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

##### 1 Answer
Mar 29, 2017

${H}_{2} C = C H - C H = C {H}_{2} + H B r \rightarrow$

$\text{H"_3"CC(Br)(H)CH=CH"_2 and "H"_3"CCH=CHCH"_2"Br}$

#### Explanation:

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

${\text{H"_2"C=CHCH=CH"_2+"HBr"rarr"H"_3"C-C"^(+)"HCH=CH"_2 +"Br}}^{-}$

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:

${H}_{3} C - {C}^{+} H - C H = C {H}_{2} \leftrightarrow {H}_{3} C - C H = C H - {C}^{+} {H}_{2}$

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.

$\text{H"_3"CC(Br)HCH=CH"_2and"H"_3"CCH=CHCH"_2"Br}$