What is the order of reactivity of halogens in electrophilic addition reaction via formation of cyclohalonium intermediate?
(A) $\ce{F2}$> $\ce{Cl2}$> $\ce{Br2}$> $\ce{I2}$
(B) $\ce{I2}$> $\ce{Br2}$> $\ce{Cl2}$> $\ce{F2}$
(C) $\ce{I2}$> $\ce{Br2}$> $\ce{F2}$> $\ce{Cl2}$
(D) $\ce{Br2}$> $\ce{Cl2}$> $\ce{I2}$> $\ce{F2}$

I tried to predict it using the mechanism, but couldn't reach a conclusion. Is this based on thermodynamic and kinetic data?

Here's the mechanism involving bromine, for reference:

enter image description here

Could someone provide a detailed explanation for the above problem, with data if possible?

  • $\begingroup$ related chemistry.stackexchange.com/questions/41130/… $\endgroup$ – Mithoron Jan 3 '18 at 16:16
  • $\begingroup$ Good to learn something new, but it didn't answer my question. $\endgroup$ – epsilon-emperor Jan 3 '18 at 16:18
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    $\begingroup$ Edit your question to remove salutation/thanks etc. Also add what you think about possible answer. This is rather important question so it should be rather improved then tempt to be closed as homework. $\endgroup$ – Mithoron Jan 3 '18 at 16:22

The rate-determining step of halogenation of alkenes is the formation of cyclic intermediate.

The cyclohalonium intermediate of bromine will be more stable than that of chlorine owing to its lower electronegativity.

The following paragraph is taken from Peter Sykes [$1$, p.$181$-$182$]:

It is not normally possible to add fluorine directly to alkenes as the reaction is so exothermic that bond fission occurs. Many alkenes will not add iodine directly either and when the reaction does occur it is usually readily reversible.

As is clear $\ce{I2}$ will be more reactive than $\ce{F2}$.

Hence, the correct answer is D.


  1. Peter Sykes; A Guidebook to Mechanism in Organic Chemistry 6th ed.; 2003 ISBN-10:8177584332

Ans is (D) Since F+ is not possible and in case. of I2 it will be reversible so In I2 &F2 the order is I2>F2 So ans is (D)

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    $\begingroup$ So how does Fluorine react ? $\endgroup$ – Maurice Mar 22 '20 at 10:20

I would say (B).

The mechanism involves the breaking of the Halogen–Halogen bond. The strength of this bond decreases down the group because atomic radius increases. The attraction of the halogen nuclei on the electrons in the covalent bond decreases as the valence electrons are further away from the nucleus.

Therefore, the I–I bond is weakest while F–F is strongest. Hence, I2 is most reactive and F2 is least.


ans is d Normally not possible to F2 because it is highly exothermic with I2 reversible reaction takes place due to more E.N of Cl2 compare to Br2. It does not give e pair to carbocation


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