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Order of Reactivity of Halogens in Electrophilic Addition
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Michelle Roy
Order of Reactivity of Halogens in Electrophilic Addition
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.
References:
Peter Sykes; A Guidebook to Mechanism in Organic Chemistry 6th ed.; 2003ISBN-10:8177584332
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.
References:
Peter Sykes; A Guidebook to Mechanism in Organic Chemistry 6th ed.; 2003ISBN-10:8177584332
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
ans is d Normally not possible to F2 because it is highly exothermic with I2 reversible reaction takes placedue to more E.N of Cl2 compare to Br2. It does not give e pair to carbocation
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.
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.
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.$181$-$182$]:
As is clear $\ce{I2}$ will be more reactive than $\ce{F2}$.
Hence, the correct answer is D.
References:
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.$181$-$182$]:
As is clear $\ce{I2}$ will be more reactive than $\ce{F2}$.
Hence, the correct answer is D.
References:
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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)
Ans is (D) Since F+ is not possible andin case. of I2 it will be reversibleso In I2 &F2 the order is I2>F2So ans is (D)
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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
ans is d Normally not possible to F2 because it is highly exothermic with I2 reversible reaction takes placedue to more E.N of Cl2 compare to Br2. It does not give e pair to carbocation
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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.
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.
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