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JEE Chemistry 2023 Question with Solution

Suitable reaction condition for preparation of Methyl phenyl ether is:

  • A

    Benzene, MeBr

  • B

    PhO−Na+\text{PhO}^-\text{Na}^+, MeOH

  • C

    Ph-Br, MeO−Na+\text{MeO}^-\text{Na}^+

  • D

    PhO−Na+\text{PhO}^-\text{Na}^+, MeBr

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: We need the suitable reaction condition for preparation of methyl phenyl ether.

Find: Which reagent pair gives Ph-O-Me\text{Ph-O-Me} by Williamson ether synthesis.

The Williamson ether synthesis involves reaction of an alkoxide or aryloxide ion with a primary alkyl halide.

R-O−+R’-X→R-O-R’+X−\text{R-O}^- + \text{R'-X} \rightarrow \text{R-O-R'} + \text{X}^-

For methyl phenyl ether, the required nucleophile is phenoxide ion and the alkyl halide must provide the methyl group.

PhO−Na++MeBr→Ph-O-Me+NaBr\text{PhO}^-\text{Na}^+ + \text{MeBr} \rightarrow \text{Ph-O-Me} + \text{NaBr}

The phenoxide ion acts as a nucleophile in an SN2\text{SN}_2 reaction and attacks methyl bromide to form the ether linkage.

Therefore, the correct option is D.

Note: the solution incorrectly labels the correct option as A, but its own reaction and explanation clearly support PhO−Na+\text{PhO}^-\text{Na}^+ with MeBr, which is option D.

Common mistakes

  • Choosing benzene with MeBr is incorrect because benzene does not directly form anisole under Williamson ether synthesis conditions. Use phenoxide ion as the nucleophile instead.

  • Using MeOH instead of MeBr is incorrect because an alcohol is not the proper electrophile for the SN2\text{SN}_2 ether-forming step. A primary alkyl halide such as MeBr is required.

  • Selecting Ph-Br with methoxide is incorrect because aryl halides do not undergo normal SN2\text{SN}_2 substitution at the aryl carbon. The alkyl halide should be methyl bromide, not bromobenzene.

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