HCl does not show the peroxide effect because | vedclass

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(B) The peroxide effect (Kharasch effect) involves a free radical mechanism.For $HCl$,the step involving the abstraction of $H$ from $HCl$ by the alky

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$I$ is exothermic and $II$ is endothermic

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(B) The peroxide effect (Kharasch effect) involves a free radical mechanism.For $HCl$,the step involving the abstraction of $H$ from $HCl$ by the alkyl radical $(CH_3-\dot{C}H-CH_2Cl + HCl 	o CH_3-CH_2-CH_2Cl + \dot{Cl})$ is endothermic because the $H-Cl$ bond is very strong $(431 \ kJ/mol)$.Additionally,the addition of the chlorine radical to the alkene is exothermic,but the overall process is not favorable due to the high energy required to break the $H-Cl$ bond.In contrast,for $HBr$,both steps are exothermic,allowing the reaction to proceed.Therefore,$I$ is exothermic and $II$ is endothermic for $HCl$.

$HCl$ does not show the peroxide effect because the reaction mechanism involves the following steps:
$CH_3-CH=CH_2 + \dot{X} \to CH_3-\dot{C}H-CH_2X$ $(I)$
$CH_3-\dot{C}H-CH_2X + HX \to CH_3-CH_2-CH_2X + \dot{X}$ $(II)$
Why does $HCl$ fail to undergo this reaction?

Similar Questions

Which of the following alkenes yield $3-$bromo$-3-$methylpentane as the major product upon addition of $HBr$?
$(I)$ $3-$methylpent$-2-$ene
$(II)$ $3-$methylpent$-2-$ene (isomer)
$(III)$ $3-$ethylpent$-1-$ene
$(IV)$ $3-$methylpent$-1-$ene

Alkenes convert into alcohols by:

The reaction of $1-$ethylcyclopentene with $BH_3 / THF$ followed by treatment with $H_2O_2 / NaOH$ produces:

What are the products obtained on hydroboration-oxidation of the given alkene?

The compound $CH_3-C(CH_3)=CH-CH_3$ on reaction with $NaIO_4$ in the presence of $KMnO_4$ gives:

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$HCl$ does not show the peroxide effect because the reaction mechanism involves the following steps:$CH_3-CH=CH_2 + \dot{X} \to CH_3-\dot{C}H-CH_2X$ $(I)$$CH_3-\dot{C}H-CH_2X + HX \to CH_3-CH_2-CH_2X + \dot{X}$ $(II)$Why does $HCl$ fail to undergo this reaction?