Question

Draw the structure of the major product in the following reaction: \( CH_{3}CH = C(CH_{3})_{2} + HBr \rightarrow \)

Hint:

Markovnikov addition is used unless peroxides are present.
With peroxides, \( HBr \) adds via an Anti-Markovnikov mechanism (Kharasch effect), but this only works for \( HBr \).

Answer 1

Concept:

• The addition of hydrogen halides (\( HX \)) to unsymmetrical alkenes follows Markovnikov's Rule.

• According to this rule, the "rich get richer": the hydrogen atom attaches to the doubly bonded carbon that already has the greater number of hydrogen atoms.

• Mechanistically, the reaction proceeds via the formation of the most stable carbocation intermediate.
Step 1: Identify the structure of the alkene
The alkene is 2-methylbut-2-ene: \( CH_{3}-CH=C(CH_{3})_{2} \).
Carbon-2 is bonded to two methyl groups (zero hydrogens).
Carbon-3 is bonded to one methyl group and one hydrogen atom.

Step 2: Determine carbocation stability
When \( H^{+} \) from \( HBr \) adds to Carbon-3, a carbocation forms at Carbon-2. This is a tertiary (\( 3^\circ \)) carbocation, which is highly stable due to inductive effects and hyperconjugation from three alkyl groups.
If \( H^{+} \) added to Carbon-2, a secondary (\( 2^\circ \)) carbocation would form at Carbon-3, which is less stable.

Step 3: Attack of the nucleophile
The bromide ion (\( Br^{-} \)) attacks the more stable tertiary carbocation at Carbon-2.

Step 4: Final Product Structure
The product is 2-bromo-2-methylbutane: \[ CH_{3}-CH_{2}-C(Br)(CH_{3})_{2} \] The final answer is 2-bromo-2-methylbutane.