Question

Reactions in which ether are obtained as a major produet is: (A) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\stackrel{\ominus}{\mathrm{O}} \mathrm{Na}^{\oplus}+\mathrm{CH}_{3}-\mathrm{I} \longrightarrow\) (B) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}^{\prime} \frac{\mathrm{H}_{1} \mathrm{SO}_{4}}{1+10 \mathrm{C}}\) (C) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \frac{\text { (i) } \mathrm{Hg}\left(\mathrm{O}-\mathrm{C}-\mathrm{CH}_{3}\right) / \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}}{(\mathrm{ii}) \mathrm{NaBH}_{4} \mathrm{OH}^{-}}\) (D)All of these

Short answer

The correct answer is (D) All of these, as all three reactions (A), (B), and (C) produce ethers as their major products.

Step-by-step solution

Reaction (A)

: First, we look at Reaction (A): \[\mathrm{CH}_{3}-\mathrm{CH}_{2}-\stackrel{\ominus}{\mathrm{O}}\mathrm{Na}^{\oplus}+\mathrm{CH}_{3}-\mathrm{I} \longrightarrow ?\] This reaction involves the sodium alkoxide \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{O}^-\mathrm{Na}^+\) reacting with an alkyl halide, \(\mathrm{CH}_{3}\mathrm{I}\). This type of reaction is called the Williamson ether synthesis, and it results in the formation of an ether: \[\mathrm{CH}_{3}-\mathrm{CH}_{2}-\stackrel{\ominus}{\mathrm{O}}\mathrm{Na}^{\oplus}+\mathrm{CH}_{3}-\mathrm{I} \longrightarrow \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{O}\mathrm{CH}_{3} + \mathrm{NaI}\] So, Reaction (A) produces an ether as the major product.

Reaction (B)

: Now, let's consider Reaction (B): \[\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}^{\prime} \frac{\mathrm{H}_{1}\mathrm{SO}_{4}}{\mathrm{1}+\mathrm{1}\mathrm{0}\mathrm{C}} \longrightarrow ?\] In this reaction, an alcohol \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\) reacts with concentrated \(\mathrm{H}_{2}\mathrm{SO}_{4}\). This reaction results in the formation of an ether through an intramolecular dehydration: \[\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}^{\prime} \frac{\mathrm{H}_{1}\mathrm{SO}_{4}}{\mathrm{1}+\mathrm{1}\mathrm{0}\mathrm{C}} \longrightarrow \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{O}\mathrm{CH}_{2}\mathrm{CH}_{3} + \mathrm{H}_{2}\mathrm{O}\] Thus, Reaction (B) also produces an ether as the major product.

Reaction (C)

: Finally, let's analyze Reaction (C): \[\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \frac{\text { (i) } \mathrm{Hg}\left(\mathrm{O}-\mathrm{C}-\mathrm{CH}_{3}\right) /\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}}{(\mathrm{ii}) \mathrm{NaBH}_{4}\mathrm{OH}^{-}} \longrightarrow ?\] This reaction is an oxymercuration-demercuration reaction, which converts an alkene to an ether. In this specific case, the following reaction occurs: \[\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \frac{\text { (i) } \mathrm{Hg}\left(\mathrm{O}-\mathrm{C}-\mathrm{CH}_{3}\right) /\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}}{(\mathrm{ii}) \mathrm{NaBH}_{4}\mathrm{OH}^{-}} \longrightarrow \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{O}\mathrm{CH}_{2}\mathrm{CH}_{3}\] Hence, Reaction (C) produces an ether as the major product.

Conclusion

: Now that we have analyzed all reactions, it is clear that all of them (A), (B), and (C) produce ethers as their major products. Therefore, the correct answer is (D) All of these.

Key concepts

Williamson Ether Synthesis

Let's dive into the world of organic chemistry where the creation of ethers, a class of organic compounds, can be fascinating. One classical method to synthesize ethers is through the Williamson ether synthesis.

How does this reaction work? It starts with an alcohol, which is converted into an alkoxide ion by reacting it with a strong base such as sodium or potassium hydroxide. This alkoxide ion then acts as a nucleophile and attacks an alkyl halide, displacing the halide ion and forming the desired ether.

Key Pointers:

• Alkoxide formation: An alcohol is deprotonated to form an alkoxide ion.
• Nucleophilic substitution: The alkoxide ion undergoes an SN2 reaction with an alkyl halide to create the ether.
• Reactivity: Primary alkyl halides work best to avoid side reactions like eliminations.

Remember, the SN2 mechanism is highly dependent on the steric hindrance around the electrophilic carbon, so primary alkyl halides are generally preferred for smoother reactions. This method is excellent for synthesizing simple symmetrical and asymmetrical ethers.

Alcohol Dehydration

Moving on from the Williamson ether synthesis, another vital technique in ether creation is alcohol dehydration. It's a type of elimination reaction where an alcohol loses a molecule of water to form an ether or an alkene, typically when heated with a strong acid like sulfuric acid or phosphoric acid.

For ether formation via dehydration, two alcohol molecules are involved. Under acidic conditions, the hydroxyl group of one alcohol molecule is protonated, making it a better leaving group. Then, an adjacent alcohol molecule attacks this now more electrophilic carbon, leading to the formation of an ether and the release of water.

Important Aspects:

• Acid catalyst: Strong acid facilitates the leaving of the water molecule.
• Temperature control: Moderate temperatures favor ether formation, while higher temperatures favor alkene formation.

It's essential to note that this method is more suitable for the formation of symmetrical ethers since using two different alcohols can lead to a mixture of different ethers.

Oxymercuration-Demercuration

The third ether synthesis method we'll explore is oxymercuration-demercuration, primarily used to transform alkenes into alcohols or ethers in a regioselective and stereospecific manner. This two-step process involves first adding mercuric acetate to an alkene, creating a mercurinium ion intermediate. In the presence of an alcohol, the intermediate is then attacked by the alcohol, leading to the formation of an organomercurial alcohol compound.

The second step is crucial, where demercuration takes place using sodium borohydride, NaBH4, which replaces the mercury atom with a hydrogen atom, resulting in an ether.

Crucial Points:

• Regioselectivity: The reaction typically follows Markovnikov's rule, adding the nucleophile (alcohol) to the more substituted carbon.
• Stereospecificity: It usually maintains the stereochemistry of the starting alkene.
• Mild Conditions: The use of mild reagents reduces the risk of rearrangements or side reactions.

Oxymercuration-demercuration allows the conversion of alkenes to ethers in a controlled manner, and it's especially good for sensitive molecules that might not withstand harsh conditions.