🧪 Synthetic Scheme: Full Detailed Solution

Step 1: Enolate Formation (Deprotonation)

The α-hydrogen next to the ester group is acidic due to resonance stabilization of its conjugate base. A strong base like NaOC₂H₅ abstracts the α-hydrogen, forming a carbanion intermediate.

✅ The carbanion now acts as a nucleophile in the next SN2 reaction.

Step 2: SN2 Reaction (First Alkylation)

The enolate ion attacks an alkyl halide (CH₃CH₂I) via the SN2 mechanism. A C–C bond is formed at the α-position.

Mechanism: One-step backside attack with inversion of configuration (if applicable).

✅ This step builds the carbon skeleton using ethyl iodide.

Step 3: Second Deprotonation and Alkylation

A second equivalent of NaOC₂H₅ removes another α-hydrogen. The new carbanion then reacts with CH₂=CHCH₂Br (allyl bromide) via SN2.

⚠️ Both alkylations occur at the α-carbon of the ester starting material.

✅ A substituted diester is now formed.

Step 4: Hydrolysis and Decarboxylation

The diester undergoes hydrolysis using H₃O⁺ and heat, converting esters into carboxylic acids. One of the β-keto acids then undergoes decarboxylation, releasing CO₂ and forming the final acid.

✅ Final product: A substituted mono-carboxylic acid after loss of CO₂.

✅ Final Answer

Starting Material: 1

Reagents Used (in order):

i – NaOC₂H₅ (base for enolate formation)
b – CH₃CH₂I (ethyl iodide)
f – CH₂=CHCH₂Br (allyl bromide)
j – H₃O⁺, heat (for hydrolysis and decarboxylation)

🔷 Correct synthetic scheme: 1ibfj

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learnlyfly

Specify a synthetic scheme that would produce the compound shown above in the fewest steps possible. Use one of the starting materials shown

Answer