Question

The heating of phenyl methyl ether with HI produces an aromatic compound A which on treatment with con. HNO₃ gives B. A and B respectively are,

• A. Methanol, Ethanoic acid
• B. Picric acid, Phenol
• C. Iodobenzene, l-Iodo-4-nitrobenzene
• D. Phenol, Picric acid

Answer 1

The Correct Option is D

The reaction described involves the heating of phenyl methyl ether with HI (hydroiodic acid). Let's break down the reaction and the products formed:

Step 1: Heating Phenyl Methyl Ether with HI

Phenyl methyl ether \((C_6H_5OCH_3)\) reacts with HI (hydroiodic acid) in a nucleophilic substitution reaction. The methyl group (–CH₃) is replaced by an iodine atom, producing phenol (C₆H₅OH) and methanol (CH₃OH) as the products. The aromatic compound A formed in this step is phenol.

Step 2: Treatment with Concentrated HNO₃

When phenol is treated with concentrated HNO₃ (nitric acid), it undergoes nitration. This results in the formation of picric acid (2,4,6-trinitrophenol), a highly nitrated aromatic compound. The product B formed here is picric acid.

Conclusion:

The products A and B are:

• A = Phenol
• B = Picric acid

The correct answer is Option B: Phenol, Picric acid.

Answer 2

Phenyl methyl ether (also known as anisole) has the chemical structure \(\text{C}_6\text{H}_5\text{OCH}_3\), where a methoxy group (\(\text{OCH}_3\)) is attached to a benzene ring.

Step 1: Heating phenyl methyl ether with HI When phenyl methyl ether is heated with hydroiodic acid (HI), it undergoes a cleavage reaction. The HI breaks the bond between the oxygen atom of the ether group and the methyl group (\(\text{CH}_3\)), resulting in the formation of phenol (\(\text{C}_6\text{H}_5\text{OH}\)) and methyl iodide (\(\text{CH}_3\text{I}\)). \[ \text{C}_6\text{H}_5\text{OCH}_3 \xrightarrow{\text{HI}} \text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{I} \]

Thus, A is phenol (\(\text{C}_6\text{H}_5\text{OH}\)).

Step 2: Treating phenol with concentrated nitric acid When phenol (\(\text{C}_6\text{H}_5\text{OH}\)) is treated with concentrated nitric acid (\(\text{HNO}_3\)), a nitration reaction occurs. In this process, the hydroxyl group (-OH) of phenol activates the benzene ring, making it more reactive to electrophilic substitution. The nitronium ion (\(\text{NO}_2^+\)) from the concentrated nitric acid attacks the benzene ring, leading to the formation of picric acid (2,4,6-trinitrophenol). This is a well-known example of nitration of phenol. The reaction can be written as: \[ \text{C}_6\text{H}_5\text{OH} + \text{HNO}_3 \xrightarrow{\text{conc.}} \text{C}_6\text{H}_2(\text{NO}_2)_3\text{OH} \ (\text{picric acid}) \]

Thus, B is picric acid (2,4,6-trinitrophenol).

Conclusion: The correct pair of compounds formed are:
A is phenol (formed after heating phenyl methyl ether with HI)
B is picric acid (formed after nitrating phenol with concentrated nitric acid)

Hence, the correct answer is (D) Phenol, Picric acid.