HCl + Alkene (Adding HCl or HBr across double bonds)
The Basics
When HCl or HBr is allowed to react with an alkene, the alkene will attack the H. After all, the double bond contains four electrons and the H is positively charged.
So, the double bond breaks open, and one of the carbons is left without a fourth bond. This means it's a carbocation and has a positive (+) formal charge. The more substituted carbocation is favoured (Tertiary > Secondary > Primary).
The other halogen atom is left with a negative charge, and will be attracted to (and will attach itself to) the carbocation.
So, the double bond breaks open, and one of the carbons is left without a fourth bond. This means it's a carbocation and has a positive (+) formal charge. The more substituted carbocation is favoured (Tertiary > Secondary > Primary).
The other halogen atom is left with a negative charge, and will be attracted to (and will attach itself to) the carbocation.
If you care about Stereochemistry
Since the carbocation intermediate is perfectly flat, the halogen could attack from either the top or the bottom. So, depending on how bulky the side chains are, there's about a 50/50 chance it attacks from either the top or bottom, and so we end up with a mix of "syn" (same side, analogous to "cis") and "anti" (opposite side, analogous to "trans") additions.
Also note, that the original attack on H could have been from either side of the double bond, and so we could end up with with a mixture (possibly even a 50/50 mixture) of R and S stereocentres.
Also note, that the original attack on H could have been from either side of the double bond, and so we could end up with with a mixture (possibly even a 50/50 mixture) of R and S stereocentres.
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