1. In such haloalkane transformation questions, the two most common possibilities for reagent $X$ are:

   • dilute aqueous $\mathrm{NaOH}$ at low temperature $\Rightarrow$ nucleophilic substitution, forming an alcohol.
   • concentrated alcoholic $\mathrm{NaOH}$ at higher temperature $\Rightarrow$ $\beta$-elimination, forming an alkene.

2. For reagent $Y$:

   • $\mathrm{HBr}$/acetic acid adds across a double bond to give a bromoalkane.
   • $\mathrm{Br_2}/\mathrm{CHCl_3}$ also adds across a double bond, but gives a dibromo product.

3. Since the options are arranged in pairs, the intended transformation is clearly of the type: $\text{haloalkane} \xrightarrow[X]{} \text{alkene} \xrightarrow[Y]{} \text{bromoalkane}$ This means:

   • First step must be elimination $\Rightarrow$ use conc. alcoholic $\mathrm{NaOH}$, $80^\circ\mathrm{C}$.
   • Second step must convert the alkene back to a monobromo alkane, which is done by $\mathrm{HBr}$/acetic acid, not by $\mathrm{Br_2}/\mathrm{CHCl_3}$.

4. Therefore: $X=\text{conc. alc. }\mathrm{NaOH},\ 80^\circ\mathrm{C}$ $Y=\mathrm{HBr}/\text{acetic acid}$

5. Checking options:

   • A: substitution first, then $\mathrm{HBr}$ addition — not consistent.
   • B: elimination first, but $\mathrm{Br_2}$ would form dibromide — incorrect.
   • C: substitution first and then $\mathrm{Br_2}$ — incorrect.
   • D: elimination first, then $\mathrm{HBr}$ addition — correct.

Hence, the correct option is D.