2. Explain the following trend (Note: "smaller pka-more acidic" is not an explanation, its a definition. Explanations require "why.")

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## Understanding the Trend in pKa Values

### Problem Statement:
2. Explain the following trend (Note: “smaller pKa=more acidic” is not an explanation, it's a definition. Explanations require "why.")

### Graph/Diagram Description:
The provided image contains three different chemical structures along with their corresponding pKa values. The structures are described from left to right as follows:

1. **Phenol (C₆H₅OH)**
   - Structure: A benzene ring (hexagon with alternating double bonds) with an OH group attached.
   - pKa: 9.95

2. **Cyclohexane with a CH group (CH₃C₆H₁₁)**
   - Structure: A cyclohexane ring (hexagon with single bonds) where there is a carbon with a hydrogen atom bonded off of a secondary carbon.
   - pKa: 41

3. **Methane (CH₄)**
   - Structure: A single carbon atom bonded to four hydrogen atoms in a tetrahedral geometry.
   - pKa: 50

### Explanation of the Trend:
The trend in this series of pKa values shows that as the pKa increases, the compound becomes less acidic. Here’s why:

- **Phenol (pKa 9.95):** Phenol is more acidic because the phenoxide ion (the conjugate base after deprotonation) is stabilized by resonance. The negative charge on the oxygen is delocalized over the aromatic ring, lowering the energy of the phenoxide ion and thus making the deprotonation more favorable.

- **Cyclohexane with a CH group (pKa 41):** This compound is significantly less acidic than phenol. The absence of resonance stabilization in the conjugate base means the anion is less stable compared to phenol's conjugate base. Furthermore, cyclohexane is a saturated hydrocarbon, and hydrocarbons are generally very poor acids.

- **Methane (pKa 50):** Methane is the least acidic among the three. When methane loses a proton, it forms a carbanion (CH₃⁻), which is extremely high in energy due to the lack of any stabilization (no resonance, inductive effects, or electronegative atoms to stabilize the negative charge). Therefore, deprotonation is highly unfavorable, resulting in a very high pKa
Transcribed Image Text:## Understanding the Trend in pKa Values ### Problem Statement: 2. Explain the following trend (Note: “smaller pKa=more acidic” is not an explanation, it's a definition. Explanations require "why.") ### Graph/Diagram Description: The provided image contains three different chemical structures along with their corresponding pKa values. The structures are described from left to right as follows: 1. **Phenol (C₆H₅OH)** - Structure: A benzene ring (hexagon with alternating double bonds) with an OH group attached. - pKa: 9.95 2. **Cyclohexane with a CH group (CH₃C₆H₁₁)** - Structure: A cyclohexane ring (hexagon with single bonds) where there is a carbon with a hydrogen atom bonded off of a secondary carbon. - pKa: 41 3. **Methane (CH₄)** - Structure: A single carbon atom bonded to four hydrogen atoms in a tetrahedral geometry. - pKa: 50 ### Explanation of the Trend: The trend in this series of pKa values shows that as the pKa increases, the compound becomes less acidic. Here’s why: - **Phenol (pKa 9.95):** Phenol is more acidic because the phenoxide ion (the conjugate base after deprotonation) is stabilized by resonance. The negative charge on the oxygen is delocalized over the aromatic ring, lowering the energy of the phenoxide ion and thus making the deprotonation more favorable. - **Cyclohexane with a CH group (pKa 41):** This compound is significantly less acidic than phenol. The absence of resonance stabilization in the conjugate base means the anion is less stable compared to phenol's conjugate base. Furthermore, cyclohexane is a saturated hydrocarbon, and hydrocarbons are generally very poor acids. - **Methane (pKa 50):** Methane is the least acidic among the three. When methane loses a proton, it forms a carbanion (CH₃⁻), which is extremely high in energy due to the lack of any stabilization (no resonance, inductive effects, or electronegative atoms to stabilize the negative charge). Therefore, deprotonation is highly unfavorable, resulting in a very high pKa
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