Question

2 -methylbutane on reacting with bromine in the presence of sunlight gives mainly (a) 1-bromo-2-methylbutane (b) 2 -bromo-2-methylbutane (c) 2-bromo-3-methylbutane (d) 1-bromo-3-methylbutane

Short answer

The main product is (b) 2-bromo-2-methylbutane.

Step-by-step solution

Understand the Reaction Type

The reaction of 2-methylbutane with bromine in the presence of sunlight is a free radical halogenation reaction. This type of reaction involves the substitution of a hydrogen atom in an alkane with a bromine atom.

Identify Possible Products

In free radical halogenation, the bromine atom can replace a hydrogen atom at different positions on the alkane. For 2-methylbutane, potential positions for substitution are at the 1st, 2nd, 3rd, or 4th carbon.

Determine Preferred Substitution

The stability of the carbon radical intermediate formed during halogenation determines the main product. Middling carbon atoms (like tertiary or secondary carbons) lead to more stable radicals.

Focus on the Most Stable Intermediate

For 2-methylbutane, the most stable intermediate results from the hydrogen at the 2nd carbon being replaced due to the tertiary carbon's stability.

Predict the Main Product

Given the formation of the most stable tertiary radical, the predominant product is formed by substitution at the second position. Thus, the main product is 2-bromo-2-methylbutane.

Key concepts

Reaction Mechanism

In organic chemistry, understanding the reaction mechanism helps explain how and why reactions occur the way they do. Free radical halogenation is one such mechanism, where halogen atoms, like bromine, replace hydrogen atoms in organic molecules. The process generally requires an initiator such as heat or light to break the bond in the halogen molecule, forming free radicals.

• Initiation: Light or heat splits the bromine molecule (\( \text{Br}_2 \)) into two bromine radicals.
• Propagation: The bromine radicals react with 2-methylbutane, leading to radicals on carbon atoms.
• Termination: Radicals recombine to form products, ceasing the chain reaction.

The initiation stage starts the reaction, the propagation stage helps develop the main products, and the termination stage finishes the reaction by using up the radicals.

Organic Chemistry

Organic chemistry focuses on compounds containing carbon. Halogenation is a significant reaction in organic chemistry for functionalizing alkanes. Alkanes, consisting of carbon and hydrogen in a stable configuration, undergo reactions such as substitution when exposed to halogens under specific conditions. In this process, bromine is used to replace hydrogen atoms in alkanes, and in our case, 2-methylbutane serves as the substrate. The reaction is facilitated by the presence of sunlight, which provides energy to break bromine's bond and generate free radicals. The overall chemistry of alkanes involves relatively stable molecules until specific conditions, like with free radical mechanisms, are introduced to activate reactions.

Alkane Substitution

Substitution reactions in alkanes, such as the free radical halogenation, are crucial for transforming simple alkanes into more complex molecules. Alkanes have saturated carbon atoms, meaning all carbon bonds are single. Substitution allows new functional groups to replace hydrogen, expanding the functional capabilities of the molecule. In the case of 2-methylbutane, bromine replaces one hydrogen atom, producing a brominated compound. The position of the substitution majorly determines the stability of the reaction's intermediate, affecting the final product. For 2-methylbutane, substitution at the tertiary carbon (second carbon) is more favorable, resulting in the formation of 2-bromo-2-methylbutane.

• This process diversifies the chemical properties of alkanes.
• Stability of intermediates influences reaction pathways and products.
• Position of substitution is key to determining the main product.