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"#