In the following reaction sequence, the product Y is
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- $\text{NaNH}_2$ deprotonates terminal alkynes, and when an alkyl halide is present in the same molecule, it readily undergoes intramolecular cyclization to form a ring.
- Lindlar's catalyst always performs selective syn-addition of hydrogen to convert alkynes into cis-alkenes.
Step 1: Understanding the Question:
We need to determine the final product Y obtained from a two-step reaction starting with $\text{Br}(\text{CH}_2)_{12}-\text{C} \equiv \text{CH}$.
Step 2: Detailed Explanation:
Let us analyze each step of the reaction:
- Step 1: Reaction of $\text{Br}(\text{CH}_2)_{12}-\text{C} \equiv \text{CH}$ with sodium amide ($\text{NaNH}_2$):
$\text{NaNH}_2$ is an extremely strong base which selectively deprotonates the acidic terminal alkyne hydrogen to generate a highly reactive acetylide anion:
\[ \text{Br}(\text{CH}_2)_{12}-\text{C} \equiv \text{C}^- \]
Since the molecule contains an alkyl bromide group at the other end of the chain, this nucleophilic carbanion undergoes an intramolecular nucleophilic substitution ($\text{S}_\text{N}2$) reaction.
The carbanion attacks the carbon bonded to the bromine atom, displacing the bromide ion and forming a cyclic alkyne.
Let us count the ring size:
The chain contains $12$ methylene carbons and $2$ alkyne carbons, making a total of $14$ carbons in the cyclic ring.
Thus, the product X is cyclotetradecyne (a 14-membered cyclic alkyne).
- Step 2: Reaction of X with $\text{H}_2$ in the presence of Lindlar's catalyst:
Lindlar's catalyst ($\text{Pd/CaCO}_3/\text{quinoline}$) selectively reduces alkynes to alkenes.
Because the hydrogenation occurs via syn-addition on the metal surface, it selectively yields the cis-alkene isomer.
Therefore, the 14-membered cyclic alkyne X is reduced to cis-cyclotetradecene (Y).
Comparing with the options, option (B) represents the cis-alkene structure where both hydrogen atoms are on the same side of the double bond.
