Now we perform hydrohalogenation with #DBr-=""^(2)HBr# so that we can follow the process. We start with propylene...the which reacts with the protium ion....
#H_2C=CH-CH_3 + DBr rarr underbrace(H_2DC-stackrel(+)CH-CH_3)_"stabilized intermediate"+Br^(-)#
#" "underbrace(H_2stackrel(+)C-CHD-CH_3)_"primary carbocation"#
From the unsymmetrical olefin, the protium ion of #""^2HBr# can add to #"carbon 1"# to give a #2^@# #"carbocation"#, or add to #"carbon 2"# to give a #1^@# #"carbocation"#. We assume the #2^@# #"carbocation"# is stabilized with respect to the #1^@# #"carbocation"#..and so the reaction should follow the lower energy route.
And so the end result should be the secondary alkyl halide as the major product....
#H_2C=CH-CH_3 + DBr rarr underbrace(H_2DC-CHBr-CH_3)_"major product"#
#" "underbrace(BrH_2C-CHD-CH_3)_"minor product"#
Do you follow? If no, we will try again...
