Question

What is the chief product obtained when \(\mathrm{n}\) -butane is treated with \(\mathrm{Br}_{2}\) in the presence of light at \(130^{\circ} \mathrm{C}\) ? (1) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}_{3}\) (2) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{Br}\) (3) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}\) (4) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\)

Short answer

Option (1) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}_{3}\)

Step-by-step solution

- Identify the Reactants and Conditions

The reaction involves \(\mathrm{n}\)-butane \(\left(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3}\right)\) and \(\mathrm{Br}_{2}\) in the presence of light at \(130^{\circ}\mathrm{C}\). Light indicates a free radical halogenation process.

- Understand Free Radical Halogenation

Free radical halogenation typically proceeds through three stages: initiation, propagation, and termination. The most stable radicals form the major products.

- Formation of Radicals

When \(\mathrm{Br}_{2}\) is exposed to light, it splits into two bromine radicals \(\left(\mathrm{Br}\cdot\right)\). These radicals will react with \(\mathrm{n}\)-butane.

- Determine Most Stable Radical

In \(\mathrm{n}\)-butane, the secondary carbon radical \(\left(\mathrm{CH_{3}}\mathrm{CH_{2}}\mathrm{\dot{C}HCH_{3}}\right)\) is more stable than the primary carbon radical. Thus, the secondary position is the most likely site for substitution.

- Bromination at the Secondary Position

The bromine radical will most likely replace the hydrogen at the secondary carbon. This substitution forms \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}_{3}\).

- Identify the Chief Product

Compare the obtained product \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}_{3}\) with the given options. Option (1) matches the structure.

Key concepts

n-butane bromination

When we talk about brominating -butane, we are referring to a chemical reaction where bromine (Br_2) is added to -butane (CH_3CH_2CH_2CH_3). In this case, the reaction is carried out in the presence of light at a high temperature (130°C).

The presence of light indicates that this is a free radical halogenation reaction. Light provides the energy needed to break the bromine molecule (Br_2) into two bromine radicals (Br·). These highly reactive bromine radicals initiate the reaction process leading to bromination of -butane.

The major product of this reaction is determined by factors such as the stability of the carbon radicals formed during the process.

reaction mechanism

Understanding the reaction mechanism of free radical halogenation is crucial for grasping why the chief product is formed. The reaction follows three main stages: initiation, propagation, and termination.

Initiation: Light breaks the bond in bromine (Br_2) to form two bromine radicals (Br·).

Propagation: The bromine radical abstracts a hydrogen atom from -butane, creating a new radical on the -butane (CH_3CH_2CH·CH_3). The newly formed carbon radical then reacts with another bromine molecule (Br_2), producing the brominated product (CH_3CH_2CHBrCH_3) and regenerating the bromine radical to continue the cycle.

Termination: Several termination steps are possible when two radicals combine to form a stable product, thus ending the chain reaction.

chemical stability

In free radical reactions, the stability of the intermediates (in this case, carbon radicals) greatly influences the final product. A secondary carbon radical (CH_3CH_2CH·CH_3) is more stable than a primary carbon radical.Because of this stability, the secondary carbon atom is the preferred site for radical formation and, subsequently, for bromine substitution.

Why Is Stability Important?

• More stable intermediates lower the energy barrier for the reaction step, making it more favorable.
• The reaction tends to produce more of the stable product, reducing the formation of less stable products.

Hence, in the case of -butane bromination, the secondary carbon radical leads to the chief product, which is CH_3CH_2CHBrCH_3.