The major product of the following reaction is:

Step 1: Identify the substrate

Tertiary carbon with:

• Cl (leaving group)
• Ester groups ($\ce{-COOC2H5}$)

Step 2: Reaction with NaOEt

Sodium ethoxide = Strong base

With tertiary halide → E2 elimination favored over $S_N2$

Step 3: Apply Zaitsev's rule

E2 elimination removes:

• Cl (leaving group)
• β-hydrogen

Forms more substituted alkene (Zaitsev product)

Step 4: Consider ester groups

Ester groups ($\ce{-COOC2H5}$) are:

• Electron-withdrawing (inductive)
• Can stabilize adjacent negative charge
• Don't significantly affect elimination regiochemistry

Product: Alkene with ester substituents

Answer: (c)

Key Points:

E2 with polyfunctional molecules:

• Identify leaving group (Cl)
• Identify β-hydrogens
• Apply Zaitsev's rule (unless bulky base)
• Consider electron-withdrawing groups

Ester effects:

• Electron-withdrawing by induction
• Can participate in conjugation if adjacent to C=C
• α,β-Unsaturated esters are stable (conjugated)

Tertiary halides:

• Strongly favor elimination over substitution
• Even with good nucleophiles
• Steric hindrance blocks $S_N2$