[Shikimic]

I'm a bit confused about the Robinson Annelation mechanism in my textbook. In the first step there is a Michael addition of an enolate (produced by base NaOEt) to an alpha-beta unsaturated carbonyl. After forming the Michael product, more base (NaOEt) is added to produce the cyclised final product.

What I don't understand is why base has to be added twice. The ethoxide ion is acting as base catalyst by my understanding (protonated by acidic beta-keto ester hydrogen and then deprotonated in Michael addition) so why would more base be necessary to catalyse the cyclisation?

Thank you in advance.

EDIT: Just left out a minor detail. In the example, the reaction is between 3-buten-2-one and ethyl acetoacetate.

[g-bones]

your book may just be indicating where the ethoxide comes into play, not necessarily a second addition of base.  Sodium ethoxide is cheap, Im pretty sure people just chuck in some sodium ethoxide into some ethanol and heat it up, not necessarily stoichiometrically stingy. 

[Shikimic]

your book may just be indicating where the ethoxide comes into play, not necessarily a second addition of base.  Sodium ethoxide is cheap, Im pretty sure people just chuck in some sodium ethoxide into some ethanol and heat it up, not necessarily stoichiometrically stingy. 

I hope that's the case because then everything would make sense. Thanks for the fast reply.

[orgopete]

The reaction will have a number of bases present. The enolate of the intermediate Michael addition reaction is a base and probably is a source of regenerated ethoxide. The aldol condensation will generate another alkoxide. This can also exchange with ethanol because of the higher concentration and regenerate ethoxide. If the aldol looses water as hydroxide, it will be another base.

When I wrote mechanisms, I included all of the steps, such as regeneration of ethoxide, borohydrides, aluminum hydrides, etc. It was my objective to make every step of a mechanism understandable. I suspect that ethoxide simply appeared over an arrow and thus may appear to be an additional amount of ethoxide being added. That ethoxide may be the same ethoxide that is generated upon protonation of the intermediate enolate by ethanol.

[Smrt guy]

The base in the Robinson annulation need only be catalytic.  The intermediate formed after conjugate addition has the enolate on the wrong side of the carbonyl.  Thus, the enolate must be protonated then deprotonated in order to generate the enolate in the correct position necessary for formation of the six-membered ring.  It may appear that a second equivalent of base was used but really it is the result of two consecutive proton transfers.  Protonation of the aldol product followed by E2 elimination will give the final product and once again regenerate the base.