Feynman Note — Why Does Anti-Markovnikov Work Only with HBr? — Notes

Explaining Anti-Markovnikov Selectivity Like a Story

Imagine you are a chain reaction factory. To keep the factory running, every step must generate energy (be exothermic). If any step costs energy, the factory shuts down.

The factory has 2 steps:

Step A: X• attacks the alkene double bond → alkyl radical + X bonded to C

Step B: Alkyl radical takes H from HX → product + new X•

For HCl:

Step A (Cl• + alkene): Cl• is a small, highly energetic radical. Adding it to a C=C is actually endothermic (costs energy). The factory shuts down at Step A. No product.

For HI:

Step A (I• + alkene): This step IS exothermic — I• adds fine.
Step B (alkyl radical + HI → product + I•): The H–I bond is weak, so this works.
BUT: I• is so large and reactive in other ways that the chain terminates too quickly before significant product forms. Essentially the factory is too disorganized.

For HBr:

Step A (Br• + alkene): Exothermic — works smoothly.
Step B (alkyl radical + HBr → product + Br•): Also exothermic — works perfectly.
Both steps give energy. The factory runs indefinitely until reactants are consumed. Perfect product yield.

The Markovnikov rule of radicals: Br• adds to the less substituted carbon NOT because of electron density, but because this placement gives the more stable (more substituted) alkyl radical at the adjacent carbon. Radical stability follows the same trend as carbocation stability (3° > 2° > 1°).