Animated Solution for Chemistry - Organic Compounds Containing Halogens: Excess of isobutane on reaction with $B{r}_2$ in presence of light at $125^{\circ }C$ gives which one of the following, as the major product?

A Classic Organic Encounter

Imagine you are in a chemistry lab, and you mix isobutane with bromine gas. You shine a bright UV light on the flask and heat it to $125^{\circ }C$. What happens next is a beautiful, chaotic dance of atoms known as a Free Radical Substitution reaction.

The presence of light ($h u$) is the ultimate trigger. It pumps enough energy into the system to homolytically cleave the relatively weak $Br-Br$ bond. This means the bond breaks evenly, and each bromine atom walks away with one electron, creating two highly reactive bromine free radicals ($B{r}^{\bullet }$). This is our Initiation Step.

A Radical Journey

Now, these bromine radicals are hungry for stability, and they look towards the isobutane molecule. Isobutane, or 2-methylpropane, has a central carbon atom bonded to three methyl groups and one lone hydrogen atom.

This gives us two distinct types of hydrogen atoms to choose from: 1. Primary ($1^{\circ }$) hydrogens: There are nine of these, located on the three methyl groups. 2. Tertiary ($3^{\circ }$) hydrogen: There is only one, located on the central carbon.

The bromine radical must abstract one of these hydrogens to form $HBr$, leaving behind an alkyl free radical. But which one will it choose?

Selectivity is Key

Here is where the magic of chemical selectivity comes into play. Bromination is an endothermic process, meaning it is highly selective and deeply cares about the stability of the intermediate it forms.

If it takes a primary hydrogen, it forms a primary free radical. If it takes the tertiary hydrogen, it forms a tertiary free radical. According to the laws of organic chemistry, the stability of free radicals follows the order: $3^{\circ }>2^{\circ }>1^{\circ }$.

The tertiary radical is significantly more stable because it benefits from hyperconjugation. The nine alpha-hydrogens from the adjacent methyl groups donate electron density into the electron-deficient p-orbital of the central carbon, stabilizing it immensely. Therefore, the bromine radical almost exclusively abstracts the tertiary hydrogen.

This stable tertiary butyl radical then attacks a fresh $B{r}_2$ molecule, capturing a bromine atom to form our product and releasing a new bromine radical to keep the chain reaction going.

Why "Excess" Matters

You might wonder, why doesn't the reaction keep going and add more bromine atoms? The question holds the key: "Excess of isobutane".

Because there is a massive ocean of unreacted isobutane molecules compared to the brominated product, a newly formed bromine radical is statistically far more likely to collide with a fresh isobutane molecule than with a molecule of 2-bromo-2-methylpropane. This effectively shuts down any chance of dibromination or polybromination.

Thus, the reaction cleanly stops at the monobrominated stage, yielding 2-bromo-2-methylpropane as the undisputed major product.