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I understand why cysteine has a $\mathrm pK_{\mathrm a}$ of 8.3. But the $\mathrm pK_{\mathrm a}$ of thiol in N-acetylcysteine is 9.27. Since there is an acetyl group attached to the nitrogen, doesn't this lower the $\mathrm pK_{\mathrm a}$ via the inductive effect?

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Shown below are the relevant ionization events of cysteine and N-acetylcysteine. The question is asking about the second ionization of each, the deprotonation of the thiol groups, but we see a similar difference in pKas for the carboxylic acid groups as well. The difference in each case is an ammonium group vs. an acylated nitrogen. It's true that the acetyl group is an electron withdrawing group, but it turns out that the cationic ammonium group is even more inductively electron withdrawing. For both the first and second ionizations, the difference in electron withdrawing power is enough to lower the pKa by >1 pKa unit.

ionization of cysteine and N-acetylcysteine

iad22agp's explanation is a good way of rationalizing this. Looking at the first ionization, cysteine is cationic while N-acetyl cysteine is neutral. Since the rest of the structures are similar, we would assume that it's easier to take a positive charge away from a positively charged compound than a neutral compound. Similarly looking at the second ionization, it's easier to take a positive charge away from a neutral compound than a compound that's already negatively charged.

pKa refs: cysteine, N-acetylcysteine

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Assume you are titrating the substance with base. You are already at pH 7 and going up. Removing a proton from a zwitterion (cysteine) to form an anion will require less energy than removing a proton from an already anionic species (N-acetylcysteinate anion) to form a dianion.

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    $\begingroup$ N-acetylcysteine is a zwitterion, too. $\endgroup$ – gsurfer04 Aug 12 '15 at 16:09

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