Consider the given reaction, the product 'X' is:

The reaction sequence involves a crossed aldol addition followed by the haloform reaction.

First, $$\mathrm{NaOH}$$ deprotonates the only available $$\mathrm{\alpha}$$-position of $$\mathrm{2,2\text{-}Dimethylcyclopentan\text{-}1\text{-}one}$$ to form an enolate ion.

This enolate attacks the carbonyl carbon of acetaldehyde, producing intermediate $$\mathrm{P}$$, a $$\mathrm{\beta}$$-hydroxy ketone containing the group $$\mathrm{(-CH(OH)-CH_3)}$$.

Intermediate $$\mathrm{P}$$ is then treated with $$\mathrm{I_2/NaOH}$$, initiating the iodoform reaction.

Sodium hypoiodite first oxidises the secondary alcohol into a methyl ketone.

The methyl group undergoes tri-iodination followed by cleavage to form yellow precipitate:

$$\mathrm{CHI_3}$$

and a sodium carboxylate salt.

Filtration removes the iodoform precipitate.

Acidification with $$\mathrm{HCl}$$ protonates the carboxylate ion to form the corresponding carboxylic acid.

The final major product $$\mathrm{X}$$ is:

$$\mathrm{2,2\text{-}Dimethyl\text{-}5\text{-}carboxycyclopentan\text{-}1\text{-}one}$$