Problem 4 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The major product formed in the following reaction is : \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}\left(\mathrm{CH}_{3}\right)_{2} \stackrel{\mathrm{HBr}}{\longrightarrow}\) [Main Sep. 05, 2020 (II)] (a) \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) (c) \(\mathrm{Br}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C}(\mathrm{Br})\left(\mathrm{CH}_{3}\right)_{2}\)

Step-by-Step Solution

Answer

The major product is (a) \(\mathrm{CH}_3 \mathrm{CH}( ext{Br})\mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\).

Step 1: Identify Reactants and Reaction Type

The given reactant is a 3-methyl-1-butene molecule \( \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}\left(\mathrm{CH}_{3}\right)_{2} \). The reaction involves the addition of HBr, indicating this is an electrophilic addition reaction.

Step 2: Apply Markovnikov's Rule

According to Markovnikov's rule, the hydrogen atom from HBr will add to the carbon with the greater number of hydrogen atoms, and the bromine atom will add to the other carbon in the double bond. In this case, the hydrogen will add to the terminal carbon of the double bond (\(\mathrm{CH}_{2}=\)).

Step 3: Determine the Major Product

The hydrogen from HBr will attach to the CH adjacent to the terminal carbon (converting \(\mathrm{CH}=\) to \(\mathrm{CH}_{2}\)), while Br will add to the other carbon, forming \(\mathrm{CH}_3 \mathrm{CH}( ext{Br})\mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\).

Key Concepts

Markovnikov's Rule

Markovnikov's Rule is a principle in organic chemistry that helps predict the outcome of certain addition reactions, particularly those involving alkenes and an electrophilic addition, like HX (where X is a halogen). According to Markovnikov's Rule, when a protic acid (HX) is added to an asymmetric alkene, the hydrogen atom will attach to the carbon with more hydrogen atoms, while the halide will attach to the other carbon.
- It's about understanding the stability of carbocations. - The rule suggests that the more stable carbocation intermediate will form first. - This leads to the major product of the reaction.
In the given reaction, applying Markovnikov's rule simplifies the process of determining the major product.

Organic Chemistry

Organic chemistry is the study of the structure, properties, composition, reactions, and preparation of carbon-containing compounds, which essentially includes not just hydrocarbons but also compounds with any number of other elements, including hydrogen (most compounds), nitrogen, oxygen, the halogens, sometimes sulfur, phosphorus, silicon, and more.
- It focuses on molecules and their transformations. - Functional groups are key players that determine molecular behavior.
In the exercise, we explore an electrophilic addition in organic chemistry, which is a typical transformation used to change the structure or functional groups of hydrocarbon chains.

Reaction Mechanism

A reaction mechanism describes the step-by-step sequence of events at the molecular level that occur during the course of a chemical reaction. Understanding the mechanism helps chemists develop predictions about reaction behavior and outcomes.
- It involves mapping out each transition state and intermediate. - In electrophilic addition, the double bond of an alkene acts as a nucleophile.
During the electrophilic addition of HBr to 3-methyl-1-butene, - The π electrons from the double bond attack the H of HBr, forming a carbocation intermediate. - The bromide ion ( Br^- ) then attacks the carbocation, leading to the final product formation.
Knowing this, the major product can be derived logically by following the sequence of electron movements.