Problem 53 - A Complete Resource Book in Chemistry for JEE Main

Question

The product formed on reaction of HBr with propene is (a) 1 -bromoproapne (b) 2 -bromopropane (c) 3 -bromopropane (d) no reaction occurs

Step-by-Step Solution

Answer

The product is 2-bromopropane (option b).

Step 1: Understand the Reactants

The reactants involved in the reaction are HBr (hydrobromic acid) and propene. Propene is an alkene with the formula \( C_3H_6 \). It contains a carbon-carbon double bond which makes it a suitable candidate for addition reactions.

Step 2: Analyze the Reaction Type

The reaction between HBr and propene is an example of an electrophilic addition reaction. In this type of reaction, the hydrogen atom from HBr adds to one of the carbon atoms in the double bond, and the bromine atom adds to the other carbon atom.

Step 3: Apply Markovnikov's Rule

According to Markovnikov's Rule, in an addition reaction of HX (where X is a halogen) to an alkene, the hydrogen will add to the carbon with the most hydrogen atoms already attached. In propene, the double bond is between the first and second carbon atoms.

Step 4: Predict the Product

Following Markovnikov's Rule, the hydrogen from HBr will add to the carbon atom that is bonded to two hydrogen atoms (the first carbon in propene), while the bromine will add to the carbon bonded to only one hydrogen (the middle carbon atom), resulting in 2-bromopropane.

Key Concepts

Markovnikov's Rule

In electrophilic addition reactions involving alkenes, Markovnikov's Rule guides us in predicting where each atom will attach itself in the final product. This rule states that when a protic acid (like HBr) adds to an alkene, the hydrogen atom will bond to the carbon with the maximum number of hydrogen atoms already connected to it. This results in a more stable carbocation intermediate, ultimately directing the formation of the final product.

In the case of propene, which consists of a double bond between the first and second carbon atoms, the hydrogen from HBr will attach to the carbon already having more hydrogen atoms, i.e., the first carbon. As a result, the bromine will join the second carbon, following Markovnikov's Rule. This leads to the formation of 2-bromopropane, which is the most stable product.

Markovnikov's Rule is crucial in predicting the major product of the reaction, ensuring that the most stable and common product is formed when dealing with asymmetric alkenes.

Reaction Mechanism

Understanding the mechanism behind electrophilic addition reactions is essential for grasping how the atoms rearrange. In our reaction, propene reacts with HBr through a two-step mechanism. Initially, the alkene's double bond acts as a nucleophile and attracts the hydrogen from HBr, leaving a bromide ion.

This step results in the formation of a carbocation intermediate on the more substituted carbon (where former double bond existed), creating a structure eager to complete its valency by being further stabilized. The resulting carbocation is why Markovnikov's Rule predicts the specific attachment of atoms.

For the final step, the bromide ion acts as a nucleophile and bonds to the positively charged carbon, completing the sequence of the electrophilic addition mechanism. These steps illustrate the reactivity of alkenes with electrophiles, highlighting the organized, step-by-step atom attachment and rearrangement.

Alkenes

Alkenes are hydrocarbons containing at least one carbon-carbon double bond which is key to their reactivity in various chemical reactions. The general formula for alkenes is \[ C_nH_{2n} \].

Their ability to engage in addition reactions arises from the presence of the double bond, increasing their reactivity compared to single-bonded alkanes. This double bond involves a pair of electrons available for reacting with other atoms or molecules.

In electrophilic addition reactions, alkenes like propene serve as an excellent example. Their nature becomes clear when they interact with electrophiles, demonstrating how they can transition from an unsaturated state to form more stable, saturated products.

Understanding the properties of alkenes, such as their reactivity and the role the double bond plays, provides insight into why these compounds are particularly important in organic chemistry and synthetic processes.