Rearrange the compounds of each of the following sets in order of reactivity towards S_(N_(3)^(2)) displacement : (i) 2- Bromo-2-methylbutane, 1-Bromopentane, 2- Bromopentaure. (ii) 1- Bromo-3-methylbutance, 2-Bromo-2-methylbutance,3-Bromo-2- methybutane. (iii)1- Bromobutane, 1- Bromo-2, 2-dimethylbutane 1-Bromo -2- methylbutane.

Rearrange the compounds of each of the following sets in order of reac

1.	Rearrange the compounds of each of the following sets in order of reactivity towards S_(N_(3)^(2)) displacement : (i) 2- Bromo-2-methylbutane, 1-Bromopentane, 2- Bromopentaure. (ii) 1- Bromo-3-methylbutance, 2-Bromo-2-methylbutance,3-Bromo-2- methybutane. (iii)1- Bromobutane, 1- Bromo-2, 2-dimethylbutane 1-Bromo -2- methylbutane.
Answer» Solution :(i) 2-Bromo-2-methybutane, 1-Bromopentane , 2Bromopentane: The reactivity is `SN^(2)` reaction depents uponsteric hindance, more thesteric hindence slower the reaction. Sincedue to stericreasionthe order of reactivityin `SN^(2)`reactionfollowsthe order: `1^(@) gt2^(@) GT3^(@)`, therefore, order of reactivityofthe givenalklyl BROMIDES is : 1-Bromopentane `gt 2`- Brompebtane `gt 2-` Bromo -2 Methylbutane. Since due to steric reason, the order of reactivity of alkyl halides in `SN^(2)` reactionfollows the order `1^(@) gt 2^(@) gt 3^(@)` therefore, the order ofreactivity of thegivenalkyl bromies is . 1-Bromo-3-Methylbutane`gt2`- Bromo -3-Methylbutane `gt 2`-Bromo -2-Methylbutane. Since in CASE of `1^(@)` alkyl halides,steric hindrancein theorder : n-alkyl halides, alkyl halide substituentsatthe `beta`-position, therefore , the reactivitydecrease in the sameorder- Thu, the reactivityof the givenalkyl bromides decrease in the order. 1-Bromobutance `gt 1`- Bromo - 3-Methybutane `gt 1`- Bromo-2- Methylbutane `GT1`- Bromo-2-Methylbutane`gt1`- Bromo-2-dimethylpropane.