A) Primary
B) Methyl
C) Secondary
D) Tertiary
Correct Answer: D
Solution :
\[\underset{\begin{smallmatrix} \\ {{3}^{o}} \end{smallmatrix}}{\mathop{C{{H}_{3}}}}\,-\underset{C{{H}_{3}}}{\overset{C{{H}_{3}}\,}{\mathop{\underset{|\,\,\,\,\,\,\,\,\,}{\overset{|\,\,\,\,\,\,\,\,\,}{\mathop{{{C}^{o}}\,\,>\,}}}\,}}}\,\underset{\,\,\,\,\,\,\,{{2}^{o}}}{\mathop{C{{H}_{3}}-\overset{C{{H}_{3}}\,}{\mathop{\overset{|\,\,\,\,\,\,\,\,\,}{\mathop{\underset{o\,\,\,\,\,\,\,\,}{\mathop{CH\,\,}}\,}}\,}}\,}}\,>\underset{\begin{matrix} {} \\ {{1}^{o}} \\ \end{matrix}}{\mathop{C{{H}_{3}}-\overset{o\,\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,}}\,\] Greater the no. of alkyl groups attached to the carbon atom carrying the odd electrons, greater is the delocalization of odd electron and hence more stable is the free radical.
You need to login to perform this action.
You will be redirected in
3 sec
