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Hydrocarbons - Aromatic Hydrocarbons - Nomenclature, Benzene Structure and Aromaticity

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Aromatic hydrocarbons, also known as arenes, are cyclic, planar compounds with delocalized π\pi electrons. Benzene (C6H6C_6H_6) is the parent member of this class.

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Structure of Benzene: Benzene is a planar molecule where all six carbon atoms are sp2sp^2 hybridized. Each carbon atom forms two C−CC-C σ\sigma bonds and one C−HC-H σ\sigma bond. The unhybridized 2p2p orbitals overlap laterally to form a delocalized π\pi electron cloud above and below the plane of the ring.

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Resonance: Benzene is a resonance hybrid of two Kekulé structures. The delocalization of π\pi electrons provides extra stability, known as resonance energy, which is approximately 150 kJ mol−1150 \text{ kJ mol}^{-1}.

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Bond Lengths: Due to resonance, all C−CC-C bond lengths in benzene are identical (139 pm139 \text{ pm}), which is intermediate between a single bond (154 pm154 \text{ pm}) and a double bond (134 pm134 \text{ pm}).

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Hückel's Rule (4n+24n + 2 rule): For a compound to be aromatic, it must be cyclic, planar, have complete delocalization of π\pi electrons in the ring, and contain (4n+2)π(4n + 2) \pi electrons, where nn is an integer (0,1,2,…0, 1, 2, \dots).

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Nomenclature: In disubstituted benzene, the relative positions are indicated by prefixes: 1,2- is 'ortho' (oo-), 1,3- is 'meta' (mm-), and 1,4- is 'para' (pp-). For poly-substituted rings, substituent positions are numbered to give the lowest possible locants.

📐Formulae

Hu¨ckel’s Rule: (4n+2)π electrons\text{Hückel's Rule: } (4n + 2) \pi \text{ electrons}

Bond Angle in Benzene Ring=120∘\text{Bond Angle in Benzene Ring} = 120^\circ

C-C Bond Length in Benzene=139 pm\text{C-C Bond Length in Benzene} = 139 \text{ pm}

General Formula for Monocyclic Arenes=CnH2n−6 (for n≥6)\text{General Formula for Monocyclic Arenes} = C_nH_{2n-6} \text{ (for } n \ge 6\text{)}

💡Examples

Problem 1:

Determine if the Cyclooctatetraene (C8H8C_8H_8) molecule is aromatic based on Hückel's Rule.

Solution:

Cyclooctatetraene has 8π8 \pi electrons. If we equate 4n+2=84n + 2 = 8, we get 4n=64n = 6 or n=1.5n = 1.5. Since nn is not an integer and the molecule is non-planar (tub-shaped), it is not aromatic.

Explanation:

To be aromatic, a molecule must be planar and satisfy the (4n+2)π(4n + 2) \pi electron rule. 8π8 \pi electrons follow the 4n4n rule, making it anti-aromatic if it were planar, but it adopts a non-planar shape to minimize strain.

Problem 2:

Identify the IUPAC name for a benzene ring substituted with a methyl group at position 1 and a chlorine atom at position 3.

Solution:

1-Chloro-3-methylbenzene (also known as mm-Chlorotoluene).

Explanation:

The substituents are Chloros (C) and Methyl (M). Following alphabetical order, Chlorine comes first. The numbering is done to give the substituents the lowest locants (1 and 3).

Problem 3:

Is the Cyclopentadienyl anion (C5H5−C_5H_5^-) aromatic?

Solution:

Yes, it is aromatic. It has 6π6 \pi electrons (4 from two double bonds and 2 from the negative charge/lone pair). 4n+2=6  ⟹  n=14n + 2 = 6 \implies n = 1.

Explanation:

Since it is cyclic, planar, has complete conjugation, and follows Hückel's rule with n=1n=1, it is aromatic.

Aromatic Hydrocarbons - Nomenclature, Benzene Structure and Aromaticity Revision - Class 11…