See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: Br—(CH2)12—C≡CH is a terminal alkyne with a bromine at the other end of a 12-methylene chain (14 carbons total including the alkyne carbons). Step 2 - Reaction with NaNH2 (Step A): NaNH2 is a strong base that deprotonates the terminal alkyne (pKa ~25) to form the acetylide carbanion: Br—(CH2)12—C≡C⁻ Na⁺. This acetylide then undergoes intramolecular SN2 displacement of the bromide by the carbanion end, forming a cyclic internal alkyne — cyclotetradec-1-yne (a 14-membered ring containing a triple bond). This is product (A). Step 3 - Reaction with Lindlar Catalyst (Step B): Lindlar catalyst (Pd/CaCO3 poisoned with lead acetate and quinoline) performs syn-selective partial hydrogenation of the internal alkyne. Syn addition of H2 across the triple bond in the cyclic alkyne gives a cis (Z) cycloalkene. In a cyclic system, syn addition means both H atoms are added to the same face, producing a cis double bond — i.e., cis-cyclotetradec-1-ene, where the two ring chain segments are on the same side (Z configuration), and both newly added H atoms appear on opposite carbons of the double bond pointing to the same face. Step 4 - Match to options: (a) Shows a cycloalkane (no double bond) — incorrect, Lindlar gives alkene not alkane. (b) Shows a cycloalkene but the H atoms are both drawn on the same carbon of the double bond, suggesting a trans arrangement — incorrect for Lindlar. (c) Shows a large ring cycloalkene with H on the left vinyl carbon and H on the right vinyl carbon of the double bond, both pointing outward/upward — this represents the cis (Z) internal cycloalkene, consistent with syn addition by Lindlar catalyst. (d) Is an open-chain bromoalkene — incorrect. Therefore, the correct answer is C.