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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 $ ext{C}_6 ext{H}_6$ compared t... 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 ΔH_a.
Answer
To calculate the enthalpy of atomization ( ΔH_a) , we can use the bond enthalpies provided in Table 1. First, we need to determine the total bond enthalpy for breaking the bonds in benzene:
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Breaking bonds in benzene:
- Benzene has 6 C–H bonds and 3 C–C bonds.
- Total bond enthalpy for C–H bonds: 6 bonds × 412 kJ/mol = 2472 kJ/mol.
- Total bond enthalpy for C–C bonds: 3 bonds × 348 kJ/mol = 1044 kJ/mol.
Thus, total bond enthalpy for breaking all bonds:
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Calculating ΔH_a:
- Using Hess's law and the values from Table 1, the heat of formation of benzene is considered along with the atomization enthalpies:
Therefore:
Step 2
Explain, in terms of structure and bonding, why C6H6 is more thermodynamically stable than C6H6(g).
Answer
C6H6 is more thermodynamically stable than C6H6(g) due to the delocalization of electrons within the benzene ring. In the benzene structure, the π electrons are not localized between individual carbon atoms; rather, they are spread out over the entire ring structure. This electron delocalization leads to lower potential energy and thus increased stability compared to the gaseous state where the bonding interactions are not as favorable.
Additionally, the resonance energy associated with benzene indicates that the actual bonded structure is lower in energy than any theoretical localized structure could ever achieve, which further contributes to its stability.
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: Add a lone pair on the nitrogen atom to show electron donation to the sulfur atom. Also, add two curly arrows from the lone pair and the double bond to create a new bond between nitrogen and sulfur.
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Step 2: A curly arrow should be added to show the movement of electrons from the oxygen to the nitrogen.
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Step 3: Add a curly arrow from the bond between the nitrogen and sulfur to the nitrogen itself, indicating bond breaking.
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Step 4: Add a curly arrow from the bond returning to the hexagon ring to show the reformation of the aromatic system.
This mechanism follows the electrophilic substitution pathway, whereby the benzene ring reacts with an electrophile, in this case, the nitronium ion (NO2+), facilitated by the sulfuric acid.