Consider the above reaction sequence, Product "A" and Product "B" formed respectively are:

Product $$\mathrm{A}$$ is formed by acetalisation of the aldehyde group using ethanol and dry $$\mathrm{HCl}$$.

The primary bromide remains unaffected under these acidic conditions.

Product $$\mathrm{A}$$ is:

$$\mathrm{Br-CH_2-CH(OEt)_2}$$

Product $$\mathrm{B}$$ is formed by $$\mathrm{E2}$$ dehydrohalogenation using potassium tert-butoxide.

The strong bulky base removes $$\mathrm{HBr}$$ to form a double bond.

Product $$\mathrm{B}$$ is:

$$\mathrm{H_2C=C(OEt)_2}$$

Correct Option: $$\mathrm{A}$$