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) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}\).

Step 1: Determine Optically Active Compounds

An optically active compound has a chiral center, meaning at least one carbon atom is attached to four different groups. We need to check each option for this feature.

Step 2: Analyze Structure (a)

The structure \((\mathrm{CH}_{3})_{2} \mathrm{CHCHO}\) is 2-butanone. This structure does not contain a carbon that is attached to four different groups, so it is not optically active.

Step 3: Analyze Structure (b)

The structure \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}\) is but-2-ene-1-ol. Checking the third carbon: It is connected to \(-OH\), \(-CH_3\), \(-CH=CH_2\), and \(-H\). This makes it a chiral center as it is bonded to four different groups.

Step 4: Analyze Structure (c)

The structure \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) is 2-butanone. Since all carbon atoms have at least two identical groups, this compound is not optically active.

Step 5: Analyze Structure (d)

The structure \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) is butanal. Butanal is a linear structure without a carbon bonded to four different groups, so it is not optically active.

Step 6: Conclusion

Option (b), \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}\), contains a chiral center and is the correct structure as an optically active compound.

Key Concepts

Optical Activity

Optical activity is a fascinating property found in certain compounds that can rotate the plane of polarized light. This rotation occurs due to the presence of an asymmetrical carbon atom within the compound, referred to as a chiral center or chiral carbon.

• This property is particularly important in understanding how certain molecules interact with light.
• Optically active compounds can be identified by their ability to rotate light, and this can be quantified using equipment like a polarimeter.

In organic chemistry, identifying whether a compound is optically active involves checking for chiral centers. Only those compounds featuring at least one carbon atom attached to four different groups can exhibit this behavior. In our exercise, the compound (b) \( \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3} \) was determined to be optically active, due to the presence of such a chiral center.

Chiral Carbon

A chiral carbon, also known as an asymmetric carbon, is crucial for the optical activity of organic compounds. It is defined by its connection to four different atoms or groups, which creates a non-superimposable mirror image.

• These non-superimposable mirror images are called enantiomers.
• A molecule needs at least one chiral carbon to be optically active.

In the exercise, structure (b) \( \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3} \) was identified as containing a chiral carbon. At the third carbon atom, there are distinct groups: -OH, -CH₃, -CH=CH₂, and -H, making it a chiral center. This is what contributes to its optical activity and differentiates it from other structures like (a), (c), and (d), which lack such chiral centers.

Organic Chemistry Structures

The architecture of organic compounds is fundamental in determining their properties, including optical activity. Organic chemistry focuses heavily on understanding these structures, which are composed of carbon atoms bonded in various configurations and to various groups.
Each organic structure can be analyzed based on its molecular formula, and understanding the nature of its bonds and configuration provides insight into possible chemical behaviors and properties.

• In evaluating the molecular formula \( \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O} \), diverse structures can be proposed.
• Option (b) features a structure where the third carbon atom forms part of an alcohol group \(-OH\), forming the key aspect of the optically active compound.

In context, having a solid grasp of the molecular structure enables chemists to predict properties and reactivities of organic compounds, crucial for applying them in various chemical and pharmaceutical purposes. With optically active compounds, these structural insights also allow for the understanding of how a molecule might react with other chiral substances in nature.