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

Question

Identify the product ' \(\mathrm{P}\) ' in the given reaction \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I} \quad \frac{\mathrm{O}^{-} \mathrm{C}_{2} \mathrm{H}_{3}}{\text { Anhy. }\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)}\) (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OC}_{2} \mathrm{H}_{5}\) (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 \(\mathrm{C}_6\mathrm{H}_5\mathrm{OC}_2\mathrm{H}_5\) (option a).

Step 1: Analyze the Reaction Components

The reactants in the given reaction are phenol \((\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{OH})\) and ethyl iodide \((\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{I})\). We also have an alkaline condition indicated by the presence of \(\mathrm{O}^{-}\) and the solvent is anhydrous ethanol \((\text{C}_2\mathrm{H}_{3}\mathrm{OH})\).

Step 2: Predict the Reaction Type

Given the presence of an alkoxide ion \((\mathrm{O}^{-})\), this reaction is likely a nucleophilic substitution reaction where the alkoxide ion will attack the carbon of the ethyl iodide, facilitating the displacement of iodide \((\mathrm{I}^- )\).

Step 3: Determine the Product

In this nucleophilic substitution reaction, the \(\mathrm{O}^{-}\) from phenol will displace \(\mathrm{I}^-\) from \(\mathrm{C}_2\mathrm{H}_5\mathrm{I}\), forming an ether product: \(\mathrm{C}_6\mathrm{H}_5\mathrm{O}\mathrm{C}_2\mathrm{H}_5\). This product matches option (a).

Step 4: Final Product Verification

Verify that the product \(\mathrm{C}_6\mathrm{H}_5\mathrm{O}\mathrm{C}_2\mathrm{H}_5\) matches the mechanism and reactants involved. Phenol forms an alkoxide ion which reacts with ethyl iodide to produce the ether \(\mathrm{C}_6\mathrm{H}_5\mathrm{OC}_2\mathrm{H}_5\), confirming this matches the product 'P'.

Key Concepts

Phenol

Phenol, chemically noted as \( ext{C}_6 ext{H}_5 ext{OH}\), is a fundamental organic compound characterized by a hydroxyl group \(-OH\) attached to a benzene ring. This makes phenol an essential starting material in many industrial processes. Despite the presence of the hydroxyl group, phenol is weakly acidic. This acidic property is vital as it allows phenol to form phenoxide ions \(\text{C}_6\text{H}_5\text{O}^- \) when reacting under basic conditions. By losing a hydrogen ion \((\text{H}^+)\), phenol transforms into a nucleophile, which can engage actively in bimolecular nucleophilic substitution reactions (also known as \(S_N2\) reactions).

• Phenol's acidic nature contributes to its reactivity with bases to form phenoxide ions.
• Phenol is crucial in the synthesis of various organic compounds, especially ethers.
• It has antiseptic properties and is often used in disinfectants and throat sprays.

Understanding phenol's properties lays the foundation for grasping how it reacts with compounds like ethyl iodide to form ethers.

Ethyl iodide

Ethyl iodide \(\text{C}_2\text{H}_5\text{I}\) is an organoiodine compound widely used as an alkylating agent in organic synthesis. It features an ethyl group \(\text{-C}_2\text{H}_5\) bound to an iodine atom, making it a primary alkyl halide known for its reactivity. The iodine atom, being a good leaving group, allows ethyl iodide to easily participate in nucleophilic substitution reactions. During such reactions, the iodine atom detaches, making way for a nucleophile to form a covalent bond with the remaining ethyl group

• Ethyl iodide's reactivity is primarily due to the presence of iodine, a large, polarizable atom.
• This reactivity is harnessed in nucleophilic substitution reactions where it acts as an electrophile.
• It is often used to introduce ethyl groups into molecules, enhancing their structural complexity.

Ethyl iodide's ability to participate in these types of reactions is pivotal in the synthesis of compounds like ethers, where it reacts with phenol-derived nucleophiles.

Ether formation

Ether formation involves a nucleophilic substitution process where an alkoxy group replaces a leaving group like iodide in an alkyl halide. In the context of phenol and ethyl iodide reaction, phenoxide ion \(\text{C}_6\text{H}_5\text{O}^-\) serves as the nucleophile. It attacks the carbon atom bonded to iodine in ethyl iodide \(\text{C}_2\text{H}_5\text{I}\). This reaction is a classic example of an \(S_N2\) mechanism, in which the nucleophile approaches the electrophilic carbon center from the opposite side of the leaving group (iodine). This results in the replacement of iodine by the phenoxide ion, leading to the formation of an ether linkage. The product of this reaction, phenyl ethyl ether \(\text{C}_6\text{H}_5\text{OC}_2\text{H}_5\), exemplifies the result of the nucleophilic attack.

• Ether formation via \(S_N2\) reaction involves one step, with simultaneous bond formation and bond breaking.
• The reaction provides ethers with extended alkoxy chains.
• The reactivity and efficiency rely heavily on the strength of the nucleophile and the quality of the leaving group.

Ether synthesis not only broadens the scope of organic synthesis but also serves crucial roles in producing solvents, pharmaceuticals, and industrial chemicals.