Textbook Question
Rank the following from strongest base to weakest base:
b.
3. Acids and Bases
pKa
5:50 minutesProblem 34c,d
Textbook Question
Given the structure of ascorbic acid (vitamin C):
(c) Predict which proton in ascorbic acid is the most acidic.
(d) Draw the form of ascorbic acid that is present in the body (aqueous solution, pH = 7.4)
Verified step by step guidance1
Identify the functional groups in ascorbic acid. Ascorbic acid contains hydroxyl (OH) groups and a lactone ring. The presence of these groups can influence acidity.
Consider the resonance stabilization of the conjugate base. The most acidic proton is typically the one whose removal results in the most stable conjugate base. In ascorbic acid, deprotonation of the hydroxyl group adjacent to the carbonyl group can lead to resonance stabilization.
Evaluate the inductive effect. The electronegative oxygen atoms in the lactone ring can withdraw electron density through the sigma bonds, increasing the acidity of nearby protons.
Predict the most acidic proton. Based on resonance and inductive effects, the proton on the hydroxyl group adjacent to the carbonyl group is likely the most acidic.
Draw the deprotonated form of ascorbic acid at physiological pH (7.4). At this pH, ascorbic acid exists primarily in its monoanionic form, where the most acidic proton has been removed, resulting in a negatively charged oxygen atom.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Acidity and pKa
Acidity in organic chemistry refers to the tendency of a compound to donate protons (H+ ions). The strength of an acid is often measured by its pKa value; lower pKa values indicate stronger acids. Understanding the pKa of functional groups in ascorbic acid helps predict which proton is most acidic, as it reflects the stability of the resulting conjugate base after deprotonation.
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Resonance Stabilization
Resonance stabilization occurs when a molecule can be represented by multiple valid Lewis structures, allowing for the delocalization of electrons. In ascorbic acid, certain protons may be more acidic due to the ability of the conjugate base to stabilize through resonance. This concept is crucial for determining which proton is most likely to dissociate in an aqueous solution.
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Physiological pH and Ionization
The physiological pH of 7.4 is significant for understanding the ionization state of ascorbic acid in the body. At this pH, certain protons may be more likely to be lost, affecting the overall charge and reactivity of the molecule. Recognizing how pH influences the ionization of functional groups in ascorbic acid is essential for predicting its form in biological systems.
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