Question

An organic compound ' \(\mathrm{P}^{\prime}\left(\mathrm{C}_{10} \mathrm{H}_{10}\right)\) on ozonolysis gives ' \(\mathrm{Q}^{\prime}\left(\mathrm{C}_{10} \mathrm{H}_{10} \mathrm{O}_{2}\right)\) which show positive test with \(\mathrm{Feh} / \mathrm{m}\) \(\mathrm{H}^{+} / \Delta /\) gives \((\mathrm{T})\) C1=Cc2ccccc2CC1 Select correct statement(s) regarding \(\mathrm{P}\) and \(\mathrm{T}\). (A) \(\mathrm{P}\) and \(\mathrm{T}\) may be same (B) \(\mathrm{P}\) and \(\mathrm{T}\) may be structural isomer (C) \(\mathrm{P}\) and \(\mathrm{T}\) both give phthalic acid on oxidation by \(\mathrm{KMnO}_{4}\) (D) \(\mathrm{P}\) can exist in two form of benzenoid isomers

Short answer

The correct statements regarding P and T are (B), (C), and (D). P and T are structural isomers with different molecular formulas. Both P and T can give phthalic acid on oxidation by KMnO4. Compound P can exist in two benzenoid isomers.

Step-by-step solution

1. Convert the SMILES string to a structural formula

Converting the given SMILES string (C1=Cc2ccccc2CC1) to its structural formula, we get a molecule of naphthalene. Naphthalene is a fused-ring aromatic hydrocarbon consisting of two benzene rings sharing two adjacent carbon atoms, and its chemical formula is C10H8.

2. Analyze the isomerism between P and T

Now, we will examine the isomerism between compound P and T. As discussed earlier, the structural formula of T is naphthalene (C10H8). Given that compound P has the molecular formula of C10H10, it suggests that P is an alkene with two hydrogen atoms less than compound T. Now let's check the given statements. (A) Since P and T have different molecular formulas - C10H10 and C10H8, respectively - they cannot be the same compound. (B) Since P and T have different molecular formulas, they can be considered structural isomers. (C) Both P and T can give phthalic acid on oxidation by KMnO4. P could oxidize to form a di-carboxylic derivative, which could be oxidized by KMnO4 to form phthalic acid. Naphthalene (T) can also be oxidized by KMnO4 to form phthalic acid. (D) Compound P can exist in two benzenoid isomers. This is possible if we assume that P is a bi-cyclic (fused-ring) compound, similar to naphthalene, but in which the two rings have an alkene linkage instead of complete aromaticity in both rings. In conclusion: - Statement (A) is incorrect. - Statement (B) is correct. - Statement (C) is correct. - Statement (D) is correct.

Key concepts

Ozonolysis

Ozonolysis is a chemical reaction that involves the cleavage of double or triple bonds in alkenes and alkynes using ozone as the oxidizing agent. The reaction produces compounds with carbonyl groups, such as aldehydes or ketones. This transformation is particularly useful in organic chemistry because it provides a method for determining the positions of double bonds in unsaturated compounds.
In the context of the problem discussed, ozonolysis is applied to the organic compound P with the formula \( C_{10}H_{10} \). When P undergoes ozonolysis, it produces a compound Q with the formula \( C_{10}H_{10}O_{2} \), which suggests the formation of oxygenated fragments, likely containing carbonyl groups from the cleavage of double bonds.
Ozonolysis is an effective tool for the fragmentation and subsequent analysis of complex hydrocarbons, providing insights into the structure and bonding of the parent compound. Its utility lies in converting unsaturated hydrocarbons into more oxidized variants, aiding in understanding their structure.

Structural Isomers

Structural isomers, also known as constitutional isomers, are compounds with the same molecular formula but different structural arrangements of atoms. This results in different chemical and physical properties. Structural isomerism is commonly seen in organic compounds where the carbon skeleton can vary significantly even with the same number of atoms.
In the scenario where compounds P and T are discussed, they possess different molecular formulas \( C_{10}H_{10} \) and \( C_{10}H_{8} \), respectively. This immediately eliminates the possibility of P and T being identical, but allows for the potential of structural isomerism. While P may have a different degree of saturation compared to T (naphthalene), they could still be structurally related in terms of carbon connectivity, allowing for varied isomers such as different placements of double bonds or other structural features.
Understanding structural isomers is crucial in organic chemistry because even though compounds may have the same molecular formula, their differing structures give rise to unique characteristics. Recognizing these differences can be vital in identifying the function and reactivity of each isomer.

Oxidation by KMnO4

Potassium permanganate (KMnO4) is a powerful oxidizing agent often used in organic chemistry for oxidizing unsaturated compounds. This reaction is typically used to convert alkenes, alkynes, and some aromatic compounds into carboxylic acids, ketones, or other oxygenated functional groups.
In the exercise, both compound P and compound T can undergo oxidation with KMnO4 to yield phthalic acid. This indicates that both compounds include reactive sites that can be oxidized into carboxylic acid groups. Naphthalene (T), being an aromatic hydrocarbon, requires oxidation conditions to cleave its rings and form phthalic acid. Similarly, compound P, having a less saturated structure, also oxidizes to a derivative which can be further oxidized to phthalic acid.
The ability of KMnO4 to break down complex organic structures into simpler oxygenated compounds makes it an indispensable tool in synthetic and analytical chemistry. By allowing conversion into identifiable substances, it helps chemists infer possible structures and functionalities of unknown compounds.