In $\mathrm{S_N1}$ mechanism alkyl halides need to be ionized, so wouldn't it make more sense if alkyl fluoride reactivity were the most, since it is the most electronegative element?
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1$\begingroup$ chemistry.stackexchange.com/questions/6808/… chemistry.stackexchange.com/questions/55829/… $\endgroup$– MithoronJan 2, 2018 at 15:28
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1$\begingroup$ chemistry.stackexchange.com/questions/76631/… $\endgroup$– MithoronJan 2, 2018 at 15:58
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$\begingroup$ Properties of a neutral fluorine atom do not necessarily extend to a fluoride anion. $\endgroup$– ZheJan 2, 2018 at 17:01
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1 Answer
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I can see why on first thoughts you would think fluorine is going to be the most reactive of the halogens, this may be due to the fact you always think of fluorine as the most electronegative; and therefore the most comfortable carrying a negative charge.
However, electronegativity is only sometimes a good indicator of the stability of the leaving group/conjugate base (i.e the F-, Cl-, Br-, I- leaving groups in a typical SN1 reaction). The reason why iodine functional groups are more reactive is purely because it's more stable when carrying that negative charge after it has left. This is because iodine is a much much larger anion and can delocalise that negative charge over a much larger area.
So normally the rule for good leaving groups, or when you are looking at acidity (HF,HCl,HBr,HI), is that the more stable the leaving group/conjugate base, the more reactive that functional group is going to be. There are other factors to consider when looking at different leaving groups, but in the case of the halogens, the increasing size of the anion, and therefore its overall stability is the determining factor in its leaving group ability..
