Question
The increasing order of stability of the following free radicals is
A.
$${\left( {C{H_3}} \right)_2}\dot CH < {\left( {C{H_3}} \right)_3}\dot C$$ $$ < {\left( {{C_6}{H_5}} \right)_2}\dot CH < {\left( {{C_6}{H_5}} \right)_3}\dot C$$
B.
$${\left( {{C_6}{H_5}} \right)_3}\dot C < {\left( {{C_6}{H_5}} \right)_2}\dot CH < $$ $${\left( {C{H_3}} \right)_3}\dot C < {\left( {C{H_3}} \right)_2}\dot CH$$
C.
$${\left( {{C_6}{H_5}} \right)_2}\dot CH < {\left( {{C_6}{H_5}} \right)_3}\dot C$$ $$ < {\left( {C{H_3}} \right)_3}\dot C < {\left( {C{H_3}} \right)_2}\dot CH$$
D.
$${\left( {C{H_3}} \right)_2}\dot CH < {\left( {C{H_3}} \right)_3}\dot C < $$ $$ {\left( {{C_6}{H_5}} \right)_3}\dot C < {\left( {{C_6}{H_5}} \right)_2}\dot CH$$
Answer :
$${\left( {C{H_3}} \right)_2}\dot CH < {\left( {C{H_3}} \right)_3}\dot C$$ $$ < {\left( {{C_6}{H_5}} \right)_2}\dot CH < {\left( {{C_6}{H_5}} \right)_3}\dot C$$
Solution :
On the basis of hyperconjugation, the order of stability of free radicals is as follows $${3^ \circ } > {2^ \circ } > {1^ \circ }.$$ Benzyl free radicals are stabilised by resonance and hence are more stable than alkyl free radicals. More the number of phenyl groups attached to the carbon atom, more is the stability of free radical.