Problem 55 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Identify, \(\mathbf{B}\left(\mathrm{C}_{4} \mathrm{H}_{8}\right)\) which adds on HBr in the presence and in the absence of peroxide to give the same product, \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br}\).

Step-by-Step Solution

Answer

The isomer is t-butylcarbyne.

Step 1: Understand the Problem

We need to identify which isomer of butene (\(\text{C}_4\text{H}_8\)) reacts with hydrogen bromide (HBr) both with and without peroxide to produce the same bromoalkane product (\(\text{C}_4\text{H}_9\text{Br}\)).

Key Concepts

Isomer Identification

The world of organic chemistry is full of interesting compounds that can exist in different forms known as isomers. Isomers are molecules with the same molecular formula but different arrangements of atoms. In this exercise, we are dealing with butene, which has the molecular formula \(\text{C}_4\text{H}_8\). There are several isomers of butene, and identifying the correct one is crucial to determine how it will react under different conditions.

• **But-1-ene** is a linear alkene where the double bond exists between the first and second carbon atoms.
• **But-2-ene** exists in two forms: *cis* and *trans*, where the double bond is between the second and third carbon atoms, but the groups attached are arranged differently.
• Another isomer is **2-methylpropene** where the double bond is within a branched structure.

Finding which of these will consistently lead to the same bromoalkane product when reacted with HBr will solve our problem.

Butene Reactions

When exploring butene reactions, the focus is primarily on the compounds' interaction with reactants to form new products. Butene contains a double bond, making it an alkene, which is reactive towards addition reactions.
During a typical addition reaction, the double bond's electrons attract other atoms, breaking the bond and allowing new bonds to form. For butene, the main addition reactions are:

• **Hydrogenation**: Adding hydrogen to convert the alkene into an alkane.
• **Halogenation**: Adding halogens like bromine to form dihaloalkanes.

However, in our scenario, we are particularly interested in the addition of hydrogen bromide (HBr), which forms a specific product: bromoalkane.

Hydrogen Bromide Addition

Adding hydrogen bromide (HBr) to an alkene like butene results in an intriguing reaction where the hydrogen atom attaches to one carbon, and the bromine atom attaches to the other in the double bond. This breaks the double bond, forming a saturated molecule. The specifics of the HBr addition to butene are guided by Markovnikov's rule, which states:

• The hydrogen atom will prefer to bond with the carbon atom in the double bond that already has more hydrogen atoms.
• The bromine atom will attach to the carbon with fewer hydrogen atoms.

For but-2-ene, whether HBr is added in the presence or absence of peroxides, the result is 2-bromobutane, a product formed consistently without rearrangement unlike in other isomers.

Peroxide Effect

The peroxide effect is a phenomenon that can alter the outcome of an addition reaction involving hydrogen bromide and an alkene. Normally, according to Markovnikov's rule, hydrogen attaches to the carbon with more hydrogen atoms already attached. But in the presence of peroxides, this trend flips, called **anti-Markovnikov addition**.
Peroxides initiate a free-radical mechanism where bromine adds first, changing the desired reaction path. Interestingly, the presence of peroxides in reactions with certain isomers such as 1-alkenes causes the bromine to attach to the less substituted carbon first, leading to a different outcome.
However, with but-2-ene specifically, the end product remains the same due to the symmetry of the molecule, creating a consistent product regardless of the reaction conditions.

Bromoalkane Formation

Bromoalkane formation through the addition of HBr to butene demonstrates how alkenes can be transformed into saturated compounds. When HBr adds to but-2-ene, it consistently forms 2-bromobutane no matter the reaction conditions applied, even in the presence of peroxides.
Understanding this involves recognizing that the double bond of but-2-ene is positioned such that both carbons of the bond equally bear the result of addition, giving rise to stability and predictable product formation.

• With **but-2-ene**, the halogen and hydrogen naturally select opposite ends of the double bond, ensuring the same outcome.
• This process clarifies why, despite experimental changes, a single bromoalkane forms consistently, offering valuable insight into chemical selectivity and reaction mechanics.