Arrange the following bromides in the order o | vedclass

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(A) The reactivity of alkyl halides towards $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed in the rate-determining

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$i > iii > ii > iv$

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(A) The reactivity of alkyl halides towards $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed in the rate-determining step.The order of stability of carbocations is $3^{\circ} > 2^{\circ} > 1^{\circ} > 	ext{methyl}$.$(i)$ $(CH_3)_3CBr$ forms a $3^{\circ}$ carbocation (most stable).(ii) $CH_3CH_2CH_2Br$ forms a $1^{\circ}$ carbocation.(iii) $(CH_3)_2CHCH_2Br$ forms a $1^{\circ}$ carbocation,but it is slightly more stable than (ii) due to the inductive effect of the alkyl group,though both are $1^{\circ}$. However,in standard textbook problems,(ii) and (iii) are both $1^{\circ}$ and (iv) is methyl.Comparing the structures: $(i)$ is $3^{\circ}$,(ii) is $1^{\circ}$ (n-propyl),(iii) is $1^{\circ}$ (isobutyl),(iv) is methyl.The order of reactivity is $i > iii > ii > iv$.

Arrange the following bromides in the order of reactivity in undergoing $S_{N}1$ reaction:
$(i)$ $(CH_3)_3CBr$
(ii) $CH_3CH_2CH_2Br$
(iii) $(CH_3)_2CHCH_2Br$
(iv) $CH_3Br$

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Arrange the following bromides in the order of reactivity in undergoing $S_{N}1$ reaction:$(i)$ $(CH_3)_3CBr$(ii) $CH_3CH_2CH_2Br$(iii) $(CH_3)_2CHCH_2Br$(iv) $CH_3Br$