Question

Which acid in each of the following pairs would you expect to be the stronger? Give your reasoning. a. \(\left(\mathrm{CH}_{3}\right)_{3} \stackrel{\oplus}{\mathrm{NCH}}_{2} \mathrm{CO}_{2} \mathrm{H}\) or \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2} \mathrm{H}\) b. \(\left(\mathrm{CH}_{3}\right)_{3} \stackrel{\mathrm{N}}{\mathrm{N}} \mathrm{H}_{2} \mathrm{CO}_{2} \mathrm{H}\) or \(\left(\mathrm{CH}_{3}\right)_{2} \stackrel{\oplus}{\mathrm{N}}(\stackrel{\ominus}{\mathrm{O}}) \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}\) c. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCO}_{2} \mathrm{H}\) or \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) d. \(\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{CO}_{2} \mathrm{H}\) or \(\mathrm{CH}_{3} \mathrm{SCH}_{2} \mathrm{CO}_{2} \mathrm{H}\) e. \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}\) or \(\mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}\)

Short answer

a. \((\mathrm{CH}_3)_3 \oplus \mathrm{NCH}_2 \mathrm{CO}_2 \mathrm{H}\); b. \((\mathrm{CH}_3)_2 \oplus \mathrm{N}(\ominus \mathrm{O}) \mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}\); c. \(\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}\); d. \(\mathrm{CH}_3 \mathrm{OCH}_2 \mathrm{CO}_2 \mathrm{H}\); e. \(\mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}\).

Step-by-step solution

Consider Inductive Effect

For pair a, - In \((\mathrm{CH}_3)_3 \oplus \mathrm{NCH}_2 \mathrm{CO}_2 \mathrm{H}\), the \(\oplus\) charge on the nitrogen will withdraw electron density, stabilizing the conjugate base.- In \((\mathrm{CH}_3)_2 \mathrm{NCH}_2 \mathrm{CO}_2 \mathrm{H}\), there is no such electron-withdrawing group.- Therefore, \((\mathrm{CH}_3)_3 \oplus \mathrm{NCH}_2 \mathrm{CO}_2 \mathrm{H}\) is the stronger acid due to better stabilization of its conjugate base.

Evaluate Resonance and Charge

For pair b,- \((\mathrm{CH}_3)_3 \mathrm{NNH}_2 \mathrm{CO}_2 \mathrm{H}\) does not have additional charge.- \((\mathrm{CH}_3)_2 \oplus \mathrm{N}(\ominus \mathrm{O}) \mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}\) has a more stabilized structure due to charge separation, making it a stronger acid because it has a more stable conjugate base.

Compare Groups' Inductive Effects

For pair c,- \((\mathrm{CH}_3)_3 \mathrm{CCO}_2 \mathrm{H}\) has bulky electron-donating groups, reducing acidity.- \(\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}\), with fewer electron-donating groups, is more acidic due to less destabilization of the conjugate base.

Examine Sulfur vs Oxygen Electronegativy

For pair d,- \(\mathrm{CH}_3 \mathrm{OCH}_2 \mathrm{CO}_2 \mathrm{H}\) has an oxygen, which is more electronegative than sulfur, stabilizing the conjugate base better.- \(\mathrm{CH}_3 \mathrm{SCH}_2 \mathrm{CO}_2 \mathrm{H}\) has sulfur, which is less electronegative.- Therefore, \(\mathrm{CH}_3 \mathrm{OCH}_2 \mathrm{CO}_2 \mathrm{H}\) is the stronger acid.

Consider Triple vs Double Bonds

For pair e,- In \(\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}\), the double bond exerts an inductive effect but not through resonance.- In \(\mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}\), the triple bond strongly withdraws electron density through inductive effect, stabilizing the conjugate base better.- Thus, \(\mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}\) is the stronger acid.

Key concepts

Inductive Effect

The inductive effect is crucial in understanding how the presence of different groups can affect the acidity of a compound. It involves the shifting of electron density along a chain of atoms, causing a change in charge distribution in a molecule.
In acids, more electronegative groups, such as \(+\) charged nitrogen or triple bonded carbon, can pull electron density towards themselves, thus stabilizing negative charges on the conjugate base.

When examining pair (a), \(\mathrm{(CH_3)_3 \oplus NCH_2 CO_2 H}\), the positive charge on nitrogen draws electrons away, providing better stabilization for the conjugate base. This makes it a stronger acid compared to its counterpart, which lacks this electron-withdrawing power.

• Electron-withdrawing groups increase acid strength.
• Electron-donating groups decrease acid strength.

Understanding this effect is a powerful tool for predicting how structural changes will impact acidity.

Acid Strength

Acid strength is determined by the ability of a molecule to donate a proton, forming a stable conjugate base. Factors such as electronegativity and molecular structure play pivotal roles.
In the case of pair (c), \(\mathrm{CH_3 CO_2 H}\) demonstrates higher acidity as it lacks bulky electron-donating groups that would otherwise destabilize the conjugate base.

• Stronger acids form more stable conjugate bases.
• Proton donation is easier when the negative charge on the base is dispersed.

In essence, any feature that enhances the stability of the conjugate base will contribute to making the original compound a stronger acid.

Conjugate Base Stabilization

Stabilization of the conjugate base is key to the strength of an acid. The more stable the conjugate base, the stronger the acid.
Conjugate base stabilization can be influenced by substituents within the molecule, such as in pair (d) where the more electronegative oxygen in \(\mathrm{CH_3 OCH_2 CO_2 H}\) stabilizes the conjugate base better than sulfur in \(\mathrm{CH_3 SCH_2 CO_2 H}\).

Factors promoting stabilization include:

• Electronegativity: More electronegative atoms better stabilize negative charge.
• Inductive Effect: Electron withdrawal by adjacent atoms or groups enhances stability.

These help spread out the negative charge over the conjugate base, making the structure less reactive and more stable.

Resonance Effect

The resonance effect plays a significant role in determining the acid strength by delocalizing electrons and stabilizing the conjugate base.
In pair (b), the presence of charge separation in \(\mathrm{(CH_3)_2 \oplus N(\ominus O)CH_2 CO_2 H}\) helps stabilize the base through resonance, making it the stronger acid.

• Resonance can distribute negative charge over multiple atoms.
• This distribution reduces the energy of the conjugate base.

By allowing the charge to be shared across different parts of the molecule, resonance decreases the localization of negative energy, which in turn boosts the acid’s ability to release a proton.