Question

# Allyl phenyl ether can be prepared by heating

A

${\mathrm{CH}}_{2}=\mathrm{CH}-{\mathrm{CH}}_{2}-\mathrm{Br}+{\mathrm{C}}_{6}{\mathrm{H}}_{5}\mathrm{ONa}$

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B

${\mathrm{C}}_{6}{\mathrm{H}}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{Br}+{\mathrm{CH}}_{3}-\mathrm{ONa}$

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C

${\mathrm{C}}_{6}{\mathrm{H}}_{5}\mathrm{Br}+{\mathrm{CH}}_{2}=\mathrm{CH}-{\mathrm{CH}}_{2}-\mathrm{ONa}$

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D

${\mathrm{CH}}_{2}=\mathrm{CH}-\mathrm{Br}+{\mathrm{C}}_{6}{\mathrm{H}}_{5}-{\mathrm{CH}}_{2}-\mathrm{ONa}$

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Solution

## The correct option is A ${\mathrm{CH}}_{2}=\mathrm{CH}-{\mathrm{CH}}_{2}-\mathrm{Br}+{\mathrm{C}}_{6}{\mathrm{H}}_{5}\mathrm{ONa}$Ethers: Ethers are a class of organic compounds that contain an oxygen atom connected to two alkyl or aryl groups. They have the general formula $\mathrm{R}-\mathrm{O}-{\mathrm{R}}^{\text{'}}$, where $\mathrm{R}$ and represent the alkyl or aryl groups.Example: Diethyl ether $\left({\mathrm{C}}_{2}{\mathrm{H}}_{5}-\mathrm{O}-{\mathrm{C}}_{2}{\mathrm{H}}_{5}\right)$Preparation of ethers: Ethers can be prepared by a number of methods. One of the most important methods is Williamson ether synthesis. Allyl phenyl ether can be prepared by this method using phenoxides $\left({\mathrm{C}}_{6}{\mathrm{H}}_{5}-{\mathrm{O}}^{-}\right)$ and allylic halides$\left(\mathrm{R}-\mathrm{CH}=\mathrm{CH}-\mathrm{X}\right)$.The reaction between phenoxide ion and allylic halides undergoes ${\mathrm{S}}_{\mathrm{N}}2$ (Nucleophilic substitution bimolecular) mechanism. This method usually does not work well for aryl halides, example- bromobenzene $\left({\mathrm{C}}_{6}{\mathrm{H}}_{5}\mathrm{Br}\right)$. Hence, option (C) is incorrect. Since, ${\mathrm{S}}_{\mathrm{N}}2$ mechanism depends on the steric hindrance, hence, option (B) is incorrect due to steric hindrance from the large alkyl group. Option (D) is also incorrect as there is no phenoxide present in the reactants. Hence, the correct option is (A) in which a primary allylic halide is reacting with sodium phenoxide to produce ether. ${\mathrm{CH}}_{2}=\mathrm{CH}-{\mathrm{CH}}_{2}-\mathrm{Br}+{\mathrm{C}}_{6}{\mathrm{H}}_{5}-\mathrm{ONa}\to {\mathrm{CH}}_{2}=\mathrm{CH}-{\mathrm{CH}}_{2}-\mathrm{O}-{\mathrm{C}}_{6}{\mathrm{H}}_{5}+\mathrm{NaBr}$

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