Decreasing order of reactivity in Williamson ether synthesis of the following is (I) M e3 C C H2 B r, (II) CH 3 CH 2 CH 2 Br, (III) CH 2= CHCH 2 Cl, (IV) CH 3 CH 2 CH 2 Cl

Question

Decreasing order of reactivity in Williamson ether synthesis of the following is (I) M e3 C C H2 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

Tardigrade Admin · Accepted Answer

Williamsons ether synthesis involves SN 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°<2°<3° As the size of halide ion increases, reactivity towards SN 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 undersetIIICH 2= CHCH 2 Cl > undersetIICH 2 CH 2 Br < undersetIVCH 3 CH 2 CH 2 Cl > undersetIMe 3 CCH 2 Br

Q. Decreasing order of reactivity in Williamson ether synthesis of the following is
(I) $M e_{3} CC H_{2} B r$ ,
(II) $C H_{3} C H_{2} C H_{2} B r$ ,
(III) $C H_{2} = C H C H_{2} Cl$ ,
(IV) $C H_{3} C H_{2} C H_{2} Cl$

$III > II > I V > I$

$I > II > I V > III$

$II > III > I V > I$

$I > III > II > I V$

Q. Decreasing order of reactivity in Williamson ether synthesis of the following is (I) Me3​CCH2​Br, (II) CH3​CH2​CH2​Br, (III) CH2​=CHCH2​Cl, (IV) CH3​CH2​CH2​Cl

III>II>IV>I

I>II>IV>III

II>III>IV>I

I>III>II>IV

Q. Decreasing order of reactivity in Williamson ether synthesis of the following is
(I) $M e_{3} CC H_{2} B r$ ,
(II) $C H_{3} C H_{2} C H_{2} B r$ ,
(III) $C H_{2} = C H C H_{2} Cl$ ,
(IV) $C H_{3} C H_{2} C H_{2} Cl$

$III > II > I V > I$

$I > II > I V > III$

$II > III > I V > I$

$I > III > II > I V$