack. The electron-rich ortho and para positions of anisole are more susceptible to attack by Br^+ ions.
   • Step 3: Formation of the sigma complex (also known as the arenium ion), followed by deprotonation to restore the aromaticity of the benzene ring.
3. Possible products: Due to the presence of the strongly activating methoxy group, bromination occurs quickly at the most activated sites (ortho and para to the -OCH_3 group). This results in two major products:
   • o-bromoanisole (ortho product)
   • p-bromoanisole (para product)
   Formation of the m-bromoanisole (meta product) is not favored due to higher energy intermediates and less stabilization by resonance.
4. Conclusion: As the methoxy group ortho-para directs, the major products of the reaction between anisole and bromine in ethanoic acid are o-bromoanisole and p-bromoanisole. Therefore, the correct answer is the option "o-bromoanisole and p-bromoanisole." Formation of meta-bromoanisole would be minimal and typically considered a minor product in this scenario.

Answer 3

The question pertains to the Williamson synthesis, a key reaction in organic chemistry for forming ethers. The process is crucial in understanding how alkoxide ions interact with alkyl halides.

Concept Explanation: The Williamson synthesis involves the reaction of an alkoxide ion (\(RO^-\)) with an alkyl halide (\(R'X\)) to form an ether (\(ROR'\)). The reaction occurs through a bimolecular nucleophilic substitution mechanism, commonly known as the SN\(_2\) mechanism.

Mechanism Details:

• Nucleophile: The alkoxide ion acts as a strong nucleophile.
• Substrate: The alkyl halide serves as the electrophile.
• Mechanism: In the SN\(_2\) mechanism, the nucleophile attacks the electrophile from the opposite side of the leaving group, resulting in an inversion of configuration. This is a concerted process occurring in a single step.

Reasoning:

The SN\(_2\) mechanism is favored because:

• Strong nucleophile: Alkoxides are typically strong, negatively charged nucleophiles, which assist in a direct backside attack on the carbon atom bearing the leaving group.
• Secondary or less hindered primary substrates: The SN\(_2\) mechanism is favored with primary or secondary alkyl halides, where steric hindrance is minimal, allowing for efficient backside attack.

Incorrect Options:

• SN\(_1\): This mechanism involves carbocation formation and is more typical with tertiary alkyl halides under different reaction conditions.
• Depends on the nature of the alkoxide ion: While the strength of the nucleophile is important, the SN\(_2\) pathway is primarily determined by the nature of the alkyl halide and its steric environment.
• Depends on the nature of the alkyl halide: Although the alkyl halide nature influences the reaction, the Williamson synthesis specifically utilizes the SN\(_2\) mechanism for effectiveness.

Conclusion: In Williamson synthesis, the attack of the alkoxide ion on the alkyl halide proceeds through the SN\(_2\) mechanism, characterized by a single, concerted step and inversion of configuration at the reaction center.

Answer 4

To determine which hydrogen halide is the most reactive for the cleavage of ethers, we need to understand the reactivity of hydrogen halides in the context of ether cleavage. Ethers can be cleaved by hydrogen halides through a nucleophilic substitution reaction mechanism. The reactivity of hydrogen halides (HX) in this reaction typically follows the order of acid strength, as stronger acids perform better cleavage.

1. The order of acid strength for hydrogen halides is: HF < HCl < HBr < HI. This order is based on the bond strength and the stability of the conjugate base.
2. Hydrogen iodide (HI) is the strongest acid amongst the options, making it the most effective at cleaving ethers. This is because the I⁻ ion is a good nucleophile, and the strong acid helps in protonating the ether, facilitating the cleavage.
3. In contrast, HF, being the weakest acid, is the least effective for the cleavage of ethers. Meanwhile, HCl and HBr show intermediate reactivity.
4. The reaction mechanism generally involves protonation of the ether oxygen to form an oxonium ion, which is then attacked by the halide ion.

Therefore, the most reactive hydrogen halide for cleavage of ethers is HI, consistent with its stronger acidic nature and better nucleophilic character of the I⁻ ion compared to other halide ions.

Answer 5

To determine which category anisole belongs to, we need to understand the different types of ethers based on their structure.

Ethers are chemical compounds that contain an oxygen atom connected to two alkyl or aryl groups. The types of ethers include:

• Dialkyl Ether: Both groups attached to the oxygen are alkyl groups. An example is ethyl ether (C₂H₅OC₂H₅).
• Diaryl Ether: Both groups attached to the oxygen are aryl groups. An example is diphenyl ether (C₆H₅OC₆H₅).
• Phenyl Alkyl Ether: One group is an aryl (e.g., phenyl) group, and the other is an alkyl group. Anisole (C₆H₅OCH₃) is an example since it consists of a phenyl group and a methoxy group.
• Alkoxy Alkyl Ether: Contains an alkyl group connected to an alkoxy group.

Analyzing the structure of anisole, we have:

• A phenyl group (C₆H₅) connected to a methoxy group (OCH₃).

Since anisole has a phenyl group and a methoxide (alkyl) group rather than two aryl groups, it fits the definition of a Phenyl Alkyl Ether, not a Diaryl Ether.

Thus, the correct classification of anisole, based on its structural properties, is Phenyl Alkyl Ether. However, there was an oversight in the question as the given correct answer is "Diaryl ether" which does not match the structure of anisole based on its definition.