Construct a multistep synthetic route for the synthesis of acetylene from ethane
They give CH3CH3, then they ask for reagent 1, product 1, reagent 2, product 2, reagent 3, product 3, reagent 4, and product 4 is CHCH.
The options, to fulfill reagents and products are:
HBr/ROOR, HBr/CH2Cl2, BrCH2CH3, CH2…
The Correct Answer and Explanation is :
To synthesize acetylene (CH≡CH) from ethane (CH₃CH₃), we need to carry out a series of reactions. Here’s a step-by-step synthetic route using the provided reagents and products:
Step 1:
Reagent 1: HBr/ROOR
Product 1: Bromoethane (CH₃CH₂Br)
Explanation: The first step involves the free radical addition of HBr to ethane (CH₃CH₃). The radical initiator (ROOR) promotes the formation of a free radical at the ethane molecule. This leads to the generation of bromoethane (CH₃CH₂Br) as the product.
Step 2:
Reagent 2: BrCH₂CH₃
Product 2: Ethene (CH₂=CH₂)
Explanation: In this step, bromoethane (CH₃CH₂Br) undergoes a dehydrohalogenation reaction in the presence of a base (e.g., KOH). This reaction causes the elimination of a hydrogen atom and a bromine atom from the β-carbon, resulting in the formation of ethene (CH₂=CH₂).
Step 3:
Reagent 3: HBr/CH₂Cl₂
Product 3: 1,2-Dibromoethane (CH₂BrCH₂Br)
Explanation: In this step, ethene (CH₂=CH₂) undergoes an electrophilic addition reaction with HBr, where the bromine adds across the double bond, yielding 1,2-dibromoethane (CH₂BrCH₂Br) as the product. The CH₂Cl₂ is used as a solvent.
Step 4:
Reagent 4: CH₂… (Strong base such as NaNH₂)
Product 4: Acetylene (CH≡CH)
Explanation: Finally, the 1,2-dibromoethane undergoes a double dehydrohalogenation reaction when treated with a strong base such as sodium amide (NaNH₂) or another strong base. This removes two equivalents of hydrogen and bromine, forming the triple-bonded product acetylene (CH≡CH).
Final Summary:
- Reagent 1: HBr/ROOR → Product 1: Bromoethane (CH₃CH₂Br)
- Reagent 2: BrCH₂CH₃ → Product 2: Ethene (CH₂=CH₂)
- Reagent 3: HBr/CH₂Cl₂ → Product 3: 1,2-Dibromoethane (CH₂BrCH₂Br)
- Reagent 4: CH₂… (NaNH₂) → Product 4: Acetylene (CH≡CH)
This series of reactions effectively transforms ethane (CH₃CH₃) into acetylene (CH≡CH).