# Does the Dieckmann Reaction proceed via the conjugated enolate of the product?

In the following mechanism of the Dieckmann Condensation, the product has been formed in the fourth step, but still the reaction proceeds and $\ce{RO^{-}}$deprotonates the product. Then $\ce{H^{+}}$ rejoins to form the same product. Won't this result in extra expenditure of energy? Also, if the reaction proceeds through this mechanism, shouldn't there be an infinite loop?

Can anyone explain why is there an extra step?

• In the depicted mechanism there is consistently a charge for the $\ce{RO-}$ omitted. – Martin - マーチン Mar 23 '16 at 7:38