Consider the following chemical reaction.
$$CH \equiv CH \xrightarrow[2) CO, HCl, AlCl_3]{1) \text{Red hot Fe tube, 873 K}}$$ Product
The number of $$sp^2$$ hybridized carbon atom(s) present in the product is ______

In the first step, acetylene ($$CH \equiv CH$$) is passed through a red hot iron tube at 873 K, which causes cyclic trimerisation to form benzene ($$C_6H_6$$).

In the second step, benzene undergoes the Gattermann-Koch reaction with CO and HCl in the presence of anhydrous $$AlCl_3$$ as a catalyst. This reaction introduces an aldehyde group ($$-CHO$$) onto the benzene ring, producing benzaldehyde ($$C_6H_5CHO$$).

Now we count the number of $$sp^2$$ hybridized carbon atoms in benzaldehyde. The six carbon atoms in the benzene ring are all $$sp^2$$ hybridized (each forms three sigma bonds and participates in the delocalised $$\pi$$ system). The carbon atom of the aldehyde group ($$-CHO$$) is also $$sp^2$$ hybridized, as it forms a double bond with oxygen and two single bonds (with H and the ring carbon).

Therefore, the total number of $$sp^2$$ hybridized carbon atoms in benzaldehyde is $$6 + 1 = \mathbf{7}$$.