Glycine is an achiral amino acid with pK_{a} values of 2.34 and 9.60. Which of the following is the predominant structure of glycine at a physiological pH?

**Explanation**

This questions asks the examinee to determine the predominate structure of the amino acid glycine. Given the pK_{a }value for the amine and carboxylic acid group, you can apply the Henderson-Hasselbalch equation to figure out whether the substituent exists as an acid or conjugate base at a physiological pH of 7.4.

COOH:

pH = pK_{a }+ log[COO^{–}]/[COOH]

7.4 = 2.34 + log[COO^{–}]/[COOH]

5.06 = log[COO^{–}]/[COOH]

10^{5.06 }= [COO^{–}]/[COOH]

[COO^{–}] = 10^{5.06}[COOH]

Thus, the conjugate base COO^{= }is the predominate form.

NH_{3}^{+}:

pH = pK_{a }+ log[NH_{2}]/[NH_{3}^{+}]

7.4 = 9.60 + log[NH_{2}]/[NH_{3}^{+}]

-2.2 = log[NH_{2}]/[NH_{3}^{+}]

10^{-2.2 }= [NH_{2}]/[NH_{3}^{+}]

10^{2.2}[NH_{2}] = [NH_{3}^{+}]

Thus, the acid NH_{3}^{+} predominates.

Therefore, the structure in answer choice D is the most common form of glycine at a physiological pH.

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Can you elaborate how can we say that [COO-] and [NH3+] is the predominant form from the equation, please?

Hi Hung, thank you for reaching out! First of all, we can actually answer this question even more rapidly by comparing its pKa values to physiological pH (7.4). The pKa of 2.34 corresponds to the carboxylic acid group. Since 7.4 (the surrounding pH) is larger than 2.34, it means our solution is relatively basic, and the carboxylic acid group will be deprotonated (COO-). Next, the pKa of 9.60 corresponds to the amine group. Since 7.4 (the surrounding pH) is smaller than 9.60, it means our solution is relatively acidic with respect to this group, and the amine will be protonated (NH3+).

Of course, we can alternatively use the Henderson-Hasselbalch equation, like this question suggests. Remember, this equation is pH = pKa + log[conjugate base]/[acid]. Let’s take our carboxylic acid group as an example. Plug in pH and pKa: 7.4 = 2.34 + log[conjugate base]/[acid]. Simplifying this equation gives us 5.06 = log [conjugate base]/[acid].

Now, do you remember how to “get rid of” a log? When we see “log,” it means log base 10. We can remove the “log” part from the equation by changing the left side to 10^5.06. Now, we have 10^5.06 = [conjugate base]/[acid]. 10^5.06 is very large, so our conjugate base (COO-) concentration must be much larger than our acid (COOH) concentration. COO- is thus the predominant form.

Let us know if we can help with anything else!