This reaction involves the Reimer-Tiemann reaction, which is commonly used to form ortho-formyl phenols from phenolic compounds using chloroform (CHCl₃) and aqueous NaOH under heating conditions.

Here's a step-by-step breakdown of the reaction process:

1. The methoxy group (\text{-OMe}) is an ortho/para-directing group due to its electron-donating nature through resonance.

2. In the presence of CHCl₃ and NaOH, the ortho position relative to the OMe group is activated due to increased electron density.

3. Chloroform is deprotonated by NaOH to form dichlorocarbene (:CCl₂), a reactive intermediate.

4. This dichlorocarbene reacts at the ortho position to the methoxy group, leading to the formation of an intermediate dichloromethyl compound.

5. Subsequent hydrolysis in basic conditions converts the dichloromethyl group into a formyl group, yielding the ortho-formyl product.

Considering these steps, the major product formed will indeed be the ortho-formyl derivative.

The correct option is:

In conclusion, the reaction conditions and the presence of the methoxy group favor the formation of the ortho-formyl product, which aligns with the given correct answer.