What is the correct order of bond dissociatio | vedclass

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(A) The bond dissociation energy $(BDE)$ is inversely proportional to the stability of the free radical formed after the homolytic cleavage of the $C-

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$IV > I > II > III$

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(A) The bond dissociation energy $(BDE)$ is inversely proportional to the stability of the free radical formed after the homolytic cleavage of the $C-H$ bond.$I. \, CH_3-H \rightarrow \cdot CH_3$ (Methyl radical,least stable)$II. \, CH_3CH_2-H \rightarrow CH_3\cdot CH_2$ (Primary radical,more stable than methyl)$III. \, CH_2=CH-CH_2-H \rightarrow CH_2=CH-\cdot CH_2$ (Allyl radical,resonance stabilized,very stable)$IV. \, C_6H_5-H \rightarrow \cdot C_6H_5$ (Phenyl radical,$sp^2$ hybridized carbon,least stable due to high $s$-character)Comparing the stability of radicals: $Allyl > Ethyl > Methyl > Phenyl$.Therefore,the order of $BDE$ is: $Phenyl > Methyl > Ethyl > Allyl$.Thus,the correct order is $IV > I > II > III$.

What is the correct order of bond dissociation energy for the following $C-H$ bonds?
$I. \, CH_3-H$
$II. \, CH_3CH_2-H$
$III. \, CH_2=CH-CH_2-H$
$IV. \, C_6H_5-H$

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What is the correct order of bond dissociation energy for the following $C-H$ bonds?$I. \, CH_3-H$$II. \, CH_3CH_2-H$$III. \, CH_2=CH-CH_2-H$$IV. \, C_6H_5-H$