Identify the correct set of reagents or reaction conditions 'X' and 'Y' in the following set of transformation: 

For an alkyl (or aryl-alkyl) halide to be converted first into an alkene and then back into a (different) halo-compound, two distinct reaction types are required:
• Step X must remove HX from the substrate (β-elimination, $$E2$$).
• Step Y must add HX to the new double bond (electrophilic addition).

Step X - β-elimination
The classical reagent for one-step dehydrohalogenation is a concentrated alcoholic solution of a strong base such as $$NaOH$$ or $$KOH$$, heated to about $$80^{\circ} \mathrm{C}$$.
Reaction: $$RCH_2CH_2Br \xrightarrow[\;80^{\circ}C\;]{conc.\,alc.\,NaOH} RCH{=}CH_2 + HBr$$

Step Y - electrophilic addition of HBr
Once an alkene is obtained, addition of hydrogen bromide across the C=C bond is required. Using dry $$HBr$$ dissolved in an anhydrous polar solvent such as acetic acid ensures straightforward Markovnikov addition without competing radical side reactions.
Reaction: $$RCH{=}CH_2 \xrightarrow[\,AcOH\,]{HBr} RCHBrCH_3$$

Eliminating the wrong choices
Option A (dil. aq. $$NaOH$$, $$20^{\circ}C$$) performs nucleophilic substitution $$\left(S_N2\right)$$ to give alcohols, not elimination.
Option B (conc. alc. $$NaOH$$, $$80^{\circ}C$$) is correct for X, but $$Br_2/CHCl_3$$ in Y would give vic-dibromides (anti-addition of $$Br_2$$), not $$HBr$$ addition.
Option C (dil. aq. $$NaOH$$, $$20^{\circ}C$$) again gives an alcohol in X, so the sequence fails.
Option D matches both mechanistic requirements: strong ethanolic base for elimination and $$HBr/AcOH$$ for addition.

Hence the correct set is
Option D which is: X = conc. alc. NaOH, 80 °C; Y = HBr/acetic acid.