Question

Decreasing order of reactivity in Williamson ether synthesis of the following is
(I) $M e_{3} C C H_{2} B r$,
(II) $CH _{3} CH _{2} CH _{2} Br$,
(III) $CH _{2}= CHCH _{2} Cl$,
(IV) $CH _{3} CH _{2} CH _{2} Cl$

• A. $ III>II>IV>I $
• B. $ I>II>IV>III $
• C. $ II>III>IV>I $
• D. $ I>III>II>IV $

Answer 1

Answer: (A)

Williamsons ether synthesis involves $S_{N} 2$ pathway, ie, involves formation of transition state and back side attack.
Thus, less hindered alkyl halides react readily in Williamsons synthesis.
The order of hindrance is Allyl halide $<1^{\circ}<2^{\circ}<3^{\circ}$ As the size of halide ion increases, reactivity towards $S_{N} 2$ reaction increases.
Thus, bromides are more reactive than their corresponding chlorides towards Williamsons ether synthesis.
Hence, the decreasing order of reactivity towards Williamsons ether synthesis is
$\underset{III}{CH _{2}= CHCH _{2} Cl} > \underset{II}{CH _{2} CH _{2} Br} < \underset{IV}{CH _{3} CH _{2} CH _{2} Cl} > \underset{I}{Me _{3} CCH _{2} Br}$