$RO-CH_2-C\equiv CH \xrightarrow{X} \xrightarrow{Y} RO-CH_2-C\equiv C-CH_2CH_2-Br$
To carry out the above conversion X and Y are respectively
Reaction Overview
The question asks for the reagents X and Y needed for the conversion of a terminal alkyne, $RO-CH_2-C\equiv CH$, to a product with an extended chain and a bromine atom, $RO-CH_2-C\equiv C-CH_2CH_2-Br$. This involves two steps: deprotonation of the alkyne and subsequent alkylation.
Step 1: Deprotonation using Reagent X
Terminal alkynes possess a weakly acidic proton attached to the sp-hybridized carbon. Strong bases are required to remove this proton. Among the options provided, $NaNH_2$ (sodium amide) is a very strong base suitable for this purpose.
The reaction is:
$ RO-CH_2-C\equiv CH + NaNH_2 \rightarrow RO-CH_2-C\equiv C^- Na^+ + NH_3 $
This generates a nucleophilic acetylide anion ($RO-CH_2-C\equiv C^-$).
Step 2: Alkylation using Reagent Y
The nucleophilic acetylide anion attacks an electrophilic alkyl halide in a nucleophilic substitution reaction ($S_N2$). The goal is to form the $RO-CH_2-C\equiv C-CH_2CH_2-Br$ structure. This requires attaching a $-CH_2CH_2-Br$ group.
Let's analyze the structure of reagent Y from the correct option (Option B): $I-CH_2-CH_2-Br$.
- This molecule has two potential leaving groups: Iodide (I$^-$) and Bromide (Br$^-$).
- Iodide is a significantly better leaving group than Bromide due to the weaker C-I bond and the stability of the iodide ion.
- Therefore, the nucleophilic attack by the acetylide anion ($RO-CH_2-C\equiv C^-$) will preferentially occur at the carbon atom bonded to Iodine.
The reaction is:
$ RO-CH_2-C\equiv C^- Na^+ + I-CH_2-CH_2-Br \rightarrow RO-CH_2-C\equiv C-CH_2CH_2-Br + NaI $
This reaction yields the desired product.
Conclusion
Based on the analysis, reagent X is $NaNH_2$ and reagent Y is $I-CH_2-CH_2-Br$. This corresponds to Option B.
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