Problem: Predict the products of the reaction of 2-methylpropene with HBr and explain the mechanism.

Given: 2-Methylpropene (isobutylene): (CH3)2C=CH2, SMILES: CC(=C)C + HBr → ?

Step 1: Identify the reaction type. Alkene + HBr = electrophilic addition. HBr provides H+ (electrophile) and Br- (nucleophile).

Step 2: Apply Markovnikov's rule to determine protonation site. The two possible protonation sites are:

• H+ adds to the terminal =CH2 (less substituted, more H atoms) → gives tertiary carbocation at C-2
• H+ adds to the internal =C(CH3)2 → gives primary carbocation at terminal carbon

Markovnikov's rule: H adds to the carbon bearing more H atoms (terminal CH2), generating the more stable carbocation.

Step 3: Identify the carbocation intermediate. Protonation at CH2 end: (CH3)2C+ -CH3 — wait, the carbocation forms at C-1 of the original molecule: (CH3)2C+-CH3. This is a tertiary carbocation (3°) — most stable (three methyl groups). SMILES: C[C+](C)C

Step 4: Nucleophilic attack by Br-. Br- attacks the electron-deficient tertiary carbocation at C-1 (the now +charged carbon): (CH3)2C+ + Br- → (CH3)2CBr-CH3...

Actually let's redo: 2-methylpropene = CH2=C(CH3)2. H+ adds to =CH2 → product is (CH3)2C+-CH3... no: (CH3)2C+- is at the sp2 carbon. After H adds to the terminal CH2=: CH3-C+(CH3)2... The cation is at the C that lost the pi bond (C-1). Product: 2-bromo-2-methylpropane: (CH3)3CBr. SMILES: CC(C)(C)Br

Step 5: Write the product and name it. Major product: 2-bromo-2-methylpropane (tert-butyl bromide). SMILES: CC(C)(C)Br Minor product (if H+ adds to the other C): 1-bromo-2-methylpropane. SMILES: CC(C)CBr

Verification using energy argument: The tertiary carbocation intermediate [(CH3)3C+] is more stable than the primary carbocation [CH2Br-CH(CH3)2 → primary]. Reaction proceeds via the more stable (lower activation energy) intermediate → major product is the Markovnikov product.

Answer: Major product = 2-bromo-2-methylpropane; Minor product = 1-bromo-2-methylpropane.