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Romashka-Z-Leto [24]
3 years ago
15

Aldol condensation of 2,5-heptanedione yields a mixture of two enone products in a 9:1 ratio. Treatment of the minor product wit

h aqueous NaOH converts it into the major product; the interconversion proceeds as follows: Hydroxide ion adds to the double bond, forming enolate ion 1; Proton transfer occurs, yielding tetrahedral intermediate 2; Ring opening occurs, yielding enolate ion 3; Protonation of enolate ion 3 occurs, yielding 2,5-heptanedione; Deprotonation at C-6 occurs, yielding enolate ion 5; Enolate ion 5 attacks C-2, yielding tetrahedral intermediate 6; Protonation occurs to yield aldol addition product 7; Dehydration yields the more stable product.

Chemistry
1 answer:
sattari [20]3 years ago
6 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The diagram of the mechanism of this reaction is shown on the second uploaded image

The structure of the enolate Ion 1 is shown on the third uploaded image

Explanation:

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