The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is :

First, phenol reacts with aqueous sodium hydroxide to form sodium phenoxide:

$$C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O$$

The phenoxide ion is far more nucleophilic than phenol. In the Kolbe-Schmitt reaction, this sodium phenoxide is treated with carbon dioxide at about 400 K and 4-7 atm. CO₂ behaves as an electrophile and is inserted predominantly at the ortho position of the activated benzene ring:

$$C_6H_5ONa + CO_2 \xrightarrow[\;4{-}7\text{ atm}\;]{\;400\,\text{K}\;} o\!-\!HO{-}C_6H_4COONa$$

The product obtained is sodium salicylate (the sodium salt of o-hydroxybenzoic acid). A final acidification step liberates the free acid:

$$o\!-\!HO{-}C_6H_4COONa + HCl \rightarrow o\!-\!HO{-}C_6H_4COOH + NaCl$$

o-Hydroxybenzoic acid is commonly called salicylic acid. Because electrophilic substitution at the ortho position is favored under these reaction conditions, salicylic acid is formed in the highest yield, with only minor para substitution.

Hence, the major product is salicylic acid.

Option C which is: salicylic acid