The major product of the following reaction is :
This question is built on drawn structures, so the options cannot be reproduced as text. The chemistry it tests is set out below.
Iodine with silver acetate in wet acetic acid, followed by hydrolysis, is the Woodward cis-hydroxylation. The answer turns entirely on stereochemistry, so the mechanism is worth following carefully.
Iodine first adds to the alkene to give a bridged iodonium ion. Acetate then opens that bridge by anti attack, giving a trans-iodo acetate, and silver ion assists by removing iodide as insoluble AgI.
The decisive step comes next. The neighbouring acetate oxygen displaces the remaining iodide intramolecularly, forming a five-membered cyclic 1,3-dioxolan-2-ylium (acetoxonium) ion. Because this ring can only form on one face, it locks both oxygens onto the same face of the original double bond.
Water then attacks the acetoxonium carbon, and basic hydrolysis with KOH cleaves the resulting monoester, delivering a cis (syn) diol.
The contrast worth remembering is with the Prevost reaction, which uses the same iodine/silver carboxylate combination but under anhydrous conditions. There, carboxylate rather than water opens the acetoxonium ion, and the product is the trans diol. Wet conditions give cis, dry conditions give trans — so the presence of water in the stem is the clue.
Per the official final answer key the answer is option (D).
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