Two moles of benzaldehyde and one mole of acetone under alkaline conditions using aqueous NaOH after heating gives $$x$$ as the major product. The number of $$\pi$$ bonds in the product $$x$$ is _____

We need to find the number of $$\pi$$ bonds in the major product when 2 moles of benzaldehyde react with 1 mole of acetone under alkaline conditions (aqueous NaOH, heated).

This is a crossed aldol condensation (Claisen-Schmidt reaction). Two moles of benzaldehyde react with one mole of acetone to form dibenzylideneacetone (dibenzalacetone):

$$2C_6H_5CHO + CH_3COCH_3 \xrightarrow{NaOH, \Delta} C_6H_5CH=CH-CO-CH=CHC_6H_5 + 2H_2O$$

The product is dibenzylideneacetone: $$C_6H_5CH=CH-CO-CH=CHC_6H_5$$.

Counting the $$\pi$$ bonds:

- First benzene ring: 3 $$\pi$$ bonds

- First $$C=C$$: 1 $$\pi$$ bond

- $$C=O$$: 1 $$\pi$$ bond

- Second $$C=C$$: 1 $$\pi$$ bond

- Second benzene ring: 3 $$\pi$$ bonds

Total $$\pi$$ bonds = $$3 + 1 + 1 + 1 + 3 = 9$$.

The answer is $$\boxed{9}$$.