1
1. The proper procedure for recrystallization is to allow the hot solution to cool to room temperature, then to chill the solution in an ice bath. Why do we not simply chill the hot solution in an ice bath initially? Ans. If the hot the hot solution is chilled rapidly, rapidly, precipitation (crystal sizes are very small) occurs instead of crystallization, and the product will probably be contaminated with impurities. 2. Explain whether or not each of the following pairs of solvents might be useful as a
for recrystallization. recrystallizati on.
(a) ethanol-me l-meth tha anol
(b) hexane-cy -cycloh lohexane
(c) (c) acetone-
ethanol (d) water-ethanol
(e) water- cyclohexane
Ans. a. ethan ethanol-met ol-methanol hanol -No. -No. There There is no no difference difference in polarity polarity (both (both are are polar and protic) b. hex hexane-cy ane-cyclohe clohexane xane -- No. There is no differ difference ence in polarity polarity (both (both are non-polar). c. ace aceton tone-e e-etha thanol nol - No. Sam Same e as (a). d. wate water-e r-ethano thanoll - Yes. Yes. Water Water and ethanol ethanol are miscible miscible and and have have differen differentt solvating characteristics (ethanol is less polar and dissolves organic compounds while water does not)
e. water water-- cyclohexane cyclohexane - No. This pair of solvents is not not miscible . miscible .
in ** Find a way to incorporate these into post-lab discussion! 1. Describe the characteris characteristics tics of a good recrystallization recrystallization solven solvent. t. Ans. A good recrystallization solvent is a. The solid to be recrys recrystallized tallized is very soluble when the solve solvent nt is near near its boiling boiling temperature and only sparingly soluble when the solvent is at room temperature or below. b. The solid is either either very very soluble in cold cold solvent solvent or insoluble insoluble in hot solv solvent. ent. 51LA W07 TA Note
2
c. The solid solid should should not not react react with with the the solv solvent. ent. d. The solvent solvent should be reasonabl reasonably y volatile, volatile, non-to non-toxic, xic, non-flamma non-flammable, ble, and inexpensive. 3. Explain how the rate rate of crystal crystal growth growth can affect the the purity of of a recrystallized recrystallized compound. Ans. If a compound is crystallized too quickly, impurities can be trapped in the crystal lattice. Slow crystallization allows for maximum selectivity (purity) in crystal formation.
1. What melting point range is typical for a pure compound? Ans. A melting range of 2 o C or less or 1.0~2.0 o C.
2. To what depth should the capillary is loaded with sample? Ans. The height of the the sample should be between between 1-2 mm.
3. When close (within 10 oC) to the melting point, how fast should you allow the temperature to rise? melting point. Ans. Rate of heating should be 1 o C/minute within 10 C to the melting o
However, However, it’s generally accepted to raise the temperature at the rate of 2-3 degree per minute.
1. Two substances, A and B, have the same melting points. How can you determine if they are the same? Ans. Check the mixed mixed melting point of A and and B mixture. If the the mixed mp is the same
as the pure one without any depression, A and B are same. If the mixed mp shows a lower mp with wider melting range than that of the pure, A and B are not same. 51LA W07 TA Note
3
2. Why is it necessary to dry the recrystallized product before taking a melting point? Ans . The residual solvent is a part of impurities that would affect the mp.
1. A student performed two melting-point determinations on a crystalline product. In one determination, the capillary tube contained a sample about 1-2 mm in height and the o
o
melting range was found to be 141 -142 C. In the other determination, the sample height o
o
was 4-5 mm and the melting range was found to be 141 -145 C. Explain the broader melting-point range observed for the second sample. The reported melting point of the o
compound is 143 C. Ans. This kind of phenomenon (the observed melting range is lower than it actually
is) is expected when the rate of heating is extremely rapid because of the thermometer lag, a condition caused by the failure of the mercury in the thermometer to rise quickly enough to accurately show the temperature of the metal heating block. The broader melting range observed for the second sample is due to the amount of sample used in the determination. The melting process is by a heat transfer process form the heater to the sample. When there are too much sample, it should take a longer time (a wider melting range) to turn the solid to liquid completely.
1. Each of the following solvents is used commonly in the experiments to extract organic compounds from aqueous solutions. a) methylene chloride (dichloromethane) b) toluene c) diethyl ether d) hexane Will the organic phase be the upper or lower layer when each of these organic solvents is mixed with water? Explain your answer for each situation. Ans. The organic layer will be the lower phase only when using methylene chloride 20
( d 1.3078). The other four solvents would be the upper phase because their densities are smaller than that of water.
51LA W07 TA Note
4
2. Explain why the following solvents are not used in the experiments to extract organic compounds from aqueous solutions. a) ethanol
b) acetone
Ans. Ethanol is not used because it is water soluble and has a high B.P. Acetone is
also not used because it is water soluble.
1. An extraction procedure specifies that the aqueous solution containing dissolved organic material be extracted twice with 10-mL portions of diethyl ether. A student removed the lower layer after the first extraction and added the second 10-mL portion of ether to the upper layer remaining in the separatory funnel. After shaking the funnel, the student observed only one phase with no interface, Explain. Ans. The student removed the wrong layer. The aqueous layer was removed leaving
the organic layer. Therefore, another addition of ether would give only one phase and no interface. 4. When the two layers form during an ether/water extraction, what would be an easy, convenient way to tell which layer is which if the densities were not available? Ans. Adding a few drops of water and observing which layer is miscible with the
drops of water is an easy and convenient way. 5. You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30 g. The distribution coefficient of benzoic acid in diethyl ether and water is approximately 10. Calculate the amount of acid that would be left in the water solution after four 20-mL extractions with ether. Do the same calculation, using one 80-mL extraction with ether, to determine which method is more efficient. Ans. For four extractions with four 20 ml portions of diethyl ether: st
1 extraction: k =10 = (0.30 - x1) g/ 20 mL
x 1= 0.10 g
x1 g/100 mL nd
2 extraction: k =10 = (0.10 – x 2) g/ 20 mL
x 2= 0.033 g
x2 g/100 mL rd
3 extraction: k =10 = (0.033 – x 3) g/ 20 mL 51LA W07 TA Note
x 3= 0.011g
5
x3 g/100 mL th
4 extraction: k =10 = (0.011 – x4) g/ 20 mL
x 4= 0.004g
x4 g/100 mL Total removed = 0.296 g Amount left in water = 0.004 g
For an extraction with one 80 mL portion of diethyl ether: st
1 extraction: k =10 = (0.30 – x) g/ 80 mL
x = 0.033 g
x g/100 mL Total removed = 0. 267g Amount left in water = 0.033 g
Therefore, the extraction using four 20 mL portions is far better than the one with 80 mL portion.
1. Why is the tea boiled with water in this experiment? Ans. Caffeine is water soluble, while some of the other components of tea such as
cellulose are not. Thus boiling water initiates the separation process.
2. Why is the aqueous tea solution cooled to 15-20 oC before the dichloromethane is added? Ans . Decreasing the temperature of the aqueous solution 1) prevents dichloromethane from boiling (evaporating) away and 2) minimizes the water solubility of caffeine so that it is more readily extracted into the dichloromethane layer. 3. Why does the addition of salt (NaCl) to the aqueous layer sometimes help to break up an emulsion that forms in an extraction? Ans . Making the aqueous layer highly ionic by dissolving NaCl will help to break up
the emulsion. 4. The distribution coefficient for caffeine in dichloromethane and water is 4.6. Assume that your 100-mL tea solution contained 0.3 g of caffeine. If you had extracted with 51LA W07 TA Note
6
only one 15-mL portion of dichloromethane, how much caffeine would have been left in the water solution? How much caffeine would be left in the water after the second 15-mL dichloromethane extraction? How much caffeine would be left in the water solution if only one extraction with30-mL of dichloromethane was performed? Ans . For the first 15 mL extraction: k = 4.6 =
conc. in dichloromethane conc. in water
=
(0.3 -x1) g/15 mL CH2Cl2 x1 g/100 mL water
Solving for x1 shows that 0.122 g is the mount of caffeine removed by one extraction of 15 mL. Therefore, x1 is 0.178 g of caffeine left in the aqueous solution. For the 2nd 15 mL extraction: 4.6 =
(0.178 - x2) g/15 mL CH 2Cl2 x2 g/100 mL water
Now x2 = 0.105 g of caffeine that is left in the aqueous solution after 2nd extraction.
If extracted once with 30 mL portion only. 4.6 =
(0.3 - x) g/30 mL CH2Cl2 x g/100 mL water
Solving for x shows that 0.126 g of caffeine will be left in the aqueous solution.
1. When 2-propanol was used as the developing solvent, two substances moved with the solvent front (Rf = 1) during TLC analysis on a silica gel plate. Can you conclude that they are the identical? If not, what additional experiment(s) would you perform? Ans. You cannot conclude that the two substances are identical because 2-propanol is a highly polar solvent and may not separate these two compounds. Further experiments should involve changing the solvent system to a solvent or solvent mixture with a lower polarity. Examples of possible solvents to be used alone or in a mixture include hexane, dichloromethane, diethyl ether, and/or ethyl acetate. 2. The Rf value of compound A is 0.34 when developed in hexane and 0.47 when developed in dichloromethane. Compound B has an R f value of 0.42 in hexane and 0.69 in dichloromethane. Which solvent would be better for separating mixture of compounds A and B? Explain.
51LA W07 TA Note
7
Ans. Dichloromethane would be a better solvent for the separation of A and B because
there is a larger difference between the R f values (0.22) than in hexanes (0.08). A larger difference between R f values translates to better separation of the compounds.
1. Why does caffeine have a larger R f value than chlorophyll? Ans. Chlorophyll is a much larger molecule with similar polarity (compare the functional groups) than caffeine. Therefore, it is naturally expected that caffeine will have a larger R f value. A larger R f value means caffeine is not as tightly bound to the polar adsorbent as is chlorophyll. Therefore, chlorophyll must have a higher polarity than caffeine. 2. Why can there be no breaks in the thin-layer surface of a TLC plate? Ans. Discontinuities in the TLC surface would disrupt passage of compounds along the surface, invalidating TLC R f value 3. Two compounds have the same R f (0.87) under identical conditions. Does this show that they have identical structures? Explain. Ans. Just because the compounds have the same R f value does not necessarily indicate
that the compounds have identical structures. The entire range of R f values is 0.0-1.0, and millions of organic compounds are known. The solvent used may not be sufficient enough to separate these compounds. High R f value is due to the choice of solvent. Choosing a less polar solvent, giving a R f of < 0.5 may separate two spots if they are not the same compound.
EXP 5
2. A crude non-acidic product mixture dissolved in diethyl ether contains acetic acid, Describe an extraction procedure that could be used to remove the acetic acid. Ans. The best way to remove acetic acid from the ether solution is to convert it to
its conjugate base, the acetate anion, by extracting the ether solution with a 5% NaHCO 3 solution. Sodium acetate is very soluble in water but not soluble in diethyl ether.
51LA W07 TA Note
8
3. What precautions need to be observed when an aqueous sodium carbonate solution is used to extract an organic solution containing traces of acid? Ans . Aqueous sodium carbonate will react with organic solutions containing
traces of acid and form carbon dioxide gas. The addition should be done slowly to minimize any exothermic reactions. CO 2 gas should be released from the separatory funnel by opening the stopcock after each shake. 6.
K2CO3 is an excellent drying agent for some classes of organic compounds. Would it be a better choice for an acid (R-COOH) or an amine (R-NH 2)? Why? Ans. Potassium carbonate is an effective drying agent that forms 2K 2 CO 3 . 3H 2O
(potassium carbonate sesquihydrate) after absorbing water molecules. However, potassium carbonate is basic in aqueous solution and, therefore, cannot be used with an acid. It can be used for amines and neutral compounds.
1. Why doesn’t the neutral organic compound dissolve in 1.5 M sodium hydroxide solution? Ans. A majority of organic compounds are insoluble in water: only those that are
ionic or are hydrogen-bonding compounds are water soluble. If an organic compound is an acid, however, NaOH will convert it to a water soluble ionic salt. 2. Why is diethyl ether a good choice for the organic solvent in this experiment? Ans. Diethyl ether is a largely water insoluble compound of intermediate polarity,
which dissolves a wide range of organic compounds. Since ether has a low boiling o
point (34 C), it is easily separated from the organic compound by evaporation or distillation. 3. What experimental difficulty would you encounter if you had neglected to include a drying step before evaporating the ether solution of the neutral organic compound?
Ans. Without including the drying step, there will be water present in the sample of
the neutral organic compound, due to some water solubility in ether and incomplete separation of the ether and water layers in the extraction. This water is very difficult to evaporate. Thus the neutral organic compound will be impure, and it may be more difficult to dissolve it in ethanol used for its recrystallization. It will also take longer to dry the solid after recrystallization. 51LA W07 TA Note
9
4. Why are the two organic compounds recrystallized before their melting points are determined? Ans. Melting points can be very sensitive to trace impurities. Presence of even a
small quantity of impurity usually depresses the melting point a few degrees and causes melting to occur over a relatively wide temperature range. 6. Describe the characteristics of a good recrystallization solvent. Ans. A good recrystallization solvent has:
1) High solubility of the compound in hot solvent Low solubility of the compound in cold solvent 2) Similar structure and polarity to the compound to be recrystallized 3) Easily evaporated/ removed from the recrystallized product
1. Why do we add boiling chips or boiling stones before the distillation starts? Boiling chips should never be added to a hot solution! Why? Ans. Solution can become superheated, i.e., heated above its boiling point without
actually boiling. When boiling does suddenly occur, it can happen with almost explosive violence. To prevent this from happening, boiling chips or stones should be added before the heating begins. If boiling ships were added to a hot solution, the solution would boil over or explode. 2. Why would it be dangerous to heat a liquid in a distilling apparatus which had no vent or opening to the laboratory? Ans . Pressure would build up as the substance vaporized. Eventually the apparatus
would blow up . 3. What is the effect of the boiling point of a solution (for example, water) produced by a soluble nonvolatile substance (for example, sodium chloride)? What is the effect of an insoluble substance such as sand or charcoal? What is the temperature of the vapor above the two boiling solutions?
51LA W07 TA Note
10
Ans. A soluble nonvolatile substance like sodium chloride will raise the boiling point
of an aqueous solution. An insoluble substance will have no effect. The bp of a solution containing a soluble substance will be higher than the pure solution. 3) Explain why refractive index values of liquids are numerically greater than one. Ans. Refractive index is defined as n=V vac /V lig , where n is refractive index, V vac =
velocity of light in a vacuum (or air), V
= velocity of light in the liquid. The velocity of
liq
light in a liquid is always less than that of light in a vacuum . 5. A solution consisting of isobutyl bromide and isobutyl chloride is found to have a o
o
refractive index of 1.3931 at 20 C. The refractive indices at 20 C of isobutyl bromide and isobutyl chloride are 1.4368 and 1.3785, respectively. Determine the molar composition (in percent) of the mixture by assuming a linear relation between the refractive index and the molar composition of the mixture.
Ans. 1.3931 x 100 = 1.4368x + 1.3785y x + y = 100 where x and y are molar % composition of isobutyl bromide and isobutyl chloride, respectively. Solution:
1.4368x + 1.3785(100-x) 1.3931 x 100 = x=
139.31 - 137.85 1.46 = = 25.04 % 0.0583 1.4368 - 1.3785
Molar % composition of isobutyl bromide is 25.04% and isobutyl chloride is 74.96%.
3. A student carried out a simple distillation on a compound known to boil at 124 oC and reported an observed boiling point of 116-117 oC. Gas chromatographic analysis of the product showed that the compound was pure, and a calibration of the thermometer indicated that it was accurate. What procedural error did the student make in setting up the distillation apparatus?
51LA W07 TA Note
11
Ans. The thermometer may not have been in correct place. It is very likely that the
thermometer was placed lower than it should be.
1. Plot a distillation curve based on your experimental results for both simple and fractional distillation as shown in Fig. 11.14 (page 143) in 2. Determine
of the "unknown" liquid mixture using the fractional
distillation curve. 3.
from the boiling points and refractive indices of the low and high boiling components.
4. Discuss the efficiency of simple and fractional distillations for separating the mixture.
EXP 7 OPTICAL ISOMERISM
1. Construct a model that has a central carbon atoms with 4 different colored spheres attached to it, representing four different atoms or groups. Draw a solid/dashed-wedge structure of this model here and answer the following questions. a blue ball Fig. 1
a.
Does the model have a plane of symmetry?
Yes ______ No __ ___
The central carbon is said to be a stereocenter, stereogenic center or chiral carbon. Change one of the colored spheres so that two of the four spheres are the same and examine the structure at different angles. . b.
Does the molecule now posses symmetry?
51LA W07 TA Note
Yes __ ___ No ______
12
What kind? c.
___
Plane of symmetry ___
In view of your answers to a and b, what condition is necessary for a carbon to be stereogenic center? A carbon is bonded to 4 different groups. d. Draw a structural formula for each of the following and indicate all stereogenic carbon with an asterisk:
a.
Looking down on the model from above, record the colors of the balls,
proceeding in a clockwise direction.
. orange, purple, green _
Construct another model that is the mirror image of the first and place it on the table with the blue ball up. b.
In what direction must you proceed to observe the same order of substituents as recorded for 2a?
c.
counter-clockwise
clockwise _____
Try to superimpose the 2 models. Are they the same?
Yes _____ No _X__
If not, how are they different? Nonsuperimposable mirror images
d.
__________________________
What is the stereochemical term that relates these two models? Enantiomers
51LA W07 TA Note
___________
13
e. What two important properties must the two molecules (models) each have to be related by the term that is the answer to 2d? Each model consists of the same colored balls but shows different spacial arrangements.
f.
On your mirror image model, switch any two groups. Is this model still the mirror image of the original model?
g.
Yes ______ No ___
Try to superimpose the two models. Are they superimposable? Yes __ ____ No______
h.
What is the stereochemical term that relates these two models? ____ Identical compounds, superimposable mirror image
_ ________
i.
Are the models still mirror images?
Yes __ ___ No ______
j.
Do the models have a plane of symmetry?
Yes __ ___ No _
If so, describe its location.
A plane bisects the model into two halves. The
____
plane contains the carbon and two balls (none blue).
k.
Are the models superimposable?
l.
Do the models represent identical or different molecules? ___ Identical________ Explain.
m.
Yes __ ___ No _____
Two models superimpose with each other.
Each of the following molecules has one stereocenter, and is therefore chiral. Draw stereorepresentations (dash-wedge formulas) for the enantiomers of each.
Br H
Br H
or
Br H Br
Cl H
Cl H
Mirror plane
Br
or
Br H Cl
Mirror plane
2-bromobutane
1-bromo-2-chloropropane . When a molecule has two or more stereogenic
centers, stereoisomers that are not mirror images can exist; these are called . Within this general class, there are special types of stereoisomers that are always optically inactive and are called 51LA W07 TA Note
.
14
Construct a model with four different colored balls about a carbon center. Construct another identical to the first and verify this by the superimposition test. Now remove the same colored balls, blue (
from Fig 1) for example, from each model and join the two carbons at
the vacant valence sites as shown in Fig 2.
Fig 2
a.
How many stereogenic carbons does this model have? Note that each carbon has the same three groups attached to it.
b.
Two __
______
Conformationally adjust (by rotation about the central C-C bond) the model in a search for symmetry. Does the molecule have a plane of symmetry? Yes _____ No ___
_
Yes _____ No ___
_
Construct the mirror image of this molecule. c.
Are the mirror images identical?
d.
What term is used to describe the relationship between the two molecules in 3c? Enantiomers____
e.
Examine each of these models to determine whether they are chiral or achiral. Chiral________
Interchange two balls (
and
for example as shown in Fig 3) on one carbon of one model.
Fig 2
Fig 3
f.
Are the two models identical now?
Yes _____ No __ _
g.
Are they mirror images?
Yes _____ No __ _
h.
Are they stereoisomers?
Yes __ __ No _____
i.
What stereochemical term describes the two molecules that are related in this way?
j.
Diastereomers
Examine various conformations of the model where the two balls were interchanged. Can you find a conformation with a plane of symmetry?
k.
Yes __ _ No _____
Does this same conformation have a plane of symmetry? Yes __ _ No _____
51LA W07 TA Note
15
Explain.
A plane bisecting the two carbons results two halves that become
the mirror immages with each other. _________________ _
l.
Predict whether this molecule's mirror image will be identical with or nonsuperimposable on the original. Explain the basis of your prediction. Test the validity of your prediction by constructing the mirror image and conducting the superimposition test. Record your observations. The mirror image is superimposable. ___
_______
m.
Is this molecule (Fig 3) chiral?
Yes _____ No __ __
n.
Is this compound optically active?
Yes _____ No __ ___
o.
What is the stereochemical term used to describe this type of stereo isomer? Meso compound
Tartaric acid, 2,3-dihydroxybutanedioic acid, can come in three different forms, corresponding to the models assembled in 2a -o: a pair of optically active enantiomers and a third optically inactive form called the
. The latter is a diastereomer of the
optically active enantiomers. HO
* C
HOOC H p.
* C
OH Tartaric acid H
COOH
Draw solid/dashed-wedge structures for the three tartaric acid forms. Label each forms and show the relationship with each other as , and the
A
.
B
C (meso)
A & B: enantiomers; A & C and B & C: diastereomers
When a molecule has two different types of stereogenic centers, it can exist in 2 n stereoisomeric forms, where n is the number of stereogenic carbons. Remove any one ball from any one-carbon center and replace it with a color not already being used. Your model should have two stereogenic centers, with two of the three colored balls on one center being the same as on the other. Construct the four stereoisomers now possible. 51LA W07 TA Note
16
q.
Record the diagrams for these 4 stereoisomers; label enantiomers and diastereomers.
A
B
C
D
A & B and C& D: enantiomers; A & C, A & D, B & C, and B & D: diastereomers
r.
Explain why 4 stereoisomers cannot be achieved with tartaric acid. Among four stereoisomers two are meso form which are superimposable mirror images each other.
. In 3q you constructed models
R/S
with two stereocenters that represented 4 stereoisomers. 2,3-Dibromobutanoic acid exemplifies such a compound. Make models for the isomers of this compound. a. Draw solid/dashed-wedge structures for the 4 possible stereoisomers and label the enantiomers and diastereomers.
A
B
C
D
A & B and C& D: enantiomers; A & C, A & D, B & C, and B & D: diastereomers
b.
Specify the sequence priority for the substituents on carbon-2 and carbon-3. Carbon 2: Br, CH(Br)CH 3, COOH, H; Carbon 3: Br, CH(Br)COOH, CH 3, H
c.
Label each structure with the proper R/S notation in the drawings above, for example (2R ,3S ).
A
d.
B
C
Draw Newman projections for all 4 possible stereoisomers.
51LA W07 TA Note
D
17
A
e.
B
C
D
Specify the sequence of priority for the substituents on C2 and C3 for tartaric acid (see 3p). Write your answer for 3p. Carbon 2: OH, COOH, CH(OH)COOH, H; Carbon 3: OH, COOH, CH(OH)COOH, H
a.
Construct a model of methanol (CH 3OH). Is methanol chiral? Why or why not? _No. It’s achiral because the carbon carries 3 Hs and, therefore, it doesn’t meet the requirement for a stereocenter.
b. Orient your model so that the plane of symmetry (indicated by the dotted rectangle) contains the H-C-O-H bonds that are in the plane of the paper, as shown below:
Now rotate the C-O bond, so that the plane no longer contains the O-H bond. All of the bonds in methanol freely rotate at room temperature, so you are just showing a different conformation of methanol. Does this conformation of methanol have a plane of symmetry? Yes _____ No _____
c.
Even though one conformation of methanol has a plane of symmetry, and one doesn’t, methanol is still considered to be achiral. It is only necessary to find one conformation with a plane of symmetry for a molecule to be achiral.
d.
Now consider cyclohexanol, shown below. Looking at the flat, hexagonal representation of cyclohexanol, does cyclohexanol contain a plane of symmetry? Yes _
__ No _____
51LA W07 TA Note
18
e. Of course, cyclohexanol never exists in a flat hexagonal form. It is just drawn this way for convenience. It exists in one of two chair conformations which are rapidly interconverting at room temperature, the conformation with the hydroxyl in the equatorial position being favored.
Make a model of cyclohexanol with the OH in the equatorial position. Now do a chair flip, so that the OH is in the axial position. Can you find a plane of symmetry for each configuration? Is cyclohexanol chiral? _______
f.
_____________________________________________________
Now consider a disubstituted cyclohexane derivative, cis-1,2-dibromocyclohexane:
Looking at the flat, hexagonal representation of cis -1,2-dibromocyclohexane, does this molecule contain a plane of symmetry?
g.
Yes _
__ No _____
Once again, we know that cyclohexane rings never exist in a flat hexagonal form. They exist as rapidly interconverting chair conformations. Only at absolute zero can you isolate a single chair conformation of any cyclohexane derivative. If you convert cis-1,2-dibromocyclohexane to a chair conformation, however, both chair conformations are chiral:
51LA W07 TA Note
19
Both chair conformations are non-superimposable mirror images of eachother and there is no plane of symmetry,
, they are interconvertible by a ring flip!
Make a model of each of the chair conformations of cis -1,2-dibromocyclohexane, and show that they are nonsuperimpoasble mirror images of each other. Is cis -1,2-dibromocyclohexane isomer optically active? The planar structure has a plane of symmetry, but the chair conformation doesn’t! Yes _____ No __
_
Although each conformer is chiral, the rapid interconversion between the two at room temperature means they cannot be separated. The compound is not chiral. Identifying chirality in cyclohexane derivatives can seem daunting. Fortunately we can greatly simplify by making using the following rule:
If you can find a plane of symmetry in the flat hexagonal drawing of a substituted cyclohexane ring, the molecule is achiral. If you cannot find a plane of symmetry, it is chiral.
51LA W07 TA Note
20
h. Using this simplification, determine whether the following substituted cyclohexane rings are chiral or achiral. Draw plane of symmetry for any achiral molecules.
Cl
HO
Chiral
Achiral
Achiral
OH
Achiral
1. Write a mechanism to illustrate what will happen if the addition of Br 2 to trans cinnamic acid is syn ; What do you expect the melting point of the syn addition product to be? Ans. d+ C6H5 C
dBr
Br
H
C
H
COOH
C6H5 H
C
H COOH 2S,3S-
and C6H5
H C
H d+
C
C
C6H5 H
H
C
COOH
C
d- COOH Br
a racemic mixture mp 93.5oC
Br 2R,3R-
syn -Addition gives a racemic mixture of (2R,3R) and (2S,3S)-2,3-dibromo-3o
phenylpropanoic acid (mp 93.5-94.5 C) 2. Write a mechanism to illustrate what will happen if the addition of Br 2 to transcinnamic acid is anti ; What do you expect the melting point of the anti addition product to be? Ans.
51LA W07 TA Note
21 C6H5
H C C
H
d+
d-
+
C6H5 C
Br
H +
C
H
COOH
Br
C 6H5
COOH
C
-
H C
H
COOH
+ Br a racemic mixture of bridged bromonuim ion intermediates is attacked by Br - and ring opening occurs
C6H5 H C
Br C
Br
H COOH
C
C 6H5
C
H COOH
H 2R,3S-
2S,3Ra racemic mixture mp 203oC
anti -Addition gives a racemic mixture of (2S,3R) and (2R,3S)-2,3-dibromo-3o
phenylpropanoic acid (mp 202-203 C) 3. Describe the melting point you expect to observe if the isolated product is the mixture of both syn and anti addition of Br 2. Ans. If the addition of bromine proceeds via both syn and anti addition, a mixture of
2,3-dibromo-3-phenylpropanoic acid will be formed and, therefore, the melting point o
will be lower than the 93 C expected from the pure syn-adduct. 4. Explain why the addition of Br 2 to trans -cinnamic acid takes place more slowly than the addition of Br 2 to a "normal" alkene. Ans. The carboxyl group of
trans-cinnamic acid is an electron-withdrawing group that slows the addition of bromine to the alkene.
1. A sample of 2-butanol shows an optical rotation of +3.25 o. Determine the % ee and the molecular composition of this sample. The optical rotation of pure (+)-2-butanol is +13.0o. Ans.
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+3.25o
x 100 %
= 25.0 % ee detrorotatory or (+)
+13.0o Molecular composition: 100 + 25 2
or
100-25 = 2
= 62.5 % dextrotoratory or (+) 37.5 % levorotatory or (-)
37.5 % of (+) and (-) to make the racemic mixture that gives 0o optical activity reading.
Therefore, (+) enantiomer of 2-butanol is 37.5 + 25.0 = 62.5 % is (+) enatiomer and 37.5 % is (-) enantiomer.
2. A sample of 2-butanol (See question 1) shows an optical rotation of –9.75 o. Determine the % ee and the molecular composition of this sample. %ee: -9.75o -13.0
x 100 %
o
= 75.0 % ee levorotatory or (-)
Molecular composition: 100+ 75 2
= 87.5 % levorotatory or (-) 12.5 %dextrotoratory or (+)
Ans.
3. An optical rotation study gives a result of a = +140 o. Suggest a dilution experiment to test whether the result is indeed +140 o and not –220 o. Ans. If the concentration of the sample solution were halved, the rotation would o o o o either be +70 or –290 if the =140 reading is originally correct. If –220 were correct o o reading, the new reading would be –110 or +250 .
Pre-Lab Question
Question 1, p 46 in “Techniques..” 1. !"#$"#li%onene has a &oilin' point o( 176 oC. )hat %a*e it possi&le to stea% *istill it at 100oC+ Ans.
Because the limonene is not miscible with water, the vapor pressures of limonene and water add together. When the combined pressure equals the
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atmospheric pressure, distillation occurs. The distillate contains limonene and water in proportion to their vapor pressures at the boiling point
51LA W07 TA Note
