Structural Representations Of Organic Compounds

Chloroform ( ) is a volatile organic compound. When chloroform is metabolized in the liver, it forms carbonyl chloride ( ), a highly toxic compound. The formation of , also known as phosgene, is harmful to liver cells and can cause damage, leading to liver toxicity. Phosgene is a potent hepatotoxic agent and can lead to liver damage upon prolonged exposure. Thus, the compound responsible for liver damage is .

The correct option is(C):

The boiling point of a compound is influenced by factors like molecular weight, intermolecular forces, and symmetry. Let's analyze the given options:

Option A: Benzene and Naphthalene

This is incorrect. Benzene (C₆H₆) has a boiling point of 80.1°C, while naphthalene (C₁₀H₈) has a boiling point of 218°C. Hence, these two compounds do not have the same boiling points.

Option B: (+)-butan-2-ol and (-)-butan-2-ol

This is correct. Both (+)-butan-2-ol and (-)-butan-2-ol are enantiomers, which are non-superimposable mirror images. These compounds have the same molecular formula and identical intermolecular forces, which results in the same boiling points despite their optical isomerism.

Option C: cis-but-2-ene and trans-but-2-ene

This is incorrect. cis-but-2-ene and trans-but-2-eneare geometric isomers with different physical properties, including different boiling points. cis-but-2-ene has stronger intermolecular forces due to its molecular shape, leading to a higher boiling point than trans-but-2-ene.

Option D: n-hexane and neo-hexane

This is incorrect. n-hexane and neo-hexane are structural isomers, and while their molecular weight is similar, neo-hexane has a more compact structure, resulting in weaker intermolecular forces and a lower boiling point than n-hexane.

Conclusion:

The correct answer is: Option B: (+)-butan-2-ol and (-)-butan-2-ol, as they are enantiomers with the same boiling point.

Sucrose is a disaccharide consisting of glucose and fructose. Sucrose itself is dextrorotatory, meaning it rotates plane-polarized light to the right. However, when sucrose undergoes hydrolysis, it breaks down into glucose and fructose. The resulting mixture shows laevorotation (rotation to the left). This occurs because of the individual optical activities of glucose and fructose.

Explanation:

The key reason the mixture shows laevorotation after hydrolysis is that fructose is more laevorotatory than glucose is dextrorotatory. Thus, after hydrolysis, the laevorotation of fructose outweighs the dextrorotation of glucose, leading to an overall laevorotatory mixture.

Review of the options:

• Option A: "Racemization is formed" - This is incorrect. The reaction does not form a racemic mixture, but rather a mixture of fructose and glucose.
• Option B: "Laevorotation of fructose is more than dextrorotation of glucose" - This is the correct answer. Fructose has a greater laevorotatory effect than glucose's dextrorotatory effect, which causes the overall mixture to exhibit laevorotation after hydrolysis.
• Option C: "Laevorotation of glucose is more than dextrorotation of fructose" - This is incorrect. Glucose does not exhibit stronger laevorotation than fructose.
• Option D: "Sucrose is a non-reducing sugar" - While sucrose is indeed a non-reducing sugar, this fact does not explain the optical rotation observed after hydrolysis.

Conclusion:

The correct answer is Option B: Laevorotation of fructose is more than dextrorotation of glucose.