Residue (A) + HCl (dil)$$\rightarrow $$ Compound (B) Structure of residue (A) and Compound (B) formed respectively is :

During the alkaline hydrolysis (saponification) of ethyl acetoacetate, the ester group is converted into its sodium salt while ethanol is set free.

Reaction: $$CH_3COCH_2COOCH_2CH_3 + NaOH \rightarrow CH_3COCH_2COONa + C_2H_5OH$$

After the reaction mixture is boiled, the volatile ethanol distils off. On evaporating the remaining liquid to dryness, the solid that remains is the sodium salt of acetoacetic acid.

Hence $$\text{Residue (A)} = CH_3COCH_2COONa$$    (sodium acetoacetate)

When this residue is subsequently treated with dilute hydrochloric acid, simple proton exchange takes place to give the free β-keto-acid (acetoacetic acid) and sodium chloride.

$$CH_3COCH_2COONa + HCl_{(dil)} \rightarrow CH_3COCH_2COOH + NaCl$$

Therefore $$\text{Compound (B)} = CH_3COCH_2COOH$$    (acetoacetic acid)

Both structures together correspond to Option D.

Final answer: Option D which is: Residue (A) - sodium acetoacetate ($$CH_3COCH_2COONa$$) and Compound (B) - acetoacetic acid ($$CH_3COCH_2COOH$$).