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 starting material is a 4-carbon alkyne with the triple bond between the second and third carbons, known as 1-butyne. ‖ ‖ 2 ⋮ **First reaction step with NaNH2**: NaNH2 is a strong base that will deprotonate the terminal alkyne, creating a negatively charged acetylide ion on the first carbon. The reaction is as follows: $ \mathrm{HC \equiv C-CH_2-CH_3 + NaNH_2 \rightarrow HC \equiv C-CH_2-CH_2^-Na^+ + NH_3} $. ‖ ‖ 3 ⋮ **Second reaction step with CH3CH2Br**: The acetylide ion is a good nucleophile and will attack the primary alkyl halide, CH3CH2Br, in an S_N2 reaction, displacing the bromide ion and forming a new carbon-carbon bond. The reaction is as follows: $ \mathrm{HC \equiv C-CH_2-CH_2^-Na^+ + CH_3CH_2Br \rightarrow HC \equiv C-CH_2-CH_2-CH_2-CH_3 + NaBr} $. ‖ ∻Answer∻ ⚹ The organic product of the following reaction is pent-1-yne (HC≡C-CH2-CH2-CH3). ⚹ ∻Key Concept∻ ⚹ Acetylide ion formation and nucleophilic substitution ⚹ ∻Explanation∻ ⚹ The acetylide ion formed from deprotonation of a terminal alkyne reacts with a primary alkyl halide in an S_N2 reaction to form a new carbon-carbon bond, extending the carbon chain. ⚹◊What is the chemical reaction mechanism involving NaNH2 and CH3CH2Br in this scenario?⍭ Generate me a similar question◊

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