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Which of the following is not a criterion for aromaticity? |
An aromatic compound is cyclic and planar Each atom in aromatic ring has p-orbital. These p-orbitals must be parallel so that a continuous overlap is possible around the ring The cyclic p-molecular orbital (electron cloud) formed by overlap of p-orbitals must contain (4n + 2) p-electrons It should decolourise \(Br_2/CCl_4\) and alkaline \(KMnO_4\) solution |
It should decolourise \(Br_2/CCl_4\) and alkaline \(KMnO_4\) solution |
The correct answer is option 4. It should decolourise \(Br_2/CCl_4\) and alkaline \(KMnO_4\) solution. To identify the statement that is not a criterion for aromaticity, we need to understand the key principles and rules that define aromaticity. Let's discuss these criteria in detail: Criteria for Aromaticity Cyclic and Planar Structure: Cyclic: The compound must be cyclic, meaning the atoms are arranged in a ring. Planar: The compound must be planar, which means all the atoms in the ring lie in a single plane. This planarity allows the p-orbitals to overlap effectively. Continuous Overlap of p-Orbitals: p-Orbitals: Each atom in the ring must have a p-orbital that can participate in the delocalization of π-electrons. Parallel p-Orbitals: These p-orbitals must be parallel to each other, enabling continuous overlap around the ring. This continuous overlap forms a conjugated π-system, which is essential for aromaticity. Hückel's Rule: (4n + 2) π-Electrons: The cyclic system must contain a specific number of π-electrons, given by the formula (4n + 2), where \( n \) is a non-negative integer (0, 1, 2, ...). This rule, known as Hückel's rule, ensures the system has a certain level of stability due to the delocalized π-electron cloud. Examples: Benzene (C₆H₆) has 6 π-electrons (n=1). Naphthalene (C₁₀H₈) has 10 π-electrons (n=2). Chemical Reactivity Aromatic compounds are distinct from alkenes and alkynes in their chemical reactivity. Aromatic compounds generally do not participate in reactions that disrupt their π-electron system because this would destroy their aromaticity. This distinction is crucial for understanding why certain tests are not applicable to aromatic compounds. Common Misconceptions Decolorization of \(Br_2/CCl_4\) and Reaction with Alkaline \(KMnO_4\): Bromine Test: Alkenes and alkynes react with bromine in carbon tetrachloride (\(Br_2/CCl_4\)), leading to the decolorization of bromine. This reaction is an addition reaction where bromine adds across the double or triple bonds. Potassium Permanganate Test: Alkenes and alkynes react with alkaline potassium permanganate (\(KMnO_4\)), which results in oxidation. This reaction also indicates the presence of carbon-carbon double or triple bonds. Aromatic Compounds: Aromatic compounds do not typically decolorize bromine in \(CCl_4\) or react with \(KMnO_4\) because these reactions would disrupt the aromatic system. Aromatic compounds undergo substitution reactions (e.g., electrophilic aromatic substitution) rather than addition reactions. Detailed Explanation of the options 3. The cyclic p-molecular orbital (electron cloud) formed by overlap of p-orbitals must contain (4n + 2) π-electrons:The option is trure. This is Hückel's rule, which is a fundamental criterion for aromaticity. The (4n + 2) π-electrons rule ensures the system has the required electron count for aromatic stabilization. 4. It should decolorize \(Br_2/CCl_4\) and alkaline \(KMnO_4\) solution: The option is false. This statement is incorrect because aromatic compounds do not typically react with \(Br_2/CCl_4\) or \(KMnO_4\) in the same way that alkenes or alkynes do. Aromatic compounds are resistant to these tests because such reactions would disrupt their aromatic π-electron system. The incorrect statement that is not a criterion for aromaticity is: (4) It should decolorize \(Br_2/CCl_4\) and alkaline \(KMnO_4\) solution |