In the given reaction 3-Bromo-2, 2-dimethyl butane $$\xrightarrow{C_2H_5OH}$$ 'A' (Major Product). Product A is:

The starting material is 3-bromo-2,2-dimethylbutane. Numbering the butane backbone from left to right (C1-C2-C3-C4), carbon C2 bears two methyl substituents and carbon C3 bears the bromine. The explicit structure is $$\text{CH}_3\text{-C(CH}_3)_2\text{-CHBr-CH}_3$$, which has a total of six carbons including the two methyl groups on C2.

When treated with ethanol (C₂H₅OH), a weak nucleophile and weak base, the reaction proceeds via an SN1 mechanism. The C-Br bond ionizes to give a secondary carbocation at C3: $$\text{CH}_3\text{-C(CH}_3)_2\text{-}\overset{+}{\text{CH}}\text{-CH}_3$$.

This secondary carbocation undergoes a 1,2-methyl shift from the adjacent C2 to the cationic C3, generating the more stable tertiary carbocation at C2: $$\text{CH}_3\text{-}\overset{+}{\text{C}}\text{(CH}_3)\text{-CH(CH}_3)\text{-CH}_3$$. In this rearranged cation, C2 now carries only one methyl group (the original one at C2 that did not migrate), the migrated methyl is at C3, and the positive charge sits at C2.

Ethanol attacks this tertiary carbocation at C2 as a nucleophile, and loss of a proton gives the ether product: $$\text{CH}_3\text{-C(OC}_2\text{H}_5\text{)(CH}_3)\text{-CH(CH}_3)\text{-CH}_3$$. This is 2-ethoxy-2,3-dimethylbutane, which corresponds to Option 3.