Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

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

X = 2–Butyne; Y = 3–Hexyne

X = 2-Butyne; Y = 2-Hexyne

X = 1-Butyne; Y = 2-Hexyne

X = 1-Butyne; Y = 3–Hexyne

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).
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.
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. $CH_3CH_2-C \equiv C^- Na^+ + C_2H_5Br \rightarrow CH_3CH_2-C \equiv C-CH_2CH_3 + NaBr$
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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