Question

Product(s) formed in the given reaction sequence is(are) 

Hint:

Br\(_2\)/H\(_2\)O gives {anti} bromohydrins via bromonium-ion opening by water. Base converts bromohydrins to epoxides by intramolecular \(\mathrm{S_N2}\) displacement of Br by the alkoxide. Attack on the bromonium from either face \(⇒\) two epoxide diastereomers.

Answer 1

Step 1: Formation of a bromohydrin (anti addition).
Br\(_2\)/KBr/H\(_2\)O adds across the \(\alpha,\beta\)-unsaturated double bond via a bromonium ion followed by attack of water. Opening occurs anti to the bromonium, giving a trans–bromohydrin with Markovnikov placement of OH (at the more substituted carbon). This creates two adjacent stereocenters.
Step 2: Base-promoted intramolecular epoxidation.
With aq. KOH in THF, the alcohol is deprotonated to an alkoxide, which displaces the anti bromide in an \(\mathrm{S_N2}\) fashion to form the epoxide. Because the bromohydrin was formed anti, intramolecular \(\mathrm{S_N2}\) ring closure preserves the overall anti relationship required for epoxide formation.
Step 3: Stereochemical outcome.
Attack of water in Step 1 can occur from either face of the planar bromonium, giving two diastereomeric bromohydrins that each cyclize to an epoxide. Thus, two epoxide diastereomers are produced—those depicted in (A) and (C). Options (B) and (D) are not obtained under these conditions (the former has an incorrect relative configuration; the latter would require oxidation/hydrolysis rather than intramolecular \(\mathrm{S_N2}\)).
\[ \boxed{\text{Products: two epoxides from anti bromohydrin formation } ⇒ \text{ (A) and (C).}} \]