Consider the above reaction and predict the major product.

The given molecule contains:

1. An ester group $$\mathrm{(-COOEt)}$$

2. An aldehyde group $$\mathrm{(-CHO)}$$

3. A carbon-carbon double bond inside the cyclopentene ring

The reagent used is:

$$\mathrm{DIBAL\text{-}H \ followed\ by\ H_2O}$$

$$\mathrm{DIBAL\text{-}H}$$ is a mild and selective reducing agent.

It partially reduces esters to aldehydes.

Unlike $$\mathrm{LiAlH_4}$$, it does not reduce esters completely to primary alcohols.

General reaction:

$$\mathrm{RCOOR' \xrightarrow[\ H_2O\ ]{DIBAL\text{-}H} RCHO}$$

The reaction occurs through formation of a tetrahedral aluminium intermediate.

At low temperature, the reduction stops at the aldehyde stage.

Also:

$$\mathrm{DIBAL\text{-}H}$$ does not reduce isolated $$\mathrm{C=C}$$ bonds.

Therefore, the double bond in the cyclopentene ring remains unchanged.

Applying this to the given compound:

$$\mathrm{-COOEt \longrightarrow -CHO}$$

while the existing aldehyde group and alkene remain unaffected.

Thus, the final product contains aldehyde groups at both ends of the molecule.

Therefore, the correct answer is:

$$\boxed{\mathrm{Option\ A}}$$