Saponification of diester with excess NaOH

Steps:

• Recognize the compound as a diester with two -COOR groups.
• Alkaline hydrolysis cleaves each ester bond using OH⁻ from NaOH.
• Each ester yields a carboxylate salt (-COONa) and an alcohol (ROH).
• Excess NaOH ensures complete conversion to salts, not acids.

Why C is correct:

• C depicts the disodium dicarboxylate and two alcohol molecules, matching the saponification formula: RCOOR' + NaOH → RCOONa + R'OH for both esters.

Why the others are wrong:

• A shows the free diacid, but excess base forms salts, not H⁺-protonated acids.
• B indicates mono-hydrolysis product, ignoring excess NaOH for full reaction.
• D includes unchanged ester or wrong byproducts, violating complete hydrolysis.

Final answer: C