The addition of HI in the presence of peroxide catalyst does not follow anti-Markovnikov’s rule because :

Question

The addition of HI in the presence of peroxide catalyst does not follow anti-Markovnikov’s rule because : (A) HI is a strong reducing agent (B) H-I bond is too strong to be broken homolytically (C) I atom combines with H atom to give back HL (D) Iodine atom is not reactive enough to add across a double bond

Tardigrade Admin · Accepted Answer

HI does not exhibit peroxide effect. HI bond although dissociates easily into iodine radicals, they being bigger in size are not much reactive but recombine together to form iodine molecule.

Q. The addition of $H I$ in the presence of peroxide catalyst does not follow anti-Markovnikov’s rule because :

$H I$ is a strong reducing agent

$H - I$ bond is too strong to be broken homolytically

$I$ atom combines with $H$ atom to give back $H L$

Iodine atom is not reactive enough to add across a double bond

HI does not exhibit peroxide effect. HI bond although dissociates easily into iodine radicals, they being bigger in size are not much reactive but recombine together to form iodine molecule.

Q. The addition of HI in the presence of peroxide catalyst does not follow anti-Markovnikov’s rule because :

HI is a strong reducing agent

H−I bond is too strong to be broken homolytically

I atom combines with H atom to give back HL

Iodine atom is not reactive enough to add across a double bond

HI does not exhibit peroxide effect. HI bond although dissociates easily into iodine radicals, they being bigger in size are not much reactive but recombine together to form iodine molecule.

Q. The addition of $H I$ in the presence of peroxide catalyst does not follow anti-Markovnikov’s rule because :

$H I$ is a strong reducing agent

$H - I$ bond is too strong to be broken homolytically

$I$ atom combines with $H$ atom to give back $H L$