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Question: Me-C=O C=O -CH3 $\xrightarrow[\Delta]{\text{OEt}^-}$ 'A' major Product A in the above reaction is...

Me-C=O C=O -CH3

→ΔOEt−\xrightarrow[\Delta]{\text{OEt}^-} 'A' major Product

A in the above reaction is :

A
\begin{array}{c}
\text{CH}_3-\text{C}=\text{O} \\
\text{CH}_2-\text{CH}_2 \\
\text{CH}_2-\text{CH}_2 \\
\text{CH}-\text{CH}_3
\end{array}
B
\begin{array}{c}
\text{CH}_3-\text{C}=\text{O} \\
\text{CH}_2-\text{CH}_2 \\
\text{CH}_2-\text{CH}_2 \\
\text{C}-\text{CH}_3
\end{array}
C
\begin{array}{c}
\text{CH}_3-\text{C}=\text{O} \\
\text{CH}_2-\text{CH}_2 \\
\text{C}=\text{C}-\text{CH}_3 \\
\text{C}-\text{C}(=\text{O})\text{CH}_3
\end{array}
D
\begin{array}{c}
\text{CH}_3-\text{C}=\text{O} \\
\text{CH}_2-\text{CH}_2 \\
\text{CH}_2-\text{CH}_2 \\
\text{CH}_2-\text{C}=\text{CH}_3
\end{array}
Answer
\begin{array}{c}
\text{CH}_3-\text{C}=\text{O} \\
\text{CH}_2-\text{CH}_2 \\
\text{C}=\text{C}-\text{CH}_3 \\
\text{C}-\text{C}(=\text{O})\text{CH}_3
\end{array}
Explanation

Solution

The reaction is an intramolecular aldol condensation of a 1,7-diketone. The reactant is 2,7-octanedione: CH3−CO−(CH2)4−CO−CH3CH_3-CO-(CH_2)_4-CO-CH_3.

The base (OEt−OEt^-) and heat (Δ\Delta) promote the formation of an enolate, which then attacks an intramolecular carbonyl group. There are two possible modes of enolate formation:

  1. Formation of a 5-membered ring: Deprotonation of a methylene group (α\alpha-carbon adjacent to a carbonyl) leads to an enolate that attacks the other carbonyl group, forming a 5-membered ring. This pathway is generally favored due to the higher stability of 5-membered rings.

    • Enolate formation at C3C_3 (or C6C_6) of CH3−CO−CH2−CH2−CH2−CH2−CO−CH3CH_3-CO-CH_2-CH_2-CH_2-CH_2-CO-CH_3.
    • The enolate attacks the carbonyl at C7C_7.
    • This forms a 5-membered ring (C3−C4−C5−C6−C7C_3-C_4-C_5-C_6-C_7).
    • The intermediate is a cyclic β\beta-hydroxy ketone.
    • Dehydration of this intermediate leads to an α,β\alpha,\beta-unsaturated ketone. The most stable product is formed when the double bond is conjugated with the ketone. This results in the formation of 2-acetyl-3-methylcyclopent-2-en-1-one, which corresponds to option (3).

  2. Formation of a 7-membered ring: Deprotonation of a methyl group (α\alpha-carbon) leads to an enolate that attacks the other carbonyl group, forming a 7-membered ring.

    • Enolate formation at C1C_1 (or C8C_8) of CH3−CO−CH2−CH2−CH2−CH2−CO−CH3CH_3-CO-CH_2-CH_2-CH_2-CH_2-CO-CH_3.
    • The enolate attacks the carbonyl at C7C_7.
    • This forms a 7-membered ring (C1−C2−C3−C4−C5−C6−C7C_1-C_2-C_3-C_4-C_5-C_6-C_7).
    • Dehydration leads to a conjugated system.

Since 5-membered ring formation is generally favored over 7-membered ring formation in intramolecular aldol reactions, the major product is the one arising from the 5-membered ring cyclization.

The structure of option (3) is: A 5-membered ring with a ketone group (C=OC=O) and a methyl group (CH3CH_3) attached to one carbon of the ring. Adjacent to the ketone is a double bond within the ring. One carbon of the double bond has a methyl group, and the other carbon of the double bond has an acetyl group (C(=O)CH3C(=O)CH_3). This structure is consistent with the product formed via the 5-membered ring pathway.