pob5e_solutions_ch03

pob5e_solutions_ch03 - 2608T_ch03sm_S26-S43 2/1/08 11:45AM...

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S-26 Amino Acids, Peptides, and Proteins chapter 3 1. Absolute Configuration of Citrulline The citrulline isolated from watermelons has the structure shown below. Is it a D- or L-amino acid? Explain. Answer Rotating the structural representation 180 ± in the plane of the page puts the most highly oxidized group—the carboxyl ( O COO ² ) group—at the top, in the same position as the O CHO group of glyceraldehyde in Figure 3–4. In this orientation, the a -amino group is on the left, and thus the absolute configuration of the citrulline is L. 2. Relationship between the Titration Curve and the Acid-Base Properties of Glycine A 100 mL solution of 0.1 M glycine at pH 1.72 was titrated with 2 M NaOH solution. The pH was monitored and the results were plotted as shown in the following graph. The key points in the titration are designated I to V. For each of the statements (a) to (o), identify the appropriate key point in the titration and justify your choice. Note: before considering statements (a) through (o), refer to Figure 3–10. The three species involved in the titration of glycine can be considered in terms of a useful physical analogy. Each ionic species can be viewed as a different floor of a building, each with a different net charge: 12 2 4 6 8 0 11.30 0.5 OH ² (equivalents) pH 1.0 1.5 2.0 (V) 9.60 (IV) (III) 2.34 (I) (II) 5.97 10 CC O ) H (CH NH 2 NH 2 22 P H C N ³ H 3 COO ² 2608T_ch03sm_S26-S43 2/1/08 11:45AM Page 26 ntt 102:WHQY028:Solutions Manual:Ch-03:
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Chapter 3 Amino Acids, Peptides, and Proteins S-27 ± H 3 N O CH 2 O COOH ± 1 ± H 3 N O CH 2 O COO ² 0 (zwitterion) H 2 N O CH 2 O COO ² ² 1 The floors are connected by steep stairways, and each stairway has a landing halfway between the floors. A titration curve traces the path one would follow between the different floors as the pH changes in response to added OH ² . Recall that the p K a of an acid (on a halfway landing) represents the pH at which half of the acid is deprotonated. The isoelectric point (pI) is the pH at which the aver- age net charge is zero. Now you are ready to consider statements (a) through (o). (a) Glycine is present predominantly as the species ± H 3 N O CH 2 O COOH. (b) The average net charge of glycine is ± . (c) Half of the amino groups are ionized. (d) The pH is equal to the p K a of the carboxyl group. (e) The pH is equal to the p K a of the protonated amino group. (f) Glycine has its maximum buffering capacity. (g) The average net charge of glycine is zero. (h) The carboxyl group has been completely titrated (first equivalence point). (i) Glycine is completely titrated (second equivalence point). (j) The predominant species is ± H 3 N O CH 2 O COO ² . (k) The average net charge of glycine is ² 1. (l) Glycine is present predominantly as a 50:50 mixture of ± H 3 N O CH 2 O COOH and ± H 3 N O CH 2 O COO ² . (m) This is the isoelectric point.
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This note was uploaded on 09/24/2011 for the course CHEM 369 taught by Professor Wang during the Spring '11 term at University of Houston.

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pob5e_solutions_ch03 - 2608T_ch03sm_S26-S43 2/1/08 11:45AM...

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