Q.(a) Write the IUPAC names of the products obtained by the addition reaction of HBr to hex-1-ene:

Markovnikov vs Anti-Markovnikov Addition

Imagine you're adding something across a double bond — say, H–Br. The double bond is like a crowded room: one carbon has more friends (alkyl groups) attached, the other has fewer. Where does the hydrogen go? That's the core question.

The Intuition: Stability Wins

A carbocation (a carbon with a positive charge) forms as an intermediate in many addition reactions. The more alkyl groups attached to that positively charged carbon, the more stable it is — alkyl groups push electrons toward the positive centre, stabilising it.

So when H⁺ from H–Br attacks the double bond, it will choose the path that leads to the more stable carbocation. That means the hydrogen goes to the carbon that already has more hydrogens (the less substituted carbon), leaving the positive charge on the more substituted carbon (which has more alkyl groups to stabilise it).

The Precise Statement

Markovnikov Addition: When H–X adds to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogen atoms already attached, and the halogen (or other group) attaches to the carbon with fewer hydrogen atoms.

Anti-Markovnikov Addition: The opposite — hydrogen attaches to the carbon with fewer hydrogen atoms, and the other group attaches to the carbon with more hydrogen atoms.

Why Does Anti-Markovnikov Happen?

Normal H–Br addition follows Markovnikov. But if you add peroxides (ROOR) to the reaction, the mechanism changes from ionic to free radical. In the radical mechanism, the bromine radical attacks first, and it prefers to form the more stable radical intermediate — which is the more substituted carbon radical. That forces the hydrogen to go to the other carbon, giving anti-Markovnikov product.

Quick Comparison Table

| Feature | Markovnikov | Anti-Markovnikov |

|---------|-------------|------------------| …