Explain why o-vanillin does not fully protonate p-toluidine. Reference appropriate pKa values and include a balanced chemical reaction and an appropriate reaction arrow in your answer.

O-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base; hence, o-vanillin does not fully protonate p-toluidine.

The pKa is defined as the negative logarithm of Ka. The dissociation constant of an acid Ka shows the extent of dissociation of an acid in solution. The higher the pKa, the lower the Ka and the weaker the acid.

The pKₐ of o-vanillin is 7.81 while the pKₐ of p-toluidine is 4.44. This means that  o-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base. Hence, o-vanillin does not fully protonate p-toluidine.

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Explanation:

pKₐ of o-vanillin = 7.81; pKₐ of p-toluidine = 4.44

The higher the pKₐ, the weaker the acid.

Thus, o-vanillin is the weaker acid and has a stronger conjugate base.

The conjugate acid of p-toluidine is the stronger and has the weaker conjugate base.

The equation for the equilibrium is

H-OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₂ ⇌ ⁻OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₃⁺

    weaker acid              weaker base          stronger  base        stronger acid

The reaction between the stronger acid and the stronger base pushes the position of equilibrium to the left.

Thus, o-vanillin does not fully protonate p-toluidine.

conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object. collisions between adjacent atoms and molecules transfer kinetic energy from the warmer to the cooler object. the objects must be in physical contact.

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