Question

Which of the following compounds is formed as major product in the following reaction?
\[ \text{2-Methylbut-2-ene} \xrightarrow{\text{HBr, Peroxide}} \text{Product} \]

• A. 3-Bromo-2-methylbutane
• B. 2-Bromo-2-methylbutane
• C. 2-Bromo-3-methylbutane
• D. 2-Bromobutane

Hint:

In anti-Markovnikov addition reactions, the halogen adds to the less substituted carbon, and the hydrogen adds to the more substituted carbon.

Answer 1

The Correct Option is C

Step 1: Identify the type of reaction.
The reaction given is a free radical addition of HBr to an alkene, where peroxide is used. This is a typical example of an anti-Markovnikov addition reaction. In this case, the bromine adds to the carbon that has fewer alkyl groups attached (the less substituted carbon), and the hydrogen adds to the more substituted carbon.

Step 2: Apply the anti-Markovnikov rule.
For 2-methylbut-2-ene, the more substituted carbon is the one at position 2, and the less substituted carbon is at position 3. In anti-Markovnikov addition, the bromine will add to the less substituted carbon (carbon 3), and the hydrogen will add to the more substituted carbon (carbon 2). This gives the product 2-bromo-3-methylbutane.

Step 3: Final conclusion.
Thus, the major product of the reaction is:
\[ \boxed{(3)\ 2\text{-Bromo-3-methylbutane}} \]