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In nucleophilic substitution reaction, order of halogens as incoming (attacking) nucleophile is:
$\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}$
The order of halogens as departing nucleophile should be :
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$\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}$
The order of halogens as departing nucleophile should be :
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Verified Answer
The correct answer is:
$\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}$
$\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}$
Since the leaving group breaks away as a base, it is easier to displace weaker bases as compared to stronger bases. Thus less basic the substituent, the more easily it is displaced.
Since the basic strength of the given groups is in order.
$\mathrm{I}^{-} < \mathrm{Br}^{-} < \mathrm{Cl}^{-}$
Thus the order of halogen leaving groups is
$\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}$
Since the basic strength of the given groups is in order.
$\mathrm{I}^{-} < \mathrm{Br}^{-} < \mathrm{Cl}^{-}$
Thus the order of halogen leaving groups is
$\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}$
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