2-Methyl propyl bromide reacts with $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}$$ and gives 'A' whereas on reaction with $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$$ it gives 'B'. The mechanism followed in these reactions and the products 'A' and 'B' respectively are :

The pair of lanthanides in which both elements have high third - ionization energy is :

In the reaction given below

'B' is

In the above reaction, left hand side and right hand side rings are named as '$$\mathrm{A}$$' and 'B' respectively. They undergo ring expansion. The correct statement for this process is: