Question

Which of the following carbonyl compounds on reduction with \(\mathrm{LiAlH}_{4}\) and subsequent acidification yields four stereoisomeric alcohols? (A) CC1CCC(=O)CC1 (B) CC1CCCCC1=O (C) O=C1CCC(=O)CC1 (D)

Short answer

The carbonyl compound that yields four stereoisomeric alcohols when reduced with LiAlH4 and subsequently acidified is (C) Cyclopentane-1,3-dione. This is because it forms two new stereocenters at the 1 and 3 positions upon reduction, leading to four possible stereoisomeric alcohols (R,R; R,S; S,R; S,S).

Step-by-step solution

Write the structures of the given carbonyl compounds

: First, let's convert the given SMILES notation for each compound to their structural formula: (A) Cyclohexanone: CC1CCC(=O)CC1 \ (B) 2-Octanone: CC1CCCCC1=O \ (C) Cyclopentane-1,3-dione: O=C1CCC(=O)CC1 \ (D) None of the above

Analyze the reduction reaction and stereochemistry of each compound

: Now let's see which compound will form four stereoisomeric alcohols when reduced: (A) Cyclohexanone: In this case, the carbonyl group is at the 2-position of the cyclohexane ring, and its reduction will form one stereocenter at the 2-position. The product alcohol will form two stereoisomers (R and S configuration). (B) 2-Octanone: This compound has a linear structure, and upon reduction of the carbonyl group, it will produce one stereocenter at the 2-position, similar to cyclohexanone, leading to two stereoisomers. (C) Cyclopentane-1,3-dione: This compound contains two carbonyl groups at positions 1 and 3 of the cyclopentane ring. Upon reduction, it will form two new stereocenters at the 1 and 3 positions. Thus, it will give a total of four possible stereoisomeric alcohols (R,R; R,S; S,R; S,S). (D) None of the above: This option indicates that none of the given compounds will produce four stereoisomeric alcohols upon reduction.

Identify the compound yielding four stereoisomeric alcohols

: Based on the analysis, we can conclude that compound (C) Cyclopentane-1,3-dione will yield four stereoisomeric alcohols when reduced with LiAlH4 and subsequently acidified. Therefore, the correct option is (C).

Key concepts

LiAlH4 Reduction

Lithium aluminium hydride, abbreviated as LiAlH4, is a powerful reducing agent frequently used in organic chemistry to reduce carbonyl compounds to alcohols. The process unfolds as the LiAlH4 donates a hydride ion (H-) to the carbonyl carbon, turning the double bond oxygen into an alcohol (OH group).

In the context of stereochemistry, this reduction is of great interest because it can create new stereocenters in a molecule, leading to the formation of stereoisomers. These are compounds that have the same molecular formula but differ in the spatial arrangement of their atoms. For students preparing for competitive exams like JEE Main and Advanced, understanding the mechanism and outcome of LiAlH4 reductions is crucial. Compounds can have one or more carbonyl groups, and the resultant stereochemistry after reduction determines the complexity of the stereochemical outcomes, which is an essential concept in organic chemistry.

Carbonyl Compounds

Carbonyl compounds are a class of carbon-containing organic molecules characterized by the presence of a carbonyl group (C=O). This group is highly reactive due to the polarization of the carbon-oxygen double bond, making these compounds amenable to a variety of chemical reactions.

Two main types of carbonyl compounds are aldehydes and ketones. Aldehydes have at least one hydrogen atom attached to the carbonyl carbon, while ketones have two alkyl or aryl groups. When these compounds are subjected to LiAlH4 reduction, they typically form primary or secondary alcohols, respectively. The reaction outcomes play a significant role in organic synthesis and are a critical topic for students preparing for exams like JEE Main and Advanced, where understanding the properties and reactivities of organic compounds is necessary.

Stereochemistry

Stereochemistry is the study of the three-dimensional structure of molecules and the effects of molecular structure on the properties and reactions of compounds. One of the central concepts in stereochemistry is the idea of stereoisomers—molecules with the same molecular formula and connectivity of atoms but differing in the three-dimensional orientation of their components.

Enantiomers and diastereomers are two types of stereoisomers. Enantiomers are non-superimposable mirror images of each other, like left and right hands, whereas diastereomers are stereoisomers that are not mirror images. When examining the reduction of carbonyl compounds, it is critical to consider the formation of potential stereocenters, which are carbon atoms attached to four different substituents. Understanding how many stereoisomers can be formed is essential for solving problems on stereochemistry in competitive exams like JEE.

Organic Chemistry JEE Main and Advanced

Organic chemistry is a significant part of the JEE Main and Advanced syllabus, challenging students to master a broad range of chemical reactions, mechanisms, and concepts. The exam tests not only the students' knowledge of organic reactions, such as reductions with LiAlH4, but also their understanding of deeper concepts like stereochemistry, and the ability to apply this knowledge to solve complex problems.

For individuals preparing for these exams, it is vital to grasp the intricacies of carbonyl chemistry and stereochemical outcomes that follow various organic reactions. Proficiency in predicting the number and types of stereoisomers from given organic reactions can be a determining factor in scoring well in these competitive exams. Students should focus on building a strong foundation in organic chemistry principles and practice extensive problem-solving to excel in this subject area.