Allyl phenyl ether can be prepared by heating

The correct answer is Option (2) → CH₂=CHCH₂Br + C₆H₅ONa

• In the Williamson synthesis an alkoxide (or phenoxide) ion attacks the carbon of an alkyl halide by an S_N2 reaction and pushes out the halide ion. The halide must be an alkyl halide. A halogen on a benzene ring is held by a bond with partial double bond character, so it is not replaced.

• Allyl phenyl ether is \( \ce{C6H5-O-CH2-CH=CH2} \). The O–phenyl bond cannot be made by substitution at a ring carbon, so the phenyl part must come in as sodium phenoxide, and the allyl part as the halide. Allyl bromide is a primary halide whose C–Br carbon is sp³, so it reacts well:

      \( \displaystyle \ce{CH2=CHCH2Br + C6H5ONa} \xrightarrow{\text{heat}} \ce{C6H5OCH2CH=CH2 + NaBr} \)

Why the other options are incorrect

Option (1) C₆H₅Br + CH₂=CH-CH₂ONa — Incorrect. Bromobenzene is an aryl halide. Its C–Br carbon is sp² and the bond has partial double bond character, so the alkoxide cannot replace the bromine.

Option (3) C₆H₅CH=CHBr + CH₃ONa — Incorrect.

• \( \ce{C6H5CH=CHBr} \) is a vinylic halide: its bromine is on an sp² carbon of a C=C, not on an alkyl carbon.

• Even a substitution would give \( \ce{C6H5CH=CHOCH3} \), a methyl ether with no phenyl–O bond.

Option (4) CH₂=CHBr + C₆H₅CH₂ONa — Incorrect. Vinyl bromide is also a vinylic halide, not an alkyl halide. The pair could at best give benzyl vinyl ether, \( \ce{C6H5CH2OCH=CH2} \), a different compound.

NCERT (2026-27 edition):
2. Williamson synthesis – “… In this method, an alkyl halide is allowed to react with sodium alkoxide. … The reaction involves S_N2 attack of an alkoxide ion on primary alkyl halide. … Phenols are also converted to ethers by this method. In this, phenol is used as the phenoxide moiety.”

6.7.2 Reactions of Haloarenes