HCl does not follow the anti-Markovnikov rule | vedclass

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Question

(C) The anti-Markovnikov addition (peroxide effect) is only observed with $HBr$ in the presence of organic peroxides. This is because the $H-Cl$ bond

Vedclass · Accepted Answer

Its bond dissociation energy is high.

Vedclass · Answer

(C) The anti-Markovnikov addition (peroxide effect) is only observed with $HBr$ in the presence of organic peroxides. This is because the $H-Cl$ bond is very strong $(431 \ kJ \ mol^{-1})$,and the enthalpy change for the homolytic cleavage of $HCl$ by free radicals is endothermic,making the reaction energetically unfavorable. In contrast,the $H-Br$ bond is weaker,allowing for the formation of bromine free radicals which facilitate the anti-Markovnikov addition.

$HCl$ does not follow the anti-Markovnikov rule because...

Similar Questions

The number of $sp^3$ hybridized carbon atoms in cyclohexene is

Arrange the following compounds in the decreasing order of their reactivity towards electrophilic addition of $HBr$:
$(a)$ $CH_3-C(CH_3)=CH_2$
$(b)$ $CH_3O-C(CH_3)=CH_2$
$(c)$ $CH_3OCH_2-C(CH_3)=CH_2$

$HCl$ 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?

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An organic compound $'A'$ $(C_4H_8)$ on treatment with $KMnO_4 / H^{+}$ yields compound $'B'$ $(C_3H_6O)$. Compound $'A'$ also yields compound $'B'$ on ozonolysis. Compound $'A'$ is:

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