The stability of the carbocations follows the order:

Arrange the following free radicals in the correct order of their stability:
$(i)$ $CH_2=\dot{C}H$
$(ii)$ $\dot{C}H_3$
$(iii)$ $CH_3-\dot{C}H-CH_3$
$(iv)$ $(CH_3)_3\dot{C}$

$CH_3-C(CH_3)_2-CH_2-CH_2^+ \to CH_3-C^+(CH_3)-CH(CH_3)-CH_3$
How many $H^-$ shifts are involved in the above rearrangement?

The major products $A$ and $B$ in the following reactions are:

Arrange the carbanions,$(CH_3)_3\overline{C}$,$\overline{C}Cl_3$,$(CH_3)_2\overline{C}H$,$C_6H_5\overline{C}H_2$ in order of their decreasing stability:

Which of the following is most stable?