Identify the substrate

2-Methyl propyl bromide means isobutyl bromide: $\mathrm{CH_3-CH(CH_3)-CH_2Br}$

This is a primary alkyl bromide.

Reaction with $\mathrm{C_2H_5O^-}$

Ethoxide ion is a strong nucleophile. With a primary alkyl halide, the reaction proceeds mainly by $\mathrm{S_N2}$ mechanism.

So substitution occurs at the carbon bearing bromine: $\mathrm{CH_3-CH(CH_3)-CH_2Br + C_2H_5O^- \rightarrow CH_3-CH(CH_3)-CH_2OC_2H_5}$

The product is isobutyl ethyl ether.

Hence,

• Mechanism for formation of $A$: $\mathrm{S_N2}$
• $A =$ iso-butyl ethyl ether

Reaction with $\mathrm{C_2H_5OH}$

Ethanol is a weak nucleophile and a polar protic solvent. In such conditions, ionization is favored. The initially formed primary carbocation is unstable: $\mathrm{CH_3-CH(CH_3)-CH_2^+}$

It undergoes 1,2-hydride shift to form the more stable tert-butyl carbocation: $\mathrm{(CH_3)_3C^+}$

Then ethanol attacks this carbocation: $\mathrm{(CH_3)_3C^+ + C_2H_5OH \rightarrow (CH_3)_3C-OC_2H_5}$

So the reaction proceeds through $\mathrm{S_N1}$ with rearrangement, giving tert-butyl ethyl ether.

Hence,

• Mechanism for formation of $B$: $\mathrm{S_N1}$
• $B =$ tert-butyl ethyl ether

Match with options

We found:

• $A$: $\mathrm{S_N2}$, iso-butyl ethyl ether
• $B$: $\mathrm{S_N1}$, tert-butyl ethyl ether

This matches Option C.

Comparison with stored answer

Stored correct answer = C

Our derived answer = C

Therefore, they agree.