Final answer:
O-vanillin does not protonate p-toluidine effectively due to high pKa values.
Step-by-step explanation:
The correct answer is a) Due to high pKa values, o-vanillin cannot protonate p-toluidine effectively.
O-vanillin and p-toluidine both contain functional groups that can act as either acids or bases. In this case, o-vanillin is acting as an acid and p-toluidine is acting as a base.
The pKa value is a measure of the acidity or basicity of a compound, with lower values indicating stronger acidity. In general, a proton transfer reaction occurs between an acid and a base when the acid has a lower pKa value than the base.
However, in this specific case, o-vanillin has a high pKa value, meaning it is a weak acid and does not easily donate a proton. On the other hand, p-toluidine has a lower pKa value, indicating it is a stronger base and readily accepts pons.
The balanced chemical reaction for the proton transfer between o-vanillin and p-toluidine would be:
C₈H₈O₂ + C₇H₉N → C₈H₇O₂⁺ + C₇H₁₀N⁺
The appropriate reaction arrow in this case would be a single-headed arrow pointing from o-vanillin to p-toluidine, indicating the transfer of a proton from o-vanillin to p-toluidine.
correct answer is a) Due to high pKa values, o-vanillin cannot protonate p-toluidine effectively.