How do we explain the reaction "1,3-butadiene" with 1*"equiv" of HBr(g) to give TWO products....?

1 Answer
Mar 29, 2017

H_2C=CH-CH=CH_2+HBrrarr

"H"_3"CC(Br)(H)CH=CH"_2 and "H"_3"CCH=CHCH"_2"Br"

Explanation:

Let's consider addition of the first equiv of H^(+), from the hydrogen bromide electrophile, and see what develops:

"H"_2"C=CHCH=CH"_2+"HBr"rarr"H"_3"C-C"^(+)"HCH=CH"_2 +"Br"^-

The proton adds to the olefin at C_1. Of course it could add at C_2, but here this would leave a primary carbocation rather than a secondary carbocation.

Moreover, addition at C_1 leaves a carbocation intermediate that is resonance stabilized:

H_3C-C^(+)H-CH=CH_2harrH_3C-CH=CH-C^(+)H_2

Now, the bromide ion (delivered from the hydrohalide electrophile can potentially add at C_1 or C_4 on the hydrocarbon chain. We would expect both substitutions to occur, and it seems they do.

"H"_3"CC(Br)HCH=CH"_2and"H"_3"CCH=CHCH"_2"Br"