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The values of \(\mathrm{p} K_{\mathrm{a}}\) for \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) and \(\mathrm{CF}_{3} \mathrm{CO}_{2} \mathrm{H}\) are 4.75 and \(0.23,\) both of which are very nearly independent of temperature. Suggest reasons for this difference.

Short Answer

Expert verified
\( \mathrm{CF}_{3} \mathrm{CO}_{2} \mathrm{H} \) is more acidic than \( \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \) due to the electron-withdrawing \( \mathrm{CF}_{3} \) group.

Step by step solution

01

Understanding pKa Values

The value of \( pK_{a} \) is a measure of the acid dissociation in water, where a lower \( pK_{a} \) represents a stronger acid. In this case, \( \mathrm{CF}_{3} \mathrm{CO}_{2} \mathrm{H} \) with a \( pK_{a} \) of 0.23 is a stronger acid than \( \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \) which has a \( pK_{a} \) of 4.75.
02

Analyzing Structural Differences

Both acids have similar structures except for the substituents on the acyl group: methyl \( (\mathrm{CH}_{3}) \) versus trifluoromethyl \( (\mathrm{CF}_{3}) \). The presence of the electron-withdrawing \( \mathrm{CF}_{3} \) group increases acidity, whereas the \( \mathrm{CH}_{3} \) group has minimal electron-withdrawing effects.
03

Role of Electron-Withdrawing Groups

The \( \mathrm{CF}_{3} \) group is a strong electron-withdrawing group due to its electronegativity, stabilizing the negative charge on the carboxylate ion once dissociation occurs. This enhances deprotonation, making \( \mathrm{CF}_{3} \mathrm{CO}_{2} \mathrm{H} \) a stronger acid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding pKa Values
When talking about acids, their strength is often measured using a scale called the pKa. The pKa value tells us how easily an acid can lose a proton, which is a tiny particle with a positive charge. The lower the pKa value, the stronger the acid because it loses its proton easily. For example, if we compare the pKa of two acids,
  • \( \text{CF}_3\text{CO}_2\text{H} \) (trifluoroacetic acid) with a pKa of 0.23, and
  • \( \text{CH}_3\text{CO}_2\text{H} \) (acetic acid) with a pKa of 4.75,
we see that \( \text{CF}_3\text{CO}_2\text{H} \) is a much stronger acid than \( \text{CH}_3\text{CO}_2\text{H} \). This is because it has a lower pKa value, meaning it can release its hydrogen ion more readily when dissolved in water.
In summary, lower pKa values correspond to acids that are stronger, because they donate their protons (or lose their hydrogen ions) more easily.
The Nature of Carboxylic Acids
Carboxylic acids are a specific type of organic compound that contain a carboxyl group \( (\text{-COOH}) \). This group is what gives these molecules their acidic properties. The carboxyl group includes both a carbonyl group \( (\text{C}=\text{O}) \) and a hydroxyl group \( (\text{-OH}) \).
  • The acidic nature comes primarily from the capability of the \( \text{-OH} \) part to lose a hydrogen ion.
  • When the hydrogen ion is released, the carboxylic acid becomes ionized to form an anion, often called a carboxylate ion.
Understanding the structure of carboxylic acids is crucial, especially since their acidic action varies depending on other elements attached, known as substituents.
Different substituents can change how easily the hydrogen ion is released, and this leads us to our next section on electron-withdrawing groups.
Influence of Electron-Withdrawing Groups
In some carboxylic acids, certain groups attached to the main structure can enhance the acidity. These are called electron-withdrawing groups. The \( \text{CF}_3 \) group is a classic example of such a substituent attached to the acyl group in trifluoroacetic acid.
  • Electron-withdrawing groups pull negative charge toward themselves due to their high electronegativity. This movement helps stabilize the carboxylate ion, formed when the acid releases a proton.
  • A stabilized carboxylate ion lowers the energy of the system, making it easier for the acid to give up its hydrogen ion.
Therefore, the presence of strong electron-withdrawing groups like \( \text{CF}_3 \) significantly boosts the acidity of the molecule.
In simpler terms, they make it easier for the acid to "let go" of the hydrogen, which increases the acid's strength. These groups essentially make the environment around the molecule's core more favorable for hydrogen ion release, explaining why \( \text{CF}_3\text{CO}_2\text{H} \) is a stronger acid compared to \( \text{CH}_3\text{CO}_2\text{H} \).

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Most popular questions from this chapter

Four \(\mathrm{p} K_{\mathrm{a}}\) values (1.0,2.0,7.0,9.0) are tabulated for the acid \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}\). Write equations to show the dissociation steps in aqueous solution and assign, with reasoning, a \(\mathrm{p} K_{\mathrm{a}}\) value to each step.

The values of \(\log K\) for the following two equilibria are 7.23 and \(12.27,\) respectively: \\[\begin{array}{l} \mathrm{Ag}^{+}(\mathrm{aq})+2 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(\mathrm{aq}) \\\ \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{AgBr}(\mathrm{s}) \end{array}\\] Determine (a) \(K_{\mathrm{sp}}\) for AgBr, and (b) \(K\) for the reaction: \\[\left[\operatorname{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{AgBr}(\mathrm{s})+2 \mathrm{NH}_{3}(\mathrm{aq})\\]

Soda-water is made by saturating \(\mathrm{H}_{2} \mathrm{O}\) with \(\mathrm{CO}_{2}\). If one titrates soda-water with alkali using phenolphthalein as indicator, one obtains a fading end-point. What does this suggest?

(a) For \(\left[\mathrm{Pd}(\mathrm{CN})_{4}\right]^{2-},\) a value of \(\log \beta_{4}\) of 62.3 (at \(298 \mathrm{K}\) in aqueous medium) has been determined. To what equilibrium process does this value refer? (b) For the equilibrium: \(\operatorname{Pd}(\mathrm{CN})_{2}(\mathrm{s})+2 \mathrm{CN}^{-}(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Pd}(\mathrm{CN})_{4}\right]^{2-}\) the value of \(\log K\) is \(20.8 .\) Use this value and the data in part (a) to determine \(K_{\mathrm{sp}}\) for \(\mathrm{Pd}(\mathrm{CN})_{2}\).

What explanation can you offer for the decrease in solubility of the alkaline earth metal sulfates in the sequence \(\mathrm{CaSO}_{4}>\mathrm{SrSO}_{4}>\mathrm{BaSO}_{4} ?\)

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