Step 1: Understanding the Concept:
The starting material is a terminal alkyne. The final product is a substituted alkyne with a two-carbon chain extension ending in a bromine atom.
Terminal alkynes can be deprotonated by strong bases to form acetylide ions, which are excellent nucleophiles. These nucleophiles can then undergo substitution reactions with alkyl halides to extend the carbon chain. The challenge here is to perform the substitution while keeping a bromine atom intact at the end of the new chain.
Step 2: Detailed Explanation:
1. Reagent X: To deprotonate the terminal alkyne, we need an exceptionally strong base. Sodium amide ($NaNH_2$) is the standard reagent for this purpose. It removes the acidic terminal proton to form the sodium acetylide: $RO-CH_2-C\equiv C^-Na^+$.
2. Reagent Y: This acetylide ion must now act as a nucleophile and attack a dihaloalkane to introduce the ethyl group.
- We want the final group to be $-CH_2CH_2-Br$.
- Therefore, we must use an unsymmetrical dihalide where one halogen is a much better leaving group than bromine.
- The leaving group ability order is $I^->Br^->Cl^->F^-$.
- If we use $I-CH_2-CH_2-Br$ (Option B), the nucleophilic acetylide will selectively attack the carbon bearing the better leaving group (Iodine), displacing $I^-$ and leaving the $-Br$ group untouched at the terminal position.
- If we used Option (A) $Br-CH_2-CH_2-Cl$, the alkyne would attack the carbon with bromine (since $Br$ is a better leaving group than $Cl$), resulting in a terminal chloride, which is not desired.
Step 3: Final Answer:
The reagents required are $NaNH_2$ for deprotonation and $I-CH_2CH_2-Br$ for selective alkylation. The correct option is (B).