Question

In the presence of platinum catalyst, hydrocarbon (A) adds hydrogen to form $\mathrm{n}$ -hexane. When hydrogen bromide is added to (A) instead of hydrogen, only a single bromo compound is formed. Which of the following is (A)? (1) ${\mathbf{C}\mathbf{H}}_2=\mathbf{C}\mathbf{H}-{\mathbf{C}\mathbf{H}}_2{\mathrm{CH}}_2{\mathbf{C}\mathbf{H}}_2{\mathrm{CH}}_3$ (2) ${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CHCH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_3$ (3) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}={\mathrm{CHCH}}_2{\mathrm{CH}}_3$ (4) ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{CH}={\mathrm{CHCH}}_3$

Short answer

Option 4: ${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{CH}={\text{CHCH}}_3$

Step-by-step solution

Understand the addition of hydrogen

When a hydrocarbon (A) adds hydrogen in the presence of a platinum catalyst, it forms n-hexane. This means that (A) must have six carbon atoms and at least one double bond which gets saturated to form n-hexane.

Analyze addition of hydrogen bromide

Hydrogen bromide (HBr) adds to the double bond in hydrocarbon (A). Since only a single bromo compound is formed, the double bond's location is such that the addition of HBr does not lead to isomeric products according to Markovnikov's rule.

Evaluate each option

Evaluate each option to find which hydrocarbon (A) is consistent with the given information.

Option 1

Option 1: ${\text{CH}}_2=\text{CH}-{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{CH}}_3$. Upon adding H_2, this forms n-hexane. However, the addition of HBr would lead to two different products (1-bromohexane and 2-bromohexane). Therefore, Option 1 is incorrect.

Option 2

Option 2: ${\text{CH}}_3\text{CH}={\text{CHCH}}_2{\text{CH}}_2{\text{CH}}_3$. Adding H_2 forms n-hexane. The addition of HBr would give two products (2-bromohexane and 3-bromohexane), making this option incorrect.

Option 3

Option 3: ${\text{CH}}_3{\text{CH}}_2\text{CH}={\text{CHCH}}_2{\text{CH}}_3$. Adding H_2 forms n-hexane. The addition of HBr would again lead to two products (3-bromohexane and 2-bromohexane), hence, this option is incorrect.

Option 4

Option 4: ${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{CH}={\text{CHCH}}_3$. Adding H_2 forms n-hexane. The addition of HBr forms only 2-bromohexane. Hence, this option is correct as it aligns perfectly with the given conditions.

Key concepts

Markovnikov's rule

Markovnikov's rule helps us predict where a hydrogen atom will attach in an alkene when an asymmetrical reagent, like hydrogen bromide (HBr), is added.

According to this rule, the hydrogen atom in HBr will attach to the carbon of the double bond that already has more hydrogen atoms, while the bromine (Br) will attach to the carbon with fewer hydrogen atoms.

This rule minimizes the formation of isomeric products, ensuring a more predictable outcome. So, in step-by-step solutions involving HBr and alkenes, Markovnikov's rule guides where each part of the molecule attaches, forming a single main product rather than multiple isomeric products.

hydrogen addition

Hydrogen addition is a fundamental reaction in organic chemistry, where hydrogen (H_2) is added across a carbon-carbon double bond in a hydrocarbon, resulting in a saturated alkane.

This process is often catalyzed by metals like platinum, palladium, or nickel to help break the double bond and allow hydrogen atoms to attach to each carbon.

In the exercise, the addition of hydrogen to hydrocarbon (A) turns it into n-hexane, a fully saturated hydrocarbon chain with no double bonds.

It is important to note that this reaction will always lead to a single, predictable product with no possibility of isomers forming.

hydrogen bromide addition

Hydrogen bromide addition to an alkene involves breaking the carbon-carbon double bond and attaching the hydrogen and bromine atoms to the carbons.

Based on Markovnikov's rule, hydrogen attaches to the carbon with more hydrogen atoms already present, and bromine attaches to the carbon with fewer hydrogen atoms.

In this case, the exercise shows that adding HBr to hydrocarbon (A) results in a single bromo compound. The placement of the double bond allows for a single product rather than multiple isomeric forms, highlighting the predictability of this reaction mechanism.

platinum catalyst

A platinum catalyst is commonly used in hydrogenation reactions due to its ability to facilitate the addition of hydrogen to alkenes efficiently.

It works by adsorbing the hydrogen gas and the alkene onto its surface, weakening the double bond in the alkene and allowing the hydrogen atoms to attach.

In the exercise, platinum speeds up the hydrogen addition to the hydrocarbon (A), converting it into n-hexane seamlessly. This ensures the reaction occurs under mild conditions with a high yield of the desired product.

isomeric products

Isomeric products arise when a reaction can produce molecules with the same molecular formula but different structural arrangements.

In the context of hydrogen bromide addition to alkenes, the formation of isomers depends on the location of the double bond and the application of Markovnikov's rule.

The exercise demonstrates that only a single bromo compound forms, indicating no isomeric products are generated.

This specificity is due to the double bond's strategic location in hydrocarbon (A), which prevents multiple structural variations and allows for a straightforward prediction of the reaction outcome.