Rank the following radicals in order of decreasing stability, putting the most stable first. $\mathrm { CH } _ { 3 } \dot { \mathrm { C } } \mathrm { H } _ { 2 }\quad$ $\mathrm { H } _ { 2 } \mathrm { C } = \mathrm { C } \dot { \mathrm { C } } \mathrm { H } _ { 2 }\quad$ $\mathrm { CH } _ { 3 } \dot { \mathrm { C } } \mathrm { HCH } _ { 3 }\quad$ $\left( \mathrm { CH } _ { 3 } \right) _ { 3 } \dot { \mathrm { C } }$
I$\quad\quad\quad\quad\quad\quad\quad$II$\quad\quad\quad\quad\quad\quad\quad$III$\quad\quad\quad\quad\quad$IV

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Quizplus · Accepted Answer

The stability of radicals increases with the number of alkyl groups attached to the radical carbon. The more alkyl groups, the more stable the radical. Therefore, the order of decreasing stability is:IV > III > II > I

Rank the Following Radicals in Order of Decreasing Stability, Putting CH3C˙H2\mathrm { CH } _ { 3 } \dot { \mathrm { C } } \mathrm { H } _ { 2 }\quadCH3​C˙H2​

Rank the Following Radicals in Order of Decreasing Stability, Putting $\mathrm { CH } _ { 3 } \dot { \mathrm { C } } \mathrm { H } _ { 2 }\quad$