Question

When 3-methyl-1-butene reacts with HBr, two alkyl halides are formed: 2-bromo-3-methylbutane and 2-bromo-2-methylbutane. Propose a mechanism that explains the formation of these two products.

Short answer

In this markoniv addition takes place.

Markoniv addition= Positive ion will go to that double bonded carbon which has more no of hydrogens attached and negative ion will go towards that carbon which has less number of hydrogens attached.

Step-by-step solution

Step 1: How the reaction proceed

In this markoniv addition takes place.

Markoniv addition= Positive ion will go to that double bonded carbon which has more no of hydrogens attached and negative ion will go towards that carbon which has less number of hydrogens attached.

Final answer

Here first double bond attack on H and HBr releases Br^- ion. Then it will form carbonation which on hydride shift form 3° carbocation. Which is more stable than 2° carbocation. Br^- reacts on both 2° and 3° carbocation and form two products i.e. 2-bromo-3-methylbutane and 2-bromo-2-methylbutane.