2-Methyl propyl bromide reacts with C$$_2$$H$$_5$$O$$^-$$ and gives 'A' whereas on reaction with C$$_2$$H$$_5$$OH it gives 'B'. The mechanism followed in these reactions and the products 'A' and 'B' respectively are:

We are given that 2-methyl propyl bromide (isobutyl bromide, (CH$$_3$$)$$_2$$CHCH$$_2$$Br) reacts with C$$_2$$H$$_5$$O$$^-$$ to give product A, and with C$$_2$$H$$_5$$OH to give product B.

Reaction with C$$_2$$H$$_5$$O$$^-$$ (ethoxide ion — strong nucleophile):

Ethoxide is a strong nucleophile. With a primary alkyl halide, the reaction follows an S$$_N$$2 mechanism. In S$$_N$$2, the nucleophile attacks the carbon bearing the leaving group directly via backside attack, with no carbocation intermediate and no rearrangement.

$$(CH_3)_2CHCH_2Br + C_2H_5O^- \xrightarrow{S_N2} (CH_3)_2CHCH_2OC_2H_5$$

Product A = iso-butyl ethyl ether.

Reaction with C$$_2$$H$$_5$$OH (ethanol — weak nucleophile/polar protic solvent):

Ethanol is a weak nucleophile and a polar protic solvent, which favors the S$$_N$$1 pathway. Although 2-methyl propyl bromide is a primary halide, the initially formed primary carbocation undergoes a 1,2-hydride shift to form a more stable tertiary carbocation:

$$(CH_3)_2CHCH_2^+ \xrightarrow{1,2\text{-hydride shift}} (CH_3)_3C^+$$

The stable tertiary carbocation then reacts with ethanol (acting as nucleophile) to form:

Product B = tert-butyl ethyl ether.

This matches Option A: S$$_N$$2, A = iso-butyl ethyl ether; S$$_N$$1, B = tert-butyl ethyl ether.