The bond dissociation energy of the C-H bond | vedclass

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

Vedclass · Accepted Answer

$IV > I > II > III$

Vedclass · Answer

(C) The bond dissociation energy is inversely proportional to the stability of the free radical formed after the homolytic cleavage of the $C-H$ bond.$1$. The radicals formed are:$(I)\ CH_3^{\bullet}$ (Methyl radical)$(II)\ CH_3-CH_2^{\bullet}$ (Primary ethyl radical)$(III)\ CH_2=CH-CH_2^{\bullet}$ (Allyl radical,resonance stabilized)$(IV)\ C_6H_5^{\bullet}$ (Phenyl radical,$sp^2$ hybridized carbon)$2$. Stability order of radicals: $(III) > (II) > (I) > (IV)$.$3$. Since bond dissociation energy is inversely proportional to radical stability,the order of energy required is: $(IV) > (I) > (II) > (III)$.

The bond dissociation energy of the $C-H$ bond for the compounds: $(I)\ CH_3-H, (II)\ CH_3-CH_2-H, (III)\ CH_2=CH-CH_2-H, (IV)\ C_6H_5-H$ follows the order:

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