$Br(CH_2)_{12}-C\equiv CH \xrightarrow{NaNH_2} X \xrightarrow[\text{catalyst}]{H_2, \text{ Lindler}} Y$
Reaction Pathway Analysis
The reaction sequence involves two steps starting from a bromoalkyne.
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Step 1: Reaction with Sodium Amide ($NaNH_2$)
The starting material is 1-bromododeca-1-yne ($Br(CH_2)_{12}-C\equiv CH$). Sodium amide ($NaNH_2$) is a very strong base. It preferentially removes the most acidic proton, which is the terminal alkyne proton.
Reaction: $Br(CH_2)_{12}-C\equiv CH + NaNH_2 \rightarrow Br(CH_2)_{12}-C\equiv C^-Na^+ + NH_3$
The intermediate, X, is the sodium acetylide: $Br(CH_2)_{12}-C\equiv C^-Na^+$.
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Step 2: Reduction with $H_2$/Lindlar Catalyst
Intermediate X is then treated with hydrogen gas ($H_2$) and Lindlar's catalyst. Lindlar's catalyst is specifically used for the partial reduction of alkynes to cis-alkenes. For a terminal alkyne ($R-C\equiv CH$), this reduction yields a terminal alkene ($R-CH=CH_2$).
Assuming the alkyne is reduced (either directly or after transient protonation):
Reaction: $Br(CH_2)_{12}-C\equiv CH + H_2 \xrightarrow{\text{Lindlar catalyst}} Br(CH_2)_{12}-CH=CH_2$
The final product, Y, is the terminal alkene $Br(CH_2)_{12}-CH=CH_2$. Terminal alkenes do not exhibit cis/trans isomerism.
Product Identification
The structure of the product Y is $Br(CH_2)_{12}-CH=CH_2$. This structure is represented by the image in Option 2 and the text in Option 3. Based on the provided correct answer, Option 2 is the correct choice.
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