The major product when 2-methylpropene reacts with HCl is

Analysis

The question asks for the major product of the addition reaction between 2-methylpropene (an alkene) and hydrogen chloride (HCl). This is a classic electrophilic addition reaction.

Concept: Markovnikov’s Rule

When an unsymmetrical reagent (like HCl) adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon atom that already has the greater number of hydrogen atoms. The halogen (chlorine) attaches to the carbon atom with fewer hydrogen atoms (the more substituted carbon).

This rule works because the reaction proceeds through the formation of a carbocation intermediate. A more substituted carbocation (tertiary > secondary > primary) is more stable and forms faster, leading to the major product.

Step-by-Step Solution

1. Identify the structure of 2-methylpropene: The structure is $CH_2=C(CH_3)_2$. • Carbon 1 ($CH_2$) has 2 hydrogens. • Carbon 2 ($C$) has 0 hydrogens and is bonded to two methyl groups.

2. The Reaction Mechanism: • The double bond breaks. The $H^+$ from HCl attacks the $CH_2$ group (because it has more hydrogens). • This leaves a positive charge on the central carbon (Carbon 2). • This forms a tertiary carbocation $(CH_3)_3C^+$, which is very stable.

3. Nucleophilic Attack: • The chloride ion ($Cl^-$) then attacks the positively charged central carbon. • The resulting structure is $CH_3-C(Cl)(CH_3)-CH_3$.

4. Naming the Product: • The longest carbon chain has 3 carbons (propane). • There is a chlorine atom on carbon 2 and a methyl group on carbon 2. • The IUPAC name is 2-chloro-2-methylpropane.

Conclusion

Following Markovnikov's rule, the chlorine atom attaches to the more substituted carbon (the central carbon), resulting in 2-chloro-2-methylpropane.

Exam Tip: Whenever you see an addition reaction with an alkene, always look for the "most substituted" carbon. The halogen will almost always end up there!