Step-by-Step Solution:

The reaction proceeds in two steps. Let's analyze each step individually.

Step 1: Reaction with DIBAL-H at -78°C

1. Identify the reactant and reagent: The starting material is methyl 3-(4-hydroxyphenyl)acrylate. It has two main functional groups: a phenolic hydroxyl group (-OH) and an α,β-unsaturated ester group ($-CH=CH-COOCH_3$). The reagent is DIBAL-H (Diisobutylaluminium hydride), a reducing agent, used at a low temperature of -78°C.

2. Determine the function of the reagent: DIBAL-H is a bulky and selective reducing agent. At low temperatures like -78°C, it is well-known for the partial reduction of esters to aldehydes. It does not typically reduce isolated carbon-carbon double bonds, and in α,β-unsaturated systems, it preferentially attacks the carbonyl carbon of the ester.

3. Predict the product of Step 1: The ester group ($-COOCH_3$) will be reduced to an aldehyde group (-CHO). The phenolic -OH group and the C=C double bond will remain unchanged. The reaction is as follows: $\text{HO} - \text{C}_6\text{H}_4 - \text{CH=CH-COOCH}_3 \xrightarrow{\text{1. DIBAL-H, -78°C}} \text{HO} - \text{C}_6\text{H}_4 - \text{CH=CH-CHO}$ The intermediate product formed is 4-(3-oxoprop-1-en-1-yl)phenol, also known as p-hydroxycinnamaldehyde.

Step 2: Reaction with (CH₃)₂CHCH₂Br and K₂CO₃

1. Identify the reactants and reagents: The reactant is the intermediate from Step 1, p-hydroxycinnamaldehyde. The reagents are isobutyl bromide ((CH₃)₂CHCH₂Br), a primary alkyl halide, and potassium carbonate (K₂CO₃), a moderately strong base.

2. Determine the function of the reagents: This set of reagents is used for Williamson ether synthesis.

   • The base, K₂CO₃, deprotonates the most acidic proton in the molecule, which is the proton of the phenolic hydroxyl group, forming a nucleophilic phenoxide ion. $\text{HO} - \text{C}_6\text{H}_4 - \text{R} + \text{K}_2\text{CO}_3 \rightleftharpoons \text{K}^+ \text{O}^- - \text{C}_6\text{H}_4 - \text{R} + \text{KHCO}_3$
   • The resulting phenoxide ion then acts as a nucleophile and attacks the primary alkyl halide (isobutyl bromide) via an Sₙ2 reaction, displacing the bromide ion.

3. Predict the product of Step 2: An ether linkage is formed at the position of the original hydroxyl group. The aldehyde group and the double bond do not react under these conditions. $\text{K}^+ \text{O}^- - \text{C}_6\text{H}_4 - \text{CH=CH-CHO} + (\text{CH}_3)_2\text{CHCH}_2\text{Br} \rightarrow (\text{CH}_3)_2\text{CHCH}_2 - \text{O} - \text{C}_6\text{H}_4 - \text{CH=CH-CHO} + \text{KBr}$ The final product is 4-(isobutoxy)cinnamaldehyde.

Evaluation of Options:

• Option A: The ester group is not reduced. This is incorrect as DIBAL-H reduces esters.
• Option B: The ester is over-reduced to a saturated primary alcohol, and the C=C double bond is also reduced. This is incorrect. DIBAL-H at -78°C selectively produces the aldehyde.
• Option C: This is the intermediate product after the first step (DIBAL-H reduction) but before the second step (Williamson ether synthesis). It is not the final major product.
• Option D: This structure matches our predicted final product. The ester has been reduced to an aldehyde, and the phenolic hydroxyl has been converted to an isobutoxy ether. This is the correct major product.

Therefore, the major product of the reaction sequence is the one shown in option D.