Write a hydrohalogenation reaction with excess HCl and name the organic product formed.

• A.
• B.
• C.
• D.

Answer: (A)

Verified step by step guidance

1. Identify the starting alkene structure. In this case, the alkene is 2-methyl-2-butene, which has a double bond between the second and third carbon atoms.
2. Understand that hydrohalogenation involves the addition of a hydrogen halide (HCl) across the double bond. The reaction proceeds via electrophilic addition, where the alkene's pi bond attacks the proton (H+) from HCl.
3. Apply Markovnikov's rule: the hydrogen atom from HCl will add to the carbon of the double bond that already has more hydrogen atoms, and the chlorine (Cl) will add to the carbon with fewer hydrogen atoms. This leads to the formation of the most stable carbocation intermediate.
4. Draw the carbocation intermediate formed after the protonation step. In this case, the carbocation forms on the more substituted carbon (the carbon with fewer hydrogens), which is stabilized by the adjacent alkyl groups.
5. Finally, the chloride ion (Cl-) attacks the carbocation, resulting in the formation of the alkyl chloride product. The product is 2-chloro-2-methylbutane, where the chlorine is attached to the more substituted carbon atom.