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Kekulé suggested this structure for benzene - AQA - A-Level Chemistry - Question 4 - 2021 - Paper 2
Question 4
Kekulé suggested this structure for benzene. Benzene is now represented by this structure. Figure 3 shows the relative stability of $C_6H_6(g)$ compared to $C_6H_6... show full transcript
Worked Solution & Example Answer:Kekulé suggested this structure for benzene - AQA - A-Level Chemistry - Question 4 - 2021 - Paper 2
Step 1
Use Figure 3 and the data shown in Table 1 to calculate \(\Delta H_{rxn}\)
Answer
To calculate (\Delta H_{rxn}), we first need to find the enthalpy changes for the bonds broken and formed.
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Bonds broken:
- 6 C–H bonds (612 kJ/mol each)
- Total for C–H:
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Bonds formed:
- 6 C–C bonds and additional hydrogen bonds.
- Total for C–C:
- Total for atoms: for carbon + for hydrogen = .
The enthalpy change for the reaction can be calculated as follows:
Thus, (\Delta H_{rxn} = -63 ; kJ/mol).
Step 2
Explain, in terms of structure and bonding, why \(C_6H_6(g)\) is more thermodynamically stable than \(C_6H_6(s)\).
Answer
The structure of benzene contributes to its stability mainly due to its resonance and delocalization of electrons. In the gaseous state, benzene exists with a planar structure that allows for resonance, leading to lower energy and increased stability. The overlapping p-orbitals in benzene facilitate delocalized pi-bonds, which distribute electron density more evenly across the molecule.
In contrast, the solid state may result in certain strain or packing arrangements that are less favorable energetically. This leads to the conclusion that gaseous benzene is more stable compared to its solid counterpart due to favorable electronic structures and minimized repulsions.
Step 3
Complete the mechanism in Figure 4 by adding any lone pairs of electrons involved in each step.
Answer
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Step 1: A lone pair from the nitrogen attacks the hydrogen on benzene, resulting in an electrophilic substitution. Draw two curly arrows showing the movement of electrons.
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Step 2: Form the nitronium ion while removing a proton from the benzene ring. Draw one curly arrow indicating the bond breaking between N and H.
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Step 3: Draw a curly arrow from the carbon in benzene back into the hexagon, signifying the reformation of the aromatic system.
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Step 4: Draw a curly arrow indicating the movement of the lone pair to bond again with nitrogen, completing the process.