Post-Lab Questions (1 pt) Is the predominant isomer of the 2,3-dibromo-3-phenylpropanoic acid that you prepared in this experiment a) achiral, b) chiral and racemic, or c) chiral and non-racemic? If it is achiral, it will not be racemic or non-racemic, just achiral. (2 pts) Recall your response to question #1. If you had access to a polarimeter, would the observed optical rotation of your purified product be zero or non-zero? Explain for full credit.

OneClass: Post-Lab Questions (1 pt) Is the predominant isomer of the 2,3-dibromo-3-phenylpropanoic ac...

Hi! I am working on a lab for Organic Chemistry and I want to make sure my answers are correct before submitting it!

Thank you in advance!

Stereoisomerism: A Model Exercise

In this experiment you will construct models with your molecular model set that illustrate the concepts of chirality, chiral center (stereogenic center, asymmetric carbon atom), enantiomers, diastereomers, and meso forms. You will also learn about two conventions, R-S and Fischer, for designating the configurations of chiral molecules.

You will be asked several questions related to each construction exercise. Record the answers and then when you have finished all of the exercises, take the assessment listed in this folder. The questions will be repeated so that you can submit your answers electronically for grading.

Chiral centers

Procedure: Construct a model* in which a tetrahedral (sp 3) carbon atom (black) has four different model atoms attached to it. Use the light blue ball, red, blue and, green polyhedrons to represent four different atoms or groups attached to the central atom (black).

1. Does the model have a plane of symmetry?

a. Yes

b. No

A carbon atom that has four different groups attached to it is a chiral center, or an asymmetric carbon atom. The carbon marked with a # in 3-methylhexane is a chiral center.

2. What groups are attached to the chiral center in 3-methylhexane?

a. Hydrogen

b. Methyl

c. Ethyl

d. n-propyl

e. n-butyl

Replace the green atom (or group) in your model with a second red atom. Now two of the groups attached to the carbon atom are identical:

3. Does the model now have a plane of symmetry?

a. Yes

b. No

4. ï¿½ï¿½ Describe it.

a. Plane through the light blue ball, black polyhedron and blue polyhedron

b. Plane through a red polyhedron, black polyhedron and blue polyhedron

c. Plane through the light blue ball, black polyhedron and red polyhedron

d. Plane through a red polyhedron, black polyhedron and red polyhedron

Draw structural formulas for the following compounds, and mark any chiral centers (asymmetric carbons) with an asterisk:

l-bromobutane, 2- bromobutane, 1,2-dibromobutane,

1,3-dibromobutane, l, 4-dibromobutane, 2,3-dibromobutane.

5. Which of the compounds contain chiral centers?

a. 1-bromobutane

b. 2-bromobutane

c. 1,2-dibromobutane

d. 1,3-dibromobutane

e. 1,4-dibromobutane

f. 2,3-dibromobutane

Chirality and Enantiomers

A center of chirality (from the Greek cheir, hand) imparts the property of handedness to a molecule. In this part of the experiment, the left- or right- handedness of molecules with a chiral center will be illustrated with models.

A molecule is said to be chiral (that is, to have the property of handedness) if its mirror image is not superimposable. The mirror image of a left hand, for example, is a right hand. A molecule that is achiral has a mirror image that is superimposible. We shall see that any molecule with a plane of symmetry is achiral.

Procedure: Reconstruct the original model (carbon (black) with light blue ball, red, blue and, green polyhedrons attached). Set the model on the desktop so that the substituent light blue ball atom points toward the ceiling.

Looking down on the model and proceeding clockwise from the green atom, record the colors of the three atoms that rest on the desktop.

Now construct a second model that is the mirror image of the first, and place it on the desktop with the light blue ball atom up:

6. In which direction, clockwise or counterclockwise, must you proceed in order to list the same sequence of colors of the three atoms resting on the desk's surface?

a. Clockwise

b. Counterclockwise

Try to superimpose the two models.

7. The models are

a. Superimposable

b. Not superimposable

The two models that you have just constructed represent chiral molecules- they lack a plane of symmetry and have mirror images that are notsuperimposable. Two substances, the molecular structures of which are related as an object and its nonsuperimposable mirror image are called enantiomers. They differ from each other only in properties that have a direction or "handedness," such as, for example, the direction(clockwise or counterclockwise) in which they rotate a beam of plane-polarized light. Because of this latter property, such substances are sometimes called optical isomers. They are optically active.

Now we will examine the consequence of having at least two identical atoms or groups attached to a tetrahedral carbon atom.

Replace the green atom in each model with a red atom, so that each model has two identical groups attached to the central carbon atom:

8. Are the models still mirror images?

a. Yes

b. No

9. Does either of the models have a plane of symmetry?

a. Yes

b. No

10. Are the models superimposable?

a. Yes

b. No

11. Do the models represent identical molecules or different molecules?

a. Identical

b. Different

Place each model on the desk so that the light blue ball substituent points up.

12. To define the same sequence of colors for the three atoms resting on the desk top, must you proceed

a. Clockwise

b. Counterclockwise

c. Either way

13. Are the models chiral (handed)?

a. Yes

b. No

The models you have just studied represent achiral molecules. Their mirror images are identical. They are optically inactive. Any molecule that has a plane of symmetry is achiral.

Diastereomers and Meso Forms

For any molecule that has two or more chiral centers, it is possible to have stereoisomers that are not mirror images. Stereoisomers that are not related as enantiomers are diastereomers. Diastereomers differ in all properties, chiral and achiral.

Procedure: Construct a model with four different groups (light blue ball, blue, red, and green) attached to a central carbon atom (black). Construct another model identical to the first. (Be sure they are identical by making sure the models are superimposable.) Now remove the green substituent from each model and connect the two carbon atoms with a bond. Use this model with the two stereogenic centers for the next series of questions (14 -26).

14. How many chiral centers (asymmetric carbons) does this model have?

a. 1

b. 2

c. 3

d. 4

e. 5

Note that there are four different groups attached to each chiral center and that each chiral center has the same four groups attached.

15. Does the model have a plane of symmetry in any of its conformations?

a. Yes

b. No

Construct the mirror image of the first model.

16. Is the mirror image identical to or different from the first model?

a. Identical

b. Different

17. What term describes the two models?

a. Enantiomer

b. Diastereomer

18. Is each model chiral or achiral?

a. Chiral

b. Achiral

Now interchange a red and blue atom on the same carbon in one of the models.

19. Are the models identical or different now?

a. Identical

b. Different

20. Are they mirror images (enantiomers)?

a. Yes

b. No

21. Are they stereoisomers?

a. Yes

b. No

22. What term describes the two models?

a. Enantiomer

b. Diastereomer

Carefully examine the conformations of the model in which you interchanged the red and blue atoms.

23. Does the model have a conformation with a plane of symmetry?

a. Yes

b. No

24. Would the mirror image of this model be identical (superimposable) or different from the model itself?

a. Identical

b. Different

Verify your prediction by constructing the mirror-image model.

25. This model is

a. Chiral

b. Achiral

26. Would a molecule corresponding to this model be optically active?

a. Yes

b. No

The last model studied here represents a meso form. The model possesses two chiral centers, but they are of equal and opposite chirality. This situation arises when a molecule has two identical chiral centers. Because the molecule has a readily accessible conformation with a plane of symmetry, it is achiral and optically inactive.

Tartaric acid is a molecule that corresponds to the models constructed in this section of the experiment.

Tartaric Acid

It exists in three forms; two are optically active enantiomers, and the third is an optically inactive meso form that is a diastereomer of the optically active forms. ,

27. Draw Newman projection formulas (looking at the bond between C-2 and C-3) for the three tartaric acids. Label pairs of enantiomers and diastereomers, as well as the meso form.

When a molecule has two different chiral centers, it may exist in four optically active forms (two pairs of enantiomers). To illustrate this with the models you just used, replace one of the colored atoms on one of the carbon atoms with a black atom. There are now four distinct ways of constructing the models: two pairs of enantiomers. Construct the two pairs of enantiomers.

28. What name is given to a pair of molecules consisting of one molecule from each of the two pairs of enantiomers you just constructed?

a. Enantiomers

b. Diastereomers

c. Meso forms

The R-S Convention

The letter R (from rectus, right) or S (from sinister, left) is used to designate the configuration at a chiral center. The four atoms or groups attached to the chiral center are arranged in a priority order according to atomic number: the higher the atomic number, the higher the priority. If two atoms have the same atomic number, we move to the next atoms out from the chiral center, or even further, until we observe a difference in atomic number. We then view the molecule from the side opposite the group with the lowest priority. If the remaining three groups in order from highest to lowest priority form a clockwise array, the configuration is R; if they form a counterclockwise array, the configuration is S.

Procedure: Construct a model of 2-chlorobutane.

29. Which carbon in the chain is a chiral center?

a. 1

b. 2

c. 3

d. 4

There are four groups attached to this chiral center

.

30. Which group has the highest priority?

a. Cl

b. Methyl

c. Ethyl

d. H

31. Which group has the lowest priority?

a. Cl

b. Methyl

c. Ethyl

d. H

32. What is the priority order of the other two groups?

a. Ethyl > Methyl

b. Methyl> Ethyl

Set the model on the desktop so that it can be viewed from the side opposite the hydrogen. Put the chlorine atom at the top.

33. When viewing the model, the three remaining groups in priority- order sequence form a

a. Clockwise array and the model has a R configuration.

b. Counterclockwise array and the model has a R configuration.

c. Clockwise array and the model has a S configuration.

d. Counterclockwise array and the model has a S configuration.

Interchange any two groups attached to the chiral center.

34. What configuration does the model have now?

a. The same as before.

b. The opposite configuration.

Note that to change configuration we must disconnect and remake bonds, whereas we can change conformation by rotating groups around single bonds.

Fischer Projection Formulas

It is sometimes convenient to have a two-dimensional representation for three-dimensional molecules, particularly when we are studying stereoisomerism. One common convention, devised by the German organic chemist Emil Fischer and named after him, is described in this section

Procedure: Construct a model of an asymmetric carbon atom [with a black, a blue, a green, and a red ball attached] that corresponds to the following three-dimensional drawing:

Set the model on the desk as shown. With your left hand, tip the model to the left so that only the red and green balls rest on the table. Viewed from the top, the black and blue balls project toward you (left and right, respectively).

The convention for Fischer projection formulas is as follows: The asymmetric carbon lies in the plane of the page or paper, horizontal groups come out of the plane of the paper toward the viewer, and vertical groups recede behind the paper away from the viewer.

Save the model you have just constructed for comparison with other models that you will construct.

Let us first consider the effect of interchanging any two groups in a Fischer projection formula. Your model corresponds to formula A below:

Consider formula A', with groups blue and green interchanged. Construct a model corresponding to A'.

35. Is A' identical to A, or is it the enantiomer of A?

a. Identical

b. Enantiomer

Now construct a model corresponding to A" (like A, but with green and red interchanged).

36. Is A" identical to A, or is it the enantiomer of A?

a. Identical

b. Enantiomer

37. What generalization can you make about the interchange of any two groups in a Fischer projection formula?

a. The new molecule is an enantiomer of the first.

b. The molecule is identical to the first.

38. If you were to make an even number of group interchanges in a Fischer projection formula, would you obtain the original molecule or its enantiomer?

a. Original molecule

b. Enantiomer of the original model

39. Do the following Fischer projections represent the same molecule or enantiomers?

a. Same molecule

b. Enantiomers

Rotation of a Fischer projection formula in the plane of the paper also affects the structure it represents.

.

Consider formula B, which corresponds to the Fischer projection formula of A rotated 90Â° clockwise in the plane of the paper. Build a model of formula B. (Remember Fischer Projection Formulas are two dimensional diagrams written on paper of three dimensional molecules.)

40. What minimum number of group interchanges in the model are necessary to convert formula B back to formula A?

a. Zero

b. One

c. Two

d. Three

41. Is model B identical to model A, or is it the enantiomer of model A?

a. Identical

b. Enantiomer

Formula C results from a 90Â° counterclockwise rotation in the plane of the paper of the Fischer projection formula of A. Build a model of formula C.

42. Is model C identical to model A, or is it the enantiomer of A?

a. Identical

b. Enantiomer

Formula D results from a 180Â° rotation in the plane of the paper of the Fischer projection formula A . Build a model of formula D.

43. What minimum number of group interchanges are necessary to convert model D back to model A?

a. Zero

b. One

c. Two

d. Three

44. Is model D identical to model A, or is it the enantiomer of A?

a. Identical

b. Enantiomer

45. What generalization can you make about the rotation of a Fischer projection formula in the plane of the paper?

a. 90Â° rotation results in an enantiomer, 180Â° rotation results in the identical molecule.

b. 180Â° rotation results in an enantiomer, 90Â° rotation results in the identical molecule.

THIS IS A LAB REPORT THAT I NEED HELP WITH I NEED HELP WITH ANSWERING THE POST LAB QUESITON THAT IS ALL THE WAY IN THE BOTTOM PLEASE HELP ME ANSWER THE POST LAB QUESTIONS THANK YOU! THE PRE LAB I HAVE ALREADY DONE BTW!

Determination of Cobalt (II) Chloride

by UV/VIS Spectroscopy

Introduction

The absorption of specific quantities of electromagnetic (light) radiation by an element or compound allows electrons to move from lower energy level to a higher energy level. We say that the energy levels are quantized. Electrons, returning to lower energy levels, will release energy within the electromagnetic spectrum. The wavelengths may be in the ultraviolet, visible, or infra red regions of the spectrum. For cobalt (II) chloride, the compound of study, the visible region, of the spectrum will be addressed.

By knowing the wavelength (l) or frequency (n) of radiation, one may determine the energy of the transition. Relationship between energy, wavelength and frequency are:

c= ln and E= hn or E= hc/l

The light radiation absorbed in a solution of this salt is proportional to its concentration through a relationship provided by the Beer-Lambert Law:

A = kC

where A= absorbance, k is Beerâs Law constant, and C is the molar concentration in solution.

The absorbance (A), a unit-less number is dependant on the quantity of light energy absorbed and transmitted by the solution through the following equation:

A= -log T or A = 2-log %T

where T= transmittance= I/I⁰ where I is the absorbed light and I⁰ is the impinging light source.

Purpose

The purpose of this laboratory experiment is to practice, learn and carefully prepare molar solutions to investigate the relationship of concentration and spectrophotometer response. A 0.150 M solution of cobalt (II) chloride will be provided. You will prepare diluted solutions through the serial dilution method, measure the transmittance and calculate the absorbance values. You will prepare a Beerâs law graph of data; molar concentration vs. absorbance. Then determine Beerâs law constant using Microsoft Excel program.

You will also be given a concentrated sample of cobalt (II) chloride of unknown concentration and you must determine the appropriate dilution to prepare and thereafter determine the unknown concentration.

Procedure

In order to carry out this analysis procedure, it will be necessary to determine the wavelength of transition. Recall, that one specific energy maximum exists for each transition. We will study the transition in the range of 400- 600 nanometers (nm). This is the visible region of the spectrum.

Step 1. Prepare the following solution concentrations using the 0.150 M cobalt (II) chloride stock solution, 18 mm x 150 mm test tubes and Mohr pipets.

TABLE 1

Solution No.	0.150 M CoCl₂ (mL)	Volume water (mL)
1	5.0	0.0
2	4.0	1.0
3	3.5	1.5
4	3.0	2.0
5	2.5	2.5
6	2.0	3.0
7	1.0	4.0

Step 2. Carefully place a cork or rubber stopper over each solution and mix the contents.

Step 3. Measure the transmittance of a 0.150M solution between 400 nm and 600 nm at 25 nm intervals.

To accomplish this, follow these steps:

Caution: instruments are delicate. If you are uncertain about instrument operation, ask your instructor for assistance. (Instructions on instrument operation- see next page.)

SPECTRONIC is a registered trademark of Spectronic Instruments, Inc.

Instrument operation- edited version of product ^*Operatorâs Manual

Transmittance and Absorbance

1. Turn on the instrument by turning the Power Switch (10) clockwise. Allow the spectro-photometer to warm up for at least 15 minutes to stabilize.

2. After the warm-up period, set the desired wavelength with the Wavelength Control Knob.

3. Set the filter lever to the appropriate position for the selected wavelength (not required for SPECTRONICÂ® 20D).

4. Adjust the display to 0%T with the Zero Control (10). Make sure that the sample compartment is empty and the cover is closed.

5. Set the display mode to TRANSMITTANCE by pressing the MODE control key until the appropriate LED is lit.

6. Fill a clean cell with water and wipe the cell with a tissue (Kimwipe^TM) to remove liquid droplets, dust and fingerprints.

7. Place the cell in the sample compartment and align the guide mark on the cell with the guide mark at the front of the sample compartment. Press the cell firmly into the sample compartment and close the lid.

8. Carefully adjust the display to 100%T with the Transmittance/Absorbance Control (9). Move the knob slowly as you approach 100%T.

9. Remove the cell from the sample compartment and empty the water.

10. Rinse the cell twice with small volumes of the solution to be measured and fill it with the solution.

11. Wipe the cell with a tissue and insert the cell into the sample compartment. Align the guide marks and close the lid.

12. Read the appropriate value (%T) from the display.

13. Remove the cell from the sample compartment and repeat steps 10 through 12 for any remaining sample solutions.

14. When all measurements are completed, turn off the spectrophotometer by turning the Power Switch counterclockwise until it clicks.

Step 4. Using solution no. 1, follow the directions for instrument operation and record (%) transmittance values for each wavelength in Table 2.

Remember to reset the instrument to 100% for each wavelength investigated.

DO NOT DISCARD THIS SOLUTION,

Step 5. The wavelength that gives the largest absorbance value will provide the greatest sensitivity to concentration change. Generate an Excel graph of wavelength (x) vs. absorbance (y) using these values. At zero absorbance, there isnât any CoCl₂ so donât forget to include the (0,0) data point. Connect your data points with a smooth curve. This is the absorption spectrum of aqueous CoCl₂. Submit this with your lab report.

Step 6. Reset your spectrometer using this wavelength (step 5), and measure this solution and the other six solutions listed in Table 1, and record your % transmittance values in Table 3.

TABLE 2

Wavelength (nm)	% Transmittance	Absorbance
400
425
450
475
500
525
550
575
600

MAXIMUM ABSORBANCE IS LOCATED AT ________________nm.

TABLE 3 -Results measured at ______nm.

Test tube No.	% Transmittance	Absorbance	Molarity(M)
1
2
3
4
5
6
7

Step 7. Plot a graph of Molarity (M) vs. Absorbance for all seven solutions using Excel plotting techniques. This is called a calibration curve, whereby one establishes a measured response to a known solution concentration. Determine the slope of the line and include the linear equation of fit, and the correlation coefficient.

Recall the relationship A = kC or y = mx + b; where b is the (0,0) data point

Then: k = A/C

Step 8. The unknown solution to be analyzed is more concentrated than any of the standard solution concentrations that were measured to prepare the calibration curve.

Points to consider:

You must achieve a dilution such that the absorbance is within the absorbance range of the standards that were previously measured.

By using serial dilutions, will multiply many errors; thus it is best to obtain a dilution is one-step if possible.

Consider reducing the concentration by Â¾, or by Â½, and by Â¼ or to 0.10 of the original concentration and test the solution to see if the absorbance response is within range. You may also consider using the dilutions in Table 1. Once this is determined, proceed to the next step. Complete the dilution table for the diluted concentrations you prepared:

Trial	mL of unknown	mL of water	(%T)	A
1
2
3

Step 9. Using the desired dilutions, determined in step 8, measure two of the diluted solutions and record the measured transmittances in Table 4.

UNKNOWN NUMBER __________

TABLE 4

Diluted concentration (M)	Transmittance (%T)	Absorbance (A)

Calculation of absorbance, for diluted Unknown solution:

Step 10. Determine the concentration of the diluted solution using the linear equation of fit, knowing the absorbance and k, the slope of the line.

C = A/ k

Step 11: Determine the concentration in your original Unknown solution. Show your

back-calculation below to support your final answer to this experiment.

Calculations:

Unknown Solution No. ________ Concentration: ___________M

Prelab Questions

1. Consider a solution of 0.400M Co(NO₃)₂ provided by your lab instructor.

You are required to make the following dilute concentrations:

0.160M and 0.240M solutions. You are provided the following glassware: 5.0 mL Mohr pipet and 18 mm x 150 mm test tubes. Describe how you would prepare each solution, and support your description with your calculations. (Recall (M_c x V_c = M_d x V_d) where M_c is the molarity of the concentrated solution, V_c is the volume of the concentrated solution, M_d is the molarity of the dilute solution, and Vd is the volume of the diluted solution.)

Given 0.400 M Co(NO3)2 solution.

Lets say we pipette out 5 mL of this solution and dilute it to get 0.160 M solution. Then

Using, Mc x Vc = Md x Vd

0.400 x 5 = 0.160 x Vd or Vd= (0.400 x 5)/0.160 = 12.5 mL

Thus we need to add 7.5mL distilled water to 5 mL of cobalt nitrate. Since we are only provided with a 5 mL pipette, we can only measure out volumes in multiple of 5.

thus pipette out (5 x 2) 10 mL of cobalt nitrate solution in the test tube and dilute it by addition of (7.5 x2) 15 mL of distilled water.

Again for preparing 0.240 M solution

Lets say we pipette out 5 mL of this solution and dilute it to get 0.160 M solution. Then

Using, Mc x Vc = Md x Vd

0.400 x 5 = 0.240 x Vd or Vd= (0.400 x 5)/0.240 = 8.33 mL

Thus we need to add 3.33mL distilled water to 5 mL of cobalt nitrate. Again taking volumes only in multiples of 5, pipette out (5 x 3) 15 mL of cobalt nitrate solution in the test tube and dilute it by addition of (3.33 x3) 10 mL of distilled water.

2. In your own words, explain, in detail, what you will be learning during this lab. Address the chemical principles and, laboratory skills, in your answer.

Post - Lab Questions:

a) Letâs say, for a given concentration of salt solution, the maximum response at a wavelength maximum of 485nm, is 0.750. If an unknown solution is analyzed at a longer or shorter wavelength would the investigator obtain the same concentration for the Unknown solution? Explain why.

b) List all possible sources of errors that may result in this experiment.

THE ONE IN ITALICS AND BOLD IS THE PRELAB QUESTION PLEASE ANSWER THAT. PLEASE HELP ME ANSWER THE POST LAB QUESTIONS. I did THE PRE LAB. I ALSO FILLED in the tables.

Determination of Cobalt (II) Chloride

by UV/VIS Spectroscopy

Introduction

The absorption of specific quantities of electromagnetic (light) radiation by an element or compound allows electrons to move from lower energy level to a higher energy level. We say that the energy levels are quantized. Electrons, returning to lower energy levels, will release energy within the electromagnetic spectrum. The wavelengths may be in the ultraviolet, visible, or infra red regions of the spectrum. For cobalt (II) chloride, the compound of study, the visible region, of the spectrum will be addressed.

By knowing the wavelength (l) or frequency (n) of radiation, one may determine the energy of the transition. Relationship between energy, wavelength and frequency are:

c= ln and E= hn or E= hc/l

The light radiation absorbed in a solution of this salt is proportional to its concentration through a relationship provided by the Beer-Lambert Law:

A = kC

where A= absorbance, k is Beerâs Law constant, and C is the molar concentration in solution.

The absorbance (A), a unit-less number is dependant on the quantity of light energy absorbed and transmitted by the solution through the following equation:

A= -log T or A = 2-log %T

where T= transmittance= I/I⁰ where I is the absorbed light and I⁰ is the impinging light source.

Purpose

The purpose of this laboratory experiment is to practice, learn and carefully prepare molar solutions to investigate the relationship of concentration and spectrophotometer response. A 0.150 M solution of cobalt (II) chloride will be provided. You will prepare diluted solutions through the serial dilution method, measure the transmittance and calculate the absorbance values. You will prepare a Beerâs law graph of data; molar concentration vs. absorbance. Then determine Beerâs law constant using Microsoft Excel program.

You will also be given a concentrated sample of cobalt (II) chloride of unknown concentration and you must determine the appropriate dilution to prepare and thereafter determine the unknown concentration.

Procedure

In order to carry out this analysis procedure, it will be necessary to determine the wavelength of transition. Recall, that one specific energy maximum exists for each transition. We will study the transition in the range of 400- 600 nanometers (nm). This is the visible region of the spectrum.

Step 1. Prepare the following solution concentrations using the 0.150 M cobalt (II) chloride stock solution, 18 mm x 150 mm test tubes and Mohr pipets.

TABLE 1

Solution No.	0.150 M CoCl₂ (mL)	Volume water (mL)
1	5.0	0.0
2	4.0	1.0
3	3.5	1.5
4	3.0	2.0
5	2.5	2.5
6	2.0	3.0
7	1.0	4.0