Let's break down the steps and fill in the table as requested: --- # Step-by-Step Solution ## a) **Account for Aromaticity with the 4n+2 Rule** The compounds are aromatic if they follow Hückel's rule: they must have a planar, cyclic, conjugated π-system with **(4n+2) π electrons** (where n = , 1, 2, ...). --- ## b) **Identify Delocalized Electrons & Non-bonding Pairs** Let's analyze each compound: --- ### **1. Indole** - **Structure:** Bicyclic, with a benzene ring fused to a five-membered nitrogen-containing ring. - **π Electrons:** - Benzene ring: 6 π electrons. - Five-membered ring: Double bond (2 π electrons) + lone pair on N can be delocalized (2 electrons). - Total: 6 (benzene) + 2 (double bond) + 2 (N lone pair) = **10 π electrons**. - **(4n+2) Rule:** 10 = 4(2) + 2; n = 2. - **Non-bonding Electrons:** The lone pair on nitrogen is **involved** in aromaticity (delocalized). --- ### **2. Azulene** - **Structure:** Fused 5- and 7-membered rings with alternating double bonds. - **π Electrons:** 10 π electrons (from 5 double bonds). - **(4n+2) Rule:** 10 = 4(2) + 2; n = 2. - **Non-bonding Electrons:** No lone pairs directly involved in aromaticity. --- ### **3. α-Pyrone** - **Structure:** Six-membered ring with an oxygen atom and three double bonds. - **π Electrons:** - 3 double bonds = 6 π electrons. - Oxygen has two lone pairs; only **one lone pair** can be delocalized in the aromatic system (the other remains as a non-bonding lone pair). - Total: 6 + 2 = **8 π electrons** if both lone pairs, but only one participates, so **6 π electrons**. - **(4n+2) Rule:** 6 = 4(1) + 2; n = 1. - **Non-bonding Electrons:** One lone pair on oxygen is **not involved** in aromaticity (remains non-bonding). --- ### **4. Adenine** - **Structure:** Purine base, two fused rings with nitrogens. - **π Electrons:** - Counted as: 3 double bonds (6 π), plus 2 from lone pair on one N, plus 2 from another N = **10 π electrons**. - **(4n+2) Rule:** 10 = 4(2) + 2; n = 2. - **Non-bonding Electrons:** Some lone pairs on nitrogens are **not involved** in aromaticity. --- ## c) **Table Filled In** | Compound | 4n+2 | Non-bonding electron explanation | |------------|------|-------------------------------------------------| | **indole** | 4n+2 | The lone pair on the five-membered ring N is delocalized and involved in aromaticity. | | **azulene** | 4n+2 | No lone pairs are involved; all π electrons are from double bonds. | | **α-pyrone** | 4n+2 | Only one lone pair on O is delocalized; the other lone pair remains non-bonding. | | **adenine** | 4n+2 | Lone pairs on some N atoms are not involved in aromaticity and remain non-bonding. | --- ## **Final Answers** ### **Non-bonding electron explanations:** - **Indole:** Lone pair on N is involved in aromaticity. - **Azulene:** No lone pairs involved; all π electrons from double bonds. - **α-Pyrone:** One of O's lone pairs is involved in aromaticity, the other is non-bonding. - **Adenine:** Some N atom lone pairs are non-bonding. --- ### **Summary Table** | Compound | 4n+2 | Non-bonding electron explanation | |------------|------|-------------------------------------------------| | Indole | Yes | Lone pair on N involved in aromaticity | | Azulene | Yes | No lone pairs involved; only double bonds | | α-Pyrone | Yes | One lone pair on O is non-bonding | | Adenine | Yes | Some N lone pairs are non-bonding | --- Let me know if you need the resonance structures or the NMR matching steps as well!