[Worth noting: bromination of alkenes is technically an oxidation reaction, because each carbon goes from being bound to another carbon (0) to bromine (–1). The oxidation state of each carbon in ethene is +2; the oxidation state of each carbon in dibromoethane is +1. ]

This is in Section 8: Some Applications of Halogenation Reactions. Oxidation number should become more positive / less negative if carbon is being oxidized (which it is here). Looks like the numbers are right but the sign is incorrect. Thank you :)

The leaving group is excellent, first of all. The C-halogen bond length is much longer than a normal C-leaving group bond length. (e.g. 2.48 A for C-I in the iodonium vs 2.13 A for normal C-I).

Secondly, because of the 3-membered ring, the geometry on the carbons is not the same as in your typical tetrahedral carbon, so there is more “room” for the nucleophile to approach.

No, what happens is that you form the bromonium ion with Br2 and then Cl(-) is the active nucleophile which attacks the resulting bromonium ion at the most substituted position.
Kind of a strange question though, because they probably don’t mention solvent. NaCl is not going to dissolve in an organic solvent. Better choice would have been Bu4N+ Cl- .

I’m studying for a quiz and the problem I came across was an alkene addition with Br-Br and NaCl. I know that the Cl should add first but I can’t seem to figure out the mechanism for it.

See this post: https://www.masterorganicchemistry.com/2017/04/11/more-on-12-and-14-additions-to-dienes/

Hi Idan. The reason the second attack occurs on the more substituted carbon, instead of the less substituted, is explained by the following. In the transition state leading to the product, one C-Br bond in the bromonium ion is already partially broken. Therefore, one of the two C atoms will bear a partial positive charge–it will be cation-like, in other words. Now, think about the relative stability of carbocations. Since more substituted carbocations are more stable, the more substituted carbon will undergo partial C-Br bond cleavage, and therefore nucleophilic attack will preferentially occur at the tertiary carbon in the above example.

While you’re right that the more substituted carbon is more sterically hindered, this effect is less important than the electronic effect of carbocation stability.