Question

An alkene reacts with a strong protic acid to form a carbocation. In Part 1 draw the curved arrow notation for the reaction between an alkene and HBr. However, an alkene will react with a halogen electrophile to form a cyclic intermediate. In Part 2 draw the curved arrow notation for the reaction between an alkene and Br2.

Answer 1

See explanation and image attached

Step-by-step explanation:

Now, we have chosen the alkene 1-propene in our example.

In the first reaction of 1-propene with HBr, the reaction proceeds by ionic mechanism leading to the formation of 2-bromo propane.

In the second reaction of 1-propene with the bromine molecule, the first step is the formation of the brominium cation which is a cyclic intermidiate followed by the addition of Br^- yielding the 1,2- dibromopropane product