Laboratory Techniques in Organic Chemistry, Fourth Edition

Instructor’s Manual

Answers for Questions

PART 1—INTRODUCTION TO THE ORGANIC LABORATORY

CHAPTER 1

SAFETY IN THE LABORATORY

1. Important safety features in the laboratory include: fire extinguisher, fume hoods, wash

station, safety shower, fire blanket, and first aid kit (which may not be in the laboratory,

depending on the institution).

2. The answer depends on the institution. At Carleton College, for example, the cut should be

washed with cold water and bandaged with a sterile bandage from the first aid kit. The

instructor must be informed of the incident and he or she will possibly encourage medical

attention.

3. Ethanol is flammable with a boiling point of 78 °C. Because its boiling point is below 100

°C, the best method for heating ethanol is to use a steam bath.

4. The answer depends on the institution. At Carleton College, for example, broken glass is

discarded in a designated broken-glass container and chemical residues are rinsed or

evaporated from glass before discarding it in the container.

5. The answer depends on the institution. At Carleton College, for example, the standard

laboratory gloves are nitrile and provide the following protection:

a. Dichloromethane: poor

b. Ethyl ether: good

c. Ethylene glycol: excellent

d. Hydrogen peroxide: good

If a student spilled some dichloromethane on the gloves, he or she should immediately remove

the gloves and wash his or her hands with soap and water.

6. CAS numbers are as follows:

a. Acetylsalicylic acid: 50-78-2

b. Salicylic acid: 69-72-7

c. Acetic anhydride: 108-24-7

d. Acetic acid: 64-19-7

e. Phosphoric acid: 7664-38-2

7. (a) Based on MSDS sheets, the chemicals that are hazardous to breathe dust/vapor/fumes are

acetylsalicylic acid, acetic anhydride, and acetic acid.

(b) Based on melting points or boiling points, the chemicals one is most likely to inhale

accidentally are acetic anhydride and acetic acid.

(c) Based on MSDS hazard information, the most dangerous chemicals to inhale are acetic

anhydride and acetylsalicylic acid.

(d) Based on both the boiling point and the inhalation hazard, acetic anhydride is the

chemical that is most important to handle in a fume hood.

CHAPTER 2

GREEN CHEMISTRY

1. The characteristics of benign solvents are low flammability and low toxicity because they

make the solvent less hazardous. A high boiling point requires more energy to evaporate and

is therefore less green. Water solubility is irrelevant to its green characteristics.

2. The one strategy from this list that is not relevant to industry is reducing the scale of a

reaction.

3. The simplest strategy to make an instructional laboratory procedure more benign is to reduce

the scale of the reaction.

4. Ethanol is lower boiling so it requires less energy to evaporate.

5. For this process, you need a boiling point of at least 110 °C and low solubility in water.

Toluene, xylenes, and isobutyl acetate all meet these criteria. Of these, isobutyl acetate has

the lowest overall score in Table 2.2 and thus is the most benign solvent across all categories.

If safety is a paramount concern, xylenes may be chosen, but this will also require more

energy to evaporate from the reaction product.

6. (a) Atom economy for the balanced equation is calculated as follows:

MW(stilbene dibromide)/[MW(trans-stilbene) + (2 MW(HBr)) + MW(HOOH)] 100

= 340/[180 + (2 81) + 34] 100 = 90%

(b) Reaction efficiency = atom economy yield = 0.90 85% = 77%

CHAPTER 3

LABORATORY NOTEBOOKS AND PRELAB INFORMATION

1. (a) True, (b) False, (c) False, (d) False, (e) True, (f) False

2. The requested data are incorporated into a sample Table of Reagents:

reagent, product, or

catalyst

MW

(g/mol)

mp or bp (°C);

d (g/mL)

quantity moles Equiv

Salicylic acid

138.12 mp = 158-161 1.0 g 0.00724 1

Acetic anhydride 102.09 bp =138-140

d = 1.08

2.0 mL 0.0212 3

85% Phosphoric acid

(catalyst)

98.00 bp = 158

d = 1.685 0.1 mL 0.00146 0.2

Acetic acid

60.05 bp = 117-118

d = 1.049

Acetylsalicylic acid

180.16 mp = 134-136 1.3 g

(theor.)

0.00724

(theor.)

1

(theor.)

3. As shown by the entries in the table, the limiting reagent is salicylic acid (phosphoric acid is

a catalyst) and the theoretical yield of acetylsalicylic acid is derived from this. These entries

were calculated from the following relationships:

moles(solid) = mass(solid) / MW(solid)

moles(liquid) = volume(liquid) d(liquid)/MW(liquid)

moles(H3PO4) = volume H3PO4 d H3PO4 0.85/MW H3PO4

equiv(reagent or catalyst) = moles(reagent or catalyst)/moles(limiting reagent)

theoretical moles(acetylsalicylic acid) = moles(limiting reagent)

theor. yield(acetylsalicylic acid) = theor. moles(acetylsalicylic acid) MW(acetylsalicylic acid)

If you isolate 0.95 g of acetylsalicylic acid, the percent yield is:

[actual yield/theoretical yield] 100 = [0.95 g/1.3 g] 100 = 73%

PART 2—CARRYING OUT CHEMICAL REACTIONS

CHAPTER 4

LABORATORY GLASSWARE

1. The first reaction, which takes place in 2 mL solvent, should use microscale glassware—

either standard taper or Williamson. The second reaction, which takes place in 20 mL

solvent, should use miniscale standard taper glassware.

2. Grease is used on the ground glass joints of miniscale standard taper glassware to prevent

chemicals from entering the joint and causing the two pieces of glassware to stick together at

the joint. It enables the two pieces of glassware to be more easily separated after a reaction is

completed; if a standard taper joint is stressed, having the joint lightly greased helps to

prevent it from becoming loose and allowing vapors to escape.

3. Grease is avoided, particularly for microscale reactions, because it contaminates the reaction

mixture, which can present challenges to purifying small quantities of reaction products. It

would be used for basic reaction mixtures, however, because when basic solutions

contaminate the joint they usually cause the joint to freeze.

4. Water can cause undesired side reactions that lower yields in most organic reactions.

5. Rinse the clean wet flask with acetone. The residual acetone coating the glassware evaporates

quickly.

CHAPTER 5

MEASUREMENTS AND TRANSFERRING REAGENTS

1. A typical top-loading balance is only accurate to ± 10 mg. In this case, it would be capable of

weighing 75 ± 10 mg, but this is too much weighing error. At 12.5 %, the relative uncertainty

is too high for careful lab work.

2. Weighing a vial or flask that will be used to collect a reaction product makes it easy to

determine the mass of the product by subtracting the mass of the empty vial or flask from the

mass of the vial or flask that contains the product.

3. The best methods for measuring the transfer of 0.45 mL of a limiting reagent are to use a

graduated pipet or a syringe.

4. To calibrate the volume of a compound delivered by an automatic pipet, use the density of

the compound to determine the expected mass of the desired delivered volume. Adjust the

volume until it delivers the correct mass.

5. To calibrate a thermometer, you need to generate a plot of measured temperatures versus

corrected temperatures. To do this, measure the melting points of a series of compounds with

the thermometer and plot the measured temperatures against the known melting points from

the literature (which will constitute the corrected temperatures). Use the resulting plot to

convert the measured temperature of an unknown into a corrected temperature.

6. (a) Because the third number in the calculation is only accurate to the tenths place, the

answer must be rounded off to the tenths place. The value before rounding off is -5085.0654

and thus, the answer is -5085.1 (because 0.06 is ≥ 0.05, it is rounded up to 0.1).

(b) For multiplication, the number of significant figures determines how to round off; in this

case, the first value has the fewest significant figures, or 2. The value before rounding off is

8.37963738 and after rounding off, the answer is 8.4.

7. 1.10 ± 0.06.

8. Use Excel or a similar program to do these calculations. The average is 10.45 and the

standard deviation is 1.7972. The standard deviation should be rounded off to one decimal

place, so the answer is 10 ± 2.

9. The appropriate G value is 2.33. None of the data points should be rejected.

CHAPTER 6

HEATING AND COOLING METHODS

1. Superheating is when a liquid is heated above its boiling point but does not boil because

bubbles are kinetically prevented from forming. It is most common in vessels with very

smooth surfaces, which do not offer as much opportunity for nucleation of bubbles as do

vessels that have scratches or other imperfections in their surfaces.

2. A boiling stone has uneven surfaces containing air pockets and sharp edges. These will help

form bubbles and prevent a heated liquid from erupting violently as it starts to boil. In other

words, it will prevent superheating of the liquid.

3. If a boiling stone is added to a hot liquid, it may cause the liquid to erupt violently because

the stone causes rapid nucleation of bubbles. If you forget to add a boiling stone before

heating, remove the heat source and allow the liquid to cool well below the boiling point

before adding a boiling stone.

4. Microwave heating enables reactions to take place faster than heating via conventional

methods.

5. An insulated Dewar flask can keep liquids cold for long periods of time and can be used as

cooling baths for reaction mixtures.

6. A lab jack is used to adjust the height of a heating mantle, a heating or cooling bath, or a

magnetic stirrer, so it can be positioned properly under a clamped reaction flask or in a

distillation.

CHAPTER 7

SETTING UP ORGANIC REACTIONS

1. This ensures that there are no air bubbles in the outer jacket of the condenser and that the

water will efficiently cool the glass wall of the condenser.

2. The heat should be adjusted so the position where the vapors condense is in the lower third of

the condenser.

3. Gas cylinders are under very high pressure and if they tip over accidentally, the valve on top

of the cylinder may be snapped off. In this case, the rapidly expanding gas would turn the

cylinder into a dangerous torpedo, which could cause bodily harm and damage to property.

4. An anhydrous atmosphere is one in which water vapor is excluded. An inert atmosphere is

composed of nitrogen or argon, in which oxygen, water vapor, and other trace components of

air are excluded.

5. A mineral oil bubbler prevents the build-up of excessive pressure inside a reaction flask that

is connected to a high-pressure cylinder of inert gas.

6. Molecular sieves are used to dry solvents or other organic liquids or to absorb water from a

reaction mixture.

CHAPTER 8

COMPUTATIONAL CHEMISTRY

FOLLOW-UP ASSIGNMENT. Page 113

The steric energies (excluding solvation energy) for equatorial- and axial-tert-butylcyclohexane,

calculated by using the Spartan ‘06 computational chemistry package and MMFF94 force field

parameter set, are:

equatorial-tert-butylcyclohexane = 78.17 kJ/mol

axial-tert-butylcyclohexane = 104.15 kJ/mol

If the difference in steric energies approximates the difference in free energy between the

conformers, the free energy difference between the axial and equatorial conformers is 25.98

kJ/mol or 6.209 kcal/mol. At room temperature (298 K), Keq = 36,800. Only one out of 36,800

molecules or less than 0.003 % of tert-butylcyclohexane would have the tert-butyl group in the

axial conformation at any one time.

1. (a) QM, (b) MM, (c) QM, (d) QM, (e) MM.

2. Quantum mechanics is the better computational method for a study of kinetic versus

thermodynamic control. Using quantum mechanics you can calculate heat of formation values to

compare the energies of the products (thermodynamic prediction) and the relative energies of

intermediates needed for understanding the kinetically formed product.

MM is used mainly for looking at the conformations of a given molecule and does not

take electrons into account; it cannot be used reliably for comparing the energies of products or

intermediates.

3. (a) NOTE: In the 1st printing of the 4th edition of Laboratory Techniques in Organic

Chemistry the numbers

1 and 2 were inadvertently omitted.

(b) MM2 would be the most appropriate.

(c) Two chair-like conformations should be submitted for compound 1, one with both the

C3H5 and OH substituents axial and one with both equatorial.

Two chair-like conformations should also be submitted for compound 2, one with the

C3H5 axial and the OH equatorial and one with the C3H5 equatorial and the OH axial.

(d)

(e)

4. Local minima steric energies were probably obtained instead of the global minima because of

hindrance of rotation around some bonds. Depending on the computational modeling package

that you are using, use of conformational searching, sequential searching, or the dihedral driver

could help find the global minimum.

5.

cis-3,3,5-trimethylcyclohexanol

88.59 kJ/mol 116.37 kJ/mol

trans-3,3,5-trimethylcyclohexanol

95.55 kJ/mol 105.96 kJ/mol

These steric energies were obtained using molecular mechanics calculations with the MMFF94

force field in the Spartan ‘06 program. The conformer with the lower steric energy is more

OH

OH

OH

OH

stable. The conformers with the greater number of larger substituents in equatorial positions are

more stable. Methyl groups have a larger effective size than hydroxyl groups.

6. Using the Spartan ’06 program, the heat of formation (ΔHf) of adamantane is −180.806

kJ/mol and ΔHf of twistane is −115.999 kJ/mol. Adamantane is the more stable compound.

PART 3—BASIC METHODS FOR SEPARATION, PURIFICATION, AND

ANALYSIS

CHAPTER 9

FILTRATION

1. Hirsh funnels come in smaller sizes than Buchner funnels, plus the plate that holds the filter

paper has a smaller area in Hirsch funnels. There is inherently less loss using a Hirsch funnel

for small-scale filtrations.

2. Fluted filter paper provides a much larger surface area than a filter paper cone, which

promotes faster gravity filtration and minimizes the chance of a solid crystallizing from a

solution during the filtration.

3. The solution would cool too rapidly during vacuum filtration and crystallize prematurely.

4. If the solvent used in the recrystallization is a volatile organic compound, it will evaporate

rapidly during the filtration. Evaporation is a cooling process.

5. If the seal is not broken in a vacuum filtration before the water flow is turned off, water may

be sucked back into the filter flask.

6. (a) Pasteur pipet packed with glass wool and a filter aid such as Celite.

(b) Hirsch funnel, adapter, filter paper, and vacuum filtration flask.

CHAPTER 10

EXTRACTION

1. Diethyl ether is less dense than water, so ether forms the upper layer of the extraction.

Adding ether to an ether layer will of course produce only one phase.

2. The best way to remove acetic acid from the ether solution is to convert the acetic acid to its

conjugate base, the acetate anion, by extracting the ether solution with a 5% NaHCO3

solution. Sodium acetate is very soluble in water but not soluble in diethyl ether, whereas

acetic acid itself is soluble both in water and in diethyl ether.

3. Reaction of Na2CO3 with acid produces gaseous CO2 by an acid-base reaction. Unless the

funnel is vented periodically, the gas will build up pressure in the separatory funnel and

could blow out the stopper, allowing the liquids in the funnel to spray out into the laboratory.

4. Remove a few drops of the questionable liquid with a Pasteur pipet and add the liquid to a

small text tube containing one mL of water. Petroleum ether is insoluble in water.

5. After the first 20-mL diethyl ether extraction, the calculation (x g/0.3 g = (100 mL/100mL +

(20 mL × 10)) shows that 0.10 g of benzoic acid would remain in the water. After two, three,

and four extractions, 0.033 g, 0.011 g, and 0.0037 g of benzoic acid would remain in the

water layer. After one 80-mL extraction with ether, 0.033 g of benzoic acid would remain in

the water. Clearly, the four 20-mL extractions with diethyl ether are more efficient than one

80-mL extraction (some 9 times more efficient) in removing the benzoic acid from water.

CHAPTER 11

DRYING ORGANIC LIQUIDS AND RECOVERING REACTION

PRODUCTS

1. CaCl2 would be the more effective drying agent because it can bind six molecules of water

per CaCl2 group, whereas CaCl2 • 6H2O has already been hydrated.

2 (a) If too little drying agent is used, the organic liquid will still contain water, which may

interfere with the subsequent use of the liquid and act as an impurity.

(b) If too much drying agent is used, a significant amount of product may be lost by

adsorption onto the surface of the drying agent, thus reducing the yield.

3. Select a neutral drying agent. Hexane is quite hydrophobic, so a solution of 2-octanone and

hexane would not likely contain very much water. Therefore, CaSO4 or K2CO3 would be a

fast and effective drying agent.

4. KOH would be a better choice for drying an amine because KOH would undergo a reaction

with the acid, producing RCO2K and H2O.

CHAPTER 12

BOILING POINTS AND DISTILLATION

1. Figure 12.5 shows that in a simple distillation without adequate fractionation, a solution

containing a 1:1 molar ratio of pentane and hexane boils just below 40 , whereas pure

pentane itself boils at 36 . The 1:1 solution boils at the higher temperature because at 36 the

partial pressure of pentane plus the partial pressure of hexane in the vapor do not add up to

the atmospheric pressure. This can be seen by considering the application of Raoult’s law

(Ppentane = P pentane Xpentane).

2. The molar composition of the vapor would be about 57% hexane and 43% pentane. After

another vaporization-condensation cycle the composition of the vapor would be about 31%

hexane and 69% pentane.

3. The thermometer bulb was probably placed too high in relation to the side arm of the

distilling head (see Figures 12.7 and 12.10). Therefore, the liquid in the thermometer was not

completely immersed in the vapor that distilled.

4. Diethyl ether is extremely flammable, and an electric heating mantle can attain a very high

temperature, well over 100 , at the mantle surface. If an accident occurred and the ether came

into contact with the hot mantle, it could burst into flames.

5. The compound would boil at about 150 when the pressure is reduced to 7 torr.

6. (a) The use of a Dean-Stark trap (Figure 12.20) under the reflux condenser allows the less

dense 1-butanol/butyl acetate solution to be returned continuously to the reaction flask where

further reaction between 1-butanol and acetic acid occurs. The water that separates from the

ternary azeotrope remains in the bottom of the Dean-Stark trap. Removal of water from the

reaction mixture shifts the equilibrium position toward the product side.

(b) 1-Butanol and butyl acetate form a binary azeotrope, so distillation will not separate

them. However, they can be separated in a straightforward manner by liquid chromatography

(Chapter 19).

CHAPTER 13

REFRACTOMETRY

1. The refractive index at 25.0 C would be 1.3169.

2. Acetone might dissolve the adhesive holding the prism, and water does not readily dissolve

most organic compounds.

CHAPTER 14

MELTING POINTS AND MELTING RANGES

1. In the second melting-point determination, using a sample height of 4 5 mm, it took longer

for the larger sample to melt. Because the temperature was rising at a steady rate, the second

melting range was greater; its high end was 2 above the actual melting point of the pure

compound.

2. Heating a melting-point sample so the temperature rises too fast can cause either a high or

low melting range. A low melting point is caused by the inability of the liquid in the

thermometer bulb to heat quickly enough to reflect the real temperature of the sample.

3. Without calibrating either melting-point apparatus, you could measure the melting point of

both compounds on one apparatus to learn if they are likely to have the same identity; a

mixture melting point would even be a better test of whether the two compounds have the

same identity. You could use another analytical technique to determine the identity of the

sample, such as IR or NMR spectroscopy.

4. When both ends of the melting-point tube are sealed, heating produces increased pressure in

the tube, which retards sublimation.

5. Upon cooling, the compound probably crystallized from the melt in a new crystalline form,

which has a higher melting point, whereas recrystallization from the same solvent

reproducibly gives the same crystal form as seen initially.

CHAPTER 15

RECRYSTALLIZATION

1. A good recrystallization solvent is one in which the compound to be recrystallized is soluble

when the solvent is near the boiling point and only sparingly soluble when the solvent is at

room temperature or below.

2. Water would be the better recrystallization solvent. It would be awkward to work with less

than 10 mL of solvent; in methanol 3 g of the compound would be lost in the

recrystallization, but only 0.022 g would be lost in water. There would be little disadvantage

to using 100 mL of water if 5 6 g of compound were being recrystallized. To look at it

another way, the differential solubility in hot compared to cold water is a factor of 30; in

methanol the factor is only 2.

3. If a compound is crystallized too quickly, impurities can be trapped in the crystal lattice.

Slow crystallization allows for maximum selectivity (purity) in the crystallization.

4. A filtration of the hot mixture is necessary if an insoluble impurity is present or if activated

carbon is used to remove colored impurities.

5. Vacuum filtration cools the solution to a much greater extent and produces excessive

crystallization of the desired compound during the filtration. Some of the solid may be lost in

the bottom portion of the Buchner funnel.

6. You can add more solvent so that the compound does not come out of solution at a

temperature above its melting point. You can also add a few crystals of the crude compound

to use for nucleation of new crystals, or you can scratch the inside of the bottom of the flask

to provide glass particles, which may serve as centers of crystallization.

7. A methanol/pentane mixture is an awkward solvent pair because when the mixture is heated

to dissolve the compound being recrystallized, the lower boiling-point pentane vaporizes

very easily, thereby making the compound much more soluble in the remaining methanol.

The use of a methanol/hexane mixture would work much better because the two solvents

have similar boiling points.

CHAPTER 16

SUBLIMATION

1. The likeliest candidate would be menthol, a solid whose vapor pressure is great enough that

the vapor above the solid can easily be detected by its odor.

2. In a sublimation one wants a substantial vapor pressure of the solid at a temperature below

the solid’s melting point. This compound has quite a substantial vapor pressure, so not too

high a temperature would have to be used for its sublimation at reduced pressure. However, it

is also necessary to recover the compound as a solid by condensation at a cold temperature.

A cold finger containing ice water, a temperature of perhaps 90 for the sublimation, and a

vacuum of 20 torr (using a water aspirator or vacuum pump) should serve.

3. As a sample of hexachloroethane is heated, it would begin to sublime and would then

recondense farther up the melting-point capillary tube, where it is at room temperature.

Therefore, it would be difficult to ascertain its melting point in an open tube. A capillary tube

sealed at both ends would be needed.

CHAPTER 17

OPTICAL ACTIVITY AND ENANTIOMERIC ANALYSIS

1. The % ee is 25%, which means that the sample is 62.5% (+)-2-butanol and 37.5% ( )-2-

butanol.

2. The optical rotation of ( )-2-butanol is –13.0 , so the % ee is 75%. The composition of the

sample is 87.5% ( )-2-butanol and 12.5% (+)-2-butanol.

3. If inversion at nitrogen is slow, the two nitrogen atoms are stereocenters. In addition, there is

one quaternary carbon atom and two tertiary carbon atoms, which are stereocenters. Lastly,

there are three secondary carbon atoms, which are stereocenters, two attached to a nitrogen

atom and one to an oxygen atom.

4. The basic site is the nitrogen atom at the top right; it is a tertiary amine. The other nitrogen

atom is an amide nitrogen, which is only very weakly basic.

5. If the concentration of the solution were halved, the rotation would either be +70 or 290 if

the +140 reading were originally correct. If 220 were the correct reading, the new reading

would be 110 or +250 .

PART 4—CHROMATOGRAPHY

CHAPTER 18

THIN-LAYER CHROMATOGRAPHY

1. If the two substances moved with the solvent front, you could say very little about their being

identical or not. The use of a less polar developing solvent would lower the Rf values and

provide a better opportunity to distinguish the two compounds from one another.

2. It is easier to evaluate a thin-layer chromatogram when the mixture of compounds being

chromatographed has a large separation in their Rf values. Diethyl ether would be the better

developing solvent.

3. Alcohols are quite polar compounds, and ketones are of intermediate polarity. Therefore a

solvent of intermediate polarity should work well, from diethyl ether to acetone.

4. The 4-tert-butylcyclohexanol is the more polar compound that can form hydrogen bonds with

the polar silica gel stationary phase. Thus, 4-tert-butylcyclohexanol will have the lower Rf

value in either developing solvent. The less polar 4-tert-butylcyclohexanone will have the

higher Rf value in both developing solvents.

CHAPTER 19

LIQUID CHROMATOGRAPHY

1. When a chromatography column dries out, channels can form, leading to poor separations.

2. The compounds usually are added in liquid form; if a solvent is needed to dissolve a solid, a

minimum amount should be used. The organic mixture should be carefully added onto the

column as the solvent level is kept just at the top of the upper sand layer, taking care not to

disturb the top of the column. After the mixture of compounds reaches the top of the sand

layer, the first elution solvent should be carefully added.

3. (a) If too strong an adsorbent is used, very polar solvents may be needed to remove the

compounds from the column and separation may be poor; (b) large elution fractions may mix

more than one band coming off the column, resulting in recovered compounds that are

impure; (c) very slow flow rate of the mobile phase can produce poor separation by diffusion

that widens the bands of compounds eluting from the column.

4. tert-Butylcyclohexane, a non-polar alkane, would elute easiest. The order of elution would be

tert-butylcyclohexane, 1,3-dichlorobenzene, 2-octanol, benzoic acid.

5. The smaller the particle size, the greater the surface area available for adsorption on the

liquid stationary phase for the compounds being separated. This produces many more

partition equilibria as the compounds move through the column, which allow for more

effective separation.

CHAPTER 20

GAS CHROMATOGRAPHY

1. A GC column has thousands of theoretical plates as a result of the huge surface area on

which the gas and liquid phases can interact. This provides much greater efficiency in

separations on a GC column compared to a fractional distillation column having less than 10

theoretical plates.

2. The liquid stationary phase must be non-reactive, nonvolatile, and be liquid over a broad

temperature range.

3. The retention time is the time from injection of the sample until it reaches the detector.

Greater flow rate and higher column temperature both produce a shorter retention time.

4. After the conditions for a successful separation have been established, some of the suspected

compound can be added to a portion of the mixture. If this produces the same chromatogram,

except for a larger peak with no shoulders for the suspected compound, it is not unlikely that

the identity of the compound is established.

5. GC uses very small sample amounts, so spectroscopic methods would have to be used.

Mass, IR, and NMR spectroscopy are likely candidates.

6. The column temperature and/or the flow rate of the carrier gas can be lowered, keeping the

compounds in the GC column for a longer time and allowing for more vapor-liquid equilibria

to separate the two components.

7. Table 20.1 can be used to answer these questions, using as a general rule that a liquid phase

provides the best separation if it is of similar polarity to the compounds being separated: (a)

polyethylene glycol; (b) any of the polysiloxanes; (c) polyethylene glycol or diethylene

glycol succinate polyester.

8. The uncorrected relative amounts are 29.2 % of compound 1 and 70.8 % of compound 2.

Correcting for the relative Mf values, the relative amounts are 32.6 % of compound 1 and

67.4 % of compound 2.

PART 5—SPECTROMETRIC METHODS

CHAPTER 21

INFRARED SPECTROSCOPY

FOLLOW-UP ASSIGNMENT. Page 327

This is an exercise on the recognition of chemical-bond stretching vibrations that would be seen

in IR spectra.

(a) cyclopentanone (C–H, 3300-2700 cm-1

; C=O, 1850-1650 cm-1

)

(b) methyl acetate (C–H, 3300-2700 cm-1

; C=O, 1850-1650 cm-1

; C–O, 1300-1000 cm-1

)

(c) methoxybenzene (C–H, 3300-2700 cm-1

; C=C, 1680-1440 cm-1

; C–O, 1300-1000 cm-1

(d) acetamide (N–H, 3550-3150 cm-1

; C–H, 3300-2700 cm-1

; C=O, 1850-1650 cm-1

)

(e) 1-aminohexane (N–H, 3550-3150 cm-1

; C–H, 3300-2700 cm-1

)

1. (a) 3311(s), 2961(s), 2119(m) cm-1

: 1-hexyne; both the C≡C and C≡C−H bond stretching

vibration bands are present in the spectrum.

3020(s), 2940(s), 1606(s), 1495(s), 741(s) cm -1

: 1,2-dimethylbenzene; the aromatic C−H

stretch, C=C stretch and C−H bend are present.

3049(w), 2951(m), 1642(m) cm-1

: 1-decene; the C=C−H and C=C stretches are present.

2924(s), 1467(m) cm-1

: dodecane; only C−H stretching and bending vibrations are present.

(b) 3060(m), 2835(m), 1498(s), 1247(s), 1040(s) cm-1

: methoxybenzene; the aromatic C−H

stretch and C=C stretch are present, as are two C−O stretches.

3370(s), 3045(m), 1595(s), 1224(s) cm-1

: phenol; the O−H and C−O stretches are present, as

are the aromatic C−H and C=C stretches.

3330(br, s), 3030(m), 2980(m), 1454(m), 1223(s) cm-1

: benzyl alcohol; the O−H and C−O

stretches and the aromatic C−H and C=C stretches are present, as is an alkyl C−H stretching

band.

(c) 3070(m), 2978(m), 2825(s), 2720(m), 1724(s) cm-1

: 2-phenylpropanal; the aromatic C−H

stretch, the H−C=O Fermi doublet, and a saturated C=O are present.

3372(m), 3290(m), 2925(s) cm-1

: octylamine; the two N−H stretching bands of a primary

amine and the alkyl C−H stretch are present.

3070(w), 1765(s), 1215(s), 1193(s) cm-1

: phenyl acetate; the aromatic C−H, saturated C=O,

and C−O stretches are present.

3300-2500(br, s), 2950(m), 1711(s) cm-1

: heptanoic acid; the broad, strong O−H stretch and

the C=O stretch of a carboxylic acid are present.

3060(m), 2985(w), 1690(s) cm-1

: acetophenone; the aromatic C−H and a conjugated C=O

stretch are present.

3352(s), 3170(s), 2960(m), 1640(s) cm-1

: 2-methylpropanamide; the two N−H stretching

bands of a primary amide and an amide C=O stretch are present.

2964(s), 1717(s) cm-1

: 2-pentanone; alkyl C−H and C=O stretches are present.

2. The product is cyclohexanol. It exhibits a strong, broad peak at 3350 cm-1

due to the O H

stretch, a C O stretch at 1080 cm-1

, characteristic of a secondary alcohol, and the normal

C H stretching and bending vibrations in the 2900 and 1400 cm-1

regions.

3. The stretching vibration of the C C in diphenylacetylene does not produce a change in the

dipole moment of this symmetrical molecule, and thus the absorption is not apparent in the

IR spectrum.

4. The liquid is benzene. Its IR spectrum shows only aromatic C−H stretch, C=C stretch and

C−H bending bands.

5. Figure 21.32: biphenyl; only aromatic C−H stretch, C=C stretch and C−H bending bands.

Figure 21.33: 4-methylbenzaldehyde; aromatic C=C, weak C−H stretches and aromatic

C−H bending, as well as a definite H−C=O Fermi doublet of an aldehyde and conjugated

C=O stretch.

Figure 21.34: ethyl propanoate; strong saturated C=O and C−O stretches, plus only alkyl

C−H stretching bands.

Figure 21.35: phenol; strong, broad O−H stretch, strong C−O stretch, aromatic C=C stretch and

C−H bending, as well as absence of strong alkyl C−H stretching bands.

Figure 21.36: benzophenone; weak aromatic C−H stretch, aromatic C=C stretch and C−H

bending, and conjugated C=O stretching.

Figure 21.37: acetanilide; strong N−H, alkyl C−H, and amide C=O stretch, as well as aromatic

C=C and C−H stretching and C−H bending.

Figure 21.38: 4-isopropyl-1-methylbenzene; aromatic C−H and C=C stretching, C−H bending

characteristic of para-substitution, and strong alkyl C−H stretching.

Figure 21.39: 1-butanol: broad, strong hydrogen-bonded O−H stretch, C−O stretch, and alkyl

C−H stretching and bending.

CHAPTER 22 1H NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY

All estimated chemical shifts in parentheses are calculated from the tables in Laboratory

Techniques in Organic Chemistry

FOLLOW-UP ASSIGNMENT. Page 362

The integration values provide a quick identification for the three NMR peaks. The peak at 2.04

ppm with relative integration of 3 is a methyl proton, and the peak at 5.10 ppm with relative

integration of 2 is a methylene proton, and the messy peak at about 7.4 ppm with relative

integration of 5 is due to the aromatic protons.

The aromatic protons are within the expected chemical shift region (6.5─9.0 ppm). The methyl

protons are consistent with being α to a carbonyl group (1.9−3.3 ppm). Finally, the methylene

protons are near the limit of the chemical shift range for protons α to an oxygen atom of an ester

(3.2−5.3 ppm) and are consistent with the chemical shift for protons α to both an oxygen atom of

an ester and a benzene ring. The unknown ester is C6H5CH2OC(=O)CH3. The ambiguity in the

correct assignment disappears when quantitative estimation of chemical shifts is used; see the

follow-up assignment on page 370.

FOLLOW-UP ASSIGNMENT. Page 368

The tert-butyl protons of (CH3)3C(C=O)OCH3 have –(C=O)OCH3 as a β-substituent.

FIRST FOLLOW-UP ASSIGNMENT. Page 370

The estimated chemical shifts (in parentheses) are much closer to the chemical shift estimates for

C6H5CH2OC(=O)CH3 than for C6H5C(=O)OCH2CH3.

C6H5CH2OC(=O)CH3 C6H5C(=O)OCH2CH3

↑ ↑ ↑ ↑

5.10 2.04 5.10 2.04

(5.4) (2.0) (4.3) (1.4)

SECOND FOLLOW-UP ASSIGNMENT. Page 370

O O

2.08(2.1) 2.46

(2. 7)

1.21(1.2)

3.16(3.0)

FOLLOW-UP ASSIGNMENT. Page 372

The measured NMR assignments are very close to the calculated ones. Changing the assignments

of Hb and Hd leads to greater errors.

Measured (ppm) Calculated (ppm)

Hb 8.42 (8.42)

Hc 7.67 (7.64)

Hd 8.38 (8.39)

PROBLEM ONE, Page 374

B is clearly the correct molecular structure. The measured chemical shifts are very far away from

the calculated chemical shifts (in parentheses) in A.

O

O

O

O

A

B

3.67(2.0)

1.20(1.4)

3.67(3.7)

1.20(1.2)

PROBLEM TWO, Page 375

The assignments of the two vinyl protons at 5.10 ppm and 5.07 ppm are too close to call with

any confidence.

PROBLEM THREE, Page 375

The protons at 1.83 and 3.75 ppm are alkyl protons, the protons at 6.07 and 6.33 ppm are vinyl

protons, and the protons at 6.80 and 7.23 ppm are aromatic protons.

FOLLOW-UP ASSIGNMENT. Page 404

The chemical shift assignments of the alkyl protons in A and B were made by taking the NMR

integrations and splitting patterns in Figure 21.34 into account. The calculated chemical shift

values (in parentheses) are clearly closer to the measured values in the case of B, which is likely

to be the correct structure. The aromatic protons were not considered, because the calculated

values would be the same for both of these compounds.

O O1.25(1.2)

2.70(2.9)

2.58(2.3)

2.58(2.6)

2.70(2.6)

1.25(1.3)

A B

1. The calculated chemical shift values are in the parentheses.

(a) C5H11Cl is 1-chloro-2,2-dimethylpropane.

(b) C5H10O2 is 3-hydroxy-3-methyl-2-butanone.

(c) C6H12O2 is 4-hydroxy-4-methyl-2-pentanone.

(d) C5H10O is 3-methylbutanal.

(e) C5H8O is 3-buten-2-ol. The assignments of the 5.19 ppm and 5.06 ppm protons cannot be

made on the basis of calculated chemical shifts; however, the relative sizes of the cis- and trans-

proton-proton coupling constants allow the assignments (see Table 22.6).

2. C5H8O2 is methyl E-2-butenoate. The methyl singlet at 3.85 ppm is the ester alkyl group. The

C-4 methyl group at 1.9 ppm is a doublet of doublets, due to splitting with each of the vinyl

protons. The C-3 vinyl proton is the doublet of quartets at 7.0 ppm, while the C-2 vinyl

proton is the doublet of quartets at 5.85 ppm. The C-3 proton is farther downfield due to the

substantial positive charge at C-3 and the effect of the cis carbonyl group.

3. The 1H NMR signal at 2.0 ppm, which exchanges with D2O to form HOD at 4.6 ppm, is the

O H of 2-propanol. This proton at oxygen splits with the methine proton at C-2 if chemical

exchange of the O H becomes slow. Increasing concentration of 2-propanol leads to greater

intermolecular H-bonding and a downfield shift. The two methyl groups comprise the

doublet at 1.2 ppm, while the septet at 4.02 ppm is the methine proton at C-2.

4. The enol O H is strongly H-bonded and appears downfield at 8.8 ppm, and the alkene proton

is at 5.5 ppm. The methyl peaks of the enol isomer appear at 1.10 ppm; the two sets of

methylene ring protons are made equivalent on the NMR time scale by rapid interconversion

of the two enol isomers and appear at 2.2 ppm. The diketone has methylene peaks at 3.4 ppm

( to two carbonyl groups) and 2.6 ppm, while its two methyl groups appear at 1.05 ppm.

Integrating the 2.2 ppm peak against the 2.6 ppm peak, the 5.5 ppm peak against the 3.4 ppm

peak, and the enol dimethyl absorption at 1.10 ppm against the methyl protons of the

diketone at 1.05 ppm, all indicate that there are 42 mol % enol and 58 mol % diketone

present in the equilibrium mixture.

5. The C8H14O compound is 6-methyl-5-heptene-2-one. The molecular formula indicates two

Double Bond Equivalents. Because one double-bond equivalent is a carbonyl group, the

other one may be a C=C. That assignment is consistent with the vinyl signal (1H) at 5.06

ppm, which is a triplet with an overlay of quartets showing the small allylic splitting across

the double bond. C8H14O has three singlet methyl groups; the chemical shift of one methyl

group (2.1 ppm) suggests that it is attached to the carbonyl carbon. The chemical shifts of the

other two methyl groups are consistent with their being allylic; there is a small allylic

splitting in the 1.65 ppm methyl peak, which suggests that it is trans to the vinyl proton

(Table 22.6). The two methylene groups appear at 2.25 ppm and 2.45 ppm.

CHAPTER 23 13

C AND TWO-DIMENSIONAL NMR SPECTROSCOPY All estimated chemical shifts in parentheses are calculated from the tables in Laboratory

Techniques in Organic Chemistry.

FOLLOW-UP ASSIGNMENT. Page 419

The estimated chemical shift for C-2 of heptane is 22.8 ppm and for C-3 is 32.2 ppm. The

measured values fit very well to these assignments.

FOLLOW-UP ASSIGNMENT. Page 422

The estimated chemical shift of C-3 of 3-methyl-1-butanol is 24.4 ppm, compared to the

measured value of 25. For C-4 and C-5 the estimated chemical shift is 22.0 ppm, compared to the

measured value of 22.5.

1. The number of signals in a 13

C spectrum of a pure compound indicates the number of

different types of carbon atoms in its molecules. If there are fewer 13

C signals than the

number of carbon atoms, there is some element of symmetry present, such as a mirror plane

or an axis of rotation.

(a) 2-pentanol (5 signals), 2,2-dimethylbutane (4), isopropyl acetate (4), 2-acetoxybutane (6).

(b) para-aminobenzoic acid (5 signals), methyl 2-hydroxybenzoate (8), 1-phenyl-2-

methylpropane (7), 1,3-cyclopentadiene (3).

(c) cyclohexane (1 signal), trans-1,4-dimethylcyclohexane (3), trans-1,2-

dimethylcyclohexane (4).

(d) (8 signals) (3 signals) (10 signals)

2. All estimated chemical shifts are in parentheses.

3. The compound is 1,2-dimethylbenzene. The 13

C spectrum shows four different types of

carbon atoms, one alkyl and three aromatic. This means that the compound is a

dimethylbenzene, whose symmetry shows that it is the ortho-isomer. The DEPT spectra are

consistent with the assignments below; the 129.6 ppm and 125.8 ppm signals are due to

carbon atoms with one hydrogen atom attached to each and the 136.4 ppm signal is due to a

carbon atom with no hydrogen atoms attached.

4. The C10H14 compound is 4-methylisopropylbenzene. The

13C spectrum shows four aromatic

carbons, indicative of para-substitution. The DEPT spectra show that the 145.8 ppm and

135.1 ppm signals are due to carbon atoms with no hydrogen atoms attached. The DEPT

spectra also show that two of the three alkyl signals are due to methyl groups and one is a

methine carbon; no methylene group is present.

The estimated 13

C chemical shift for the methine carbon of the isopropyl group is only

approximate because it is difficult to ascertain the appropriate steric additivity parameter

from Table 23.4; its estimated value was calculated using a methylene group for the benzene

ring carbon.

5. The number of 13

C signals in each spectrum provides a preliminary sorting of the seven

possible C4H10O isomers. The three spectra that have four signals must include

CH3CH2CH2CH2OH, CH3CH2CH(OH)CH3, and CH3OCH2CH2CH3. The two spectra with

three signals must include (CH3)2CHCH2OH and CH3OCH(CH3)2. The two spectra with two

signals must include (CH3)3COH and CH3CH2OCH2CH3. Knowing that the downfield

signals must be carbon atoms next to oxygen atoms and utilizing the DEPT spectra the

tentative assignments are:

(a). CH3CH2CH2CH2OH

(b) CH3CH2CH(OH)CH3

(c) CH3CH2OCH2CH3

(d) (CH3)2CHCH2OH

(e) CH3OCH(CH3)2

(f) CH3OCH2CH2CH3

(g) (CH3)3COH

Confirmation of these assignments can be made using a quantitative estimation of their chemical

shifts. The calculated values are in parentheses.

6. The methyl ester of ibuprofen has the molecular formula C14H20O2 and its 13

C NMR

spectrum has eleven signals. Notice that the DEPT (135) spectrum has both a methylene and

a methyl group that appear to be one signal at 45 ppm in the complete 13

C spectrum. The four

signals in the aromatic chemical shift region indicate a benzene ring and there is also a

carbonyl carbon signal far downfield. These account for the five Double Bond Equivalents in

the methyl ester of ibuprofen.

One could either ascertain the correct molecular structure by an initial analysis of the 1H

NMR spectrum or by beginning with the 13

C spectrum and the DEPT information.

The 13

C spectral data indicate a para-disubstituted benzene ring and the ester functional

group. The upfield alkyl signals indicate two methine carbons, one methylene carbon, and

three methyl groups among the seven remaining carbon atoms in C14H20O2.

An analysis of the 1H spectrum produces the following table of structure fragments:

1

H NMR Data for the Methyl Ester of Ibuprofen

Chemical Shift

(ppm)

1H Type Integration Splitting Pattern Structure Fragment

0.90 C-C-H 6 Doublet –CH(CH3)2

1.49 C-C-H 3 Doublet –CHCH3

1.85 C-C-H 1 Multiplet –CHn–CH–CHn–

2.45 Ar-C-H 2 Doublet ArCH2CH–

3.65 O-C-H 3 Singlet –OCH3

3.7 O=C-C-H 1 Quartet O=CCHCH3

7.1 Ar-H 2 Doublet para-disubstituted

benzene ring 7.2 Ar-H 2 Doublet

Combining the 1H and

13C NMR data, one of the substituents on the benzene ring may be

a (CH3)2CHCH2− group. This leaves one additional methine and two methyl groups, one

of which is the alkyl group of the methyl ester. A −CH(CH3)CO2CH3 group is not

unreasonable. These observations lead to the hypothesis of the following structure for the

methyl ester of ibuprofen:

The 2D (C,H) COSY spectrum shows coupling between the three protons at 1.49 ppm

with the carbon atom at 18 ppm, the six protons at 0.90 ppm with the carbon atoms at

22.5 ppm, the proton at 1.85 ppm with the carbon atom at 30.8 ppm, the two protons at

2.45 ppm, as well as the proton at 3.70 ppm, with the carbon atoms at 45 ppm, the three

protons at 3.65 ppm with the carbon atom at 52 ppm, the two protons at 7.2 ppm with the

carbon atoms at 127.5 ppm, and the two protons at 7.1 ppm with the carbon atoms at 129

ppm. The 2D (H,H) COSY spectrum shows coupling between the CH proton at 3.70 ppm

with the CH3 group at 1.49 ppm, coupling between both the CH2 group at 2.45 ppm and

the CH3 group at 0.90 ppm with the 1.85 ppm CH group, and between the ortho protons

of the benzene ring near 7 ppm with each other.

Quantitative estimation of the 13

C chemical shifts confirms the proposed structure for the

methyl ester of ibuprofen. The calculated values are in parentheses. Since the

−CH2CO2CH3 group is not included in Table 22.5, the −CH2Cl group additive parameters

were used instead for its contribution when calculating the chemical shifts of the aromatic

carbon atoms; the −CH2Cl values should be a fair mimic for the electronic influence of

−CH2CO2CH3. Most all of the calculated values are within 3 % of the measured values.

CHAPTER 24

MASS SPECTROMETRY

1. Ethanol, pyridine, and azobenzene (C6H5 N N C6H5) have MW’s of 46, 79, and 182,

respectively. The odd number MW of pyridine shows that it has one nitrogen atom, whereas

azobenzene has an even number of nitrogen atoms.

2. The m/z base peak for 1-bromopropane is C3H7+, the propyl cation.

3. The most stable cations are the base peaks. The base peak at m/z 31 for 1-pentanol is

CH2=OH+, and the base peak at m/z 45 for 2-pentanol is CH3CH=OH

+.

4. The ratio of 3:1 for the m/z 139 and 141 peaks suggests that each fragment contains a

chlorine atom. Both have the formula p-ClC6H4C O+. The base peak at m/z 105 is due to

C6H5C O+.

5. The m/z peaks at 127 and 125, in a ratio of 1:3, must both contain a chlorine atom. The likely

cation is ClC7H6+, the chlorotropylium cation (see bottom of page 454 for the structure of a

tropylium cation). The m/z 105 peak is probably the methyltropylium cation, CH3C7H6+,

which forms by a complex fragmentation process.

6. The odd number molecular-ion peak suggests an odd number of nitrogen atoms, and the

broad infrared band of medium intensity at 3300 cm-1

is diagnostic for a secondary amine.

Using the Rule of Thirteen and MW 121 leads to a molecular formula of C8H11N. With the

low C/H ratio it is likely that the compound has a benzene ring, which leaves C2H6N as a

substituent. The large m/z peak at 91 is likely C7H7+, the aromatic tropylium cation (see page

454). The compound is N-methylbenzylamine, C6H5CH2NHCH3. The m/z peak at 120

probably results from loss of H from the molecular ion to give (C6H5CH=NHCH3)+. The

base peak at m/z 44 is the (CH3NH=CH2)+ cation.

7. The molecular formula of aspirin is C9H8O4: (9 × 12.0000) + (8 × 1.00783) + (4 × 15.9949)

= 80.0422. The mass of the major nuclide is used in each case: 1H,

12C, and

16O.

8. It is likely that the major component of clove oil is eugenol, hit 6 (d). Except for the height of

the 149 fragment, the fragmentation peaks are smaller than the peaks in the MS of the major

component of clove oil (hit 0); however, they have comparable ratios of heights in all cases,

except in the mass range below 30. Hits 1 and 2 have some peaks that are too large and

others that are too small, compared to hit 0 (a).

Hit #1: No 163 peak; small 137, 122, and 94 peaks; large 91 peak; small 66, 55, and 39 43

peaks.

Hit #2: Large 136, 133, 43, and 41 peaks; small 103 and 91 peaks.

However, this is not conclusive proof. Subtle changes in the GCMS experimental conditions

can cause minor changes in mass spectral fragmentation patterns. The identity of the clove

oil component would have to be confirmed by NMR analysis, which would show definitive

differences in the alkene signals of eugenol, compared to the other two isomers.

9. M = 145.9691, M + 2 = 147.9661, M + 4 = 149.9631.

CHAPTER 25

ULTRAVIOLET AND VISIBLE SPECTROSCOPY

1. The energy gap in π−π* electronic transitions is greater than in n−π* transitions; thus, π−π*

bands occur at shorter wavelengths than do n−π* bands. Also, the intensity of π−π*

transitions is inherently greater than the intensity of n−π* transitions because of the lack of

effective overlap between non-bonding n orbitals and π* orbitals.

2. Cyclohexane and ethanol do not absorb UV radiation above 210 nm, whereas toluene has

significant UV absorption in the 230−270 nm region. Good solvents for UV spectroscopy are

transparent to UV radiation.

3. You can calculate ε by using the Beer-Lambert law, which relates absorbance to the

extinction coefficient; ε = 4166 M−1

cm−1

.

4. Benzene has a number of π and π * electronic energy levels that can be involved in UV

transitions. In addition, vibrational energy transitions can produce fine-structure features to

the electronic bands.

CHAPTER 26

INTEGRATED SPECTROSCOPY PROBLEMS

1. The compound is 2-butanone, C4H8O (MW 72). The MS m/z peak at 43 is the CH3CO+

cation. The IR peaks are the C=O stretch at 1715 cm-1

and sp3 C H stretches in the 2990-

2850 cm-1

range. In the 1H NMR spectrum the C-4 methyl group appears at 1.08 ppm, the C-

1 methyl group at 2.15 ppm, and the C-3 methylene group at 2.45 ppm.

2. The IR spectrum has a strong, broad O−H stretch of an alcohol. The five-proton NMR signal

at 7.3 ppm is indicative of a monosubstituted benzene ring. Consistent with the NMR singlet

at 2.5 ppm and the broadened singlet at 4.65 ppm, the compound is benzyl alcohol

(C6H5CH2OH). In addition to the characteristic O−H stretch at 3350 cm−1

, there is aromatic

C−H stretching at 3030 cm−1

, alkyl C−H stretching at 2880 cm−1

, and a strong C−O

stretching vibration of a primary alcohol at 1030 cm−1

.

3. The MS peaks at m/z 92 and 94 in a 3:1 ratio show the presence of a chlorine atom. With a

MW of 92, it is likely that the compound has four carbon atoms; thus, its molecular formula

would be C4H9Cl. The 1H NMR singlet at 1.65 ppm is consistent with (CH3)3CCl. In the

13C

NMR spectrum the weak peak at 67 ppm is the quaternary carbon and the peak at 35 ppm is

due to the methyl groups. The compound is 2-methyl-2-chloropropane.

4. The broad, strong absorption at 3300 cm-1

is indicative of O H stretch, and the C O stretch

at 1230 cm-1

indicates a phenol. Application of the Rule of Thirteen to the molecular ion

peak at m/z 108 leads to the molecular formula C7H8O. From the symmetry of the aromatic

region of the 1H NMR spectrum, the compound is a para-disubstituted benzene compound.

The 2.3 ppm methyl singlet is perfectly consistent with the compound being 4-methylphenol

(CH3C6H4OH).

5. The 1H NMR spectrum indicates four types of hydrogen atoms in C3H6Cl2. One is a methyl

group adjacent to a methine hydrogen atom. The 13

C DEPT spectra indicate that the three 13

C

signals are methyl, methylene, and methine carbons. The 1H chemical shifts and coupling

pattern are consistent with CH3CHClCH2Cl. The two hydrogen atoms of the methylene

group, which is adjacent to a stereocenter, are diastereotopic protons that couple with one

another and with the proton at C-2.

6. The C7H16O compound has no double-bond equivalents and most importantly seven different

types of carbon atoms, so there are no symmetry elements that make any of the carbon atoms

equivalent. The 13

C DEPT spectrum indicates that six of the carbon signals are assignable to

methylene groups. The IR spectrum shows an alcohol O H stretch, which could be primary

or secondary from the position of the C O stretching frequency at 1080 cm-1

. The 2H triplet

at 3.6 ppm in the 1H NMR shows that the compound is a primary alcohol, with an adjacent

CH2 group at 1.6 ppm. There is also a methyl group at 0.9 ppm adjacent to a CH2 group. All

of the information points to the compound being 1-heptanol. The close correspondence

between the measured and calculated 13

C chemical shifts confirms it.

7. Except for the base peak at m/z 30, the mass spectrum has very small peaks. If the molecular

ion has an m/z of 101, there is probably an odd number of nitrogen atoms in the molecule.

Using the Rule of Thirteen, and knowing that a nitrogen atom is equivalent to the mass of

CH2, C6H15N would be the molecular formula, which indicates that there are zero Double-

Bond Equivalents. The intense m/z 30 peak suggests a CH2=NH2+ fragmentation peak. This

also fits with the two medium-intensity peaks at 3376 cm−1

and 3300 cm−1

in the N H

stretching region of the IR spectrum and the broad singlet 2.5 ppm 1H NMR spectrum peak

with an integration of 2H, all of which indicate a primary amine. The two hydrogen atoms of

the methylene group, which is adjacent to a stereocenter, are diastereotopic protons that

couple with one another and with the proton at C-2.

The 13

C NMR spectrum shows only four different types of carbon atoms, which indicates

symmetry in the molecule, also established by the 6H 1H NMR methyl triplet at 0.9 ppm, the

triplet splitting pattern of which indicates that the methyl groups are adjacent to methylene

groups. The 13

C DEPT spectra indicate one methine signal, two methylene signals, and one

methyl signal. It is not unlikely that the most deshielded 45 ppm 13

C signal is a methylene

group attached to the electronegative nitrogen atom; the most deshielded peak in the 1H

NMR spectrum is a doublet, which suggests a CHCH2NH2 grouping.

A reasonable hypothesis for the structure of C6H15N is 2-ethyl-1-aminobutane,

(CH3CH2)2CHCH2NH2. The 1H NMR multiplet signals in the 1.2−1.4 ppm region indicate

the multiple couplings of the methine and CH2 portion of the ethyl groups.

8. When two pieces of data seem inconsistent, it is important to decide which one to trust for

making an initial hypothesis. The MS suggests that the MW of the unknown compound is

100, but fortunately the problem states that the molecular ion peak is not discernible in the

mass spectrum. The IR spectrum shows an intense C=O stretching vibration at 1740 cm−1

and a strong peak at 1190 cm−1

, which likely indicates a C−O stretch and perhaps an ester

functional group. The 1740 cm−1

peak is consistent with a nonconjugated ester carbonyl

group. There is no O−H stretch apparent, but there is a C=C stretch at 1650 cm−1

.

The 13

C NMR spectrum indicates seven different kinds of carbon. The 173 ppm signal is

consistent with an ester functional group. The DEPT spectra lead to the conclusion that the

molecules of the compound contain one methyl group, four methylene groups, one methine

group and one carbon with no protons attached. This is consistent with the 1H NMR

spectrum, although one CH2 group appears far downfield in the vinyl proton region. The

methyl group is farthest upfield in the 1H NMR spectrum and its triplet pattern suggests that

it is attached to a methylene group. There is another methylene group at 2.3 ppm which is

also a triplet, and an additional methylene group at 1.65 ppm which shows a multiplet

pattern. This part of the 1H NMR spectrum suggests that a CH3CH2CH2− group is present.

The chemical shift of the 2.3 ppm CH2 suggests that it is attached to a carbonyl carbon atom.

Thus we have a CH3CH2CH2(C=O)− present in the molecule, probably as a

CH3CH2CH2CO2− group. The 1H NMR spectrum indicates three vinyl protons, and the

13C

spectrum indicates two alkene carbon atoms. If we attach the remaining CH2 group at 65 ppm

to the C=C and to the ester oxygen (not unreasonable considering its chemical shift), we have

CH2=CHCH2O−. The CH2 group attached to the ester oxygen atom is also suitably downfield

at 4.58 ppm in the 1H NMR spectrum. Attaching the two pieces, we have

CH3CH2CH2(C=O)OCH2CH=CH2, allyl butanoate, C7H12O2 (MW 128).

The MS fragmentation peak at m/z 71 fits CH3CH2CH2CO+, the m/z 43 peak is likely

CH3CH2CH2+, and the m/z 41 peak fits CH2=CHCH2

+. The small MS peak at m/z 100 results

from the loss of CH2=CH2 from the molecular ion. If you feel that the quantitative estimation

of the NMR signals is necessary to clinch the compound’s identity, they can be shown to do

so.