Question

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

Answer 1

`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"`