A. Using ARIO, predict the position of equilibrium in the following intramolecular proton transfer reaction. Briefly explain your answer. B. One proton below has a surprisingly low pk, value and is the most acidic proton in the structure. Indicate which proton it is and using ARIO provide an explanation for its low pK2. ole
Electronic Effects
The effect of electrons that are located in the chemical bonds within the atoms of the molecule is termed an electronic effect. The electronic effect is also explained as the effect through which the reactivity of the compound in one portion is controlled by the electron repulsion or attraction producing in another portion of the molecule.
Drawing Resonance Forms
In organic chemistry, resonance may be a mental exercise that illustrates the delocalization of electrons inside molecules within the valence bond theory of octet bonding. It entails creating several Lewis structures that, when combined, reflect the molecule's entire electronic structure. One Lewis diagram cannot explain the bonding (lone pair, double bond, octet) elaborately. A hybrid describes a combination of possible resonance structures that represents the entire delocalization of electrons within the molecule.
Using Molecular Structure To Predict Equilibrium
Equilibrium does not always imply an equal presence of reactants and products. This signifies that the reaction reaches a point when reactant and product quantities remain constant as the rate of forward and backward reaction is the same. Molecular structures of various compounds can help in predicting equilibrium.
**Reaction Description:**
- The left structure depicts a resonance-stabilized carboxylate ion (R-COO⁻) and an alcohol group (-OH).
- The right structure shows a protonated carboxyl group (R-COOH) and an alkoxide ion (R-O⁻).
**Explanation:**
To predict the position of equilibrium, apply the ARIO factors:
1. **Atom**: Oxygen is more electronegative compared to hydrogen and can stabilize negative charge better.
2. **Resonance**: The left structure benefits from resonance stabilization due to the carboxylate ion, which can delocalize the negative charge across two oxygen atoms.
3. **Induction**: Electron-withdrawing groups near the proton can stabilize negative charges, making deprotonation favorable.
4. **Orbital**: The hydroxyl group's lone pair is in an sp³ hybrid orbital, which is less effective at stabilizing a negative charge compared to the sp² hybrid orbitals in carboxylate.
Given these factors, the equilibrium position favors the resonance-stabilized carboxylate ion and alcohol group (left structure).
#### B. Identifying the Most Acidic Proton
One proton in the given structure has a surprisingly low \(pK_a\) value and is the most acidic proton. Indicate which proton it is and, using ARIO, provide an explanation for its low \(pK_a\).
**Structure Description:**
- The molecule consists of a benzene ring, a ketone group (C=O), and an additional carbonyl group.
**Identification and Explanation:**
The most acidic proton is the α-proton (adjacent to the carbonyl group).
**Rationale:**
- **Atom**: The proton is attached to a carbon atom adjacent to the carbonyl.
- **Resonance**: Deprotonation of the α-proton can lead to resonance stabilization, where the negative charge](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe4d0ca9f-229e-40d6-ac53-f10954775736%2F80d74c67-9421-40c2-b716-5dabefc32d18%2Fx7lfnp_processed.jpeg&w=3840&q=75)
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