Draw the organic product of the following reaction.
- You do not have to consider stereochemistry.
- You do not have to explicitly draw $\mathrm{H}$ atoms.

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∻Solution∻ ‖ 1 ⋮ **Identify the starting material**: The organic compound described is 1-butyne, which has the structure $CH_3C\equiv CCH_3$. ‖ ‖ 2 ⋮ **First reaction step with NaNH2**: Sodium amide, NaNH2, is a strong base that will deprotonate the terminal alkyne, creating a sodium acetylide anion. The reaction is $CH_3C\equiv CCH_3 + NaNH_2 \rightarrow CH_3C\equiv C^-Na^+ + NH_3$. ‖ ‖ 3 ⋮ **Second reaction step with CH3CH2Br**: The sodium acetylide anion will act as a nucleophile and attack the primary alkyl halide, bromoethane, in an SN2 reaction, displacing the bromide ion and forming a new carbon-carbon bond. The reaction is $CH_3C\equiv C^-Na^+ + CH_3CH_2Br \rightarrow CH_3C\equiv CCH_2CH_3 + NaBr$. ‖ ∻Answer∻ ⚹ The organic product of the following reaction is pent-1-yne, $CH_3C\equiv CCH_2CH_3$. ⚹ ∻Key Concept∻ ⚹ Alkyne deprotonation and subsequent nucleophilic substitution ⚹ ∻Explanation∻ ⚹ A terminal alkyne is deprotonated by a strong base to form an acetylide anion, which then acts as a nucleophile in an SN2 reaction with a primary alkyl halide to form a new carbon-carbon bond. ⚹◊What is the chemical equation for the reaction involving NaNH2 and CH3CH2Br as reagents?⍭ Generate me a similar question◊

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