Question

Select the pair in which second acid is stronger than the first. (A) \(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) (B) \(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (C) \(\mathrm{CH}_{2} \mathrm{FCO}_{2} \mathrm{H}, \mathrm{CH}_{2} \mathrm{ClCO}_{2} \mathrm{H}\) (D) \(\mathrm{CHF}_{3}, \mathrm{CHCl}_{3}\)

Short answer

The correct answer is (D) because Pair D, \(\mathrm{CHF}_{3}\) and \(\mathrm{CHCl}_{3}\), is the only pair in which the second acid is stronger than the first.

Step-by-step solution

Pair A: \(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)

Their conjugate bases are \(\mathrm{CCl}_{3} \mathrm{CO}_{2}^{-}\) and \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\). Since chlorine is more electronegative than carbon, it stabilizes the negative charge in \(\mathrm{CCl}_{3} \mathrm{CO}_{2}^{-}\) more effectively than the methyl group stabilizes the negative charge in \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\). Thus, \(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}\) is a stronger acid than \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\).

Pair B: \(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{OH}\)

Their conjugate bases are \(\mathrm{CCl}_{3} \mathrm{CO}_{2}^{-}\) and \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\). Again, the \(\mathrm{CCl}_{3} \mathrm{CO}_{2}^{-}\) is more stable due to the effect of the electronegative chlorine atoms. The \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\) is less stable due to the negative charge being on a less electronegative oxygen. Thus, the first acid in this pair is stronger.

Pair C: \(\mathrm{CH}_{2} \mathrm{FCO}_{2} \mathrm{H}, \mathrm{CH}_{2} \mathrm{ClCO}_{2} \mathrm{H}\)

Their conjugate bases are \(\mathrm{CH}_{2} \mathrm{FCO}_{2}^{-}\) and \(\mathrm{CH}_{2} \mathrm{ClCO}_{2}^{-}\). The fluorine atom is more electronegative than chlorine, stabilizing the electron cloud to a greater extent in \(\mathrm{CH}_{2} \mathrm{FCO}_{2}^{-}\) compared to \(\mathrm{CH}_{2} \mathrm{ClCO}_{2}^{-}\). Thus, \(\mathrm{CH}_{2} \mathrm{FCO}_{2} \mathrm{H}\) is a stronger acid than \(\mathrm{CH}_{2} \mathrm{ClCO}_{2} \mathrm{H}\).

Pair D: \(\mathrm{CHF}_{3}, \mathrm{CHCl}_{3}\)

Their conjugate bases are \(\mathrm{CHF}_{2}^{-}\) and \(\mathrm{CHCl}_{2}^{-}\). In this case, the chlorine's lack of electronegativity compared to fluorine stabilizes the electron cloud to a greater extent in \(\mathrm{CHCl}_{2}^{-}\) than \(\mathrm{CHF}_{2}^{-}\). Thus, \(\mathrm{CHCl}_{3}\) is a stronger acid than \(\mathrm{CHF}_{3}\). Based on the analysis above, Pair D is the only pair in which the second acid is stronger than the first. Therefore, the correct answer is (D).

Key concepts

Conjugate Bases

When an acid donates a proton, it transforms into its conjugate base. This change is central to understanding acid strength. Conjugate bases are the species left behind after an acid loses a hydrogen ion. For example, the conjugate base of trichloroacetic acid (\(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}\)) is \(\mathrm{CCl}_{3} \mathrm{CO}_{2}^{-}\).

A stable conjugate base usually results in a stronger acid. This is because a stable base more easily remains without the proton, making its parent acid more willing to give it away. Comparing conjugate bases can help you determine which acid is more potent. Essentially, the more stable the conjugate base, the stronger the acid from which it forms.

Electronegativity

Electronegativity refers to an atom's ability to attract electrons in a bond. It's a significant factor in comparing acid strength. Atoms like fluorine are incredibly electronegative, pulling electrons toward themselves. This action affects how the negative charge is held around the conjugate base.

When an electronegative atom is involved, it stabilizes the conjugate base by holding onto negative charge tightly. Let's take \(\mathrm{CH}_{2} \mathrm{FCO}_{2}^{-}\) as an example. Here, fluorine's electronegativity stabilizes the negative charge effectively, making \(\mathrm{CH}_{2} \mathrm{FCO}_{2} \mathrm{H}\) a stronger acid than \(\mathrm{CH}_{2} \mathrm{ClCO}_{2} \mathrm{H}\), where chlorine is less electronegative.

Stabilization of Negative Charge

The stabilization of negative charge in a conjugate base is crucial to determining acid strength. This stabilization often defines how aggressive an acid can be in donating its proton. When a conjugate base holds a negative charge more stably, the parent acid is stronger.

Factors that play into this stabilization include the surrounding atoms' electronegativity and any stabilization mechanisms, such as resonance or inductive effects.

• An example is \(\mathrm{CCl}_{3} \mathrm{CO}_{2}^{-}\), where the chlorine atoms help stabilize the negative charge, enhancing the acid strength of \(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}\).
• Therefore, looking at how the conjugate base can accommodate and stabilize its charge helps in identifying which acid is stronger.