Predict the structure of the major product formed by 1,2 -addition of HCl to 2-methyl1,3 -butadiene (isoprene).

In this case, the reaction taking place is the addition of HCl to 2-methyl-1,3-butadiene (isoprene), which is an example of an electrophilic addition reaction. In electrophilic addition reactions, an electrophile (e.g. HCl) is added to a nucleophile (e.g. the isoprene molecule).

First, let's examine the isoprene molecule, 2-methyl-1,3-butadiene: C5H8. This molecule contains two double bonds (C=C), making it a diene. Now, let's look at HCl, which is a strong acid. In this reaction, HCl will dissociate into a proton (H+) and a chloride ion (Cl-). The proton will act as an electrophile, while the chloride ion will act as a nucleophile later in the reaction.

Markovnikov's rule states that when an electrophile is added to an alkene or a diene, the hydrogen atom will bond to the carbon with the most hydrogens already connected, while the other part of the electrophile (in this case, Cl-) will bond to the carbon with fewer hydrogens connected. This rule helps to form the most stable carbocation intermediate during the reaction. In our case, when the proton (H+) from HCl is added to the isoprene molecule, it will form a carbocation intermediate. The most stable carbocation will have the positive charge on a tertiary (3°) carbon, if possible. By following Markovnikov's rule, the proton (H+) will bond to one of the terminal carbons (C1) of the double bond, since the terminal carbon has two hydrogens already connected. This creates a carbocation at the neighboring carbon (C2), which is a secondary (2°) carbon with a positive charge.

Now that we have our carbocation intermediate, the chloride ion (Cl-) will act as a nucleophile and attack the positively charged carbon (C2) to form a new C-Cl bond. This step completes the 1,2-addition of HCl to isoprene.

After the nucleophilic attack by the chloride ion, we can now predict the structure of the major product. The molecule will have a new C-Cl bond at C2, and the double bond between C3 and C4 will remain intact. The resulting product will be 2-chloro-3-methyl-2-butene (C5H9Cl).