instructor’s resource guide for the laboratory...

Instructor’s Resource Guide for the Laboratory Manual

Organic Chemistry

A Short Course Thirteenth Edition

Hart/Craine/Hart/Hadad

Prepared by Leslie E. Craine, Central Connecticut

State University &

T.K.Vinod, Western Illinois University Brooks/Cole Cengage Learning Boston, MA

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Preface The laboratory manual is independent of the text package and can be used successfully with virtually any “short-course” text. The level of the experiments is designed primarily for the short, or at least nonmajors’, organic course, although the manual has been used successfully in second-year high school chemistry courses up to chemistry majors’ courses in universities. The range covered by the 31 multipart experiments is quite broad. Each experiment has tear-out perforated sheets for prelab exercises and postlab reports, which simplify the grading of reports by yourself or your teaching assistants. Answers to all of the prelab exercise questions and most of the questions in the report sheets are in this manual. Each experiment is taken in order. In some instances, additional comments about the experiment itself are included, although in most cases essential comments to the instructor are in the Laboratory Manual itself as footnotes. An interesting feature in this laboratory manual is a series of eight “Chemistry at Home” experiments, explorations that students can do safely outside the laboratory setting. Each activity is placed at the end of an appropriate experiment as a separate tear-off sheet. These activities may be formally assigned or used as a springboard for other laboratory or classroom activities, or students may simply be encouraged to explore them on their own. Safety Note: It is assumed throughout the laboratory manual that each instructor is thoroughly familiar with ordinary safety precautions involved with solutions he or she prepares for students. If students, lab assistants, or untrained personnel are used to prepare these solutions, the instructor is solely responsible for training and supervising these persons so that no injuries occur. The instructor is also responsible for the safety of the students who use the laboratory manual; even though safety precautions are featured as required in each experiment, they are not intended as a substitute for the instructor’s directions and supervision. The “Chemistry at Home” experiments are as safe as any normal activity carried out in the home. However, students should be reminded to pay attention to the cautions on the labels of the household products they use and in general to behave sensibly around kitchen appliances. In two of the experiments, cautions are explicitly given because of the nature of the household products used.

© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or

duplicated, or posted to a publicly accessible website, in whole or in part.

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Experiment 1 Melting-Point Determination: Purity and Identity of Crystalline Organic Compounds Comments. The possible unknowns listed in Table 1.1 range in melting point (mp) from 70-166o

If there is a shortage of equipment, Experiments 1 (Melting-Point Determination) and 2 (Recrystallization) can be combined, with the class divided into two groups that alternate between the two experiments. Following are some suggestions for combining the experiments.

C. Compounds with higher melting points have not been listed because in general the higher the mp, the longer it takes the student to determine it. Keep this fact in mind when grading.

Melting-Point Experiment. Divide half the class into groups of three students, and have each student in a group prepare one of the three samples-urea, cinnamic acid, and a 50:50 mixture-by powdering a microspatula tipful of compound in a labeled 10-mL beaker. Each student can prepare her or his own three melting point samples using the common beakers. If there is time at the end of the lab, students can do 75:25 and 25:75 mixture melting points for extra credit. Each student prepares individual samples of an unknown and a 50:50 mixture of the unknown and the suspected known. Using microspatula tipfuls of compound will avoid congestion at balances. Recrystallization Experiment. Start the other half of the class on the microscale recrystallization of acetanilide. To save time, the melting point of the crude material can be eliminated and a representative value given to the students. Students can mix crude acetanilide with decolorizing charcoal before dissolving the acetanilide. While the acetanilide is being dissolved, the filtration apparatus (a 10-mL Erlenmeyer flask with a few milliliters of water in it and the funnel) can be warmed on a steambath. The hot acetanilide solution is decanted quickly through the filter and followed with a 0.5-mL hot water rinse. The resulting filtrate should be clear and will not need to be concentrated. Note: Sand and methyl orange can be substituted for sawdust and brown sugar as contaminants in the impure mixture. Answers to Questions (Prelab Exercise)

1. Melting point range is the temperature range between the point when liquefaction begins and the point when it is complete.

2. The desired rate of heating during a melting point determination is about 1-

20C/min to ensure uniform heating of the entire sample. A determination of the rough melting point of the unknown sample by quickly heating the sample allows the experimenter to identify the approximate melting point range so that the


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sample can be quickly heated to about 15-200

C below the expected melting point range and the actual melting point range precisely determined by slowly ramping up the heating from that point.

3. a. They broaden the mp range

b. They lower the mp range

Answers to Questions (Report)

1. Tight packing facilitates heat transfer throughout the sample. Slow steady heating also allows time for heat transfer and prevents too high a mp from being recorded.

2. Too large a sample leads to a larger mp range because of the time needed for heat

transfer throughout the sample.

3. Determination of the mp range of the known sample allows the researcher to verify whether the thermometer (mp apparatus) used for the mp determination requires calibration.

4. a. They are used to help identify crystalline compounds. b. They are used to get an indication of purity from the mp range.

5. B and C would melt broadly, 130–139o

C. A and C are probably identical, whereas B, although it has the same mp, has a different structure from A and C.

Experiment 2 Recrystallization: Purification of Crystalline Organic Compounds _______________________________________________________________________ Comments. If time forces a choice between the two parts of this experiment, the second part (p-dibromobenzene) gives the more general experience regarding techniques that may have to be used later in the course. However, if conditions are crowded and safety is a serious factor, the first part (acetanilide) avoids the use of flammable solvents and gives excellent, satisfying results (beautiful white crystals from the rather messy looking material used at the start of the experiment). If the microscale procedures are used, both recrystallizations can be accomplished within a 3-hour period. See “Comments” for Experiment 1 for a suggested combination of Experiments 1 and 2 if shortage of equipment is a factor.



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Answers to Questions (Prelab Exercise)

1. The use of a minimum amount of the required solvent will ensure that the yield of the isolated crystals will be high as loss of material due to the solubility of the compound in the solvent is minimized.

2. The solubility of a compound in a solvent increases with temperature and thus

yield of the isolated crystals increase when the hot solution is cooled to room temperature where the solute is less soluble in the chosen solvent.

3. The compound being purified is dissolved in a hot solvent. Insoluble impurities

will not dissolve and are removed by filtration.

4. Finely divided charcoal effectively removes colored impurities during a recrystallization step.


1. The purpose of recrystallization is to purify an impure crystalline substance. 2. If the amount of soluble impurity is small relative to the amount of substance

being recrystallized, then the impurity will remain dissolved in the mother liquor when the pure crystals are collected.

3. The fine glass particles produced by scratching the inside of the vessel may act as

nuclei on which crystallization may begin or the solution drawn onto the sides of the vessel during scratching will evaporate quickly to produce tiny solute particles that can act as nuclei for crystallization to start.

4. Four properties are listed in the first paragraph under “General Principles.”

5. It is faster. It helps prevent premature deposition of crystals.

6. The solubility of acetanilide is considerably greater in hot than in cold water:

3.5g/100mL at 80oC, 0.56g/100mL at 25o

C.

7. A cold funnel might cause the acetanilide to crystallize prematurely.

8. Charcoal is insoluble in water and would be removed at the first filtration. Sugar is very soluble in water and would not crystallize out from even cold water.

9. p-Dibromobenzene is insoluble in water but soluble in ethanol. When the amount

of water present is too great to completely dissolve the sample, the solution becomes turbid. If just enough ethanol to clarify the hot solution is then added,


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the solubility of p-dibromobenzene that remains dissolved in the cooled mother liquor will be minimized.

10. Because water and hexane are not miscible. Mixed crystallizing solvents must be

mutually soluble.

Experiment 3 Distillation: Separation and Purification of Organic Liquids _____________________________________________________________ Comments. Even a simple Vigreux column will give reasonably good results in the hexane-toluene fractionation. Student results vary quite a bit depending on how quickly (or how patiently) students carry out the distillations. However, student curves plotted in the answer to Question 1 should show rather clearly an improved separation with the column. If time or an equipment shortage is a factor, have half the class do the simple distillation and the other half use a column and then share results. Other liquid pairs that have been used successfully for simple and fractional distillation are cyclohexane and toluene, 2-propanol and water, and 1-chlorobutane and 1-bromobutane. For a nonflammable alternative, try dichloromethane and 1,1,1-trichloroethane. Answers to Questions (Prelab Exercise)

1. The bp is the temperature at which the vapor pressure of the liquid equals the pressure of its surroundings. By applying a vacuum to the distillation set-up, one is reducing the pressure of its surroundings and thus lowering the bp of the high boiling liquid.

2. Boiling chips should be added during distillation to prevent bumping.

3. The pressure exerted by the vapor in equilibrium with its liquid is called the vapor

pressure. The bp is the temperature at which the vapor pressure of the liquid equals the pressure of its surroundings.

4. The vapor pressure of a liquid increases with temperature as there is more vapor

in equilibrium with its liquid as the temperature is increased. Answers to Questions (Report) The answers to Questions 1 and 2 will vary from student to student depending on results.

3. Fractionating column.



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4. If the bp range is broad, the compound is impure. If the range is narrow, the

compound may be pure, or it may contain impurities with the same bp. Mixtures of some liquids in definite proportions form azeotropes, which have a constant bp.

5. The bp is the temperature at which the vapor pressure of the liquid equals the

pressure of its surroundings. Lowering the pressure lowers the bp.

6. Because the heat of vaporization (the heat necessary to turn liquid to vapor at the bp) must be supplied to the entire sample, and this takes time.

7. Pressure would build up as the substance vaporized. Eventually, the apparatus

would blow up.

8. If water flowed the other way, it would run down the bottom wall of the condenser and out, without filling the condenser and surrounding the inner tube with cold water.

9. Because the liquid expands on heating, and when boiling begins, some of the

liquid may bump or foam over instead of distilling. Experiment 4 Extraction: A Separation and Isolation Technique ______________________________________________________ Comments. This is a very nice, practical experiment that teaches the principles of extraction, which are so useful later in preparative experiments. It also teaches the technique of using a separatory funnel and gives students a chance to run melting points on their own products. The following suggestions apply to the microscale procedure. A test tube works well as a separatory funnel, but a conical centrifuge tube works better. Dispensing diethyl ether from an autopipet directly into the sample test tube avoids waste. If each student is given six long disposable pipets, to be labeled, contamination in solution transfer operations is minimized. Samples can be dried more rapidly by attaching the flasks to a vacuum line and evacuating at room temperature. You can add zest to the experiment by varying the percentages of mixture components (say 40:40:20 or 50:30:20) and checking on students’ technique by their yield of each component. An alternative two-component microscale extraction procedure that works well is as follows: Place a mixture of benzoic acid and camphor (100 mg each) in a 5-mL conical vial. Add 2 mL of diethyl ether (the solids dissolve), followed by 1 mL of 10% aqueous sodium bicarbonate, cap the vial and shake with occasional venting for 5


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minutes. Upon separation of the layers, transfer the lower layer to a 25-mL Erlenmeyer flask. Repeat the extraction of the ether layer with 0.5 mL of 10% aqueous sodium bicarbonate, adding ether as needed to maintain the organic phase at a volume of 2 mL. Organic phase: Dry the organic phase with anhydrous sodium sulfate, transfer it to another 5-mL conical vial, and remove the solvent on a heating source. Equip the vial with a sublimation apparatus (Experiment 5), and sublime the camphor. (A sublimation apparatus can be made from a multipurpose adapter and a glass rod sealed at one end and filled with ice-cold water.) Camphor sublimes rapidly when the temperature of the heating source reaches 1600C. Recovery is about 50%. The mp should be taken in a sealed capillary and should be approximately 1700

Aqueous phase: Acidify the aqueous phase by adding 10% HCl (about 0.5 mL). Crystalline benzoic acid will crash out and can be recrystallized directly, adding hot water as needed. Collect crystals on a Hirsch funnel, dry, and weigh. Recovery: 50 to 75%. Alternatively, extract the acidic aqueous phase with 2 mL of diethyl ether or dichloromethane, dry the organic extract, and remove the solvent. Recrystallize the product from water if necessary.

C. (Trace contaminants lower the mp drastically.)

If you wish, you can let students omit the calculation questions on the report sheet (numbers 6 and 7). Answers to Questions (Prelab Exercise)

1. Four desired criteria are listed in the first paragraph under “Practical Considerations.”

2. A carboxylic acid can be readily converted into its water-soluble salt form by

reacting with aqueous sodium bicarbonate while phenol does not react with aqueous bicarbonate and remain unionized and in the organic solvent.

3. Chlorinated solvents are toxic and some are even suspected carcinogens. The

body does not rid itself easily of chlorinated solvents and thus one should minimize exposure to these solvents by wearing gloves when handling and working in properly ventilated fumehoods when handling these solvents

4. Dichloromethane is heavier and immiscible with water and thus forms the bottom

layer during extractions using a separatory funnel allowing one to drain the organic layer without having to empty the aqueous layer from the separatory funnel.


1. Advantages: Good solvent for organic compounds, and low bp; therefore, easily removed. Disadvantages: Flammable, develops explosive peroxides on standing, and is partially miscible with water.



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2. Can remove solvent as lower layer in separatory funnel; then add more for the

next extraction, without having to remove the aqueous layer from the funnel.

3. Otherwise, as liquid is withdrawn, a vacuum will be created in the funnel, and air bubbles will be sucked in through the stopcock and mix the layers.

4.

O2N NH3Cl+ _

m-nitroanilinium chloride

CO2H

benzoic acid napthalene

CO2Na

sodium benzoate

_ +

napthalene

O2N NH2

m-nitroaniline

CO2H

benzoic acid 5.


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O2N NH2

+ HClO2N NH3Cl

+ _

O2N NH3Cl+ _

+ NaOH

O2N NH2

CO2H+ NaOH

CO2Na

_ +

CO2Na+ HCl

CO2H_ +

(water layer)

+ NaCl + H2O

+ H2O

+ NaCl

6. For the first extraction, 5 = (x/50) ÷ (60-x)/60 x = 48.4 mg extracted; 60-x = 11.6 mg remaining

For the second extraction, 5 = (x/50) ÷ (11.6-x)/60 x = 9.4 mg extracted Total for the two extractions = 48.4 mg + 9.4 mg = 57.8 mg 7. KD 4 = (90/x) ÷ (10/100)

= 0.56/0.14 = 4

x = 225 mL Experiment 5 Isolation of Natural Products: Caffeine and Eugenol ________________________________________________________________________ Comments. You can ask each student to bring his or her own tea leaves or No-Doz tablets for Part A and cloves for Part B, saving you some expense and adding to the interest. If we were to have to select only one of the two parts to do, we would choose Part A because it teaches another technique, sublimation. But both parts work well, despite the fact that beginners are handling, in the end, rather small amounts of materials. The “Chemistry at Home: Spice Essences” exploration found at the end of this experiment is a nice extension. It could be substituted for Part B if time or equipment constraints are a consideration.



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1. Sodium carbonate removes tannins from caffeine.

2. Anhydrous sodium sulfate at the apex of the funnel removes moisture from the dichloromethane extract of caffeine.

3. A mixture of immiscible liquids will boil, and can thus be distilled, at a lower

temperature than the boiling points of the separate components. When steam is used to provide one of the immiscible phases, the process is called steam distillation and thus distillation would occur below 1000

C.

4.

H3CO

OH

Eugenol

+ Br2

H3CO

OH

Br

Br

Answers to Questions (Report: Part A)

1. In sublimation, a solid is converted directly to a vapor and then recondensed as a solid, without going through the liquid phase.

2. The vacuum lowered the pressure and thus the temperature at which the caffeine

would sublime. The lower temperature is desirable to prevent decomposition of the caffeine.

Answers to Questions (Report: Part B)

1. Because eugenol is not miscible with cold water.

2. The product should decolorize bromine and give a pink-violet color with ferric chloride.

Answers to Questions: Chemistry at Home: Spice Essences


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1. The odor of the extracts will resemble the odors of the plants.

2. Eugenol and isoeugenol are constitutional isomers. They differ by virtue of the position of the carbon-carbon double bond in the three-carbon chain. The double bond is conjugated to the aromatic ring in isoeugenol and isolated from the ring in eugenol. The difference in placement of the double bond leads to differences in shapes of the molecules, which leads to differences in how our odor-detecting systems (receptors) interact with each molecule, which leads to differences in their odors.

3. Many fragrance molecules are “small,” so they have some vapor pressure at room

temperature. They must get “airborne” to be detected at a distance, for example, by your nose, a bird, or a bee.

4. Isopropyl alcohol (common name) or 2-propanol (IUPAC name).

5. Isopropyl alcohol is moderately polar and thus will dissolve the moderately polar

organic compounds such as eugenol, isoeugenol, and cinnamaldehyde.

6. Some desirable characteristics of a good extraction solvent are polarity needed to dissolve the desired solutes, low boiling point so it is easy to remove from the solutes, not toxic, inexpensive.

Experiment 6 Chromatography: Column, Thin-Layer, Gas-Liquid, and paper ________________________________________________________________________ Comments. Each part of this experiment requires a full laboratory period, so it is usually not possible, in a short course, to do all four parts. Your choice of which part(s) to do will depend on the time and equipment available, and the number of students in your class. For large classes, you may wish to omit Part C, which requires access to a gas chromatograph. In Part D (paper chromatography), the use of Petri dishes may be a bit expensive for large classes. Nevertheless, the circular chromatograms that are obtained are extremely satisfying, and the experiment works very well indeed. But the experiment can be performed just as successfully using more conventional techniques, as indicated in the last footnote. Another alternative is to make the “Chemistry at Home: Food Dyes and Marking Pens” exploration at the end of this experiment an outside-of-the-lab assignment. In the sets of food coloring dyes commercially available now, the yellow dye appears to be homogeneous. The blue dye is also homogeneous. The green dye appears to be a mixture of a blue dye and a yellow dye, with the blue having the larger Rf. The



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yellow dye appears to be identical with the yellow in the green dye, and the blue dye appears to be the same as the blue component of the green dye. Answers to Questions (Prelab Exercise)

1. The different components of the mixture adsorb differently onto the stationary phase (solid support) during column chromatography resulting in the separation of these components.

2. Pentane, dichloromethane, methanol, water

3. The dyes present in the ink will elute along with the components of the sample

making the identification difficult after the TLC analysis.

4. Rf value is defined as the ratio of the distance a compound has traveled from origin to the distance the developing solvent has traveled from origin. Rf

value of a given compound is a characteristic constant under the given conditions of the TLC analysis (the adsorbent, its thickness and developing solvent used) and can be compared to that of the reference (standard) compound for identification purposes.

5. The separation of the components in a mixture between the stationary and mobile phases in gas-liquid chromatography occurs while the components are in their vapor phase and hence the term vapor-phase chromatography.


1. In methyl orange, the negative charge is localized on the sulfonate ion (RSO3 _

), whereas in methylene blue, the positive charge is delocalized, as in the following structures:

SNH3C

CH3

NCH3

CH3

+SN

H3C

CH3

NCH3

CH3

+

Thus methyl orange is the more polar dye and is more firmly absorbed on the alumina.


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2. Channels would develop, and the lines of demarcation that separate the two dyes would not be as clear.


1. The layers could be scraped off the slides separately and then extracted to isolate the pigments.

2. The sample would dissolve in the solvent instead of moving up the plate.

Answer to Question (Report: Part D)

1. The blue. Because it moves farther (has a larger Rf

) in the chromatogram.

Answers to Questions: Chemistry at Home: Food Dyes and Marking Pens

1. Blue food color will appear to be pure compound. Yellow food color frequently appears as a pure compound but sometimes contains a trace of a red dye that can be detected by paper chromatography.

2. The green food color appears to be a combination of blue and yellow food colors.

It has the same chromatographic behavior as the green material prepared by mixing the blue and yellow food colors.

3. The blue food color travels faster than the yellow food color under the suggested

conditions. Differences in polarity account for the different rates at which the colored compounds rise on the filter paper.

4. If two substances rise at different rates, they are different. If they rise at the same

rate, they may be the same, but may be different. More experiments are needed.

5. If a substance does not separate into two or more compounds with rubbing alcohol, you cannot be too confident that it is a pure compound. Try some different solvents, for example turpentine, ethanol, or various mixtures of rubbing alcohol with water, vinegar, or household ammonia.



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Experiment 7 Conformations of Alkanes and Cycloalkanes: An

Exercise with Molecular Models and Computer Modeling

________________________________________________________________________ Comments. In this “experiment,” students construct models of ethane, propane, butane, cyclohexane, chlorocyclohexane, and certain di- or trichlorocyclohexanes and examine their conformations. The directions are written so that they can be followed using any good set of molecular models. In Part C of this “experiment” students will minimize the energies of acyclic and cyclic compounds for which they have made models in Part A and B and compare the predictions made about the energies of the various conformations. The questions, which should be answered directly on the report sheets as the experiment is performed, are numbered consecutively but separately for Parts A, B and C so that any part can be performed independently. Answers to Questions (Prelab Exercise)

1. The various molecular shapes that result from the rotation of groups or atoms about single bonds are called conformations.

2.

Br CH

HCH

HBr BrH2CCH2Br

Dash Formula Condensed Formula

3. One plane of symmetry passing through the two carbon atoms and the pair of anti bromine atoms. One two-fold axis of symmetry bisects the carbon-carbon bond perpendicular to the plane of symmetry. The center of symmetry is the midpoint of the C-C bond.

4.

Br

HH

H

Br

H


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1. Yes. Three planes through the two carbons and a pair of anti hydrogens.

2. The C__

C bond is a threefold rotational axis. Also, axes perpendicular to this bond and through its midpoint are twofold axes (interchanging “front” and “back” of the molecule).

3. Midpoint of the C__

C bond.

4.

HH

H

H

H

H

H

H H

H

H

H

H H

H HHH

5. Three planes pass through the two carbons and each pair of eclipsed hydrogens. Also, one plane perpendicularly bisects the C__

C bond.

6. The C__

C bond is a threefold rotational axis. The perpendicular bisector of this bond is a twofold rotational axis (front-to-back).

7. No center of symmetry.

8. Eclipsed.

9. Eclipsed.

10. Rotation around the C__

C bond.

11. The arrangement of atoms with respect to rotational motions.

12. The curve should show equal minima of 60o, 180o, and 300oand equal maxima of 0o, 120o, 240o, and 360o

. The minima represent staggered conformations, and the maxima represent eclipsed conformations.

13.



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H

HH

H

CH3

H

14. The plane that passes through all three atoms of the CH2 group and the plane that passes through both C__

C bonds are mutually perpendicular symmetry planes.

15. The axis that bisects the H__C__H bond of the CH2

group is a twofold symmetry axis.

16.

H

H

HH

CH3

H

17. Again, the plane that passes through all three atoms of the CH2 group and the

plane that passes through all three carbons are mutually perpendicular planes of symmetry. The bisector of the H__C__H bond of the CH2

group is a twofold symmetry axis.

18.

HHHH

HHH

H

19. The plane through all three carbons is a plane of symmetry. There are no symmetry axes.

20. The shape of the curve should be the same as for ethane (Question 12).

21. The energy difference is larger; the two staggered conformations (ethane and

propane) are nearly equal in energy, but the energy of the eclipsed conformation of propane is much higher (methyl-hydrogen vs. hydrogen-hydrogen) than that of the eclipsed conformation of ethane.

22.


18

HH

HHH

HHH

H H

23. The plane through all four carbon atoms. 24. The midpoint of the C2__C3 bond is a center of symmetry. There is one twofold

axis of symmetry passing through the midpoint of the C2__

C3 bond, perpendicular to the plane of symmetry.

25.

CH3

HH

H

CH3

H

A

26. Assuming that the “back” carbon is C3, we get

CH3

HH

H

H

CH3

H CH3

HHHCH3

H3C CH3

HHH

H

B C D

27. A and C

28. B and D

29. Center of symmetry at the midpoint of the C2__C3 bond; symmetry plane through all four carbon atoms; C2 symmetry axis passing through the midpoint of the C2__

C3 bond, perpendicular to the plane of symmetry.

30. A C2 axis, the bisector of the angle between the two C__CH3

bonds.

31. B has a C2 axis in the plane through the two eclipsed C__H bonds, passing through the midpoint of the C2__C3 bond. D has two mirror planes, one passing



19

through the eclipsed C__Ch3 bonds and another that perpendicularly bisects the C2__

C3 bond; it also has a C2 axis at the intersection of these two planes.

32. A > C > B > D

33. The curve has minima at 60o, 180o, and 300o, with the one at 180o lower in energy than those at 60o and 300o, which have equal energies. The curve will have equal energy maxima at 120o and 240o and a higher energy maximum at 0o (360o). The minimum at 180o is conformation A; those at 60o and 300o are C and its mirror image. The maxima at 120o and 240o are conformation B and its mirror image, and the one at 0o (360o

) is conformation D.


1. It is puckered

2. Staggered

3.

H

CH3H

H

H

CH3

in cyclohexane, these are CH2 groups

4. The planes are mutually perpendicular

5. Six

6. Three above and three below

7. C3 and C5. Those on C2, C4 and C6 are on the opposite side of the mean

molecular plane

8.

a

e

a

e

a

e

a

ea

ea

e


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9. Eclipsed

10.

in boat cyclohexane, these are CH2 groups

H H

HCH3H

CH3

11. Chair, because all interactions are staggered

12. Note: The conformation shown below is quite difficult for students to draw.

13.

Cl

14. Axial

15.

Cl

16. Cl equatorial; avoids Cl-H axial interaction

17. They are trans; one chlorine is attached to the “upper” bond of a carbon atom,

whereas the other chlorine atom is attached to the “lower” bond of the carbon



21

atom. If the ring were flattened, one Cl would be above the plane and the other

would be above the plane and the other Cl would be below it.

18. Both axial

19. On opposite sides of the mean ring plane

20. Trans; flipping the ring does not interconvert cis and trans (configurational)

isomers.

21. The conformation in which both chlorines are equatorial; avoids Cl__

22. Cis; on the same side of the mean ring plane.

H axial

interactions

23. The C1 chlorine becomes axial, and the C2 chlorine becomes equatorial.

24. Still cis.

25. They have equal energies; each has one equatorial one axial chlorine.

26. Trans, because it has one conformation in which both chlorines are equatorial.

27. Cis, because it has one conformation in which both chlorines are equatorial.

28. Trans, for the same reason as in question 27.

29.

Cis

Trans

1, 2 1, 3 1, 4

ea ae

ee aa

ee aa

ea ae

ea ae

ee aa

30. All cis.

31. They all become axial.


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32. Equatorial bonds become axial and axial bonds become equatorial when ring is

“flipped”.

Answers to Questions (Report: Part C)

1.

H

HH

H

H

H

2. Staggered.

3. 1.500 Å

4. 1.117 Å

5. A bond distance is defined by the radii of the two bonded atoms. The H atom is smaller

than the C atom (it has a shorter atomic radius), so the C–H bond is shorter than the C–C

bond.

6. 108.20°

7. 110.72°

8. These angles are very close to the ideal tetrahedral angle.

9. 60° and 180°

10. Staggered

11. Two: 60° and 180°

12. –30.4 kcal/mol

13. 74.60°

14. –31.13 kcal/mol



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15. 180°

16. The conformation with the CCCC dihedral angle of 180° is more stable.

17. The methyl groups at C1 and C4 are farthest away from each other in the

conformation in Question 16. In the conformation with a CCCC dihedral angle of 74.60°,

even though it is staggered, the two methyl groups are close enough to experience steric

crowding compared to the methyl groups in the conformation with the CCCC dihedral

angle of 180°.

Answers to Questions (Report: Part D)

1. Puckered

2. Staggered

3. Bond Angles: HCH 107.49°; HCC 109.57° (eq) and 109.41° (ax); CCC 111.31°. These

values are close to the ideal tetrahedral angle.

4. CCCC dihedral angle: 55.18°. HCCC dihedral angle for Hax 65.86°. HCCC dihedral

angle for Heq

5. Energy is –44.87 kcal/mol. The C–Cl bond is eclipsed with respect to the C–H axial

bonds on the same side of the ring.

176.54°. The equatorial H atoms appear to be in the plane of the ring.

6. Energy is –45.70 kcal/mol. The C–Cl bond is not eclipsed with any other bond.

7. The conformation in which Cl is equatorial.

8. The conformation in which Cl is equatorial avoids Cl-H axial interactions.

9. They are trans to each other; if the ring were flattened, they would be on opposite sides

of the mean ring plane.

10. Energy is –50.82 kcal/mol.

11. They are trans to each other.


24


13. They are cis to each other.


15. 1,2Cleqeq and 1,2Claxax are conformational isomers of each other. 1,2Claxeq

represents a different structural isomer, because the Cl atoms are cis to each other, while

in the other two structures the Cl atoms are trans. Trans and cis isomers cannot be

interconverted by rotation around bonds.

16. The trans isomer is more stable. One of the two conformations of this isomer does

give rise to C–Cl and C–H axial interactions (the other does not), while the cis isomer has

C–Cl and C–H axial interactions in both of its conformations. The trans isomer is slightly

lower in energy than the cis isomer because it has a conformation in which no C–Cl and

C–H axial interactions occur.

17. cis, because this isomer has a conformation in which there are no C–Cl C–H axial

interactions.

18. trans, because this isomer has a conformation in which there are no C–Cl C–H axial

interactions.

19. 1,2– and 1,4–

Experiment 8 Preparation of Alkenes from Cyclohexanol or 2-Methyl-2-butanol, and Tests for Unsaturation

____________________________________________________________ Comments. In the macroscale dehydration of cyclohexanol (Part A), use of phosphoric acid results in less charring, but the reaction takes appreciably longer. For a 2-hour laboratory period, use of sulfuric acid for the dehydration is recommended. Either Part A



25

(macroscale or microscale) or Part B can be completed, along with Part C, in one 2- or 3-hour laboratory period. Answers to Questions (Prelab Exercise)

1.

OHOH

OHOH OH

OH

2. Alcohols can form strong hydrogen bonds to each other making their bp higher than the corresponding alkenes.

3.

major product minor product

4. No; both compounds, due to the presence of the double bonds in them, will decolorize bromine water and yield brown MnO2

precipitate when reacted with aqueous potassium permanganate.


26


1.

H OH

+ H+

H O H

H+

slow, rds

H

+ HH

fast+ H+

+ H+

slow, rds

fast+ H+

H3COH

CH3

CH2CH3 H3CO

CH3

CH2CH3

HH +

H2C

H3CCHCH3

+

HaHb

loss of Ha

CH2CH3

CH3

H

H

fast

loss of Hb+ H+

CH2CH3

H

H3C

H3C

2. a. Mechanical losses (the alkene sticks to the condenser walls, some losses in transfer from one vessel to another, and so on)

b. Chemical losses (some of the alcohol distills over without dehydration, some of the alkene is polymerized or oxidized by the sulfuric acid)

3. To remove any acid present (either mechanically through entrainment or through

distillation of SO3

)

2H2SO4 + Na2CO3 → 2NaHSO4 + CO2 + H2

O

4. Get a mixture

H OHHCH3

H+ CH3 HCH3+



27

5. 2-Methyl-2-butene

6.

+ Br2

H BrHBr

+ 2KMnO4 + 4H2O

H OHHOH + 2MnO2 + 2KOH

+ H2SO4

HOSO3H

cold

For Part B, substitute each of the two alkenes for cyclohexene rings. If regiochemistry of electrophilic additions has not been discussed, the third reaction equation should be evaluated with this consideration in mind. Experiment 9 Cis-Trans Isomerism in Alkenes: Models and

Experiments ____________________________________________________________ Comments. The model experiment, as presented here, is a good introduction to the experimental parts of the experiment. Part A of this experiment can be performed with molecular model kits or can be adapted for use with molecular modeling software. Part B2 is quite short and is easily performed with Part A in a 2-hour laboratory period.


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1.

H3C Cl

H H

cis-1-chloropropene

H3C H

H Cl

trans-1-chloropropene

H3C H

Cl H

2-chloropropene(no cis-trans forms)

H3CH2CH2C

H

H

H

1-pentene(no cis-trans forms)

H3CH2C

H

CH3

H

cis-2-pentene

H3CH2C

CH3

H

H

trans-2-pentene

2. Stereoisomers that are not

3. Pi-bonds are weaker than sigma bonds and are therefore easier to break.

mirror images of each other are diastereomers.

4. Bromine provides the necessary bromine radical needed for the isomerization as shown in the reaction mechanism below.

hνBr2 2Br

H

H

CO2CH3

CO2CH3

H

H

CO2CH3

CO2CH3

BrBr

H

H3CO2C

CO2CH3

H

Br-Br

H

H3CO2C

CO2CH3

H


1. No.

2. They are stereoisomers, and since they are not mirror images, they are diastereomers.

3. Cis-1,2-dichloroethene (Z-1,2-dichloroethene).

4. Trans-1,2-dichloroethene (E-1,2-dichloroethene).



29

5. Restricted rotation around the double bond.

6. Break the pi bond, rotate by 1800

around the sigma bond, and remake the pi bond.

7. Yes.

8. Chloroethene (vinyl chloride).

9. One of the double bond carbon atoms has two identical groups attached

10. There must be two different groups attached to each carbon of the double bond

11.

H3C Cl

H H

cis-1-chloropropene

H3C H

H Cl

trans-1-chloropropene

H3C H

Cl H

2-chloropropene(no cis-trans forms)

H3CH2CH2C

H

H

H

1-pentene(no cis-trans forms)

H3CH2C

H

CH3

H

cis-2-pentene

H3CH2C

CH3

H

H

trans-2-pentene

Only one isomer is possible for 2-chloropropene and 1-pentene.

12.

O

O

O

13. No; a five membered ring is too small to accommodate a trans double bond. The two carboxyl groups are too far apart to react to form an anhydride.


30

Answers to Questions: Part B

1.

H

H

C

C

H

H

C

C

H+

OHOOH

OOH

OH+ H

H

C

C

OH

OH

+O

OHO

OH

H

H

C

C

OH

OH+

OHO

H

H

C

C

OH

OH

OHO +

- H+H

C

C

H

OOH

OHO

2.

H

H

CO2CH3

CO2CH3

H

H

CO2CH3

CO2CH3

H

H3CO2C

CO2CH3

H

H

H3CO2C

CO2CH3

H

hν

(break pi-bond)

(reform pi bond)



31

Experiment 10 Cycloadditions: The Diels-Alder Reaction ____________________________________________________________ Comments. Each part of this experiment actually illustrates two reactions: a cycloreversion and a cycloaddition. In Part A and C, a retro-Diels-Alder reaction is used to generate cyclopentadiene from its commercially available dimer. The diene then reacts with maleic anhydride (Part A) or with 1,4-benzoquinone (Part C) to give the Diels-Alder adduct. Part B begins with a chelotropic reaction, the elimination of sulfur dioxide from 3-sulfolene; the resulting butadiene then undergoes cycloaddition to maleic anhydride. You may wish to supplement the brief stereochemical discussion presented in the lab manual. The microscale procedures for Part A and B of this experiment can be modified to avoid the use of special glassware. In Part A, a small test tube or centrifuge tube can be used as a reaction vessel. The mother liquor from recrystallization can be removed by Pasteur pipet or by centrifugation. In Part B, a test tube can be used as a reaction vessel, and a gas trap can be fashioned from a one-hole stopper and a drying tube. Part C requires no special glassware.

Part C is a short experiment that can easily be done with Part A in the same laboratory period.


1. Cyclopentadiene prepared by cracking the dimer must be kept cold and used fairly soon in order to prevent it from dimerizing.

2. 1,3-Butadiene is a gas at room temperature and thus difficult to handle and use in

an undergraduate experiment. 3-Sulfolene, on the other hand, is a solid at room temperature.

3.

O

O

O

O

mono adduct bis adduct

4. Petroleum ether (60-900C) is, in fact, a mixture of hydrocarbons that boil between 60-900C and is not an “ether” as the name might suggest. Three possible


32

compounds that might be present in this solvent are: hexane (bp 690), 2-methylpentane (bp 620) and 3-methylpentane (bp 640

)


1. The diene is a conjugated 1,3-diene. The dienophile is an alkene with a suitably activated double bond. In the Diels-Alder reaction, the diene supplies four pi electrons and the dienophile supplies two electrons to the transition state.

2. Two new sigma bonds and one new pi bonds are created.

3. Three new pi bonds are created in a retro-Diels-Alder reaction; one pi bond and

two sigma bonds are destroyed.

4.

O

O

O

+a.

+b.

O

O Answers to Questions (Report: Part B)

1. Both are cycloreversion reactions, both involve six electrons in a cyclic transition state.

2.



33

CH3

H3C CH2

heatCH3

2

limonene

2-methyl-1,3-butadiene(isoprene)

Answers to Questions (Report: Part C)

1. The reaction solution turns yellow within 10 minutes and is colorless within 24 hours. The control solution remains yellow.

2.

+

O

O

O

O

or two equivalents of cyclopentadiene might add to the quinone to give a bis-adduct.

3. Answers will vary depending on what related topics have been covered in class and the ingenuity of the student. A reasonable answer is that upon cycloaddition, 1,4-benzoquinone loses part of its extended conjugated pi-system. Organic dye molecules have extended conjugated pi systems (Experiment 24), so it is reasonable to think that loss of this system is related to loss of color.


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instructor’s resource guide for the laboratory...

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