The major product when 2-methylpropene reacts with HCl is

Answer Details

When 2-methylpropene reacts with HCl, the reaction follows Markovnikov's rule: the hydrogen atom adds to the carbon of the double bond that already has more hydrogen atoms, and the halide (Cl) adds to the more substituted carbon.

2-Methylpropene has the structure CH₂=C(CH₃)₂. The double bond is between C-1 (which bears two hydrogens) and C-2 (which bears no hydrogens but has two methyl groups). According to Markovnikov's rule:

• H adds to C-1 (the less substituted carbon)
• Cl adds to C-2 (the more substituted carbon)

This produces 2-chloro-2-methylpropane, (CH₃)₃CCl. The reaction proceeds via a tertiary carbocation intermediate at C-2, which is the most stable carbocation possible in this molecule. This stability drives the regioselectivity.

The other options are incorrect: 1-chloro-2-methylpropane would result from anti-Markovnikov addition; 3-chloro-2-methylpropane does not correspond to a valid position on a three-carbon chain with a methyl branch; and 2-chloro-3-methylbutane would require a five-carbon skeleton that is not present in the starting material.