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

Step 1: Look at where the hydrogens already sit on the double bond.
The alkene is $CH_3-CH=C(CH_3)_2$. One carbon of the double bond carries one hydrogen, the other carbon carries none, it has two methyl groups sitting on it instead. Markovnikov's rule tells us the incoming hydrogen from $HBr$ heads for the carbon that already has more hydrogens.
Step 2: Add the proton and see what cation forms.
So $H^+$ adds to the $CH$ carbon, and the positive charge is left on the carbon that used to hold the two methyl groups. That carbon is now attached to three carbon groups, so we get a tertiary carbocation, and tertiary cations are the most comfortable, most stabilised kind there is, thanks to hyperconjugation and the electron push from three alkyl groups.
Step 3: Let bromide close the loop.
$Br^-$ simply attaches itself to this stable tertiary carbon, giving the final product.
\[ CH_3-CH=C(CH_3)_2 + HBr \rightarrow CH_3-CH_2-\underset{\displaystyle CH_3}{\overset{\displaystyle CH_3}{C}}-Br \]
\[ \boxed{\text{2-bromo-2-methylbutane}} \]