Question

Allyl phenyl ether can be prepared by heating

• A. ${\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{Br}+{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{ONa}$
• B. ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{CH}=\mathrm{CH}-\mathrm{Br}+{\mathrm{CH}}_3-\mathrm{ONa}$
• C. ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{Br}+{\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{ONa}$
• D. ${\mathrm{CH}}_2=\mathrm{CH}-\mathrm{Br}+{\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{CH}}_2-\mathrm{ONa}$

Answer 1

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 $({\mathrm{C}}_2{\mathrm{H}}_5-\mathrm{O}-{\mathrm{C}}_2{\mathrm{H}}_5)$

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 $({\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{O}}^{-})$ and allylic halides$(\mathrm{R}-\mathrm{CH}=\mathrm{CH}-\mathrm{X})$.
• 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.

${\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{Br}+{\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{ONa}\to {\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{O}-{\mathrm{C}}_6{\mathrm{H}}_5+\mathrm{NaBr}$