Excess of isobutane on reaction with $Br_2$ in presence of light at $125^\circ C$ gives which one of the following, as the major product ?
Show Hint
Bromination is highly regioselective. Always look for the most substituted carbon (tertiary > secondary > primary) to place the bromine atom for the major product.
Step 1: Understanding the Concept:
The reaction of an alkane with bromine in the presence of light is a free radical substitution reaction. The selectivity of bromine is very high compared to chlorine, favoring the substitution of hydrogen atoms on more stable radical intermediates. Step 2: Key Formula or Approach:
Stability of free radicals follows the order: $3^\circ>2^\circ>1^\circ>\text{methyl}$.
The rate of bromination for $3^\circ : 2^\circ : 1^\circ$ hydrogens is approximately $1600 : 82 : 1$. Step 3: Detailed Explanation:
Isobutane is 2-methylpropane, having the structure $(CH_3)_3CH$.
It contains nine $1^\circ$ hydrogens and one $3^\circ$ hydrogen.
During bromination, two types of radicals can form:
1. A primary radical $(CH_3)_2CH\dot{C}H_2$ by abstracting a $1^\circ$ hydrogen.
2. A tertiary radical $(CH_3)_3\dot{C}$ by abstracting the $3^\circ$ hydrogen.
Because the tertiary radical is significantly more stable and the selectivity of bromine for $3^\circ$ hydrogens is extremely high (1600 times that of $1^\circ$), the reaction occurs almost exclusively at the tertiary carbon.
The resulting major product is 2-bromo-2-methylpropane (tert-butyl bromide). Step 4: Final Answer:
The major product is $(CH_3)_3C-Br$.
