Biochemistry Assignment

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Name: _____________________________

 

CHM 332: Fall 2020

Homework #4

1. The following experimental data were collected during a study of the catalytic activity of an intestinal peptidase

capable of hydrolyzing the dipeptide glycylglycine.

 

Glycylglycine + H2O → 2 glycine

 

 

[S] (mM)

Product formed

(µmol/min)

 

[S] (mM)

Product formed

(µmol/min)

1.5 0.21 4.0 0.33

2.0 0.24 8.0 0.40

3.0 0.28 16.0 0.45

 

From these data determine by graphical analysis the values of Km and Vmax for this enzyme preparation and

substrate.

 

2. Many enzymes are inhibited irreversibly by heavy-metal ions such as Hg2+, Cu2+, or Ag+, which can react with essential sulfhydryl groups to form mercaptides:

 

Enz−SH + Ag+ → Enz−S−Ag + H+

 

The affinity of Ag+ for sulfhydryl groups is so great that Ag+ can be used to titrate –SH groups quantitatively. To

10 mL of a solution containing 1.0 mg/mL of a pure enzyme was added just enough AgNO3 to completely

inactivate the enzyme. A total of 0.432 µmol of AgNO3 was required. Calculate the minimum molecular weight

of the enzyme. Why does the value obtained in this way give only the minimum molecular weight?

 

3. The enzymatic activity of lysozyme is optimal at pH 5.2 (see graph below). The active site of lysozyme contains two amino acid residues essential for catalysis: Glu35

and Asp52. The pKa values of the carboxyl side chains of these two residues are 5.9

and 4.5, respectively. What is the ionization state (protonated or deprotonated) of

each residue at the pH optimum of lysozyme? How can the ionization states of these

two amino acid residues explain the pH-activity profile of lysozyme shown at right?

4. Two different enzymes are able to catalyze the same reaction, A → B. They both have the same Vmax, but differ in

their Km for the substrate A. For enzyme 1, the Km is 1.0 mM; for enzyme 2, the Km is 10 mM. When enzyme 1

was incubated with 0.1 mM A, it was observed that B was produced at a rate of 0.0020 mmoles/minute.

 

a. What is the value of the Vmax of the enzymes? b. What will be the rate of production of B when enzyme 2 is incubated with 0.1 mM A? c. What will be the rate of production of B when enzyme 1 is incubated with 1M (i.e., 1000 mM) A?

 

5. An enzyme can catalyze a reaction with either of two substrates, S1 or S2. The Km for S1 was found to be 2.0 mM, and the Km for S2 was found to be 20 mM. A student determined that the Vmax was the same for the two substrates.

Unfortunately, he lost the page of his notebook and needed to know the value of Vmax. He carried out two

reactions: one with 0.1 mM S1, the other with 0.1 mM S2. Unfortunately, he forgot to label which reaction tube

contained which substrate. Determine the value of Vmax from the results he obtained:

 

Tube Number Rate of formation of product

1 0.5

2 4.8

Organic Chem Lab Reports (Purification – Recrystallization Of Benzoic Acid)

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CHEM 2423 Recrystallization of Benzoic Acid Dr. Pahlavan

1

EXPERIMENT 4 – Purification – Recrystallization of Benzoic acid

Purpose:

a) To purify samples of organic compounds that are solids at room temperature b) To dissociate the impure sample in the minimum amount of an appropriate hot solvent

Equipment / Materials: hot plate 125-mL Erlenmeyer flask ice stirring rod spatula

Büchner funnel impure benzoic acid weighing paper digital scales rubber tubing (hose) benzoic acid boiling stones (chips) filter paper

25 mL graguated cylinder 50 mL beaker Mel-temp apparatus Discussion: The products of chemical reactions can be impure. Purification of your products must be performed to remove by-products and impurities. Liquids are customarily purified by distillation, while solids are purified by recrystallization (sometimes called simply “crystallization”). Recrystallization is a method of purifying a solid. There are two types of impurities: those more soluble in a given solvent than the main component and those less soluble. (If there are any impurities that have the same solubility as the main component, then a different solvent needs to be chosen.) When organic substances are synthesized in the laboratory or isolated from plants, they will obviously contain impurities. Several techniques for purifying these compounds have been developed. The most basic of these techniques for the purification of organic solids is recrystallization, which relies on the different solubilities of solutes in a solvent. Compounds, which are less soluble, will crystallize first. The crystallization process itself helps in the purification because as the crystals form, they select the correct molecules, which fit into the crystal lattice and ignore the wrong molecules. This is of course not a perfect process, but it does increase the purity of the final product. The solubility of the compound in the solvent used for recrystallization is important. In the ideal case, the solvent would completely dissolve the compound to be purified at high temperature, usually the boiling point of the solvent, and the compound would be completely insoluble in that solvent at room temperature or at zero oC. In addition the impurity either would be completely insoluble in the particular solvent at the high temperature, or would be very soluble in the solvent at low temperature. In the former case, the impurity could be filtered off at high temperature, while in the latter case the impurity would completely stay in solution upon cooling. In the real world, this will never happen and recrystallization is a technique that has to be practiced and perfected. Regardless of crystallization method, the purity of the solid can be verified by taking the melting point. A good (suitable) recrystallization solvent will dissolve a large amount of the impure compound at temperatures near the boiling point of the solvent. Small amount of compound being purified should remain in solution at low temperatures, between approximately 25 and –5 oC. Low solubility at low temperatures minimizes the amount of purified compound that will lose during recrystallization.

 

 

CHEM 2423 Recrystallization of Benzoic Acid Dr. Pahlavan

2

A suitable recrystallization solvent should also be partially volatile in order to be easily removed from the purified crystals. The solvent should not react with the compound being purified and it should have the boiling point below the melting point of the compound being purified because solid melts before dissolves (oiling out). In selecting a good recrystallization solvent one should also consider flammability, toxicity, and expense. In selecting a solvent consider that like likes like. Polar compounds dissolve polar compounds and non-polar compounds dissolve non-polar compounds. The most commonly used recrystallization solvents are presented in the following table.

solvent formula polarity boiling point (0C) water H2O very polar 100 ethanol CH3CH2OH polar 78 methanol CH3OH polar 65 dichloromethane CH2Cl2 slightly polar 40 diethyl ether (CH3CH2)2O slightly polar 35

Organic compounds with one polar functional group and a low number of carbon atoms such as methanol, ethanol, and n-propanol are highly soluble (miscible) in water. These alcohols form hydrogen bond with water due to the polar –OH functional group. As the number of carbons per polar functional group increase, solubility decreases. The solubility of alcohols with four to five carbons is given in the following table. alcohol formula Solubility (g/100 ml H2O) n-butanol CH3CH2CH2CH2OH 8

n-pentanol CH3CH2CH2CH2CH2OH 2 n-hexanol CH3CH2CH2CH2CH2CH2OH 0.5 n-pentanol CH3CH2CH2CH2CH2CH2CH2OH 0.1

Compounds with six or more carbons for each polar group will not be very soluble in polar solvents but will be soluble in non-polar solvents such as benzene and cyclohexane. If a single solvent cannot be found that is suitable for recrystallization, a solvent pair often used. The solvents must be miscible in one another. Some commonly used solvent pairs are water-ethanol, acetic acid – water, ether-acetone. Typically, the compound being recrystallized will be more soluble in one solvent than the other. The compound is dissolved in a minimum amount of the hot solvent in which it is more soluble. The following formulas used in solubility problems. % lost in cold solvent = (solubility in cold solvent/solubility in hot solvent) x100 % recovery of solid = [g (solid ) – g (solid lost)] x 100 / g (solid)

 

 

CHEM 2423 Recrystallization of Benzoic Acid Dr. Pahlavan

3

Example (1)- The solubility of solid “X” in hot water (5.50 g/100 ml at 100 oC) is not very great, and its solubility in cold water (0.53 g/100ml at 0 oC) is significant. What would be the maximum theoretical percent recovery from crystallization of 5.00 g of solid “X” from 100 ml water? Assuming the solution is chilled at 0 oC. Percent solid lost in cold water = (solubility in cold water/ solubility in hot water) x100 = (0.53/5.50) x100 = 9.64% grams solid lost in cold water = grams mass of original solid x percent lost = 5.00 g x 9.64% = 0.482 g g (solid recovered) = g (solid) – g (solid lost) = 5.00 – 0.482 = 4.52 g % recovery = g (solid recovered) x100 / g (solid) = (4.52/5.00) x100 = 90.4 % Example (2) – The solubility of compound “X” in ethanol is 0.80 g per 100 ml at 0 oC and 5.00 g per 100 ml at 78oC. What is the minimum amount of ethanol needed to recrystallize a 12.00 g sample of compound “X”? How much would be lost in the recrystallization, that is, would remain in the cold solvent? amount of ethanol needed at 78 oC = (12.00 g)( 100 ml/5.00 g) = 240 ml amount of sample remaining in the cold solvent at 0 oC = (240 ml)(0.80 g/100 ml) = 1.9 g or % lost = (0.80/5.00) x100 = 16 % 12.00 x 16% = 1.92 g The actual laboratory we will do is the recrystallization of benzoic acid from water using the temperature gradient method. Benzoic acid is not very soluble in cold water, but it is soluble in hot water. The purpose of this experiment is to learn the technique of recrystallization by purifying benzoic acid.

 

 

CHEM 2423 Recrystallization of Benzoic Acid Dr. Pahlavan

4

 

Experimental Procedures Using a weighing paper, weigh out about 1.00 g of “impure Benzoic acid for recrystallization” and transfer it to a 125-ml Erlenmeyer flask. Add about 20 ml distilled water, using a graduated cylinder, to the flask and bring the mixture to the boiling point by heating on a hot plate, while stirring the mixture and boiling gently to dissolve benzoic acid completely. (Fig 1)

 

benzoic acid solution

Erlenmeyer flask

hot plate

 

Fig 1- Dissolving benzoic acid

Remove the flask from the hot plate and examine the solution. If there are particles of benzoic acid still undissolved, then add an additional amount of hot or cold water in small increments and resume heating the solution. The objective is to dissolve the entire solid in only as much as hot or near boiling solvent (water) as is necessary. Do not add too much water or the solution will not be saturated and the yield of purified benzoic acid will be reduced. Keep adding water in small amounts (several drops at a time from a Pasteur pipette) until all of the benzoic acid is dissolved and the solution is boiling. If the solution is completely clear (though not necessarily colorless) and no solid benzoic acid is visible, then add additional 10-15 ml water to the mixture and place the Erlenmeyer flask on a countertop where it will not be disturbed and cover with an upside-down small beaker (to prevent dust contamination). Allowing the flask to cool slowly will give the best-shaped crystals after about 5-10 minutes. If crystallization does not occur after 10 minutes, scrape the sides of the flask above the level of the solution with the sharp end of a glass rod hard enough to audibly scratch the interior surface of the flask. This may dislodge some undetectable, small crystals that will drop into the solution and “seed” the solution, helping to induce crystallization. A seed crystal can serve as a nucleation point for the crystallization process. Cooling the solution in an ice bath may also help at this point.

 

 

CHEM 2423 Recrystallization of Benzoic Acid Dr. Pahlavan

5

When the crystals have formed completely (may required ice bath), collect your solid chemical by setting up a vacuum (suction) filtration on a properly fitted filter paper in a clean Büchner funnel apparatus as described by your instructor. (Fig 2)

 

vacuum(suction)

filtrate

benzoic acid

Buchner funnel

 

Fig. 2 – Büchner funnel and suction flask

Pour the chilled mixture into the Buchner funnel. The water should filter quickly – if not, check for vacuum leaks. Get all the crystals out of the flask using a spatula or stirring rod. Rinsing with 1 or 2 mLs of cold water helps get the crystals out of the flask, and rinsing helps remove impurities. Let the aspirator run for a few minutes to start air-drying the crystals. Then use a spatula to lift the filter paper and crystals out of the Buchner funnel, then press them as dry as possible on a large clean paper towel (hand dry), allow them to dry completely, and transfer the dry sample to a pre-weigh weighing paper. Determine the weigh the DRY crystals of recovered benzoic acid. Calculate the percent recovered using the following written formula and determine the melting point of your recrystallized benzoic acid. Weight of benzoic acid obtained after recrystallization % Recovered = x100 Weight of benzoic acid before recrystallization Note: Submit product to the instructor in a properly labeled container.

 

 

CHEM 2423 Recrystallization of Benzoic Acid Dr. Pahlavan

6

EXPERIMENT 4 – Recrystallization of Benzoic Acid

Data and Results (Recrystallization) REPORT FORM Name _______________________________ Instructor ___________________________

Date ________________________________ 1. Sample name ____________________________ 2. Data on the impure Benzoic acid a. Mass of the benzoic acid + weighing paper ________ g b. Mass of weighing paper ________ g c. Mass of impure benzoic acid ________ g 3. Data for recrystallized benzoic acid a. Mass of recrystallized benzoic acid + weighing paper ________g b. Mass of weighing paper ________ g c. Mass of recrystallized benzoic acid ________g d. Calculation of percentage recovery (show calculation) ________% d. Melting point of recrystallized benzoic acid ________ oC e. Structural formula of the benzoic acid

 

 

CHEM 2423 Recrystallization of Benzoic Acid Dr. Pahlavan

7

Pre-Laboratory Questions–EXP 4 Name:

Due before lab begins. Answer in space provided. 1. What is the ideal solvent for crystallization of a particular compound? What is the primary consideration in choosing a solvent for crystallizing a compound? 2. Impure benzoic acid was dissolved in hot water. The container of solution was placed in an ice-water bath instead of being allowed cooling slowly. What will be the result of cooling the solution in this manner? 3. Outline the successive steps in the crystallization of an organic solid from a solvent and state the purpose of each operation. 4. Compound X is quite soluble in toluene, but only slightly soluble in petroleum ether. How could these solvents be used in combination in order to recrystallize X? 5. 0.12 g of compound “Y” dissolves in 10 ml of acetone at 25 oC and 0.85 g of the same compound dissolves in 10 ml of boiling acetone. What volume of acetone would be required to purify a 5.0 g sample of compound?

 

 

CHEM 2423 Recrystallization of Benzoic Acid Dr. Pahlavan

8

Post-Laboratory Questions–EXP 4 Name:

Due after completing the lab.

1. Give some reasons why Suction filtration (vacuum) is to be preferred to gravity filtration.

2. A student recrystallized some impure benzoic acid and isolated it by filtration. He scraped the purified benzoic acid off the filter paper after it had dried and took the melting point as a test for purity. He was surprised that most of the white solid melted sharply between 121 and 122oC but that a small amount remained unmelted even at temperatures above 200oC. Explain this behavior.

3. What does the term “oiling out” mean? How can one prevent oiling out?

3. What are the purposes of the following in recrystallization of solids? I) boiling stones –

II) activated carbon – III) seed crystals –

4. Give one reason why we cannot reuse boiling chips? 5. 0.12 g of compound “Y” dissolves in 10 ml of acetone at 25 oC and 0.85 g of the same compound dissolves in 10 ml of boiling acetone. If 5.0 g of compound “Y” were to be recrystallized from 75 ml acetone, what will be the next maximum amount of “Y” that will be recrystallized?

 

  • EXPERIMENT 4 – Purification – Recrystallization of Benzoic acid
    • Discussion:
  • EXPERIMENT 4 – Recrystallization of Benzoic Acid
  • Pre-Laboratory Questions–EXP 4 Name:
  • Post-Laboratory Questions–EXP 4 Name:

Formal Report For Organic Chemistry

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Multistep Synthesis

Step 1: The Benzoin Condensation: Cyanide Ion Catalyzed

The reaction of two moles of benzaldehyde to form a new carbon-carbon bond is known as the benzoin condensation. It has been catalyzed by two rather different catalysts, cyanide ion and the vitamin, thiamine. Both of which on close examination appear to function in exactly the same way.

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Benzaldehyde MW 106.12

bp 178°C

Benzoin

MW 212.24

mp 135°C

 

 

 

 

Caution: Do not handle the potassium cyanide if you have open cuts on your hands. Wash your hands immediately after handling the crystals. Never acidify a cyanide solution, HCN gas (the killer) would be given off !

 

 

 

Procedure:

In a round bottom flask place 20 mg of potassium cyanide (POISON!), dissolve the solid in 0.25 mL of water, add 0.50 mL of 95% ethanol and from a micropipette 0.400 mL of pure benzaldehyde. Introduce a stirring bar, and while stirring, reflux the solution gently with the heating block for 30 minutes. Remove the round bottom flask and cool it in an ice bath, and if no crystals appear within a few minutes scratch the inside of the reaction flask to induce crystallization. When the crystallization is complete, remove the solvent by filtration using a Hirsch funnel. Wash the crystals twice with separate 0.4 mL portions of a cold, one to one mixture of 95% ethanol to water. The product is usually colorless crystals with a melting point of 134-135°C. Weigh the dry product and save some solid to take a melting point. 150 mg of dry product is needed for the next step. If your synthesis does not yield over 100 mg, your instructor will provide additional benzoin to make up the difference (a few points will be deducted).

 

 

Clean up: This is the most important step in the whole experiment. Discard all filtrate into a waste bottle labeled CYANIDE WASTE in the fume hood. The waste bottle contains 1 % sodium hydroxide and bleach solution (5.25% sodium hypochlorite) that oxidize the cyanide ion.

 

 

 

 

Step 2: Oxidation of Benzoin to Benzil

 

Benzoin can be oxidized to the a -diketone, benzil, by a variety of mild oxidizing agents. Too vigorous oxidation by sodium dichromate or potassium permanganate yields chiefly benzaldehyde by cleavage of the bond between the two oxidized carbon atoms, activated by both of the attached phenyl groups. In this procedure, benzoin is oxidized with two equivalents of copper (II) acetate in an acetic acid­-water mixture. Purportedly, this procedure offers the advantage that the benzil obtained is well crystallized and of high purity.

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Procedure:

 

Weigh out ~150 mg of the benzoin product and calculate the stoichiom

My Topic Is Racemization, Organic Chemistry

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Write a 2- 3 page report

-require only the final report

-Full Instruction is on file

•Provide answers to the case study questions in Part 1 (a–e) at the start of your report.

• Critical Analysis Report: Using the answers to the case study questions in Parts 2–4, describe the major points of the case in essay format. Summarize the outcomes and concepts discussed in your lab session. Incorporate the answers to the case study questions into your report essay—do not write separate individual answers.

• Provide a short description of the outcomes of the experiment by Noorduin, et al. How does the experiment by Noorduin, et al., provide a “proof of principle” to the theory of chirality enhancement?

• Provide a critical analysis of the experiment. Does this experiment provide the answer to our question (i.e., how did amino acids become 100% “L” from achiral starting materials)? Describe the positive and negative contributions of this paper to the theory.

• In your conclusion, please address the fi nal point from Part 4. Noorduin’s paper suggests a reason for the enhancement of chirality from a small excess (1–2%) to complete asymmetry (100%). It does not address how that initial excess appeared in the fi rst place. You have been given one theory about how chiral molecules appeared on earth—do you believe it? Are there weaknesses in that theory? Th is section is intended to be open— ended—points will be given for your analysis, not just whether your answer is “correct.” (Th ere is no proven answer to this question.)

Page 1“Chirality and the Origins of Life” by Young and Hooley

by Michael C. Young and Richard J. Hooley Department of Chemistry University of California, Riverside

NATIONAL CENTER FOR CASE STUDY TEACHING IN SCIENCE

Before Class

1. Read the case scenario provided below.

2. In addition, read the following in advance of class:

a. Noorduin, et al. “Emergence of a Single Solid Chiral State from a Nearly Racemic Amino Acid Derivative.” J. Am. Chem. Soc. 2008, 130, 1158–1159.

b. Chapter 5 and section 26.1 (“Stereochemistry,” pages143–173 and “Amino Acid Structures,” pages 1045– 1049) of your assigned organic chemistry textbook (McMurry, 8th edition).

c. Th e defi nition of “Ostwald ripening” (from Wikipedia; attached to the end of this worksheet).

3. Answer the pre-case question and be prepared to submit your response to your discussion group when you arrive to class.

4. Read and familiarize yourself with the case study questions; these will be discussed and answered in groups during the discussion sessions.

In Class

1. You will be divided into collaborative groups for the pre-case discussion in which you identify the major issues of the case, and determine what types of questions need to be answered in order to resolve the case.

2. You will work in your collaborative groups and arrive at answers to the case study questions.

After Class

Complete the post-case study individual report. Th e due date for this fi nal report will be announced in class.

Chirality and the Origins of Life: A Case Study in Organic Chemistry

•

Case copyright held by the National Center for Case Study Teaching in Science, University at Buff alo, State University of New York. Originally published November 6, 2014. Please see our usage guidelines, which outline our policy concerning permissible reproduction of this work. Title block detail from Michelangelo’s “Creation of Adam.”

 

 

NATIONAL CENTER FOR CASE STUDY TEACHING IN SCIENCE

Page 2“Chirality and the Origins of Life” by Young and Hooley

The Case Scenario

Carsten and Charles were having a philosophical discussion on the origins of life in the universe. Carsten had been reading some literature on the topic. Th is stated that the fact that amino acids are chiral, but naturally exist in only one form, means that evolution cannot be responsible for the initial formation of proteins and other building blocks of life. Charles disagreed with this, saying that the proportion of each form can alter over time—four billion years should be suffi cient time for anything to occur. After a spirited debate, they decided to do some reading, and came across a paper in the Journal of the American Chemical Society by Noorduin, et al., that discussed their topic in more detail.

Background

Natural amino acids form the basis for the structure of all proteins and enzymes in living organisms. Th e key to this is that each amino acid contains one stereocenter in its structure. All natural amino acids exist as a single enantiomer (the L isomer); this enantiomeric purity allows proteins (amino acid polymers) to form stable three dimensional structures.

When determining how life on this planet began, the question of how enantiomerically pure amino acids were created from achiral building blocks is still a signifi cant puzzle. A plausible “prebiotic” synthesis of amino acids that can be performed in a laboratory setting called the Strecker reaction is shown below. While this addresses the issue of how amino acids could have originally been formed, it does not address the chirality issue—achiral reactants always give a racemic mixture of products.

Th ere are many questions to ask about how life began: from a chemist’s point of view, the most interesting is “how could amino acids have been converted from achiral starting materials to 100% of a single (L) enantiomer, allowing the creation of proteins and other building blocks of life as we know them?” In this case study, you will address that issue.

Th is question can be split into two parts: (a) how was the enantiomeric purity enhanced to 100% over time from a small excess, and (b) how did that small excess occur in the fi rst place? Th e fi rst question is addressed by your assigned reading (Noorduin, et al.). Th e second is much trickier…

Question

1. Defi ne the following terms: racemic mixture, enantiomer, achiral, enantiomeric excess, and diastereomer (see McMurry, Chapter 5).

 

H3C H

O + NH3 HCN+

H2O

CH3

OH

O NH2

H CH3

OH

O NH2

HAchi ral Reactant s

+

Racemic M ixt ure of P roducts

NH2

CH3

HO

O

Natural L-Alanine: single enant iomer

 

 

NATIONAL CENTER FOR CASE STUDY TEACHING IN SCIENCE

Page 3“Chirality and the Origins of Life” by Young and Hooley

Part 1 – Amino Acids, Chirality, and Racemization

In this fi rst section, you will answer questions about the chiral nature of amino acids and how they are susceptible to interconversion between enantiomers (“racemization”).

Questions

a. Four natural amino acids are shown below. Determine the chiral centers in each amino acid and assign those centers R or S, according to the Cahn-Ingold-Prelog rules.

b. In Noorduin, et al., amino acid surrogate 1 (a molecule that mimics the reactive behavior of natural amino acids) was used to illustrate enantiomeric enhancement upon crystallization. Th e fi rst experiment was to illustrate racemization in solution upon treatment with added base, as shown below. Determine the position of the stereocenter in 1, and use the Cahn-Ingold-Prelog rules to confi rm the R/S stereochemistry.

c. DBU (1,8-diazabicycloundec-7-ene) is an organic amine base that can reversibly deprotonate 1. Th ere are two lone pairs in DBU. Which is most basic? Draw the two protonated species, and see which one is most stabilized by resonance.

a. Drawing DBU as B: (i.e., “base”), draw the mechanism for racemization of 1 upon treatment with base. Pay attention to the geometry of the charged intermediate.

b. Why did Noorduin, et al, use the amino acid surrogate 1 instead of a natural amino acid? (Hint: What happens when you treat L-alanine with a base?)

NH2

CH3

HO

O NH2HO

O

NH2HO

O

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NH2HO

O

NHL-Alanine

L-Phenylalanine L-Cysteine

L-Tryptophan

N

NH2O(S)-1 MeOH

solution

“DBU” H

N

NH2O(R)-1

H

N

N 1,8-diazabicyclo-

undec-7-ene (DBU)

 

 

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Page 4“Chirality and the Origins of Life” by Young and Hooley

Part 2 – Chirality Enhancement via Crystallization

In this section, you will discuss the results of Noorduin, et al., and describe why the phenomenon is observed.

Questions

a. Th e concept of the paper is shown below. Upon addition of a small excess of one enantiomer of 1 or a diff erent chiral species (such as phenylglycine), crystals of 1 with 100% enantiomeric excess can be obtained over time. Th e key to this theory and experiment is selective crystallization. Read the Ostwald ripening attachment (P-1) and explain why a large crystal is more thermodynamically favorable than a small one.

b. Th e experiment also relies on the fact that individual enantiomers crystallize together in a single crystal (i.e., the crystals are either all-S or all-R, not a racemic crystal). Explain why this occurs; consider the solid state interaction of two molecules of the same handedness and diff erent handedness. See McMurry, pages161–163.

c. If there is only a small excess of one enantiomer in solution, can selective crystallization alone (i.e., in the absence of base) cause chirality enhancement?

d. Discuss why the combination of added base (DBU in this case) and selective crystallization allows the enhancement of chirality of the system. Consider equilibrium eff ects and Le Chatelier’s principle in your answer.

e. Th is is a proof-of-principle experiment (i.e., it mimics the natural process, but several changes were made to allow study in a laboratory setting). To maximize the speed of the enhancement, glass beads and magnetic stirring were employed. How does this speed up the process? Is this a plausible method of mimicking natural evolution? Why did the researchers not simply perform the experiment as it would occur in prebiotic systems?

 

S

SS

S

R

R R

R

crystallizationdissolution crystallizationdissolution

N

NH2O(S)-1 MeOH

solution

H N

NH2O(R)-1

HDBU

 

 

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Page 5“Chirality and the Origins of Life” by Young and Hooley

Part 3 – Critical Analysis

You have discussed the results of Noorduin, et al. Th is is a controlled experiment designed to show the proof of principle that a slight excess of one enantiomer can lead to complete conversion to a single enantiomer over time, under the correct conditions. Here, in this question set, you will discuss whether this applies to prebiotic systems and is a good model for the origin of life.

Counterpoints:

a. Is base-catalyzed racemization the most favorable acid-base reaction for natural amino acids? Determine the most acidic H atom in 1 and in L-alanine. Are they the same?

b. Th is theory relies on crystallization. Are there large solid deposits of amino acids in nature? Are amino acids more or less soluble in water than 1? In question 2b you described why crystals of one enantiomer are more favorable than crystals of a racemic mixture. Is this always the case (see McMurry, pages161–163)?

c. Provide a critical analysis of this experiment. Discuss whether or not you believe the results of this experiment. Discuss whether this is a good “proof-of-principle” experiment, and whether it can apply to natural amino acids in a prebiotic environment.

 

 

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Part 4 – Extraterrestrial Amino Acids

http://en.wikipedia.org/wiki/Murchison_meteorite

So far, you have discussed chirality amplifi cation. Th is theory still requires a small excess of one enantiomer to begin the amplifi cation process. In this question set you will discuss how that original excess appeared.

One (disputed) theory of the original genesis of “chiral molecules” on Earth comes from outer space. Th e Murchison meteorite, which fell to earth in Australia in 1969, was shown to contain traces of amino acids and other simple organic materials. Th ese materials displayed a “slightly enhanced” enantiomeric ratio, i.e., more of one enantiomer than the other (Engel & Macko 1997; Cronin & Pozzarello 1997). Th e theory was posited that similarly chiral materials crashed to earth during the prebiotic period (Pizzarello & Groy 2011). (Note: Th ere are a lot of scientifi cally dubious theories on this topic. We won’t discuss how the enhanced chirality appeared on the meteor while in outer space; ask a physicist.)

a. Th e presence of natural amino acids with enantiomeric excess from the meteorite is still a subject of debate. Discuss why this is—what are the experimental issues with stating that alanine found on the Murchison meteorite has an excess of one (L) enantiomer? How might “false positives” be observed in the analysis of these meteorite samples?

b. An argument put forward to corroborate the meteorite theory is the presence of unnatural -alkyl amino acids such as isovaline, which was observed to exist in the meteorite with a slight enhancement of the L isomer. Discuss why the observation of unnatural amino acids such as these is less susceptible to false positives than the presence of alanine above.

c. Assume that there was an excess of (L)-isovaline on the Murchison meteorite. Discuss the plausibility of (L)-isovaline conferring chirality on species such as 1, based on your reading of Noorduin, et al.

NH2

CH3 HO

O

L-Isovaline CH3

 

 

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Page 7“Chirality and the Origins of Life” by Young and Hooley

Final Report

Write a two- to three-page report on the case study, following the structure below

• Provide answers to the case study questions in Part 1 (a–e) at the start of your report.

• Critical Analysis Report: Using the answers to the case study questions in Parts 2–4, describe the major points of the case in essay format. Summarize the outcomes and concepts discussed in your lab session. Incorporate the answers to the case study questions into your report essay—do not write separate individual answers.

• Provide a short description of the outcomes of the experiment by Noorduin, et al. How does the experiment by Noorduin, et al., provide a “proof of principle” to the theory of chirality enhancement?

• Provide a critical analysis of the experiment. Does this experiment provide the answer to our question (i.e., how did amino acids become 100% “L” from achiral starting materials)? Describe the positive and negative contributions of this paper to the theory.

• In your conclusion, please address the fi nal point from Part 4. Noorduin’s paper suggests a reason for the enhancement of chirality from a small excess (1–2%) to complete asymmetry (100%). It does not address how that initial excess appeared in the fi rst place. You have been given one theory about how chiral molecules appeared on earth—do you believe it? Are there weaknesses in that theory? Th is section is intended to be open— ended—points will be given for your analysis, not just whether your answer is “correct.” (Th ere is no proven answer to this question.)

References

McMurry, J.E. 2012. Organic Chemistry, 8th edition, Cengage Learning.

Noorduin, et al. 2008. Emergence of a single solid chiral state from a nearly racemic amino acid derivative. Journal of the American Chemical Society 130: 1158–1159.

Wikipedia. Murchison meteorite. http://en.wikipedia.org/wiki/Murchison_meteorite

Wikipedia. Ostwald Ripening. http://en.wikipedia.org/wiki/Ostwald_ripening

Further Reading

Th ese references below are optional and only for students who want a deeper understanding of the material. Th ey are not necessary for the case or fi nal report.

Engel, M. H, and Macko, S. A. 1997. Isotopic evidence for extraterrestrial non-racemic amino acids in the Murchison meteorite. Nature 389: 265–268.

Cronin, J. R., and Pizzarello, S. 1997. Enantiomeric excesses in meteoritic amino acids. Science 275: 951–955.

Pizzarello, S.; Groy, T.L 2011. Molecular asymmetry in extraterrestrial organic chemistry: An analytical perspective. Geochimica et Cosmochimica Acta 75: 645–656.

 

 

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Page 8“Chirality and the Origins of Life” by Young and Hooley

Assigned Reading

Ostwald Ripening

From Wikipedia (http://en.wikipedia.org/wiki/Ostwald_ripening), last accessed October 14, 1014:

Th is thermodynamically-driven spontaneous process occurs because larger particles are more energetically favored than smaller particles. Th is stems from the fact that molecules on the surface of a particle are energetically less stable than the ones in the interior.

Consider a cubic crystal of atoms: all the atoms inside are bonded to 6 neighbors and are quite stable, but atoms on the surface are only bonded to 5 neighbors or fewer, which makes these surface atoms less stable. Large particles are more energetically favorable since, continuing with this example, more atoms are bonded to 6 neighbors and fewer atoms are at the unfavorable surface. As the system tries to lower its overall energy, molecules on the surface of a small particle (energetically unfavorable, with only 3 or 4 or 5 bonded neighbors) will tend to detach from the particle, as per the Kelvin equation, and diff use into the solution. When all small particles do this, it increases the concentration of free molecules in solution. When the free molecules in solution are supersaturated, the free molecules have a tendency to condense on the surface of larger particles. Th erefore, all smaller particles shrink, while larger particles grow, and overall the average size will increase. As time tends to infi nity, the entire population of particles becomes one large spherical particle to minimize the total surface area.