Homolytic fission of the following alkanes fo | vedclass

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(B) The stability of free radicals is determined by the number of hyperconjugative structures and inductive effects. Primary $(1^\circ)$ radicals are

Vedclass · Accepted Answer

$CH_3-\dot{C}H_2 < CH_3-\dot{C}H-CH_3 < (CH_3)_2\dot{C}-CH_2-CH_3 < (CH_3)_3\dot{C}$

Vedclass · Answer

(B) The stability of free radicals is determined by the number of hyperconjugative structures and inductive effects. Primary $(1^\circ)$ radicals are less stable than secondary $(2^\circ)$ radicals,which are less stable than tertiary $(3^\circ)$ radicals. $1$. $CH_3-\dot{C}H_2$ (Primary) $2$. $CH_3-\dot{C}H-CH_3$ (Secondary) $3$. $(CH_3)_2\dot{C}-CH_2CH_3$ (Secondary,but more substituted) $4$. $(CH_3)_3\dot{C}$ (Tertiary) Stability order: $1^\circ Thus,the increasing order is $CH_3-\dot{C}H_2 < CH_3-\dot{C}H-CH_3 < (CH_3)_2\dot{C}-CH_2CH_3 < (CH_3)_3\dot{C}$.

Homolytic fission of the following alkanes forms free radicals: $CH_3-CH_3$,$CH_3-CH_2-CH_3$,$(CH_3)_2CH-CH_3$,and $CH_3-CH_2-CH(CH_3)_2$. The increasing order of stability of the resulting free radicals is:

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