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

${\text{S}}_{\mathrm{N}}2,\mathrm{A} =$iso – butyl ethyl ether; ${\mathrm{S}}_{\mathrm{N}}\mathrm{1,}\mathrm{B}=$ tert – butyl ethyl ether

${\mathrm{S}}_{\mathrm{N}}\mathrm{1,}\mathrm{A} =$tert-butyl ethyl ether;  ${\mathrm{S}}_{\mathrm{N}}\mathrm{2,}\mathrm{B} =$iso-butyl ethyl ether

${\mathrm{S}}_{\mathrm{N}}\mathrm{1,}\mathrm{A} =$ tert-butyl ethyl ether; ${\mathrm{S}}_{\mathrm{N}}\mathrm{1,}\mathrm{B} =$ 2- butyl ethyl ether

${\mathrm{S}}_{\mathrm{N}}\mathrm{2,}\mathrm{A} = 2-$$\mathrm{butyl} \mathrm{ethyl} \mathrm{ether};$${\mathrm{S}}_{\mathrm{N}}\mathrm{2,}\mathrm{B}=$iso- butyl ethyl ether