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 (b) \(\text{CH}_2=\text{CH}-\text{CH}(\text{OH})\text{CH}_3\).

Step 1: Understanding Optical Activity

An optically active compound must have a chiral center, meaning one carbon atom is bonded to four different groups. Determine which structures have a chiral center.

Step 2: Evaluating Option (a) \((\text{CH}_3)_2 \text{CHCHO}\)

Visualize the structure: this compound is \(\text{CH}_3 - \text{CH}(\text{CH}_3) - \text{CHO}\). The carbon atom bonded to the \(\text{CHO}\) group is also bonded to two identical \(\text{CH}_3\) groups, meaning this carbon is not a chiral center. Thus, this compound is not optically active.

Step 3: Evaluating Option (b) \(\text{CH}_2=\text{CH}-\text{CH}(\text{OH})\text{CH}_3\)

Considering its structure, \(\text{CH}_2=\text{CH}-\text{CH}(\text{OH})\text{CH}_3\), the carbon atom in the \(\text{CH}(\text{OH})\) position is attached to four different groups: \(-\text{CH}_3\), \(-\text{CH}=\text{CH}_2\), hydroxyl \(\text{OH}\), and a hydrogen atom, thus serving as a chiral center.

Step 4: Evaluating Option (c) \(\text{CH}_3 \text{COCH}_2 \text{CH}_3\)

The structure \(\text{CH}_3 \text{COCH}_2 \text{CH}_3\) is a ketone with no carbon bonded to four different groups, so it cannot be optically active.

Step 5: Evaluating Option (d) \(\text{CH}_3 \text{CH}_2 \text{CH}_2 \text{CHO}\)

The molecular structure \(\text{CH}_3 \text{CH}_2 \text{CH}_2 \text{CHO}\) contains no chiral carbon since none of its carbon atoms are bonded to four different groups.

Step 6: Conclusion and Correct Structure

Among the given options, only option (b) \(\text{CH}_2=\text{CH}-\text{CH}(\text{OH})\text{CH}_3\) possesses a chiral center, which makes it optically active. Therefore, this is the correct structure for the given compound.

Key Concepts

Chiral Centre

In organic chemistry, the concept of a "chiral centre" is central to understanding optical activity. A carbon atom in a molecule becomes a chiral centre when it is bonded to four different groups. This unique configuration leads to molecules that are non-superimposable on their mirror images. To visualize a chiral centre, imagine a carbon atom with four different colored spheres connected to it, each representing a different atom or group.

When examining molecular structures, identifying a chiral centre involves checking each carbon atom’s neighboring groups. If you find one with distinct attachments all around, this carbon is likely chiral. A molecule with at least one chiral centre can have stereoisomers, which are pairs of molecules that are mirror images of each other and not identical, much like left and right hands. These molecules may have different optical properties, such as rotating polarized light in different directions.

This is where optical activity comes into play, making the understanding of chiral centres fundamental for determining the optically active nature of a compound.

Organic Chemistry

Organic chemistry is the study of the structure, properties, composition, reactions, and synthesis of organic compounds, which contain carbon. It is a branch of chemistry that has a significant focus on carbon-containing compounds due to carbon's unique ability to form strong covalent bonds with itself and other elements. This creates an immense variety of organic molecules with diverse characteristics and uses.

In organic chemistry, the versatility of carbon is showcased through numerous functional groups, isomers, and geometric arrangements that create fascinating molecules. These molecules can range from the simplest hydrocarbons to complex natural compounds like proteins and DNA. Concepts such as chirality and optical activity are essential in organic chemistry because they help explain how molecules interact with biological systems and their environments.

Whether you’re looking at industrial chemical processes or the intricate chemistry of life itself, the study of organic compounds, like the optically active compound in this exercise, plays a crucial role.

Optically Active Compounds

Optically active compounds have the distinctive ability to rotate the plane of polarized light. This property is intrinsically linked to the presence of chiral centres within the molecule. When light passes through a solution of an optically active compound, it can be rotated either to the left (levorotatory) or to the right (dextrorotatory), depending on the specific configuration of the molecule.

The optical activity of a compound is measured using an instrument called a polarimeter, which can help determine the direction and degree of rotation. The understanding of optical activity is crucial in various fields, such as pharmaceuticals, where the optical isomer of a drug can affect its efficacy and safety.

In the context of this exercise, option (b) \(\text{CH}_2=\text{CH}-\text{CH}(\text{OH})\text{CH}_3\)is confirmed as having a chiral centre, making the compound optically active. This illustrates the vital relationship between molecular structure and optical properties in organic chemistry.