CHEM-2221 Lecture Notes - Lecture 23: Anomer, Fischer Projection, Cyclic Form

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champagnemarten225

27 May 2016

School

Vanderbilt University

Department

Chemistry

Course

CHEM-2221

Professor

Dr.Alissa Hare

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Document Summary

Carbohydrates: carbohydrates were considered to be hydrates of carbon because their molecular formulas correspond to cn(h2o)m, a carbohydrate is a polyhydroxy aldehyde or polyhydroxy ketone, carbohydrates have several things to consider , number of monomers. Fischer projections and d, l notation: fischer projections are the two crossed lines that represent a tetrahedral carbon, to assign r or s, with the h pointing toward you, solve the configuration and assign the opposite. Allowed manipulations: fischer projections can be rotate by 180 in the plane of the page. If one group is held steady, the other 3 groups can be rotated clockwise or counterclockwise. Fischer projections: assign priorities to the four substitutents according to the cahn-ingold- Fischer projections and d, l notation: remember for most molecules, we can"t correlate absolute configuration with d or l = (+) or (-, fischer arbitrarily assigned (+)-glyceralderhyde to be d and (-)- glyceraldehyde to be l, d-carbohydrates.

Related Questions

Â¥Write a brief (less than 150 words, 3 or 4 sentences) of the aim of the experiment

Â¥Address question of what was done, how it was done, and what compounds were used.

Â¥Include Chemical Reaction where relevant.

See text below. Please provide Purpose following the steps above. Thank you.

Alcohols R-O-H are characterized by the presence of the hydroxyl group

Because of intermolecular co-association, alcohols show higher boiling points

than aldehydes, ethers, or ketones of similar mass. Among alcohols, boiling

points increase with increased surface area. Increased surface area occurs as

more carbons are added to the alkyl portion or with decreased branching of th

alkyl group. Thus, 1-propanol has a higher boiling point than ethanol

(increased C#). 1-propanol also has a higher boiling point than 2-propanol

(decreased branching)

The solubility of alcohols in water decreases as the surface area of the

alkyl (nonpolar) portion increases. Thus, alcohols become less soluble in water

as the number of carbons in the alkyl group increases. When comparing

alcohols that contain the same number of carbons, branching of the alkyl group

decreases its surface area and thus increases the water solubility of the alcohol.

1-Propanol is slightly soluble in water while 2-propanol is quite soluble

The limit of water solubility is reached at about 4 or 5 carbons in the

alkyl group. However, it is possible to "salt out" or separate a water-soluble

alcohol from the water it is mixed with by the addition of certain salts such as

K2CO3. If an alcohol-water mixture is saturated with such a salt, the anion

selectively forms strong ion-dipole bonds to water, leaving the alcohol as a

separate layer. Thus, water is separated from the alcohol because the ionic salt

bonds more strongly with water than with the alcohol. There are twice as many

ion-dipole bonds possible with water than with the alcohol.

Alcohols are relatively acidic organic molecules, being weaker acids than

water, yet stronger than acetylene.

Acidity: water > alcohol > acetylene > ammonia > alkanes

Alcohols may be converted to sodium alkoxides by the action of metallic sodium

under anhydrous conditions. Primary alcohols react more readily tharn

secondary or tertiary alcohols due to the relative stability of the alkoxide ions in

the solvent alcohol.

R-o:-

Alcohols undergo nucleophilic substitution reactions with ease under

acidic conditions. The relative ease of such substitution for primary, secondary

and tertiary alcohols serves as a means of classification. Upon treatment with

concentrated HC1, tertiary alcohols are rapidly converted to alkyl chlorides

Secondary alcohols require warming in a boiling water bath for 15 minutes

Primary alcohols do not react at all under these conditions. The alkyl chlorides

formed by this reaction are insoluble in the concentrated HCl and form an

easily detectable second layer

R-O-H HC1 >

Ð-Ð-Ð

Complications arise from the fact that benzylic and allylic alcohols also

react immediately, but yield products that are soluble in the concentrated HCI

Alcohols are easily oxidized by any of these strong oxidizing agents:

dilute KMnO4, NajCr.07 in acid, or CrO3 (chromic anhydride)

acidified sodium dichromate is somewhat helpful in the classification of

alcohols. Prim

ary and secondary alcohols are oxidized rapidly within several

minutes, while tertiary alcohols do not react under these conditions

Oxidation occurs more easily when the carbon bearing the -OH group

bonded to at least one atom of hydrogen. The complex mechanism of

involves the transfer of a hydride ion H:- as a critical step.

hydrogen atom is absent (as in tertiary alcohols), no oxidation can occur

directly. Tertiary alcohols may be oxidized indirectly through the formation of

Alcohols which contain the partial structure C- CH3 give positive

Ð¾Ð½

results in the iodofora reaction. When the alcohol is dissolved in water and

KOH, the addition of a solution of 12 in KI yields a yellow precipitate of

iodoform, CHI3, along with the oxidized portion of the alcohol

Hypoiodite ion is produced by the reaction between molecular iodine and

hydroxide ion:

I2 + 4(OH) -

The hypoiodite ion in turn oxidizes an alcohol with the proper structure

Once this oxidation has occurred, additional hypoiodite ion continues to

react with the aldehyde or ketone produced:

CH,-C

3 (0I)-IC-c3 (OH)-

resulting in the final substitution reaction to give iodoform and the anion of a

carboxylic acid containing one carbon atom less than the original alcohol:

128

lodoform precipitates as fine yellow crystals. This precipitate, a result of a

specific oxidation followed by a substitution, is an indication of the partial

structure of the original alcohol.

Simple alcohols of up to ten carbon atoms may be detected by the

ammonium hexanitrocerrata(lV) reagent, (NH4Ce(NO,). The hexanitro-

cerrate(IV) ion has a distinctive orange-red color. When an alcohol, ROH is

added, one or more OR- groups replace the nitro groups as ligands in the

complex ion. The new complex ion formed has a distinctive color. The general

equation for this reaction is:

This test is positive for aliphatic alcohols up to approximately ten carbons in

length. Phenols or larger alcohols yield insoluble mixtures

Alcohols react with acids or with acid derivatives to form esters. This

characteristic reaction is useful in the formation of specific esters such as ethyl

acetate, acetylsalicylic acid (aspirin) and methyl salicylate (oil of wintergreen). In

this experiment, methyl salicylate will be prepared by reacting methyl alcohol

with salicylic acid in the presence of a trace of sulfuric acid as the catalyst

Lower molecular weight alcohols such as ethanol, methanol, and

isopropyi alcohol absorb water on exposure to moist air or are prepared as

azeotropic mixtures containing water. The presence of water in alcohols may be

detected in several ways. If a small lump of CaC2 is added to any alcohol that

contains a trace of water, a reaction occurs that produces immediate bubbles of

acetylene gas:

Thus, evolution of acetylene gas upon the addition of calcium carbide indicates

the presence of water in the alcohol. Another way to test for the presence of

water in alcohols involves the use of anhydrous cupric sulfate powder (CuSO4)

Anhydrous cupric sulfate is a very pale, almost white powder. In the presence of

water, deep blue crystals of cupric sulfate pentahydrate,

CuSO4-5H20, are formed. Thus, anhydrous cupric sulfate indicates the

presence of water in an alcohol by a simple color change.

Commercial ethanol is produced as an azeotropic mixture that contains

95% ethanol and 5% water. Absolute alcohol contains no water. Many reagent

grade alcohols are composed of a nonpotable mixture of 95% ethanol and 5%

isopropyl alcohol. Thus, absolute ethanol and reagent alcohols are both water

free and give negative results when tested for the presence of water

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.

CHEM-2221 Lecture Notes - Lecture 23: Anomer, Fischer Projection, Cyclic Form

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