When an ether is treated with ${{\rm{P}}_2}{{\rm{S}}_5}$ we get:

NEET UG CHEMISTRY MCQ: ALCOHOLS & PHENOLS & ETHERS

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Understanding +4 | -1

When an ether is treated with

{\mathrm{P}_2}{\mathrm{S}_5}

we get:

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Correct option: C

Thio-ether

Option C (Thio-ether) is correct because

{\mathrm{P}_2}{\mathrm{S}_5}

replaces oxygen with sulfur in ethers, forming thioethers.

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{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH}}\xrightarrow{{{\text{PCC}}}}{\text{A}}\xrightarrow[{\left( {{\text{ii}}} \right){\text{ }}{{\text{H}}_2}{\text{O/}}{{\text{H}}^ + }}]{{\left( {\text{i}} \right){\text{ HCN}}}}{\text{B}}

\mathrm{O}_2

(X)

Ph - CH = C{H_2}\xrightarrow[{\begin{array}{cccccccccccccccccccc} {\left(2 \right)\,HBr} \\ {\left(3 \right)\,Mg,{\text{ dry ether}}} \\ {\left(4 \right)\,HCHO} \end{array}}]{{\left(1 \right){B_2}{H_6}/{\text{THF/}}{H_2}{O_2}/O{H^ - }}}X\left( {{\text{major}}} \right)