UNIVERSITY OF B AHRAIN
COLLEGE OF SCIENCE
DEPARTMENT OF CHEMISTRY
LABORATORY MANUAL
ORGANIC CHEMISTRY I
CHEMY 221
Taxol, anticancer drug.
2015
Edited by : Dr. Awatef Mahdi
1
Table of Content
Experiment 1
Melting Point Determination
Experiment 2
Purification by Recrystallization
Experiment 3
Separation of an Unknown Mixture by Acid/Base Extraction
Experiment 4
Distillation: Separation and Purification of Organic Liquids
Experiment 5
Thin Layer Chromatography (TLC).
Experiment 6
Elimination: Synthesis of Alkenes from Alcohols
Experiment 7
Bromination of alkene: Preparation of stilbene dibromide.
Experiment 8
Conformations and Configurations using Molecular Models
Tutorial on use of ChemDraw.
Experiment 9
Optical activity of chiral molecules
Experiment 10
Nucleophilic Substitution: Relative Reactivity of Organic Halides.
2
INTRODUCTION
Laboratory work is an integral and essential part of any chemistry course. Chemistry is an
experimental science – the compounds and reactions that are met in lecture and classroom
work have been discovered by experimental observations. Organic compounds exist as
gases, liquids, or solids with characteristic odors and physical properties.
They are
synthesized, distilled, crystallized and chromatographed, and then transformed by reactions
into other compounds. The purpose of laboratory work is to provide and opportunity to
observe the reality of compounds and reactions and to learn something of the operation and
techniques that are used in experimental organic chemistry and in other areas in which
organic compounds are encountered.
LABORATORY SAFETY
Most organic compounds are flammable, and they are toxic or irritating to a greater or lesser
degree.
Many organic reactions are potentially violent.
Laboratory work in organic
chemistry is not a dangerous occupation, provided that some simple precautions and safety
rules are followed. There are some potential hazards that must be recognized and avoided:
accidents can and do occur when these hazards are ignored.
RULES FOR PERSONAL SAFETY
1- Eye Protection. Approved eye protection must be worn at all times in the laboratory,
regardless of what is being done. In many locations, chemical safety goggles are
required by law. Safety glasses (either prescription or plain) with side shield to
protect from splashes also offer good eye protection.
2- Never work in a laboratory without another person being present or within calling
distance. Minor accidents can become disasters if help is not available.
3- Do not carry out any reaction that is not specifically authorized by the instructor.
4- Never taste a compound ; never pipet a chemical by mouth; do not eat, drink, or
smoke in the laboratory.
5- Avoid contact of the skin with any chemical. If a substance is spilled on your hands,
wash them thoroughly with soap and water. Do not rinse them with a solvent, since
this may cause more rapid absorption.
3
6- Long hair should be tied back. Shoes must be worn to prevent injury from spilled
chemicals or bits of glass.
Avoid loose-fitting sleeves and clothing that leave
expanses of skin unprotected.
7- Never heat a flask or any apparatus that is sealed or stoppered (i.e., a closed system)
make certain that there is an opening to the atmosphere.
8- Some experiments require the use of a well-ventilated hood. These experiments
should not be attempted in the open laboratory.
4
Typical glassware found in Organic Chemistry Laboratory
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Hardware and other nonglass items
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Experiment 1
Melting Point Determination.
1.1
Introduction
The melting-point of a pure organic compound is a highly reproducible physical
constant characteristic of the substance.
However, in order to establish the
identity of an unknown, it is not sufficient to show that it melts at the published
temperature from the suspected structure – many hundreds of organic compounds
will share the same melting-points. The chemist makes use of the principle that a
mixture of two different substances with fairly close melting-points usually melts
below the value of either singly – a “depression of m.p.”
If a sample of A is available, you can quickly deduce whether is identical to X by
determining the mixed melting point. A mixture of X and A should have the
same melting point as either substance alone, provided the two substances are
identical. If X and A are not the same substance (even though they separately
have the same melting point), then a mixture of the two will usually have a lower
melting point and a broader melting-point range than either substance alone.
This phenomenon occurs because each substance acts as an impurity in the other.
Miscible or partially miscible impurities, even when present in small amounts,
usually lower the melting point and broaden its range.
Conversely, a wide
melting-point range usually indicated that a substance is impure.
In this experiment you will determine the melting points of two different pure
solids that have approximately the same melting-point range. You will then
prepare a mixture of the two substances and determine its melting-point range.
Finally, you will obtain a “unknown” from your instructor. After you determine
its melting point, you will identify it by a mixed melting point.
1.2.1
Determination of the melting point of a substance.
Place a microspatula-full of urea on a small piece of pours tile and press out to a
fine powder. This also serves to remove any traces of solvent or moisture. Take
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a clean m.p. capillary, seal one end in a hot flame, and press the open end into the
powder until about 1 mm has been collected. Tap this gently down to the closed
end and fix the capillary to its position, raise the temperature fairly rapidly the
first time (20º a minute) to get a rough melting-point. Then repeat once with a
fresh capillary to obtain a more accurate value, slowing the rate of heating to 2º/
min as the m.p. is approched.
Record the melting-point range of urea on the report sheet. Then, in a similar
way determine and record the melting point of a sample of cinnamic acid.
1.2.2.
The effect of impurities on the melting point if a pure substance.
Thoroughly mix equal amounts of urea and cinnamic acid (50mg of each
compound) with a microspatula on a porous tile. Determine the melting range of
the 50-50 mixture. Repeat with 75-25 and 25-75.
Using the midpoints of the melting ranges, plot the data on the graph on the
report sheet.
1.2.3.
Identification of an unknown by mixed m.p.
Obtain an unknown sample (which will be one of the compounds listed in Table
1.1) and record its number on your report sheet.
Introduce a few mm of your unknown into a capillary tube and determine the
melting point. Compare your m.p. with the data in Table 1.1. and make a
preliminary identification of your unknown. Confirm its identity by the mixed
m.p. technique: mix a bout 50mg of the unknown with an equal weight of the
suspected known compound taken from the side-bench and determine the mixed
m.p.
If the mixed m.p. is about the same as the melting points of your unknown and
the known compound you took from the side-bench then your unknown must be
the same compound. If , however, your mixed m.p. is lower and wider than the
melting points of your unknown and the known compound then your unknown
must be a different compound. In this case, take the next most likely known
compound from the side bench and repeat the mixed m.p. with your unknown to
determine its identity. Record your results and conclusions on the report sheet.
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Table 1.1 Melting Points of some organic compound.
Compound
M.P.ºC
1.
2.
3.
4.
5.
6.
70-71
97-99
113-114
121-122
132-133
156-158
Biphenyl
Glutaric acid
Acetanilide
Benzoic Acid
Urea
Salicylic acid
If the acetanilide was recrystallized from water and not thoroughly dried, it may melt at
83-84ºC.
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Experiment 1 data sheet
Melting-Point Determination:
Identity and Purity of
Organic Compounds
DETERMINATION1.
OF MELTING
POINTS
Record the observed melting points in the table.
Melting Points. ºC
Finish
Mid-point
Compound
Start
Urea
---------------
----------------- -----------------
Cinnamic acid
---------------
----------------- -----------------
50-50 mixture
---------------
----------------- -----------------
25-75 mixture
---------------
----------------- -----------------
---------------
----------------- -----------------
urea-cinnamic acid
75-25 mixture
urea-cinnamic acid
2. Graph the midpoints of the melting-point ranges of urea,
cinnamic acid, and their mixtures.
140
130
Temperature ºC
120
110
100
100%
urea
50-50
Percentage composition
10
100%
cinnamic acid
Answer the following questions:
1-
Which will give index of purity? A narrow M.P. range or a wide M.P. range?
2-
What is the effect of increasing molecular weight on M.P.? give example.
3-
The following table lists the observed melting points for several compounds isolated
in the student laboratory.
Compound
Observed melting point (°C)
naphthalene
benzophenone
p-anisic acid
salicylic acid acetate
3-chlorobenzoic acid
sulfanilamide
ferrocene
79° - 80°
45° - 47°
178° - 182°
135°
157° - 158°
165° - 166°
157.5° - 161.5°
(a) Look up the melting point for each compound in a suitable resource.
(b) Which of the compounds are pure? Which are impure?
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Experiment 2
Purification by Recrystallization
2.1
Introduction:
When a solid organic compound is prepared in the laboratory or isolated from
a natural source, it is almost always impure. A simple technique for the
purification of such a solid compound is crystallization.
Crystallization is the slow formation of a crystalline solid, as opposed to
precipitations, which is the rapid formation of an amorphous solid. If a hot,
saturated solution is cooled too quickly, the compound may precipitate instead
of crystallizing. A precipitated solute may contain many impurities trapped in
the rapidly formed amorphous mass on the other hand, when a solution is
allowed to crystallize slowly, impurities tend to be excluded from the growing
crystal structure.
2.1.1.
Solvent Selection
The most crucial aspect of the recrystallization process is selecting the
appropriate solvent. The best solvent for recrystallization is one in which the
material is insoluble at room temperature but completely soluble at elevated
temperature.
In considering what solvent to use, the rule “like dissolves like” should
always be kept in mind. A solvent in which the substance is very soluble will
be a poor one for recrystallization.
Likewise, a solvent in which the
compound is almost totally insoluble, even at elevated temperatures, will also
be a poor solvent. The best compromise is usually a solvent in which a
compound is relatively insoluble at low temperatures but soluble at high
temperatures.
The solvent chosen for crystallization should also have the following criteria:
 The solvent should be readily volatilized after collection of the purified
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crystals.
 The solvent should be unreactive towards the sample.
2.1.2.
Mixed-Solvent Systems
It sometimes happens that no single solvent will meet all the requirements of a
good crystallization solvent. In such cases, one must resort to so-called mixed
solvent systems.
The choice of a mixed solvent system is usually predicted from the mutual
solubility of two (or more) solvents in each other and the high affinity of the
compound to be purified for one of the two solvents. It is also necessary that
the other solvent in the pair have a low affinity for the compound to be
purified. Other requirements to be kept in mind are:(1) the boiling points of the two solvents.
(2) the two solvents should be completely miscible so that no new phase
appears on mixing.
Some solvent pairs which are often used for recrystallization of organic
compounds are listed below.
Methanol – water
Ethanol - water
Acetone - petroleum ether (or hexane)
Ethyl acetate - petroleum ether (or hexane or cyclohexane)
Ether - petroleum ether (or pentane)
Dichloromethane - petroleum ether (or pentane)
Toluene - petroleum ether (or heptane)
Acetone - water.
In this experiment three solid organic compounds are tested for their hot and
cold solubilities in five different solvents and a good recrystallization solvent
is then chosen for one of the compounds.
Compounds
Benzoic acid
Cinnamic acid
urea
Solvents
Hexane
Toluene
Acetone
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Water
2.1.3
Steps of Crystallization
The general procedure for recrystallization involves the following steps:
1. Selection of a suitable solvent through experiment or from data on
solubility
2. Dissolution of the material in the hot solvent (near the boiling point).
3. Filtration of the hot solution to remove insoluble impurities or
impurities adsorbed on activated carbon. (This step is sometimes
omitted – see below.)
4. Crystallization of the solute from the cool solution.
5. Collection of the purified crystals.
6. Washing and drying the product.
Figure 2.1 Steps in crystallization.
2.2
Procedure
2.2.1
Choosing a solvent
Label each of the five test tubes with the initial letter of the solvent to be used
in it (H , T, E, A & W). Place a small spatula-full of benzoic acid in each of
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the test tubes. Just cover the sample in each tube with the corresponding
solvent. Shake each tube and observe the solubility at room temperature (If
the sample dissolves completely in a particular solvent, that solvent is useless
for recrystallization). Enter the results of the cold solubility test in the table
given, using the terms: soluble, partially soluble or insoluble, as appropriate.
Heat the tubes containing undissolved solid in a hot water bath (CAUTION:
extinguish any flames). Be careful not to evaporate the more volatile solvents
(acetone, hexane and ethanol). If necessary, add a little more solvent and reheat. Report the hot solubility in the table. If any sample is completely
soluble in the hot solvents, let the tube stand at room temperature to see if any
solid crystallizes out. (Only solvents which give a good recovery of the solute
on cooling are useful for recrystallization).Repeat the same procedure with the
other solutes (cinnamic acid and urea).
2.2.2.
Recrystallization
Place 1g of the Bonzoic acid in an Erlenmeyer flask. Add approximately
50ml of water to the flask and warm the mixture on a hot plate. Add more
solvent, a little at a time, if necessary, until the solid has just dissolved in the
boiling solvent. Remove the clear solution from the hot plate, cover the
mouth of the flask with watchglass to prevent entry of dust and allow the
solution to cool gradually, at room temperature, until crystals deposit. The
slower the cooling, the larger are the crystals obtained. Then cool the solution
in an ice/water bath until no more crystals are formed when crystallization is
complete , filter off the crystals using a Buchner funnel (or Hirsch funnel) and
filter flask as shown in Figure 2.2.
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Figure 2.2 A vacuum filtration apparatus
When the crystals have been collected on the filter paper, dry them as much as
possible by pressing them down with a spatula and drawing air through them.
If necessary, the crystals may be dried more thoroughly by pressing them
between two pieces of filter paper or by placing them on a watch glass in a
drying oven. When your sample is dry, put it in a labeled sample tube which
has been weighed empty first. Reweigh the tube plus sample and hence
obtain the mass of the sample by difference.
Record the mass of the
recrystallized compound and its melting point on your report sheet. Finally,
calculate the percentage recovery. In recrystallization, the maximum amount
of material that can be recovered is the same as the weight of the crude
sample. However, considerably less is usually obtained as pure crystals.
Percentage recovery - = Weight of recrystallized product X 100%
Weight of crude product
Take the melting point of the crude and pure benzoic acid.
Show your results and calculation on the report sheet and hand it to your
instructor together with your sample.
Safety Notes Summary:
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 Toxic solvents should be heated only in a hood.
 A Bunsen burner should be used only for aqueous solutions – and only
when no flammable solvents are being used in the vicinity.
 Some hot-plate surfaces are capable of igniting flammable solvents.
 The safest source for heating a solvent, especially a low-boiling
solvent, is a steam bath.
Solute
Solvent
Condition
Hexane
Toluene
Ethanol
Acetone
Water
Cold
Benzoic
acid
Hot
Recovery
Cold
Cinnamic
acid
Hot
Recovery
Cold
Urea
Hot
Recovery
Cold
Unknown
Hot
Recovery
Conclusion : The unknown is
% recovery of Benzoic acid =
mp of crude benzoic acid =
mp of pure benzoic acid =
Questions:1-
Compounds, A and B are equally soluble in the three solvents listed. In each
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case, which solvent would you choose? Give reasons for your answer. (More
than one answer may be correct.)
(a) Compound A: benzene, acetone, or chloroform
(b) Compound B: methanol, ethanol, or water.
2-
A student was recrystallizing a compound. As the hot solution cooled to
room temperature, no crystals appeared. The flask was then placed in an icewater bath. Suddenly a large amount of solid material appeared in the flask.
The student isolated a good yield of product however, the product melting
point was lower than expected. Explain.
3-
A chemist crystallizes 17.5g of a solid and isolates 10.2g as the first crop and
3.2g as the second crop.
(a) What is the percent recovery in the first crop?
(b) What is the total percent recovery?
4-
The solubility of acetanilide in hotwater (5.5 g/100 mL at 100°C ) is not very
great, and its solubility in cold water (0.53g /100 mL at 0°C ) is significant.
What would be the maximum theoretical percent recovery (first crop only)
from the crystallization of 5.0g of acetanilide from 100 mL of water
(assuming the solution is chilled to 0°)?
5-
During a crystallization, while heating a solution of a compound to dissolve it
in hot solvent, you boil it so long that a substantial amount of the solvent
evaporates. What is likely to happen to some of the solute? What should you
do if this occurs?`
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Experiment 3
Separation of an Unknown Mixture by Acid/Base Extraction
3.1
Experimental Aims: The objective of this exercise is to separate a two-component
mixture using extraction techniques and then to identify the isolated components by
determining their melting points.
Each student will be given a mixture of two substances, which belong to two of the three
categories listed below.
Possible carboxylic acids benzoic acid
3.2
2-chlorobenzoic acid
Possible phenols
4-tert-butylphenol
2-naphthol
Possible neutrals
1,4-dimethoxybenzene
fluorene
Introduction:
Extraction is a particularly useful means of separating organic compounds if one compound
in the mixture can be chemically converted to an ionic form. The ionic form is soluble in an
aqueous layer and can be extracted into it. Other, non-ionic organic compounds in the
mixture will remain dissolved in the organic solvent layer. Separation of the two layers
results in the separation of the two compounds.
The extent to which an acid-base reaction proceeds to completion depends upon the relative
acidity of the reactants and products. Reactions occur so that stronger acids and bases are
converted into weaker conjugate base and conjugate acids, respectively. The pKa value of
the acids gives a measure of the acidity of each compound. Stronger acids have smaller pKa
values and their conjugate bases are weaker. The position of an acid-base equilibrium can
then be predicted from knowledge of the pKa values of the acids involved.
Take a look at the following acid-base reactions in Figure 1, paying particular attention to
the position of the equilibrium and its relationship to the pKa values given.
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Figure 1:
The reactions of a carboxylic acid and a phenol with bicarbonate ion. Note that the
carboxylic acid has a lower pKa than the conjugate acid of bicarbonate ion (carbonic acid).
The reaction, therefore, proceeds to products. The reaction of a phenol, however, favors the
reactants since the pKa of phenol (10) is larger than that of the carbonic acid (6.4). Acidbase reactions favor the side with the weaker acid (that is, they favor the side with the larger
pKa). So, extracting a mixture of these two compounds with bicarbonate results in the
ionization and extraction of a carboxylic acid in the presence of phenol thus separating the
two compounds from one another.
Now, look at the reaction in Figure 2 where we use a stronger base to do the reaction:
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Figure 2:
The reactions of a carboxylic acid and a phenol with hydroxide ion. Note that in both cases,
the reactions favor the formation of products. Therefore, extracting with hydroxide ion
would result in the ionization and extraction of both compounds at the same time. A close
look at these two figures indicates that separating a mixture of a carboxylic acid and a
phenol would best be done using bicarbonate ion since only the carboxylic acid is converted
into its conjugate base by bicarbonate. The conjugate base of the carboxylic acid, being an
ionic species, is soluble in the aqueous layer while the phenol (left unionized) would remain
dissolved in the organic layer. However, if we were to extract with hydroxide ion, both the
carboxylic acid and the phenol would be converted into their conjugate bases (see figure2).
The conjugate bases, again are both ionic species and therefore soluble in the aqueous layer.
This means that both compounds would be extracted at the same time, resulting in no
separation.
A neutral compound will not react with either bicarbonate ion or hydroxide ion since a
neutral compound does not have protons acidic enough to be removed by these bases.
Therefore, such a compound will remain dissolved in the organic layer, no matter which
base is added. For example, a mixture of neutral compound and a carboxylic acid can be
separated using bicarbonate ion since only carboxylic acid will be ionized by the
bicarbonate ion. Once extracted, the carboxylic acid and phenol can both be recovered by
adding HCl to the aqueous solutions. The carboxylate ion and phenoxide will both be
protonated by HCl, resulting in the formation of the original carboxylic acid and phenol,
21
neither of which is soluble in water so they precipitate from solution. The solid can then be
isolated by filtration. Figure 3 shows this chemistry for you.
Figure 3:
The reactions of a carboxylate ion and a phenoxide ion with HCl. Since HCl is stronger acid
than either of the conjugate acids, the products are favored in both cases. The products, a
carboxylic acid and a phenol, are insoluble in aqueous solutions and precipitate from
solution. The resulting solids can be isolated and their melting points determined.
The procedure you will use in this exercise exploits the difference in acidity and solubility
just described.
1. you will dissolve your unknown in ethyl acetate (an organic solvent). All of the
possible compounds are soluble in ethyl acetate.
2. you will extract with sodium bicarbonate to remove any carboxylic acid that is
present.
3. you will extract with sodium hydroxide to remove any phenol that is present.
4. you will acidify both of the resulting aqueous solutions to cause any compounds that
were extracted to precipitate.
These solids will be isolated by vacuum filtration, dried, and then their melting point ranges
determined to identify them. If a neutral compound was present in your unknown, it will
remain in the organic layer throughout the extraction procedure. To isolate it, you will
simply evaporate the ethyl acetate to leave a solid. The melting point ranges of all solids
will be determined. You will also weigh each solid you obtain to determine the percent
22
recovery of your procedure. Remember, though, that you only have two compounds in your
unknown mixture so that you should not isolate solids from all of the extracts.
3.3
Extraction procedure:
General Notes: To measure the small volumes called for in this procedure, it is convenient
to measure them in a graduated measuring cylinder. Make sure you label everything so
that you can identify which layer you are putting into each flask correctly - label one
125 mL Erlenmeyer flask "bicarbonate", a second one as "hydroxide", and a 50 mL
Erlenmeyer flask "ethyl acetate". We are using ethyl acetate in this lab, so avoid
excessive exposure. Be sure you are familiar with the procedure below before starting the
lab.
1. Collect an unknown and record the unknown number. Without this number, we
cannot grade your report. Label three Erlenmeyer flask as directed above.
2. Dissolve approximately 1.0 g of your unknown mixture in 10 mL of ethyl acetate.
3. Pour the solution into a clean separatory funnel and add 10 mL of 10% aqueous
sodium bicarbonate found on your bench.
4. Stopper the funnel and invert it. Slowly open the stopcock to release any built up
pressure, then close the stopcock (Figure 4).
5. Gently shake the separatory funnel to allow intimate mixing of the solutions and
effect extraction of the compound from the organic mixture. (Caution: When
shaken, the mixture may develop pressure; be sure to vent it periodically).
6. Clamp the separatory funnel to a retort stand and allow the mixture to separate into
two layers (Figure 5).
7. Remove the stopper and collect the aqueous layer (the lower layer) in the 125 mL
Erlenmeyer flask labeled "bicarbonate".
8. Repeat steps 3-7 two more times draining each portion successively into the same
flask. At the end of this sequence you will have extracted the organic solution with
three 10 mL portions of 10% aqueous sodium bicarbonate.
9. Put the Erlenmeyer flask labeled "bicarbonate" aside in a safe place. Later you will
isolate any compound that was extracted by the bicarbonate. Do you remember
which functional group that would be?
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10. Add 10 mL of 5% aqueous NaOH to the separatory funnel with the remaining ethyl
acetate.
11. Stopper the funnel and invert it. Slowly open the stopcock to release any built up
pressure, then close the stopcock.
12. Gently shake the separatory funnel to allow intimate mixing of the solutions and
effect extraction of the compound from the organic mixture.
13. Clamp the separatory funnel to a retort stand and allow the mixture to separate into
two layers.
14. Remove the stopper and collect the aqueous layer in the 125 mL Erlenmeyer flask
labeled "hydroxide".
15. Repeat steps 10-14 two more times draining each portion successively into the same
flask. At the end of this sequence you will have extracted the organic solution with
three 10 mL portions of 5 % aqueous sodium hydroxide.
16. Put the Erlenmeyer flask labeled "hydroxide" aside in a safe place. Later, you will
isolate any compound that was extracted by the hydroxide. Do you remember which
functional group that would be?
The remaining steps described in this section will allow you to isolate any compound
remained in the ethyl acetate layer. Recall, this would be a neutral compound, if you
have one.
1. Add 5 mL of saturated aqueous NaCl and 5 mL of distilled H2O to the ethyl acetate
layer in the separatory funnel.
24
2. Separate and set aside the lower, aqueous layer.
3. Pour the organic layer in the 50 mL Erlenmeyer flask and dry with anhydrous
Na2SO4.
4. Filter the dried organic solution into a dry pre-weighed 50 mL round bottom flask
and remove the ethyl acetate on a rotary evaporator. If a solid remains after
evaporation of the ethyl acetate, it is a neutral substances and you will determine its
weight and melting point.
Instructions follow for isolating the carboxylic acid and / or phenol from aqueous
layers you put into the Erlenmeyer flasks labeled "bicarbonate" and "hydroxide",
respectively.
1. Take the Erlenmeyer flask labeled "bicarbonate" and carefully acidify the aqueous
solution by the dropwise addition of 6M HCl. (CAUTION: The bicarbonate solution
will vigorously liberate carbon dioxide when neutralized with HCl - that is, it will
bubble a lot). Check to make sure the solution is acidic with blue litmus paper.
2. If a solid precipitates, add a boiling stone and then gently heat the solution to bring
most of the solid back into solution. Cool slowly to room temperature and then use
an ice/water bath to complete the precipitation. If no solid precipitates, your
unknown did not contain a carboxylic acid. In that case, skip steps 3-4.
3. When the solution is ice cold, isolate the solid precipitate by suction filtration.
4. Filter, rinse the solid with ice-cold water, and determine the weight and melting point
range of the carboxylic acid.
Now, we will follow the same procedure to isolate the phenol from the Erlenmeyer flask
labeled "hydroxide".
1. Take the Erlenmeyer flask labeled "hydroxide" and carefully acidify the aqueous
solution in the centrifuge tube by the dropwise addition of 6M HCl. (CAUTION:
The hydroxide solution will become hot when neutralized with HCl). Check to make
sure the solution is acidic with blue litmus paper.
2. If a solid precipitates, add a boiling stone and then gently heat the solution to bring
25
most of the solid back into solution. Cool slowly to room temperature and then use
an ice/water bath to complete the precipitation. If no solid precipitates, your unknown did
not contain a phenol. In that case, skip steps 3-4.
1. When the solution is ice cold, isolate the solid precipitate by suction filtration.
2. Filter, rinse the solid with ice cold water, and determine the weight and melting point
range of the phenol next week.
Safety Notes
You must wear eye protection at all times. In the event that any reagent used in this
investigation comes in contact with your skin or eyes, wash the affected area immediately
with lots of water. Notify your instructor. Avoid excessive exposure to all organic solvents.
Acids and bases can cause severe burns. No flames should be present in the laboratory
during this experiment.
Acid/Base Extraction Flow chart :
26
3.4
REFERENCES
L. M. Harwood and C. 1. Moody, Experimental Organic Chemistry- Principles and
Practice, Blackwell Scientific Publications.
C.A. MacKenzie, Experimental Organic Chemistry, Prentice-Hall. 4th Edition
J. A. Moore and D. L. Dalrymple, Experimental Methods in Organic Chemistry, Saunders
Golden Sunburst Series, W. B. Saunders Company.
C. F. Wilcox and M. F. Wilcox, Experimental Organic Chemistry- A Small-scale Approach,
Prentice-Hall. 2nd Edition.
O. R. Rodig, C. E. Bell Jr. and A. K. Clark, Organic chemistry Laboratory- Standard and
Microscale Experiments, Saunders College Publishing.
J.R. Mohrig, C.N. Hammond and P.F. Schatz, Techniques in Organic Chemistry, Freeman
Publishers, 2nd Edition.
27
Experiment 4
Distillation: Separation and Purification of Organic Liquids
Distillation is a technique used to separate and purify liquids. It consists of heating a liquid
to its boiling point, conducting the vapors to a cooling device where they are allowed to
condense, and collecting the condensate
General Principles:
In a sealed container partially filled with a liquid, some molecules escape from the liquid’s
surface into the space above. In their random motion, molecules that have escaped to the
vapor may strike the liquid surface again and stick to it. At equilibrium, the number of
molecules that leave the liquid surface equals the number of vaporized molecules that strike
the liquid surface and stick. The molecules in the vapor also strike the walls of the container
and exert a pressure, which is called the vapor pressure of the liquid. If the temperature of
the liquid is raised, more molecules escape to the vapor until equilibrium is once again
established. The vapor pressure of a liquid, therefore, increases with increasing temperature
The Boiling Point: is that temperature at which the vapor pressure of the liquid becomes equal
to the pressure exerted by its surroundings. If the liquid is open to the atmosphere, the boiling
point is the temperature at which the vapor pressure of the liquid becomes equal to the
atmospheric pressure. The vapor pressure of a pure liquid rises steadily as the temperature is
increased until the boiling point is reached.
A thermometer placed in the vapor of a boiling pure liquid registers that liquid's boiling
point. The temperature remains constant throughout the distillation of a pure liquid. This is
because at the boiling point, vapor and liquid are in equilibrium, if the phase composition
of the vapor and liquid remains constant throughout the process, the temperature also
remains constant. The boiling point (at a given pressure) is a characteristic property of a
pure liquid, just as the melting Point is a characteristic property of a pure crystalline solid.
A Mixture of Ideal Liquids: When a mixture of two miscible liquids with different boiling
points is heated, the vapor does not have the same composition as the liquid. Instead, the
vapor is richer in the morning volatile component
A fractionating column: is a device used to increase the efficiency of distillations It
consists of a vertical tube that either is packed with inert material (such as glass beads or
glass helices) or has some, other device (such as indentations) to increase the surface on
which the rising vapors can condense. As the hot vapors rise through the column, they
condense, and the liquid flows back down the column. As this liquid reaches the lower,
hotter portions of the column, it is revalorized, and the more volatile components Once
again proceed up the column. If the column is efficient, this process is repeated many
28
times inside it and the distillate consists of the lowest-boiling component of the mixture in
nearly pure form.
Objectives of this Experiment:
In this experiment you will first distill a pure liquid and observe that it has a constant
boiling point. Then you will distill a two-component mixture twice, first using a simple
distillation apparatus (Figure 1.) and then using a fractionating column (Figure 2.). You
will be able to compare the efficiency of these two types of apparatus by evaluating the
separation of the mixtures.
Distillation of Hexane:
Procedure: Read the CAUTION below. You will first distill a pure liquid in a simple
distillation apparatus. Assemble the apparatus (distilling flask, thermometer, condenser,
adapter, and graduated receiver) as illustrated in Figure 1., using a 50-mL distilling or
round-bottomed flask. Be sure that the thermometer bulb is positioned just below the side
arm so that it can measure the temperature of the vapor as that vapor passes out of the flask
into the condenser and receiver.
Caution:
Hex ane and tolue ne are fla mmable . Do not use a Bunsen burner as
heat sourc e. Heat i ng mantl es attac hed to rheo stats a r e Preferred. Be
sure that y our apparatus is put to gether snug ly in all parts of this
ex periment , and be sure t hat or gani c vapors and liqui d d o not co me
near the heat sour ce
Place 15 mL of hexane in the flask, add a boiling stone (to prevent "bumping" due to
superheating), put the thermometer in place, and start water circulating through the
condenser. Have the instructor check your apparatus at this point.
Caution:
Always slop a distillation before the flask becomes completely dry. When the flask 'is dry,
its temperature can rise sharply. Some organic substances, especially alkenes and
ethers, may contain peroxide impurities that become concentrated and can explode at
dryness.
29
30
Start heating the flask and increase the heat gradually until the hexane boils. When liquid
begins to drip into the receiver, adjust. the heat so that the drops come steadily at a rate of
about one per second. Record the temperature after you have collected 1, 3, 5, 7, 9 and 11
mL of distillate. Then stop the distillation. The temperatures you have recorded should not
vary by more than 2° C and should collectively represent the boilingpo.int range of hexane.
Discard the residue as advised by your instructor.
Separation of a binary Mixture by simple distillation:
Procedure: Use the same apparatus (Figure 1.) to distill a mixture of 10 ml. of hexane and
10 ml of toluene Adjust the heat during the distillation so that the distillate drips slowly and
steadily, into the receiver. Record the temperature after every 1-2 ml. as the distillation
proceeds until you have collected 16 mL of distillate. Save distillate and residue for the
next part of the experiment.
Separation of a Binary Mixture by Using a Fractionating Column:
Procedure: Assemble the apparatus. Shown in Figure 2; Place the combined distillate and
residue from Sec. above (or, if it has been accidentally lost, use a fresh mixture containing
10 mL each of hexane and toluene) in a 50-mL round-bottomed flask, connect this flask to
the distilling column, and proceed as before. Again record the temperature every 1-2 ml. as
the distillation proceeds until you have collected 16 mL of distillate. Save the second, fifth,
and last fractions for use in the experiment on gas-liquid chromatography. Discard the
other fractions as advised by your instructor.
Waste Disposal: Pour any hexane or toluene distillate that is not to be saved into a
nonhalogenated organic liquids waste bottle provided by your instructor.
31
Data Sheet
Distillation of Hexane:
Quantity Distilled, mL
1
3
1
2
5
7
9
11
Temperature
Separation of a Binary Mixture:
Quantity Distilled, mL
4
6
8
10
12
14
Temperature without Column
Temperature with Column
Boiling point
Questions :
1. Plot the boiling point against the volume of distillate for the distillation of the hexanetoluene mixture both with and without a column. Label each curve.
115
110
105
100
95
90
85
80
75
0
2
4
6
8
10
12
Volume of distillate, ml
14
16
18
20
2. From the distillation curves, estimate the volume of liquid boiling below
85°C with and without a column.
3. Which procedure was more efficient at separating the mixture into its components?
4. A pure liquid has a constant boiling point, but a liquid with a constant
32
16
boiling point is not necessarily pure. Explain.
5. What effect does a reduction in the atmospheric pressure have on the boiling
point of .a liquid?
6. Why doesn't a pure liquid in a distilling flask vaporize all at once when the boiling
temperature is reached?
7. Why is it dangerous to heat a liquid in a distilling apparatus that is closed tightly at
every joint and has no vent to the atmosphere?
8. Why is it important that the cooling water in a distillation apparatus enter the condenser
jacket at the lower end and exit at the upper end, and not vice versa?
9. Why should a distilling flask be filled to not more than two-thirds of its capacity at the
beginning of a distillation procedure?
33
Experiment 5
Thin Layer Chromatography (TLC).
Introduction:
Chromatography is a technique used to separate the components of a mixture
and to purify substances. Although first developed in connection with the
separation of coloured compounds (hence the name after the Greek chroma,
colour), chromatography is now used to separate all types of substances,
including those that are colourless.
Separation may depend on selective adsorption on a highly porous surface
(adsorption chromatography).
Such separations may be carried out in a
vertical column packed with the adsorbing solid (column chromatography)
Alternatively, the adsorbing solid may be spread in a thin layer on an inert
surface (thin layer chromatography).
Figure 1 Thin-layer chromatography
34
The spotted plate is placed in the developing jar with a piece of filter paper, which acts
as a wick to saturate the atmosphere with solvent. Different compounds move up the
plate at different rates: the less polar compounds move the fastest and are found close
to the solvent front.
The Rf, value for a compound is constant, but only if all variables are also held
constant: temperature, solvent, adsorbent, thickness of adsorbent, and amount of
compound on the plate.
The Rf, value for compound A is the ratio of the distance it has travelled to the
distance the solvent has travelled.
In partition chromatography, separation depends on the partition of the mixture’s
components between two solvents, of which one (the stationary phase) is adsorbed on
a solid support. The solid support may be packed in a column, as in adsorption
chromatography.
Another technique involves passing the vapors of the mixture
(mobile phase) through a heated tube containing a nonvolatile liquid (stationary phase)
adsorbed on a finely divided solid support. This is called gas-liquid or vapor-phase
chromatography.
In paper chromatography filter paper serves as the solid support, and the water
adsorbed on the cellulose of the paper is the stationary phase. The mixture to be
separated is placed on the paper as a small spot, and solvent is allowed to move by
capillary action through the spot and along the paper.
Thin-Layer Chromatography (TLC)
Thin-layer chromatography involves: (1) preparing a thin-layer plate of a smoothly
spread layer of some adsorbent on glass, aluminum or plastic, (2) spotting the plate
with a small amount of the materials to be separated, dissolved in an appropriate
35
solvent, (3) developing the chromatogram, AND (4) dying or otherwise treating the
chromatogram in some way that allows observation of the separated components.
The solid adsorbents most often used in thin-layer chromatography are silica gel ( SiO2
. x H2O) and aluminum oxide. A slurry of the adsorbent in an organic liquid or water
is spread on to the glass plate to obtain a layer of uniform thickness. The plate is dried
to remove most of the liquid. The commercial adsorbent often contains a binder
(usually CaSO4) to ensure that it sticks to the surface.
Table 4.1. shows typical solvents. Mixtures of these solvents are also commonly used
to develop chromatograms. The greater the developing power of the solvent, the faster
the adsorbed compounds will move on the plate. For example, nonpolar hydrocarbons
should be eluted with hydrocarbon solvents, but a mixture of an alcohol and an ester
might be developed with a toluene-methylene chloride mixture.
Table 1
Common chromatographic solvents, listed in order of increasing
elutive or developing power
Aliphatic hydrocarbons (pentane, hexane)
Carbon tetrachloride (tetrachloromethane)
Aromatic hydrocarbons (benzene , toluene)
Methylene chloride (dichloromethane)
Chloroform (trichloromethane)
Diethyl ether (ethoxyethane)
Esters of organic acids (ethyl acetate)
Acetone (propanone)
Ethanol
Methanol
Water
The chromatogram is recorded in terms of a number called the Rf value (Fig. 5.2). The
Rf value is characteristic of a specific compound under specified conditions (adsorbent
and developing solvent) and is defined by
Rf = distance compound has traveled from origin
distance developing solvent has traveled from origin
Detection of spots on the chromatogram is easy for colored materials, and a
number of procedures are available for locating spots of colorless materials.
36
For example, irradiation of the plate with ultraviolet light will permit location
of the spots of compounds that fluoresce. Alternatively, the solid adsorbent
may be impregnated with an otherwise inert, fluorescent substance. Spots of
materials that absorb ultraviolet light but do not fluoresce will show up as
black spots against the fluorescing background when the plate is irradiated
with ultraviolet light.
Other detecting agents are more often used. These
agents may be sprayed onto the chromatograms, causing the spots to become
readily apparent. Examples of detecting agents used in this way are sulfuric
acid, which causes many organic compounds to char, and potassium
permanganate solution. Iodine is another popular detecting agent. In this case
the plate is placed in a vessel whose atmosphere is saturated with iodine vapor.
Iodine is adsorbed by many organic compounds, and their spots on the
chromatogram become colored (usually brown). Since these spots usually fade,
it is a good idea to circle the spots with a pencil while they are still visible in
order to have a permanent record of the chromatogram.
Thin-Layer Chromatography in Analysis of a Commercial Analgesic
Most
non-prescription
pain-relieving
remedies
contain
aspirin
(o-
acetylsalicylic acid) or acetaminophen - and some contain both. Many also
have some auxiliary compounds with analgesic properties.
In addition,
caffeine is included in some preparations. Caffeine is not an analgesic, but
aids in the relief of certain types of headaches because of its stimulant effect
(CNS and cardiac).
You will receive as an "unknown" a tablet of a commercial analgesic, to be
analyzed by TLC for the presence or absence of three specific constituents:
aspirin, ibuprophen, and caffeine. Some of the products also have other active
ingredients, but no attempt will be made to identify these other substances.
Procedure:1- Obtain a 250-ml beaker and carefully pour 10 ml of ethylacetate :
hexane (3:1) solution. It is important that the solvent level is below the
37
spots on the TLC plate. Line the beaker with a piece of filter paper and
cover it.
2- On a TLC plate draw a light line with a pencil 1 cm above. This is the
starting line. Mark 6 dots on the line equally spaced. Label the dots
from 1-6.
3- Obtain the 1% solutions of analgesic tablets and standards:
1. Aspirin
2. Profen
3. Panadol extra
4. Caffeine
5. Aspirin standard
6. Ibuprofen
standard
standard
4- Using a clean capillary spot a tiny amount of each substance on a dot.
Take care that the spots are no larger than 2mm diameter.
5- Dry the spots in air at room temperature for 2 minutes.
6- Place the plate in the tank and observe the development of the
chromatogram. When the solvent front has migrated about 5cm remove
the TLC sheet and air dry.
7- After the TLC plate has dried, observe the plate under a UV lamp and
lightly outline or mark the spots with the pencil. Measure the distance
and calculate the Rf value for each spot.
Questions:-
1-
Calculate the Rf values for the following compounds
(a) Spot, 5.0 cm; solvent front, 20.0 cm
(b) Spot, 3.0 cm; solvent front, 12.0 cm
(c) Spot, 9.8 cm; solvent front, 12.0 cm
2-
If two compounds have Rf values of 0.50 and 0.61, how far will they be
separated from each other on a plate when the solvent front is developed to:
(a) 5 cm?
38
(b) 15 cm?
3-
A single spot with an Rf of 0.55 showed up on the plate after developing the
sample in hexanes – ethylacetate 50:50. Does this indicate that the unknown
material is a pure compound? What can be done to verify the purity of the
sample?
39
Experiment 6
Elimination: Synthesis of Alkenes from Alcohols
Prelab Exercise:
Prepare a detailed flow sheet for the preparation of cyclohexene, indicating at
each step which layer contains the desired product.
Introduction:
One of the more general and synthetically useful methods of obtaining alkenes
involves dehydration of an alcohol:
OH
Cl
C
C
C
Alcohol
C
+
H2O
Alkene
Most commonly, a strong and high-boiling mineral acid, such as sulfuric or
phosphoric acid, is used as a catalyst for the reaction. The acid protonates the
alcohol. Subsequently, a molecule of water and a proton are eliminated:
If the alcohol is tertiary or secondary, or has other structural features that
stabilize the corresponding carbocation, the elimination may proceed stepwise
(E1 mechanism). Alternatively, the loss of H2O and H+ may occur in one step
(E2 mechanism).
In cases where the carbocation intermediate can rearrange to a more stable ion,
rearranged products are obtained.
40
In this experiment the secondary alcohol cyclohexanol will be dehydrated to
cyclohexene:
Experimentally, we take advantage of the fact that alkenes boil at much lower
temperatures than the alcohols from which they are prepared. The alcohol is
heated with acid to a temperature above the boiling point of the alkene but
below that of the alcohol. The alkene and water distill out of the reaction flask
as they are formed, whereas the unchanged alcohol remains behind to be
further acted upon by the acid. In the present case, the dehydration is carried
out at 130-140˚C, which is above the boiling point of cyclohexene (83˚C) but
below that of the starting alcohol (161˚C).
Ether formation competes with elimination. The amount of ether produced is
small in the case of secondary alcohols. Therefore in the present reaction, the
ether is only a minor product:
Procedure
CAUTION!
1. Phosphoric acid is a corrosive liquid. Avoid any contact of this acid with
your skin or clothing. If you have an accidental spill, wash immediately with a
large amount of water.
2.
A flame is used for the distillation, but your product cyclohexene is
flammable. Keep flames away from the receiver.
41
Preparation of cyclohexene
Set up a simple distillation apparatus , (Fig.4 in experiment 4). Add 20g
(0.20mol) of cyclohexanol and 5 ml of 85% phosphoric acid to a 100 ml
round-bottomed flask.
Mix the contents thoroughly by swirling before
connecting the flask to the distillation set up. Add two anti-bumping granules
and heat the flask gently over a low flame so that the temperature of the
distilling vapor does not exceed 100C. Continue the distillation until only a
few milliliters of high boiling residue remain in the flask and at least 17 ml
distillate have been collected.
Note that the distillate consists of two layers. Transfer the distillate to a small
separatory funnel and add 5 ml of saturated sodium chloride solution (to
decrease the solubility of cyclohexene in the water layer). Then add 5 ml of
10% sodium carbonate solution to neutralize any traces of acid. Shake the
mixture gently. Allow the layers to separate, then draw off and discard the
lower, aqueous layer. Pour the upper layer (crude cyclohexene) out of the top
of the separatory funnel into a small, dry conical flask containing anhydrous
calcium chloride to remove traces of water from the cyclohexene. The product
should be clear, not cloudy.
Decant or filter the cyclohexene from the calcium chloride into a pear shaped
flask, add two anti-bumping granules, connect the flask to the distillation
apparatus and distill. Collect the product that boils between 79 and 84C in a
receiver of known weight. Weigh the product and calculate the percentage
yield. Transfer the product to a labeled sample tube.
a- Use some of the product for the following tests for unstauration then
hand in the remainder to the laboratory instructor.
b- Check the purity of the product by gas chromatography.
42
Tests for Unsaturation
Bromine is frequently added to alkenes as a simple qualitative test for
unsaturation. Bromine is a dark re-brown liquid, whereas alkenes and alkene
dibromides are colorless.
Thus a dilute solution of bromine in some inert colorless solvent, such as
carbon tetrachloride or dichloromethane, is rapidly decolorized when it is
added to an alkene. In contrast, most saturated compounds do not decolorize
bromine solutions.
Oxidizing agents also readily attack carbon-carbon double bonds. You can use
this property to distinguish alkenes from alkanes.
In the Baeyer test the
reagent is alkaline permanganate. When the alkene is oxidized, the reagent is
reduced to manganese dioxide and the color changes fro purple to brown:
a.
Most alkanes, on the other hand, do not react with permanganate under the test
conditions.
Procedure for Bromine Test
CAUTION!
Avoid skin contact with or inhalig the vapors of the
bromine solution.
Dissolve 5 drops of the cyclohexene you prepared in 1 ml dichloromethane.
Add to this solution, drop wise, a 3% solution of bromine in dichloromethane.
Report the result. For comparison, repeat the test using cyclohexane in place
of cyclohexene.
b.
Procedure for Baeyer Test
Add 5 drops of the cyclohexene you prepared to 1 ml of 0.5% potassium
permanganate solution, shake the tube well for 1-2 min, and report the results.
For comparison, repeat the test using cyclohexane.
43
Experiment 6 report.
Reagents
MW
Reagent
C6H11OH
H3PO4
100.16
97.99
Density
B.p.
Mass
Moles
(g/mL)
0.96
-
(C )
161
-
Used (g)
20.0
(5 ml)
Theory
-
Used
0.20
Products and By-products
Compound
C6H10
(C6H11)2O
MW
82.14
182.30
Density
(g/mL)
Given
Found
0.81
-
83
Calculations
Theoretical yield of the desired product
0.20 (mole of alcohol) x 82.14 (MW of product) = 16.4g cyclohexene
LIMITING REAGENT
Percentage yield = Actual yield (g)
Theoretical yield (g)
Yield,g(%)
B.p. (C)
x 100% =
44
Theory
Found
16.4(100)
g %
Experiment 7
Bromination of alkenes: Preparation of stilbene dibromide.
Introduction
Alkenes (olefins) are hydrocarbons containing C-C double bond which can be halogeneated
to form alkyl halides which are capable to undergo various organic chemical reactions.
Bromination is an addition chemical reaction in which Br2 is added across the double bond
to yield a vicinal (1,2) dibromide product. The mechanism for the addition of Br 2 to the C-C
double bond is shown in scheme 1.
Br

Br
Br
Br 
Br
Br
Scheme 1= Mechanism for bromination of an alkene
In this experiment Br2 will be added to (E)-stilbene to afford 1,2-dibromo-1,2diphenylethane equation1 Typically, bromination of alkene is carried out using bromine
(Br2) in chlorinated solvent such as carbon tetrachloride (CCl4) or dichloromethane
(CH2Cl2). These solvents are suspected to be carcinogenic therefore in this experiment
ethanol will be used as a safer alternative solvent.
Br
H
+
H
Br2
H
Br
(E) -stilbene
H
1,2=dibromo-1,2 -diphenylethane
Equation1 = Bromination of (E) –stilbene
Bromine (Br2) is volatile and highly corrosive, causing severe burns when contact with the
skin and extremely irritating upon inhalation. An alternative reagent is pyridinium
tribromide. This reagent exists in rapid equilibrium with pyridinium hydrobromide and
bromine equation2, thus slowly provides Br2 into the reaction media. Compared to liquid
bromine, pyridinium tribromide is a solid thus it is easily weighed and handled ( When a
reagent is generated in the reaction medium rather than added to it, it is said that the reagent.
N_ H
Br3
N_ H
45
Br
+ Br2
Equation2: In site generation of bromine from pyridinium tribromide
Reaction mechanism involves interaction of Br2 with π orbital of the alkene to form a cyclic
brominium ion which then reacts with bromide ion to give a vicinal dibromoalkane
(bromine atoms are on the neighboring carbon atoms) product. The addition of Br 2 across
the double bond proceeds with anti stereochemistry which means that two bromine atoms
add to the opposite sides of the double bond.
Chemicals
 Pyridinium tribromide (corrosive and lachrymator so a void contact and clean up any
spills immediately, particularly on the balance as the metal parts will quickly be
corroded.
 Ethanol is volatile and flammable.
Experimental procedure
1. Place 2.0 g of (E)-stilbene, 40 mL of ethanol and a magnetic bar in a 125 mL
Erlenmeyer flask.
2. Heat with stirring the mixture on a hot plate to dissolve all the stilbene.
3. Using gloves add 4.0 g of pyridinium tribromide to the above mixture. If solid material
adheres to the interior wall of the flask add little ethanol to rinse it down.
4. Heat the mixture for 5 min then remove the flask from the hot plate. The product
dibromide should quickly precipitate.
5. Allow the reaction mixture to cool to room temperature than collect the product by
vacuum filtration and wash the collected solid with small amount of ice-cold methanol to
remove any adsorbed pyridinium salts then allow your sample to dry in vacuum.
6. Determine the mass of your product and calculate the percentage yield.
7. Measure the melting point of the product.
Test for alkyl halides
Beilstein test
It is a test to identify alkyl halides. A red-hot copper wire reacts with alkyl halide to form
the corresponding copper halide salt. When this copper –halide salt is placed in a flame, a
blue-green color indicates that the organic compound contains a halogen (Cl, Br, I).
The chemical procedure for this reaction is as following:
1. Wrap a copper wire (1mm diameter or greater) around one end of glass rod leaving
3-4 cm long straight piece.
2. Hold the end of the copper wire in a Bunsen burner flame until the flame becomes
colorless.
46
3. Dip the hot wire in a small amount of your product.
4. Hold the copper wire in the flame and record your observation.
Questions
In your Discussion provide Answers to the following questions:
1. Report the mass and the percent yield of your product.
2. Describe the color and the state of your product.
3. Write a detailed mechanism of the reaction between a stilbene and bromine.
4. What isomers do you expect to be the reaction product? Explain your answer.
5. Would you expect that the steric and electronic effects would influence reactivity of
bromine towards electrophilic addition to the double bond? How would substituent on
the double bond effect the reactivity of it?
Reference
Green Organic Chemistry – Strategies, Tools, and Laboratory Experiments by K.M. Doxsee
and J.E. Hutchison, Thompson Brooks/Cole,2004: pp 120-124.
47
Pre-Lab Preparation
Preparation of Dibromostilbene
Date:…………………………………………………………
Name:……………………………………………………....
Draw a mechanism and predict what will be the product of bromoniation of Cis-stilbene and
what will be the product of bromination of trans-stilbene.
48
Experiment 8
Conformations and Configurations using Molecular Models
3.1.
Introduction:
Molecular models can help you visualize the three-dimensional structures of
molecules.
In this experiment, we will use models to examine the
conformations and configurations of acyclic molecules (Part A) and cyclic
molecules (Part B). For construction of models of the molecules listed
below, you are provided with the following parts:
6 black, tetrahedral (sp3) carbon centers
12 white hydrogen centers
2 green chlorine centers
12 short , green sticks for carbon-hydrogen bonds
8 long, green sticks for all other bonds except
3 white, flexible sticks for “bent bonds” in cyclopropane
1 ruler for measuring
The models you make generally show the correct angles between the
various bonds in a molecule but they do not accurately represent the relative
sizes of the atoms or their precise inter-nuclear distances.
Your drawings of the molecules and answers to questions must be written
clearly on the report sheet at the time of construction of the models. The
completed report sheet should be handed in for marking at the end of the
laboratory period.
49
Experiment 8 , Part A report
8.2
Procedure.
Part A : Acyclic Molecules
Construct a model of each of the following molecules and make a drawing
of each using dashed line – wedge formulas. Carry out any additional
instructions given.
1- methane
Structural Formula
Dashed line wedge formula
2- chloromethane
Structural Formula
Dashed line wedge formula
3- ethane – staggered and eclipsed conformations.
between an eclipsed pair of hydrogens.
Measure the distance
Structural Formula
Line Formula
Sawhors projection
staggered
eclipsed
Newmann projection
staggered
eclipsed
50
4-
Chloroethane
Structural Formula
Newmann projection staggered
Line bond formula
5-
Propane
Structural formula
Newmann projection staggered
Line bond formula
6.
Butane – anti and gauche conformations. Measure the distance between
the closest pairs of hydrogens in each conformation and show this on your
drawings.
Structural Formula
Linebond Formula
Newmann projection
0°
7-
60°
120°
Dihedral angle
180°
240°
300°
360°
2-chlorobutane – enantiomers (mirror images). Confirm that the two
enantiomers are nonsuperimposable mirror images.
Structural formula
Line bond formula
Enantiomers:-
Experiment 8 , Part B, report.
51
Part B: Cyclic Molecules.
Construct a model of each of the following molecules and make a
perspective drawing of each. Carry out any additional instructions given.
1-
Cyclopropane. Use the flexible white sticks for the carbon-carbon bonds.
2-
Cyclobutane – “bent” conformation.
3-
Cyclopentane – envelope conformation.
4-
Cyclohexane (a) chair
(a)
Label the axial and equatorial hydrogens in your drawing of the chair
conformation. Confirm that the axial and equational hydrogens change
places on “flipping” the ring to the other chair conformation.
Chair
(b) boat
boat
52
chair
4(b)
In the boat conformation measure the distance between the transannular
hydrogens and between a pair of eclipsed hydrogens and show these on your
drawing.
5-
Trans-1,3-dichlorocyclohexane – enantiomers. Show that there are no
conformations in which the two isomers are superimposable.
6.
Cis-1,4-dichlorocyclohexane – both chair conformations. What relationship
do the two chair conformations have? Is there any plane of symmetry?
7-
Trans-1,4-dichlorocyclohexane – both chair conformations.
symmetry plane in each chair conformation.
8-
Cis-1,2-dichlorocyclopropane. Use the three white, flexible sticks for the
53
Show the
carbon-carbon bonds. Show the internal symmetry plane on your drawing.
9-
Trans-1,2-dichlorocyclopropane. Use the three white, flexible sticks for the
carbon-carbon bonds. Show that the two trans isomers have a mirror image
relationship..
54
Experiment 9
Optical activity of chiral molecules
OBJECTIVES
In this experiment, you will
Become familiar with the use of the use of Polarimeter.
Measure the angle of optical rotation for known sugar solutions.
Calculate the specific rotation for known compounds and compare them with literature
values.
Measure the angle of optical rotation for the two enantiomeric forms and racemic mixture of
tartaric acid.
Measure the angle of optical rotation for a meso compound.
Background:
A special class of stereoisomers, called enantiomers, are mirror image pairs that cannot be
superimposed on each other. These stereoisomers are called enantiomers. An example of a
pair of enantiomers is provided by 2-butanol.
OH
*
C2H5
OH
H
CH3
H
H3C
*
C2H5
Molecules that have two enantiomeric forms are called chiral and they generally contain at
least one atom that is bonded to four different groups. This atom is called a stereogenic
centre. The carbon atom marked with an asterisk in 2-butanol above is a stereogenic centre it is bonded to four different groups.
Unlike diastereoisomers, enantiomers have identical chemical properties except when
interacting with other chiral molecules. They also have identical physical properties except
when interacting with polarized light. Why bother with distinguishing such molecules when
they are identical in so many ways? The reason is that Nature is chiral - many of the
molecules in living things (e.g. proteins and sugars) can form enantiomers, but only one of
the possible enantiomers is actually found. This very particular selection of enantiomers in
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biological systems means that all enantiomers are distinguishable when introduced into a
living cell. A classic example is that of thalidomide, a drug prescribed from 1957 to 1962 as
a sedative and painkiller and as a treatment for morning sickness during pregnancy.
Thalidomide is a chiral molecule and it was sold as a 50:50 mixture of both enantiomers. It
was subsequently discovered that the (R)-enantiomer was effective against morning
sickness, but the (S)-enantiomer was responsible for horrific birth defects. The tragedy of
thalidomide stresses the importance of being able to distinguish enantiomers. On the bright
side, the bitter experience gained with thalidomide has led to much more stringent testing of
drugs before they enter the market.
The physical similarity between enantiomers makes separating them very difficult.
Synthetic chiral molecules are, as a result, generally expensive to produce from non-chiral
starting materials. As a living organism will only produce one of the possible enantiomers,
natural products remain the most important source of chiral molecules. In some cases, such
as the terpenes limonene and carvone, both enantiomers can be obtained from natural
sources, although from different species. In the case of limonene, one enantiomer, (R)limonene, is extracted from citrus rind while the other enantiomer, (S)-limonene, is obtained
from pine resin.
Optical Activity
Visible light consists of electromagnetic waves that oscillate in all possible planes
perpendicular to the direction in which the light is travelling. Certain filters (e.g. calcite and
Polaroid film) permit only light waves that are vibrating in one particular plane to be
transmitted. After passage through such a filter, the light is called plane-polarized light.
Jean-Baptiste Biot, a French scientist, discovered in 1815 that the plane of polarization was
rotated by passing plane-polarized light through solutions of certain organic compounds.
Molecules that possess this property are said to be optically active. Optical activity is a
property of each individual molecule - the more molecules encountered by the light, the
greater the amount of rotation. Consequently, the degree of rotation is dependent on the
molecule, its concentration and the sample path length.
To determine the specific rotation of the sample, use Biot’s law:
α = [α] ℓ c
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where α is the observed optical rotation in units of degrees,
[α] is the specific rotation in units of degrees dm-1 mL g-1 ( scientific literature uses just
degrees),
ℓ is the length of the cell in units of dm
c is the sample concentration in grams per milliliter.
Measurement of optical rotation is performed using an instrument called a polarimeter.
Figure 1: A diagram of a polarimeter.
As shown in Figure 1, incident non-polarized light is transmitted through a fixed polarizer
that only allows a certain orientation of light into the sample. The sample then rotates the
light at a unique angle. As the analyzer is turned, the rotated light is maximally transmitted
at that unique angle, allowing the user to determine properties of the sample.
Optically active compounds that rotate plane-polarized light to the right are given the
symbol (+) and, conversely, compounds that rotate plane-polarized light to the left are given
the symbol (–). A pair of enantiomers will rotate plane-polarized light by the same amount,
but in opposite directions. The amount and direction of rotation of polarized light is a
physical property of each and every compound and needs to be determined experimentally.
If you take a 50:50 mixture of two enantiomers, the (+) rotation from one enantiomer will
exactly equal the (–) rotation from the other enantiomer. Such mixtures are called racemic
mixtures or racemates . Racemic mixtures display an
optical rotation of zero.
This experiment allows you to use a Polarimeter to measure the specific angle of rotation for
different solutions.
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Procedure:
Part 1:
1. Calibrate the Polarimeter:
a. Pour distilled water in the Polarimeter cell.
b. Place the cell in the Polarimeter.
c. Zero the instrument.
2. You are now ready to add the optically active sample into the Polarimeter cell.
a. Pour the sucrose solution in the Polarimeter cell.
b. Place the sample cell in the Polarimeter.
c. Record the angle of optical rotation in the table below.
3. Repeat Step 2 for solutions of 0.1M glucose, 0.1M galactose and 0.1M fructose solutions.
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Part 2:
1. Measure the angle of optical rotation of 20% (-)- tartaric acid.
2. Measure the angle of optical rotation of 20% (+)- tartaric acid.
3. Measure the angle of optical rotation of 1:1 mixture of the two tartaric acid solutions.
4. Measure the angle of optical rotation of stilbene dibromide (.5g/10ml DMSO).
5. Record the angles of optical rotation in the table below.
Results:
Table 1: Measuring the optical activity of sugars.
compound
Conc. g/ml
α
[α]
[α]lit
% error
0.1 M sucrose
Table 2: Measuring the enantiomers, racemic mixture and meso compounds.
compound
Conc. g/ml
α
[α]
[α]lit
% error
20% (-)- tartaric acid
Calculations:
1. Calculate the exact concentrations of each sample in g/mL.
2. Using the observed angles of rotation and Biot’s law calculate the specific angles of
rotation.
3. Compare with literature values and calculate the % error.
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Experiment 10
Nucleophilic Substitution:
Relative Reactivity of Organic Halides.
Introduction:
A large number of organic reactions involve substitution of one group by another
at a saturated carbon atom:
Nu:-
+
R
Nucleophile
-
L
R - Nu
substrate
product
L:-
+
leaving group
Nu:- represents an anion or a molecule with one or more unshared pairs of
electrons,e.g.,
CN  , RC  C  , N3 , NH 3 , OH  , RO  , CH 3CO2 , H 2O, ROH , RS  , SCN  , Cl  , Br  , I  .
Substrates are molecules or cations such as RC1 , RBr , RI , CH3OSO2OCH3 ,

ROSO2R , R O H 2 ,
The nature of the product and leaving group follows from the structures of the
nucleophile and substrate used in the reaction.
The rate and mechanism of nucleophilic substitution depend on:
1. The strength of the nucleophile (nucleophilicity).
2. The structure of the alkyl group (R)
3. The nature of the leaving group
4. The reaction conditions (polarity of the solvent)
The most common mechanisms of nucleophilic substitution are SN1 and SN2
which stand for “Substitution Nucleophilic Unimolecular” and – Bimolecular” ,
respectively, i.e., the slow rate determining step involves one and two molecular
species, respectively.
SN1 rate = k1 [R-L]
SN2 rate = k2 [R-L] [Nu:-]
The SN2 mechanism is favored by : Strong nucleophiles, less substituted
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substrates (methyl and primary) and polar aprotic solvents, e.g., CH3COCH3
The SN1 mechanism is favored by: Weak nucleophiles , more substituted
substrates (tertiary) and hydroxylic solvents, i.e., water and alcohols.
For a given nucleophile and set of reaction conditions the mechanism and rate are
determined by the structure of the alkyl group of the substrate:
SN1 rate increases
CH3-L
RCH2-L
R2CH-L
R3C-L
methyl
1
2
3
SN2 rate increases
SN2 Mechanism (One Step)
Iodide ion is an effective nucleophile in SN2 displacements. In acetone solution,
other alkyl halides can be converted to alkyl iodides:
acetone
-
I + R-X
R-I + X-
Although you might expect such a reaction to be reversible, it can be made to proceed in
the forward direction by using anhydrous acetone as the solvent because sodium iodide is
soluble in this solvent but sodium chloride and sodium bromide are not. We can detect
whether a reaction has occurred by the formation of a precipitate of sodium chloride or
bromide.
The mechanism involves a one-step, concerted SN2 displacement displacement.
Therefore reaction will occur most quickly when attack at the carbon that bears the
halogen X is least hindered sterically. For alkyl halides, the order is primary > second >
tertiary. Tertiary halides rarely react by the SN2 mechanism.
In cases where the reaction is at a chiral carbon atom (stereocenter), the configuration of
the product is opposite to that of the substrate since the nucleophile attacks from the
opposite side of the molecule to the leaving group, e.g.,
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For cyclic halides (Secondary), the reactions rate depend on the ring size:
SN1 Mechanism (two steps)
A solution of silver nitrate in ethanol represents essentially the limiting
conditions for the SN1 mechanism, viz., weak nucleophile and hydroxylic solvent,
i.e., the solvent molecule (CH3CH2OH) contains an acidic hydrogen atom. A
protic (hydroxylic) solvent can solvate both cations, e.g., carbocations, and
anions, e.g., charged nucleophiles, thereby stabilizing them but reducing the
nucleophilicity of the latter. In cases where the reaction is at a stereocenter, the
product will be a racemate since attack by the nucleophile occurs to an equal
extent on both sides of the planar carbocation, e.g.
The reaction has been considerably simplified above.
Nitrate is not the only nucleophile. Ethanol and, probably, water are also present
and compete to give additional products as well as solvating the ions present.
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Primary and methyl halides do not usually react by the S 1N mechanism.
Alkenyl (vinyl) are aryl (aromatic) halides are much less reactive than alkyl
halides and do not usually react under SN1 or SN2 conditions. In the following
experiment, the relative rates of SN1 and SN2 reactions will be estimated for a
series of organic halides by measuring the time for appearance of a visible
precipitate in the reactions with silver nitrate and sodium iodide respectively.
Procedure
SN2 Reaction:Label 8 dry tubes from 1 to 8. Then place 0.2 cm3 of each of the following
halides into the tube indicated.
1. bromoethane
5. 1- chlorobutane
2. 1- bromobutane
6. 2- chlorobutane
3. 2- bromobutane
7. 2- chloro-2-methylpropane
4. bromobenzene
8. chlorobenzene
To each tube add 2 cm3 of 15% sodium iodide in propanone (acetone) in one
portion, shake and stand. Note the time and record any rapid reactions. After 5
minutes, place any tubes that do not contain a precipitate into a beaker of water at
50C. Note the time and record any reactions that occur. On completion of this
part, clean the test tubes, dry and go on to the next experiment.
SN1 Reaction:Place 0.2 cm3 of the same bromides and chlorides into test tubes 1 to 8 as in the
previous experiment. Then label four more tubes from 9-12. Place 0.2 cm3 of
each of the following iodides into the tube indicated:
9. 1- iodobutane
11. 2-iodo-2-methylpropane
10. 1- iodo-2-metylpropane
12. iodocyclohexane
To each tube add 2 cm3 of 1% silver nitrate in ethanol in one portion, shake and
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stand. Note the time and record the time both for appearance of the first turbidity
and for a definite precipitate. After 5 minutes, place any tubes that do not contain
any precipitate into a beaker of water at 50C. Note the time and record any
reactions that occur.
Results and Conclusions
Summarize your observations and conclusions in the table on the report sheet.
Account for the observed differences in reactivity and draw conclusions about the
correlation of relative rate of reaction with the structure of the hydrocarbon group
(alkyl or aryl) and the nature of the leaving group (Cl- , Br- or I- ) by answering
the questions on the results sheet.
Questions
1.
Arrange the halides tested in order of decreasing reactivity towards I-
2.
Explain the order observed for the alkyl halides towards I-
3.
Account for the behavior of bromobenzene and chlorobenzene towards I- and Ag+
4.
Arrange the halides tested in order of decreasing reactivity towards Ag+
5.
Explain the order observed for the alkyl halides with Ag+
6.
Why must the acetone used in the reactions with NaI be anhydrous?
7.
Which, if any, of the substrates in this experiment is/are chiral?
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Experiment 9 report
Condensed
Tube
No.
Structural
Nature of R
Formula (RX)
NaI in acetone
Ag NO3 in ethanol
Time (min/s) for reaction (ppt.)
Time (min/s) for turbidity & ppt.
Room temp.
1-
C2H5Br
1
2-
CH3 (CH2)3Br
1
3-
C2H5CHBrCH3
2
4-
C6H5Br
5-
CH3(CH2)3Cl
1
6-
C2H5CHClCH3
2
7-
(CH3)3CCl
3
8-
C6H5Cl
9-
CH3(CH2)3I
10-
(CH3)2CHCH2I
11-
(CH3)3Cl
12-
C6H11l
Aryl
Aryl
1
1 branch
3
2cyclic
65
50C
Room temp.
50C
66