$RCH _2 Br + I ^{-} \stackrel{\text { Acetone }}{\longrightarrow} \underset{\text { }}{ RCH _2 I + Br ^{-}}$
સાચું વિધાન શોધો.
Clearly, the transition state is less polar than free anions. $Br ^{-}$and $I ^{-}$
Acetic acid is protic which does not support $S _{ N } 2$
Acetone does not solvate anion
$Br ^{-}$gets precipitated and hence can not compete with $I$



$\mathop I\limits_{({C_3}{H_6}C{l_2})} \xrightarrow{{KOH(aq)}}II\xrightarrow[{(ii){H_2}O/{H^ + }}]{{(i)C{H_3}MgBr}}III\xrightarrow{{Anhy.ZnC{l_2} + Conc.HCl}}$ તરત જ terbidity મળે છે