Hydrocarbons Part 2

Concept:The reaction converts a terminal alkyne into an internal alkyne via deprotonation followed by alkylation.Explanation:The starting material is 3‑methylbut‑1‑yne ($\mathrm{HC≡C–CH(CH_3)_2}$).It is treated with sodium amide ($\mathrm{NaNH_2}$) which removes the terminal hydrogen atom, forming an alkynide ion.The alkynide ion then undergoes an $S_N2$ reaction with ethyl bromide ($\mathrm{CH_3CH_2Br}$).The new triple bond is formed between the original terminal carbon and the ethyl group.The alkyl group adds to the end of the chain, giving a six‑carbon backbone with a triple bond at the 3‑position.The methyl branch from the starting material remains on carbon 2.The final IUPAC name of the product is 2‑methylhex‑3‑yne ($\mathrm{CH_3CH(CH_3)C≡CCH_2CH_3}$).Answer:A. 2‑methylhex‑3‑yne