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Bromination of alkene occurs by the mechanism as shown (source: Master Organic Chemistry — Bromination of Alkenes: The Mechanism):

Bromination of alkene

Here $\ce{Br^-}$ attacks bromonium ion from a side which is more hindered to form Markovnikov product. But shouldn't it attack on the less hindered carbon as we do in case of reaction following SN2 mechanism to form anti-Markovnikov product?

For example, in ring opening of epoxide in basic medium attack by $\ce{CH3O^-}$ occurs on the less hindered carbon (source: Lumen's Organic Chemistry 1: An open textbook — 9.6. Epoxide reactions):

Ring opening of epoxide in basic medium

Why does the same doesn't happen in case of bromination of alkenes and oxymecuration of alkenes?

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    $\begingroup$ While the process involves the formation of a cyclic bromonium intermidiate, with positive charge residing on bromine, it may not be completely ignored that during the opening of the ring, a partial positive charge will develop on either carbon atom. Since, the partial positive charge is more stable on the -CH3 attached carbon, formation of that is favoured. Similarly, it is formed when attack happens on the more hindered sight. Moreover the rate determining step is the bromonium ion intermidiate formation. See Hammond Postulate. $\endgroup$
    – Solid - NMR
    Commented Aug 21, 2020 at 5:58
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    $\begingroup$ @AdityaRoychowdhury the partial positive charge is also more stable on tertiary position in case of ring opening of epoxide then why the does the same does not apply to epoxides $\endgroup$
    – Param_1729
    Commented Aug 21, 2020 at 6:14
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    $\begingroup$ chemistry.stackexchange.com/q/100788/16683 and generally the same explanation is invoked for acid-catalysed epoxide opening too as the transition state has cationic character. Base-catalysed / promoted epoxide openings are different. $\endgroup$
    – orthocresol
    Commented Aug 21, 2020 at 7:04
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    $\begingroup$ Generall, epoxide opening occurs by acid and basic means. Since the nucleophile is neutral the ring openong step is the rate determining step and more stable carbocation forms like Sn1. But in basic medium the nucleophile is stronger negatively charged. Hence the rate determining step invoves simultaneous ring opening and necleophilic attack like an Sn2 reaction. Here steric effect prdominates. The bromination is completely different. In that the bromine is leaving. It is a much better leaving group and is also larger than Oxygen. $\endgroup$
    – Solid - NMR
    Commented Aug 21, 2020 at 7:23
  • $\begingroup$ Comparing epoxide and bromonium is a bit ambiguous. $\endgroup$
    – Solid - NMR
    Commented Aug 21, 2020 at 7:24

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