Problem 87 - A Complete Resource Book in Chemistry for JEE Main

Question

An organic compound \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\) is found to be optically active. Which of the following is correct structure of the given compound? (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCHO}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\)

Step-by-Step Solution

Answer

The correct structure is option (b): \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}\).

Step 1: Understand Optical Activity

Optically active compounds have molecules that can rotate the plane of plane-polarized light. For a molecule to be optically active, it must be chiral. This generally means that the molecule must have a carbon atom attached to four different groups, making it asymmetric.

Step 2: Analyze the Compound's Formula

The molecular formula given is \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\). We are to determine which of the provided options corresponds to a structure that could be optically active, which means identifying a chiral center.

Step 3: Evaluate Each Structure for Chirality

- Option (a): \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCHO}\) has a carbon bonded to two identical methyl groups. This cannot be chiral.- Option (b): \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}\) has a carbon with an OH group. Check if it's chiral.- Option (c): \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) has no carbon bonded to four different groups.- Option (d): \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) has no asymmetric carbon atom and cannot be chiral.

Step 4: Identify the Chiral Center

In option (b), the structure \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}\) has the second carbon atom bonded to a hydroxyl group (\(\mathrm{OH}\)), hydrogen, an ethyl group (\(\mathrm{CH}_{2}=\mathrm{CH}-\)), and a methyl group (\(\mathrm{CH}_{3}\)). This carbon is bonded to four different groups, indicating it is a chiral center.

Key Concepts

Chiral Center

A chiral center in a molecule is a critical aspect of understanding optical activity. Imagine standing at a car's governor wheel—your location in relation to the vehicle changes what you see and how you interact with the surroundings. Similarly, a chiral center in an organic compound provides a unique spatial arrangement, significantly affecting how the molecule behaves.

This uniqueness is attributed to a carbon atom being attached to four different groups. In the compound from our exercise, the configuration of these groups varies, giving the molecule a property called chirality. This configuration makes it impossible for the molecule to be superimposed on its mirror image, leading to optical activity as these structures can rotate plane-polarized light.

Organic Compound

Organic compounds are molecules that are mainly composed of carbon and hydrogen atoms. They often include other elements such as oxygen, nitrogen, and more. The backbone of these structures is typically carbon, allowing for versatile bonding and structural complexity.

In the case of the compound (\(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\)), this is an organic compound comprising four carbon atoms, eight hydrogen atoms, and one oxygen atom. This simple combination can form various structures, as seen in the options considered in the solution. Organic chemistry focuses a lot on understanding how changes in these molecular structures impact physical properties like optical activity.

Asymmetric Carbon

An asymmetric carbon atom is known as a stereocenter—a key player in determining the chirality of the molecule. Imagine a glove: it fits one shape perfectly but not its mirror image. Similarly, an asymmetric carbon makes its host molecule chiral and allows it to rotate plane-polarized light.

In our compound example, the carbon bonded to an OH group, hydrogen, ethyl, and methyl groups forms an asymmetric center. This particular configuration ensures there's no identical mirror image, thus contributing to the optical activity of the compound.

Molecular Structure Analysis

Molecular structure analysis helps us understand the arrangement and connection of atoms within molecules. It is vital in determining the chemical and physical properties of a compound. This analysis allows chemists to understand why certain molecules act the way they do, such as being optically active.

For this reason, evaluating each molecular structure against others becomes essential. In our exercise, a careful inspection of bond connections and spatial orientation helped us identify the presence of a chiral center. Such analyses are crucial to predict or manipulate this property in synthetic and medicinal chemistry, ultimately affecting how a compound might interact in biological systems or react in chemical processes.