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

Complete Question

The complete question is shown on the first uploaded image

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:

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.