Why is Ti(bpy) (CH3)4 Much more thermally stable than Ti(CH3)4?

In Ti(CH3)4 the metal is coordinatively unstaurated and the coordination sites required for decomposition reaction are avaliable (vacant) . There is easy pathway for thermodynamically possible decomposition reaction to occour. Hence Ti(CH3)4 is thermaly unstable, But in Ti(bpy)(CH3)4 has high thermal stablity due to the reasons that the coordination sites required for decomposition reaction to proceed are blocked.