Amongst the following, the major product of the given chemical reaction is

Solution

Chemistry ma’am · 4 steps

Methanol opens the bromonium at the carbon next to the ring O, from the back face: OMe and Br end up trans.

1. Bromonium ion. Number the ring oxygen as 1, so the double bond is between C2 (bonded to O) and C3. Br₂ adds to this C=C to give a cyclic bromonium ion bridging C2 and C3. Methanol is the solvent and is present in huge excess, so methanol, not Br⁻, is the nucleophile that opens the ring.
2. Which carbon. The ring oxygen's lone pair stabilises positive charge on C2 (oxocarbenium character). So the C₂-Br bond is the weaker one, and methanol attacks C2. The OCH₃ goes on the carbon beside the ring O, and Br stays on C3.
3. Which face. Methanol must come in from the face opposite the bridging bromine. This anti addition leaves OCH₃ and Br trans: one on a wedge, the other on a hash.

   Option	OCH₃ on	Br on	OMe and Br	Verdict
   (a)	C2, next to O	C3	trans (wedge, hash)	correct
   (b)	C2, next to O	C3	cis (both hashed)	wrong stereochemistry
   (c)	C3	C2, next to O	trans	wrong regiochemistry
   (d)	C3	C2, next to O	trans	wrong regiochemistry

4. So. The major product is trans-3-bromo-2-methoxytetrahydropyran, which is option (a).

The idea to keep. In a halonium opening, the solvent nucleophile goes to the carbon that best holds positive charge, here the one beside the ring O, and it always attacks anti to the halogen.

Answer from the NTA final answer key. Solution by PrepWiser.