Which of the following has the highest pK_a value?

Iodoacetic acid

The correct answer is option 4. Iodoacetic acid.

The acidity of carboxylic acids, represented by their \(pK_a\) values, is influenced by several factors, including the electron-withdrawing effects of substituents attached to the carboxyl group. In haloacetic acids (acetic acids substituted with halogen atoms), the acidity generally increases as the electronegativity of the halogen increases. This trend occurs because more electronegative halogens stabilize the negative charge on the conjugate base formed after deprotonation, making the acid more acidic (lower \(pK_a\)).

Let us break down the reasoning for each compound:

Bromoacetic acid \((CH_2BrCOOH)\):

Bromine \((Br)\) is more electronegative than iodine but less electronegative than chlorine and fluorine (electronegativity: 2.96). Bromoacetic acid has a lower \(pK_a\) compared to iodoacetic acid due to the stronger electron-withdrawing effect of bromine compared to iodine.

Chloroacetic acid \((CH_2ClCOOH)\):

Chlorine \((Cl)\) is more electronegative than bromine (electronegativity: 3.16). Chloroacetic acid has a lower \(pK_a\) compared to bromoacetic acid due to the stronger electron-withdrawing effect of chlorine compared to bromine.

Fluoroacetic acid \((CH_2FCOOH)\):

Fluorine \((F)\) is the most electronegative halogen (electronegativity: 3.98). Fluoroacetic acid has the lowest \(pK_a\) among the haloacetic acids because fluorine strongly stabilizes the conjugate base (fluoroacetate ion, \(CH_2FCOO^-\)) through its strong electron-withdrawing effect.

Iodoacetic acid \((CH_2ICOOH)\):

Iodine \((I)\) is the least electronegative halogen among the options provided (electronegativity: 2.66). The weak electron-withdrawing effect of iodine results in a less stabilized conjugate base (iodoacetate ion, \(CH_2ICOO^-\)), leading to a higher \(pK_a\) value compared to other haloacetic acids.

Conclusion:

Iodoacetic acid has the highest \(pK_a\) among the haloacetic acids listed because iodine is the least electronegative halogen, resulting in the least stabilization of the conjugate base. Therefore, the correct answer remains: Iodoacetic acid.