Arrange the following alkyl chlorides in order of decreasing reactivity in a $S_{N} 1$ reaction.
(I) isopropyl bromide
(II) propyl bromide
(III) tert-butyl bromide
(IV) methyl bromide

(III) > (I) > (II) > (IV)

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(I) > (III) > (IV) > (II)

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(IV) > (III) > (II) > (I)

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(I) > (II) > (III) > (IV)

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Solution

The correct option is A (III) > (I) > (II) > (IV)
tret+-buty 1 bromide > isomeroply 1 bromide > propyle bromide > methyl bromide
i.e. $(I I I) > (I) > (I I) > (I V)$
Reason:
Since we know that in $S N^{'}$ reaction first is $10 s s$ of a leaving group to give a carbocation, then nuclophile is attack
So carbocation must be stable
Carbocation Stability increases with increasing substitution of carbon
i.e. tertiary > secondry >> primary

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