Lactide formation from alpha-hydroxy acids

Steps:

• Identify lactide as the cyclic diester dimer of 2-hydroxypropanoic acid (lactic acid).
• Recognize that alpha-hydroxy acids (OH on carbon adjacent to COOH) form six-membered lactide rings via acid-catalyzed dehydration.
• Eliminate beta-hydroxy acids, which form five-membered lactones instead.
• Confirm the structure: lactide requires the alpha-position for proper cyclization.

Why B is correct:

• 2-Hydroxypropanoic acid (CH3CH(OH)COOH) is lactic acid, which dehydrates to form the specific six-membered lactide ring (3,6-dimethyl-1,4-dioxane-2,5-dione).

Why the others are wrong:

• A: 2-Hydroxybutanoic acid (CH3CH2CH(OH)COOH) forms a homolog of lactide, not the standard unsubstituted lactide.
• C: 3-Hydroxypropanoic acid (HOCH2CH2COOH) is a beta-hydroxy acid that forms gamma-butyrolactone, a five-membered ring.
• D: Identical to C, so same reason.

Final answer: B