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15240672380's Question
Posted 2 months ago

Draw the organic product of the following reaction.
- You do not have to consider stereochemistry.
- You do not have to explicitly draw H\mathrm{H} atoms.
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Answer from Sia
Posted 2 months ago
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
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: HCCCH2CH3+NaNH2HCCCH2CH2Na++NH3 \mathrm{HC \equiv C-CH_2-CH_3 + NaNH_2 \rightarrow HC \equiv C-CH_2-CH_2^-Na^+ + NH_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: HCCCH2CH2Na++CH3CH2BrHCCCH2CH2CH2CH3+NaBr \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}
The organic product of the following reaction is pent-1-yne (HC≡C-CH2-CH2-CH3).
Key Concept
Acetylide ion formation and nucleophilic substitution
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.

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