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Consider the structure of the amino acid isoleucine. a. How many chiral centers does isoleucine have? b. How many optical isomers does isoleucine have? c. Draw perspective formulas for all the optical isomers of isoleucine.

Short Answer

Expert verified
Isoleucine has 2 chiral centers, 4 optical isomers.

Step by step solution

01

Identify Chiral Centers in Isoleucine

A chiral center (or chiral carbon) in a molecule is a carbon atom that is attached to four different groups. To determine the number of chiral centers in isoleucine, draw its structure and evaluate each carbon atom to see if it is bonded to four different substituents. Isoleucine has two chiral centers: one at C2, which is attached to NH2, H, CH(CH3)CH2CH3, and COOH; and the other at C3, which is attached to C2, CH2, CH3, and H.
02

Calculate Optical Isomers using Chiral Centers

For a molecule with 'n' chiral centers, the number of possible optical isomers is given by the formula \(2^n\). Since isoleucine has 2 chiral centers, substitute n = 2 into the formula: \(2^2 = 4\). Therefore, isoleucine can have 4 optical isomers.
03

Draw All Optical Isomers

Perspective formulas for each optical isomer must be drawn to show different spatial arrangements. Draw each chiral center to show its possible 'R' or 'S' configurations. Each of the four optical isomers can be represented by these configurations: (2R,3R)-isoleucine, (2S,3S)-isoleucine, (2R,3S)-isoleucine, and (2S,3R)-isoleucine, ensuring to switch the positions of substituents around each chiral center to generate each specific isomer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Optical Isomers
Optical isomers are essential concepts in chemistry, especially in the field of stereochemistry. These are molecules that have the same molecular formula and sequence of bonded atoms, but they differ in the 3D orientation of their atoms. Optical isomers are mirror images of each other that cannot be superimposed, much like our left and right hands. This lack of superimposability results from the presence of chiral centers in the molecule, which are carbon atoms bonded to four different groups.

When a molecule, such as isoleucine, contains chiral centers, it can have multiple optical isomers. These isomers have varying properties, such as how they rotate plane-polarized light. Some will rotate the light clockwise, others counterclockwise. This characteristic is crucial in fields such as pharmaceuticals, where the orientation of atoms can significantly impact a drug's efficacy and safety.

Optical isomers are a fascinating part of studying molecules because they challenge us to visualize molecules' spatial arrangements. They also illustrate the impact of chirality on chemical properties, which can profoundly affect their biological interactions.
R/S Configuration
The R/S configuration system is used to define the absolute configuration of chiral centers in molecules. This system is critical for distinguishing between the two possible optical isomers (enantiomers) that arise from each chiral center. The idea is to assign an order of priority to the groups attached to the chiral carbon and determine the spatial arrangement of these groups around the carbon.

To assign the R/S configuration:
  • Identify the chiral center and label the groups attached to it by atomic number, with the highest being first priority.

  • Orient the molecule so that the group with the lowest priority is pointing away from you.

  • Observe the remaining groups: If they appear in a clockwise order, the configuration is "R" (rectus, right). If counterclockwise, it is "S" (sinister, left).

For isoleucine, which has two chiral centers, the R/S configuration helps define its four optical isomers: (2R,3R)-isoleucine, (2S,3S)-isoleucine, (2R,3S)-isoleucine, and (2S,3R)-isoleucine. The configurations of these isomers play a significant role in their biochemical interactions and functionalities.
Isoleucine Structure
Isoleucine is one of the essential amino acids, which means that it cannot be synthesized by the body and must be obtained from diet. Its structure is distinct due to the presence of two chiral centers, giving it unique properties and enabling the formation of four optical isomers.

The structure of isoleucine includes a branched side chain, making it part of the aliphatic group of amino acids. Specifically, it consists of a carbon (C2) linked to an amino group (NH2), a hydrogen atom (H), a carboxyl group (COOH), and a complex branch consisting of another chiral center.

This complex branch attached to C3 includes another set of groups (CH, CH2, CH3, and H) making C3 another chiral center. As a result, these two centers create a situation where isoleucine can exist in different spatial arrangements, contributing to its unique role in proteins. Understanding these structures is vital in fields such as protein chemistry and molecular biology.
Biochemistry Education
Biochemistry education is an exciting field that bridges biology and chemistry, providing insights into the molecular processes of life. It involves understanding molecules like amino acids and their significance in biochemistry. Learning about chiral centers, optical isomers, and amino acid structures is foundational for students of biochemistry.

Developing an understanding of concepts such as R/S configuration and optical isomerism improves students' skills in stereochemistry, an essential part of molecular biology and pharmacology. Being able to draw and visualize molecules in 3D can deepen one's ability to predict the behavior of biomolecules in complex biological systems.

To engage effectively in biochemistry education, students should:
  • Practice drawing molecular structures and identifying chiral centers.

  • Explore the relationship between molecular orientation and biological function.

  • Incorporate knowledge about amino acids into broader studies of proteins and enzymes.

Ultimately, biochemistry education provides the tools and knowledge needed to understand the microscopic chemical processes that sustain life. This knowledge is critical for anyone pursuing careers in biotechnology, medicine, or pharmaceutical sciences.

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Most popular questions from this chapter

Extracts from the bacterium Bacillus brevis contain a peptide with antibiotic properties. This peptide forms complexes with metal ions and seems to disrupt ion transport across the cell membranes of other bacterial species, leading to bacterial death. The structure of the peptide has been determined from a series of observations. a. Complete acid hydrolysis of the peptide, followed by amino acid analysis, yielded equimolar amounts of Leu, Orn, Phe, Pro, and Val. Orn is ornithine, an amino acid not present in proteins but present in some peptides. Ornithine has the structure b. The molecular weight of the peptide is approximately 1,200 Da. c. The peptide failed to undergo hydrolysis when treated with the enzyme carboxypeptidase. This enzyme catalyzes the hydrolysis of the carboxyl- terminal residue of a polypeptide unless the residue is Pro or, for some reason, does not contain a free carboxyl group. d. Treatment of the intact peptide with 1-fluoro-2,4dinitrobenzene, followed by complete hydrolysis and chromatography, yielded only free amino acids and the derivative shown here. e. Partial hydrolysis of the peptide followed by chromatographic separation and sequence analysis yielded these di- and tripeptides (the amino-terminal amino acid is always the first amino acid): Leu-Phe Phe-Pro Orn-Leu Val-Orn Val-Orn-Leu Phe-Pro-Val Pro-Val-Orn Given this information, deduce the amino acid sequence of the peptide antibiotic. Show your reasoning. When you have arrived at a structure, demonstrate that it is consistent with each experimental observation.

A peptide with the primary structure Lys-Arg-Pro-Leu-Ile-Asp-Gly-Ala must be synthesized by the methods developed by Merrifield. Calculate the percentage of the peptides synthesized that will be full length and have the correct sequence if the addition of each amino acid residue is \(96 \%\) efficient. Do the calculation a second time but assume a \(99 \%\) efficiency for each cycle.

A purified protein is in a Hepes ( \(N\)-(2-hydroxyethyl)piperazine- \(N^{\prime}\)-(2-ethanesulfonic acid)) buffer at pH 7 with \(500 \mathrm{~mm} \mathrm{NaCl}\). A dialysis membrane tube holds a \(1 \mathrm{~mL}\) sample of the protein solution. The sample in the dialysis membrane floats in a beaker containing \(1 \mathrm{~L}\) of the same Hepes buffer, but with \(0 \mathrm{~mm} \mathrm{NaCl}\), for dialysis. Small molecules and ions (such as \(\mathrm{Na}^{+}, \mathrm{Cl}^{-}\), and Hepes) can diffuse across the dialysis membrane, but the protein cannot. a. Calculate the concentration of \(\mathrm{NaCl}\) in the protein sample, once the dialysis has come to equilibrium. Assume that no volume changes occur in the sample during the dialysis. b. Calculate the final \(\mathrm{NaCl}\) concentration in the protein sample after dialysis in \(250 \mathrm{~mL}\) of the same Hepes buffer, with \(0 \mathrm{~mm} \mathrm{NaCl}\), twice in succession.

Amino Acid Constituents of Glutathione Glutathione is an important peptide antioxidant found in cells from bacteria to humans. Identify the three amino acid constituents of glutathione. What is unusual about glutathione's structure?

Suppose a column is filled with a cation-exchange resin at \(\mathrm{pH}\) 7.0. In what order would the given peptides elute from the column if each has the same number of residues? Peptide A: Ala \(30 \%\), Asp \(10 \%\), Lys \(10 \%\), Ser \(15 \%\), Pro \(25 \%\), Cys \(10 \%\) Peptide B: Ile \(25 \%\), Asp \(20 \%\), Arg \(5 \%\), Tyr \(15 \%\), His \(5 \%\), Thr \(30 \%\) Peptide C: Ala \(40 \%\), Glu 5\%, Arg 20\%, Ser 5\%, His 5\%, Trp \(25 \%\)

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