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I am a little confused on the electrophilic addition of hydrogen bromide, HBr, to propene, CH3CHCH2, producing 2-bromopropane, CH3CHBrCH3. I don't understand why the 2nd carbon gets a partial positive charge. I also don't get how we know which carbon the H from the HBr gets bonded to. Could someone please explain this reaction?
If you draw it out, you will see why it gets a partial positive charge. Look at the bonds. Our middle carbon (the one you're talking about with the positive charge) is bonded to a carbon on the left (1 bond), + a carbon on the right (now we have 2 bonds), and then a hydrogen (that's 3 bonds). But our formal charge is going to be +1 (you can prove this using the equation for formal charge) since our our carbon has 4 valence electrons and only 3 bonds.
In this class we won't give examples in which you need to rationalize which carbon of the double bond gets the H and which gets the Br. If you are given the product, then your addition of H and Br should match the product drawn. If you are not shown the product, then you can choose either one.
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