X and Y in the above-mentioned reaction are respectively:

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Accepted Answer

1. **Missing Data Analysis:** The specific reaction scheme is not provided in the text, but the question asks to identify reactants/products X and Y. Based on the options and standard alkyne chemistry, the reaction likely involves the conversion of a lower alkyne (X) to a higher alkyne (Y).
2. **Chemical Principle (Acidity of Alkynes):** Terminal alkynes (like 1-Butyne) possess an acidic hydrogen atom attached to the $sp$ hybridised carbon atom. They react with strong bases like sodamide ($NaNH_2$) to form stable sodium alkynides [NCERT 11th, Hydrocarbons, Sec 13.4.4; Source 136, 303].
   $$ R-C \equiv C-H + NaNH_2 \rightarrow R-C \equiv C^- Na^+ + NH_3 $$
   Internal alkynes (like 2-Butyne) do not have this acidic hydrogen and do not undergo this reaction.
3. **Synthesis of Higher Alkynes:** The sodium alkynide formed can react with primary alkyl halides (e.g., Ethyl bromide, $C_2H_5Br$) via nucleophilic substitution to form higher alkynes.
   $$ CH_3CH_2-C \equiv C^- Na^+ + C_2H_5Br \rightarrow CH_3CH_2-C \equiv C-CH_2CH_3 + NaBr $$
4. **Deduction:** 
   - For X to react, it must be a terminal alkyne. This suggests X is **1-Butyne** ($CH_3CH_2-C \equiv CH$), eliminating 2-Butyne options.
   - If 1-Butyne is alkylated with an ethyl group (to increase carbon count from C4 to C6), the product is **3-Hexyne** ($CH_3CH_2-C \equiv C-CH_2CH_3$).
   - Option 4 matches this sequence.

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