The proton on the right is more acidic.
Work Step by Step
Draw the conjugate bases: In the first one, I am going to remove the proton on the left: Remove the highlighted hydrogen on the left, and give a lone pair to the nitrogen that was connected to it (It will get a negative charge). In the second one, I am going to remove the proton on the right: Remove the highlighted hydrogen on the right, and give a lone pair to the carbon that was connected to. it (It will get a negative charge). Now, check which one is more stable, use the ARIO rules: 1. Atom: - The first conjugate base has its charge on a nitrogen, and the second one, on a carbon. Therefore, the first base should be more stable. 2. Ressonance: - There are no other structures in both case, so this doesn't make a difference. 3. Induction: - There are no electronegative atoms or excess of carbons next to them. 4. Orbital: - The first conjugate base has its charge on a $sp^3$ hybridized atom, and the seconde one has its charge on a $sp$ hybridized atom. - Therefore, the second conjugate base should be more stable. ** Normally, the first rule is more important than the fourth, but, this is an exception, when the structure has a $sp$ hybridized carbon, and a $sp^3$ nitrogen, the carbon one is more stable. - Since the second conjugate base is more stable, the proton on the right is more acidic.