#"H"_3"C"-"C"="CH" + "HBr" -> #?

1 Answer
Jun 16, 2018

This may be what you mean:

#"1-propene"#, #"H"_3"C"-"CH"="CH"_2#, reacts with #"HBr"# in an addition reaction.

  1. The #"H"^(+)# from #"HBr"#, a strong acid, simply protonates a carbon on the #pi# bond by responding to the #pi# electrons coming in towards the #"H"# on #"HBr"#.

    This alkene follows Markovnikov addition, meaning that the #"H"^(+)# goes onto the carbon with more hydrogens than the other one. Carbons 1 and 2 were choices, and we would then choose carbon 1.

  2. Hence, carbon 2 becomes the carbocation when the secondary carbocation intermediate forms. At that point, #"Br"^(-)# acts as a nucleophile to add onto carbon 2 to yield #color(blue)"2-bromopropane"#.