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When a substitution reaction takes place in a substrate having halogen,

Why is it that iodine reacts the fastest and fluoride is the slowest ?

For sure the iodine has a weaker bond but it is also very less electronegative and the carbon would be less electrophilic whereas in the case of fluorine, highly electrophilic centre is created.

$S_N1$ where the RDS is breaking of a bond to form carbo cation, I think it should be easiest with the fluoride because of its electronegative character, the bond is already quite weak as all the electrons are closer to the fluorine and it would be easier to break the bond.

$S_N2$ which has only a single step the approach of the nucleophile is the most important which should be easiest in fluoride because the carbon would be highly electrophilic

Also my teacher told that fluoride is a strong base so it does not act as a good leaving group but I had read in a book that when aqueous medium is present the fluoride ion gets heavily solvated and thus its nucleophilic character is significantly reduced.

In aqueous medium the trend followed is $\ce{I}^->\ce{Br}^->\ce{Cl}^->>\ce{F}^-$.

So I could not understand how even if it is a strong base ,would affect the leaving capacity of $\ce{F}$ because it would quickly get solvated and could not act as a nucleophilic again. So should be a good leaving group, where is iodine being a strong nucleophile attack again and would not be a good leaving.

Please help.

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    $\begingroup$ Please avoid pointless editing (such as adding extra space before punctuation marks) made solely for the purpose of bumping your question up. $\endgroup$ – andselisk Jun 17 at 16:25

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