Answer:
The reaction of one equivalent of HBr with 1,3-butadiene gives different products at under different conditions, at 0ºC gives the following major product (See the drawing)
Explanation:
The low-temperature product is formed via an electrophilic addition, with Markovnikov regioselectivity.
The first step is the protonation of one of the C=C double bonds. The protonation occurs regioselectively to give the more stable carbocation.
The most stable cation is not only secondary, but also allyl, and therefore has resonance stabilization. Observing the resonance forms of the carbocation in the drawing.
This carbocation is reorganized to the most stable carbocation.
In the second step the nucleophilic bromine attacks the carbocation, resulting in the largest product being 3-bromo-1-butene.
