Question

1. benzene (c6h6) consists of a six membered ring of carbon atoms with one hydrogen bonded to each carbon. write lewis structures for benzene, including resonance structures. 2

Explanation:

1. Step 1: Determine Valence Electrons

• Carbon (C) has 4 valence electrons, and Hydrogen (H) has 1. For benzene (\(C_6H_6\)):
• Total valence electrons from C: \(6\times4 = 24\)
• Total valence electrons from H: \(6\times1 = 6\)
• Total valence electrons: \(24 + 6 = 30\)

1. Step 2: Arrange Atoms

• Benzene has a six - membered carbon ring, with one H atom bonded to each C atom. The carbon atoms are arranged in a hexagonal ring (each C - C - C bond angle is approximately \(120^{\circ}\) in the planar structure).

1. Step 3: Form Single Bonds and Distribute Remaining Electrons

• First, form single bonds between the C atoms in the ring and between each C and its attached H atom. Each C - H single bond uses 2 electrons, and each C - C single bond also uses 2 electrons.
• For the 6 C - H bonds: \(6\times2 = 12\) electrons used.
• For the 6 C - C single bonds: \(6\times2 = 12\) electrons used.
• Electrons used so far: \(12+12 = 24\) electrons.
• Remaining electrons: \(30 - 24=6\) electrons. These remaining electrons are the delocalized \(\pi\) electrons.

1. Step 4: Draw Lewis Structure and Resonance Structures

• The Lewis structure of benzene shows a six - membered carbon ring. Each carbon is \(sp^2\) hybridized. In the Lewis structure, we can represent the double bonds in two equivalent ways (resonance structures).
• In one resonance structure, the double bonds are between C1 - C2, C3 - C4, and C5 - C6. In the other resonance structure, the double bonds are between C2 - C3, C4 - C5, and C6 - C1. The actual structure of benzene is a resonance hybrid, where the \(\pi\) electrons are delocalized over the entire ring. The Lewis structure is often represented with a hexagon and a circle inside (to represent the delocalized \(\pi\) electrons), but the two resonance structures with alternating double bonds are also used to show the resonance.
• The two resonance structures (using the Kekulé structures) are:
• Structure 1: \( \ce{H - C = C - C = C - C = C - H}\) (with the carbons in a ring, so it's a hexagonal ring with double bonds between C1 - C2, C3 - C4, C5 - C6)
• Structure 2: \( \ce{H - C - C = C - C = C - C = H}\) (with double bonds between C2 - C3, C4 - C5, C6 - C1) (again, the carbons are in a hexagonal ring)

Answer:

The Lewis structure of benzene (\(C_6H_6\)) has a six - membered carbon ring with one hydrogen atom attached to each carbon. The two main resonance (Kekulé) structures are:

1. A hexagonal ring of carbon atoms with alternating single and double bonds: In the first resonance structure, the double bonds are between carbon atoms 1 - 2, 3 - 4, and 5 - 6 (with each carbon also bonded to a hydrogen atom).
2. A hexagonal ring of carbon atoms with alternating single and double bonds: In the second resonance structure, the double bonds are between carbon atoms 2 - 3, 4 - 5, and 6 - 1 (with each carbon also bonded to a hydrogen atom).

The actual structure is a resonance hybrid, and it can also be represented with a hexagon and a circle inside the hexagon to show the delocalized \(\pi\) electrons.