Decreasing order of reactivity in Williamson synthesis of the following :
I. $M e_{3} C C H_{2} B r$ II. $C H_{3} C H_{2} C H_{2} B r$
III. $C H_{2} = C H C H_{2} C l$ IV. $C H_{3} C H_{2} C H_{2} C l$

Question

Decreasing order of reactivity in Williamson synthesis of the following :
I.

M e_{3} C C 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} C l

IV.

C H_{3} C H_{2} C H_{2} C l

Toppr Admin · Accepted Answer

C-Br bond is weaker than C-Cl bond- therefore- alkyl bromide -II- reacts faster than alkyl chlorides- -III- and -IV- - Since CH2-CH-x2212- is electron withdrawing therefore- CH2 has more positive charge on III than on IV- In other words- nucleophilic attack occurs faster on III than on IV- Further- since Williamson synthesis occurs by SN2 mechanism- therefore- due to steric hindrance alkyl bromide -I- is the least reactive- Thus- the decreasing order of reactivity is II-gt-III-gt-IV-gt-I- III- CH2-CH-x2190-x3B4-x3B4-CH2-x2212-ClIV- CH3-x2212-CH2-x2192-x3B4-CH2-x2212-Cl

Decreasing order of reactivity in Williamson synthesis of the following :I.Me3CCH2Br II.CH3CH2CH2BrIII. CH2=CHCH2Cl IV. CH3CH2CH2ClI>II>IV>IIIIII>II>IV>II>III>II>IVII>III>IV>I

Decreasing order of reactivity in Williamson synthesis of the following :
I. $M e_{3} C C H_{2} B r$ II. $C H_{3} C H_{2} C H_{2} B r$
III. $C H_{2} = C H C H_{2} C l$ IV. $C H_{3} C H_{2} C H_{2} C l$
$I > I I > I V > I I I$
$I I I > I I > I V > I$
$I > I I I > I I > I V$
$I I > I I I > I V > I$