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.

See explanation and image attached

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

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).

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