Problem 132 - A Complete Resource Book in Chemistry for JEE Main

Question

Identify the product ' \(\mathrm{P}\) ' in the given reaction \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I} \quad \frac{\mathrm{O}^{-\mathrm{C}}_{2} \mathrm{H}_{5}}{\text { Anhy. }\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)}\) (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OC}_{6} \mathrm{H}_{5}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}\).

Step-by-Step Solution

Answer

The product \(\mathrm{P}\) is (b) \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OC}_{2}\mathrm{H}_{5}\).

Step 1: Understand the reaction components

The reactants in the given reaction are phenol (\(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{OH}\)) and ethyl iodide (\(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{I}\)). The solvent or catalyst listed is anhydrous ethyl alcohol (\(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{OH}\)). The reaction conditions suggest a Williamson ether synthesis.

Step 2: Identify the type of reaction

This is a Williamson ether synthesis reaction where an alkoxide ion attacks an alkyl halide to form an ether. In this case, the ethoxide ion (formed from ethanol and a base) will displace the iodine in ethyl iodide.

Step 3: Predict the product formation

The ethoxide ion (\(\mathrm{O^{-}C_{2}H_{5}}\)) is a good nucleophile and will attack the electrophilic carbon atom in ethyl iodide (\(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{I}\)), forming diethyl ether (\(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OC}_{2}\mathrm{H}_{5}\)). This matches with option (b).

Step 4: Verify the side product

Given the side conditions and reactants, verify which product results. In the Williamson synthesis, phenol can generate alkoxide ions that will act as nucleophiles, but since ethoxide ion is explicitly provided, it directly forms diethyl ether without involving phenol in forming the ether product.

Key Concepts

Phenol

Phenol, chemically known as \(\mathrm{C}_6\mathrm{H}_5\mathrm{OH}\), is a widely studied organic compound. It's a colorless crystalline solid that possesses a strong scent. Phenol is often found in the form of a white crystalline solid, but it can appear slightly pink when in contact with impurities.

One of phenol's most notable chemical properties is its acidic nature. Although it is a weak acid, it is more acidic than alcohols due to the aromatic ring, which stabilizes the phenoxide ion after deprotonation. This ability to form ions is crucial in reactions like the Williamson ether synthesis.

• Phenol’s acidity means it can lose its hydrogen ion \(\mathrm{H}^+\) easily.
• This forms the phenoxide ion \(\mathrm{C}_6\mathrm{H}_5\mathrm{O}^-\).

In the context of the exercise, phenol is a potential participant in ether synthesis, but in the given reaction setup, it does not form the ether product. The ethoxide ion takes precedence.

Ethyl Iodide

Ethyl iodide, \(\mathrm{C}_2\mathrm{H}_5\mathrm{I}\), serves as a crucial component in the Williamson ether synthesis. It is recognized for its role as a primary alkyl halide. Characteristically, ethyl iodide is a colorless liquid but can appear slightly yellowish due to light exposure and formation of iodine.

The functionality of ethyl iodide in synthesis primarily relies on its ability to act as an electrophile. This means it can be attacked by nucleophiles, making it highly reactive in substitution reactions.

• Iodine is a good leaving group, assisting in the progress of the reaction.
• This makes ethyl iodide highly suitable for an alkoxide ion attack.

In the exercise, ethyl iodide partners with the ethoxide ion to successfully synthesize diethyl ether. This transformation underscores its pivotal role as a reactive alkylating agent.

Diethyl Ether

Diethyl ether, often simply referred to as ether, is a widespread organic compound with the formula \(\mathrm{C}_2\mathrm{H}_5\mathrm{OC}_2\mathrm{H}_5\). It plays a significant role not only in chemical syntheses but also historically as an anesthetic.

In chemical reactions, diethyl ether functions as an important solvent due to its relatively low polarity and excellent solubilizing capability. In Williamson ether synthesis, it represents an ideal ether product.

• It is formed when ethoxide ion attacks ethyl iodide, displacing iodine.
• The reaction demonstrates a typical nucleophilic substitution process.

In the context of this synthesis, diethyl ether is the product resulting from the specific reaction between the reactants, showcased in the exercise. This confirms the choice of option (b) in the problem set.