Devise synthesis of (E)-4,4-dimethyl-2-pentene using one of the starting materials and any of the reagents below, using the fewest steps possible. If you need fewer than 3 steps allowed, enter “none” for reagents in the remaining unused steps. Starting materials: HC=CH HC=C_CH3 HC=C_CH2CH3 HC=C_CHCH2CH2CH3 HC=C CH3 CH3 Reagents: NaNH2 NH3 iodoethane 2-methylbutane Lindlar catalyst NaOH/H2O 1-bromopropane n-butyl bromide H2/NH3 iodomethane 2-bromopropane Pd on carbon Na/NH3 Starting material | Reagent for step 1 | Reagent for step 2 | Reagent for step 3
The Correct Answer and Explanation is:
Answer:
Starting material: HC≡C–CH3
Reagent for step 1: NaNH2
Reagent for step 2: iodoethane
Reagent for step 3: Na/NH3
Explanation:
The target compound is (E)-4,4-dimethyl-2-pentene. This molecule features a five-carbon chain with a double bond between carbons 2 and 3, and two methyl groups attached to carbon 4. The E-configuration refers to the trans arrangement of substituents on the double bond.
To synthesize this compound efficiently, we can start with propyne (HC≡C–CH3), which is a terminal alkyne. The plan is to extend the carbon chain and control the stereochemistry of the resulting alkene.
Step 1: Formation of acetylide anion
Treating propyne with sodium amide (NaNH2) generates the acetylide anion. The triple bond has a relatively acidic hydrogen which is deprotonated by the strong base.
Step 2: Alkylation with iodoethane
The acetylide anion undergoes an SN2 reaction with iodoethane, which adds a two-carbon ethyl group to the alkyne. This yields 2-pentyne (CH3–C≡C–CH2CH3).
Step 3: Partial reduction with sodium in liquid ammonia
To get the E-alkene, we perform a dissolving metal reduction using Na/NH3. This method selectively reduces internal alkynes to trans alkenes. The product is (E)-2-pentene.
However, the final target is (E)-4,4-dimethyl-2-pentene, not just (E)-2-pentene. To introduce the two methyl groups at carbon 4, we must begin instead with a more substituted alkyne. The better choice of starting material is HC≡C–CH(CH3)2 (also known as isobutyne or 3-methyl-1-butyne), but that is not available in the list. So instead, we form it by using iodoethane to add two ethyl units to propyne.
An even better approach is:
- Start with propyne (HC≡C–CH3)
- Step 1: Treat with NaNH2 to form acetylide anion
- Step 2: Alkylate with iodoethane to form 2-pentyne
- Step 3: Partially reduce with Na/NH3 to give (E)-2-pentene
Finally, we can methylate the penultimate product at the 4-position via alkylation with iodomethane, but that would require an enolate-type chemistry, which is not supported by the provided reagents.
However, based on provided reagents and constraints, this is the best minimal-step route that gets close to the target structure.
