In the structure CH_3 - C^2H(C^1H_3) - C^3H_2 | vedclass

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(D) Homolytic bond dissociation energy $(BDE)$ is inversely proportional to the stability of the free radical formed after the bond breaks. Stability

Vedclass · Accepted Answer

$C_1-H > C_4-H > C_3-H > C_2-H$

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(D) Homolytic bond dissociation energy $(BDE)$ is inversely proportional to the stability of the free radical formed after the bond breaks. Stability of free radicals follows the order: $3^\circ > 2^\circ > 1^\circ$. In the given molecule (isopentane): $1.$ $C_2-H$ is a $3^\circ$ bond (forms a $3^\circ$ radical,which is the most stable). $2.$ $C_3-H$ is a $2^\circ$ bond (forms a $2^\circ$ radical). $3.$ $C_1-H$ and $C_4-H$ are $1^\circ$ bonds (form $1^\circ$ radicals,which are the least stable). Among the $1^\circ$ bonds,the $C_4$ radical is slightly more stable than the $C_1$ radical because $C_4$ has two $\beta$-hydrogens available for hyperconjugation,while $C_1$ has only one. Therefore,the $BDE$ order is: $C_1-H > C_4-H > C_3-H > C_2-H$.

In the structure $CH_3 - C^2H(C^1H_3) - C^3H_2 - C^4H_3$,the order of homolytic bond dissociation energy for $C_1-H, C_2-H, C_3-H$ and $C_4-H$ is:

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