[AnkurGel]

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[OC pro]

Any ideas by yourself?

[AnkurGel]

Initially, I thought first step of the problem to be the condensation in intermolecular ketones as base is also present to facilitate the aldol type reaction. Then, going by my approach will yield two major intermolecular-products and this must go intermolecular-aldol once again as per next step.
But answer is different. Seeing the answer I can easily deduce what happened in reaction, but basic touch of this problem would not leave any imprint. So, I posted it just like it is, so as to get the real reason behind the product.

[OC pro]

I assume deprotonation of the most acidic proton (will be the H at R). Then attack onto ketone, leading to a tertiary alcohol which upon basic treatment will eliminate.

[AnkurGel]

Hey... the given compound has double bonded carbon in beginning. Though obvious, but it may have been gone unnoticed while seeing.

[OC pro]

Sorry my mistake. Sum formulas are more difficult to read then structure formulas  .
Now correct: the deprotonated diketone will attack at the double bond to give the first intermediate A (Michael Addition). Then deprotonation will take place at the methyl group of the ketone. Then attack onto a carbonyl leading to a Wieland-Miescher Ketone type (http://en.wikipedia.org/wiki/Wieland%E2%80%93Miescher_ketone). The overall process is a Robinson-Annulation (from the famours Sir Robert Robinson). This reaction has been used for Steroid synthesis.

[AnkurGel]

Thanks a lot. Michael reaction was the key to this. Thanks again for all info... specially for Wieland-Miescher ketone