$RCH _2 Br + I ^{-} \stackrel{\text { Acetone }}{\longrightarrow} \underset{\text { }}{ RCH _2 I + Br ^{-}}$
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 \(I\)
$(I)\,\,CH_2 =CH-CH_2Cl$ $(II)\,\,CH_2=CH-CH(CH_3) Cl$
$(III)\,\,CH_2 =C(CH_3)CH_2Cl$ $(IV)\,\, CH_3CH = CH-CH_2Cl$
$2-$ મીથાઈલબ્યુટેન $\xrightarrow{{B{r_2},\,hv}}$ $2-$ બ્રોમો $-3-$ મીથાઈલબ્યુટેન
(not the major product)