Crystallization and Determination of Melting and Boiling Points

Experiment 

1 

Crystallization and Determination of Melting and 

Boiling Points 

by Anawat Ajavakom 

Objectives 

1) To perform a purification of organic compounds by crystallization. 

2) To determine melting point and boiling point of unknown compounds. 

3) To become familiar with general apparatus in organic chemistry laboratory. 

Principles 

1. Crystallization 

Solid organic compounds are usually purified by crystallization. This general 

technique involves dissolving the impure material in a minimum amount of hot solvent 

and cooling the solution slowly. The dissolved material has lower solubility at lower 

temperature and a solid material will form as the solution is cooled down. This process is 

called either ‘crystallization’ if the crystal growth is relatively slow and selective or 

‘precipitation’ if the process is rapid and yields smaller crystals. 

In microscale organic laboratory, there are two common methods for crystallization; 

1) Semi-microscale crystallization (for solid material of more than 0.1 g). 

2) Microscale crystallization (for solid material of less than 0.1 g) 

In this experiment, you will perform a semi-microscale crystallization using 0.5 g of 

solid unknowns, which could be either benzoin or benzoic acid. 

O 

COOH 

OH 

Benzoin 

Benzoic acid 

Four main steps in semi-microscale crystallization are; 

1) Dissolving the solid 

2) Removing insoluble impurities (if necessary) 

3) Crystallization 

4) Isolation of crystals 

2. Melting Points and Boiling Points 

Physical properties that are useful for distinguishing compounds include color, 

melting point, boiling point, density, and reflective index. In this experiment, we will learn 

how to determine melting and boiling points. 

- 1 -

Melting Points 

The melting point is used by organic chemists not only to identify compounds, but 

also to verify their purity. To measure the melting point, a small amount of sample is 

heated in an apparatus equipped with a thermometer while two temperatures are noted. 

The first is the point at which the first drop of liquid forms among the crystals; and the 

other is the point at which the whole mass of crystals turns to a clear liquid. The melting 

point is recorded as a range of temperatures, for example 180-182 °C. Pure compounds 

usually have narrow melting temperature ranges (within 1-2 degree). 

Melting point can be used as supporting evidence in identifying a compound as well. 

Not only can we compare melting points of two individual compounds, a procedure called 

‘mixed melting point’ may be used with the same compound taken from different 

sources. Since melting point is a unique physical property, the mixture of two samples 

would have a narrow melting range if the mixture composes of just one compound. On 

the other hand, the observed melting ranges would be broad if the mixture composes of 

two different compounds. 

Boiling Points 

When a liquid is heated, its vapor pressure increases until it equals the atmospheric 

pressure. And at this temperature, the liquid starts to boil. The normal boiling point is 

measured at 760 mmHg or 1 atm. However at lower pressures, the vapor pressure 

needed for boiling is also lower, and so the liquid boils at lower temperature. (Look at the 

reference, fig 6-8 p.606.) 

Experimental Procedure 

Part A 

1) Obtain a sample from your instructor. Record sample number. 

2) Keep 10 mg (or about 2 grains of rice) in a watch glass for melting point 

determination and place the same amount in each of two test tubes. 

3) Add 10 drops of ethanol in one test tube, and 10 drops of water in another. 

4) Gently shake both test tubes and observe the solubility at room temperature. 

a) If the sample completely dissolves in just one tube, your solvent for crystallization 

is the one that cannot dissolve your sample at room temperature. 

b) If the sample does not dissolve in both tubes, place the tubes in a steam bath for 

30 seconds and observe the solubility. Your solvent for crystallization is the one 

that can dissolve your sample at high temperature. 

c) Report the solvent of choice before proceeding to the next step. 

5) Place the rest of your sample in a 50 mL Erlenmeyer flask, follow by 5 mL of your 

solvent. 

6) Heat the mixture on a hotplate and swirl occasionally. Do not allow the mixture to 

boil vigorously. If sample doesn’t completely dissolve when the solvent starts to boil, 

add a small amount of solvent until a clear solution is obtained. However, impurities 

or some black particles may still exist. 

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7) Setup a hot filtration kit as shown in the picture below. Filter paper is available in the 

supply room. Consult your instructor on how to transfer and filter a hot solution. 

Glass funnel 

with filter paper 

Empty 

flask 

Solution to be 

filtered 

Figure 1 

8) When the hot filtration is finished, remove the flask from the hot plate and set aside. 

Wait for the crystals to grow and do not move the flask around. You may measure 

the melting point of unpurified sample while waiting (see step 10). 

9) Vacuum filter and dry the crystals by placing them on a watch glass heated on a 

steam bath. Use a glass rod to gently turn the crystals, ensuring thorough drying. 

10) Put the unpurified sample into a capillary tube (about 4-5 mm in height). Attach the 

tube to a thermometer with sewing thread and set the apparatus as in Figure 2. 

Clamp 

Thermometer 

50 mL Beaker 

Paraffin Oil 

Sewing thread 

Sample in 

capillary tube 

Hotplate 

Figure 2 

- 3 -

11) Turn on the hotplate and observe the melting point. Use a clean glass rod to stir the 

paraffin oil to ensure a uniform heat distribution. 

12) Record the temperatures 

a) when the sample starts to melt. 

b) when the melting process is finished. 

13) Allow the paraffin oil to cool down and repeat step 10) to12) with the purified sample. 

Part B 

1) Obtain an unknown liquid sample from your instructor. Record the sample number. 

2) Attach a clean and empty test tube to a thermometer with sewing thread. Put an 

empty capillary tube in the test tube so that the open end of capillary is down. Set up 

the apparatus as in Figure 3. 

Test tube 

Clamp 

Thermometer 

Capillary 

Close end 

50 mL Beaker 

Capillary tube 

Capillary 

tube 

Paraffin Oil 

Sewing thread 

Capillary 

Open end 

Hotplate 

Sample in 

test tube 

Figure 3 

3) Ensure that the paraffin oil is not hot and place 2-3 mL of sample in the test tube. 

4) Turn on the hot plate and use a clean glass rod to stir the paraffin oil to ensure a 

uniform heat distribution. 

5) Record the temperatures when rapid air bubbles come out from the capillary 

6) Turn off the hot plate and carefully insert a ceramic tile between the beaker and the 

hotplate. 

7) As the temperature decreases, air bubbling will gradually slow down. Record the 

temperature when you see the last bubble comes out and the liquid goes into the 

capillary tube 

8) Report the boiling point to your instructor. If a repetition is needed, allow the paraffin 

oil to cool, then replace the capillary tube with a new one. Add more liquid sample if 

necessary and repeat step 4) to 7). 

- 4 -

Safety Precautions 

Wear safety goggles and lab coat at all times. 

Reference 

D. L. Pavia, G. M. Lampman, G. S. Kriz, R. G. Engel, Introduction to Organic Laboratory 

Techniques, A Microscale Approach, Part 5, 577-616. 

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Experiment 

2 

Extraction and Simple Distillation 

by Anawat Ajavakom 

Objectives 

1) To separate neutral, acidic, and basic compounds in a mixture by extraction. 

2) To become familiar with separatory funnel and other general apparatus in organic 

chemistry laboratory. 

3) To perform a simple distillation of organic compounds. 

Principles 

To isolate pure components from a mixture, many practical techniques can be 

chosen depending on the difference in physical and chemical properties. Extraction is a 

very simple way for isolation. It can be used with mixtures containing neutral, acidic, and 

basic compounds, for example, the isolation of benzoin (a neutral component) and 

benzoic acid (an acidic component) from a mixture given to you in this experiment. 

O 

COOH 

OH 

Benzoin 

Benzoic acid 

Both benzoin and benzoic acid are white solid substances which can dissolve in 

dichloromethane (CH 2 Cl 2 ). With the difference in acidity, these two compounds can be 

separated by acid-base extraction method as outlined in the flow chart. 

Mixture of benzoin and benzoic acid 

dissolved in CH 2 Cl 2 

Shake with diluted NaOH 

Organic layer 

Benzoin 

in CH 2 Cl 2 

Benzoin 

Dry and evaporate 

CH 2 Cl 2 

Aqueous layer 

Sodium benzoate 

in water 

White Precipitate 

Add cold conc. HCl 

Filter 

Benzoic acid 

- 6 -

Acidic compounds (HA) react with bases (OH - ) to give their conjugated bases (A - ) 

and water, as described in equation (i). 

HA + OH - A - + H 2 O 

(i) 


Part A 

1) Obtain a mixture from your instructor. This mixture consists of approximately equal 

weights of benzoin (neutral) and benzoic acid (acidic). 

2) Weigh 1g of the mixture and dissolve with 20 mL of CH 2 Cl 2 in a 50-mL separatory 

funnel. Swirl until solid dissolves completely. 

3) Add 8 mL of 10% NaOH into the separatory funnel. Shake the funnel carefully for 10 

seconds as demonstrated by your instructor. (*Remember to hold the separatory 

funnel with both hands and to vent it frequently with the lower end pointed upward 

and away from other people.) Settle for a few minutes and split the layers into two 

separate flasks. 

4) Transfer the organic layer into the separatory funnel. Add a new portion of 8 mL of 

10% NaOH into the separatory funnel. Shake, settle, and split the layers. Transfer 

the lower organic layer in a flask and keep the upper aqueous layer in the funnel. 

5) Combine the first aqueous portion with the solution in the funnel. Add 15-mL of 

CH 2 Cl 2 . Shake, settle, and split the layers. Combine the CH 2 Cl 2 layers with that from 

step 4). Transfer the aqueous layers into another flask for experiment in Part B. 

6) Wash the organic solution by shaking with 10 mL of water. Discard the aqueous 

layer. Dry the organic layer over anhydrous sodium sulfate (Na 2 SO 4 ) (roughly about 

2-3 tea spoons). After standing for a few minutes, this organic solution should be 

clear. If not, consult your instructor. 

7) Gravity filter the CH 2 Cl 2 solution into a 50-mL round bottom flask and evaporate the 

solvent by distillation. The distillation set should be prepared as shown in Figure 1 

(Don’t forget to place a lab jack underneath the hot plate so that the heating source 

can be instantly removed in case of EMERGENCY.) Consult your instructor before 

heating the distillation set. 

8) Observe the temperature of the distillate (solvent) and collect in a conical flask. 

9) Remove the heating source when the content remaining in the flask is about 2-3 mL. 

Allow the solvent to evaporate while the flask is cooling to room temperature. 

10) Collect and weigh the solid residue of benzoin. 

- 7 -

Thermometer 

Thermometer 

Adapter 

Water out 

50 mL Round 

Bottom Flask 

Condenser 

Parafin Oil 

Boiling Stone or 

Ceramic Pieces 

Oil Bath 

Water in 

Hot Plate 

Conical Flask 

Part B 

Figure 1 

1) Acidify the combined aqueous portions from Part A step 5) by slowly add 6M HCl 

until the solution is acidic to litmus paper. 

2) Cool the resulting suspension in ice bath for a few minutes. 

3) Vacuum filter and air dry the benzoic acid for a few minutes. 

4) Collect and weigh the solid benzoic acid. 


Concentrated HCl is highly corrosive. Safety goggles and lab coat must be worn at all 

times. 

Reference 

D. L. Pavia, G. M. Lampman, G. S. Kriz, R. G. Engel, Introduction to Organic Laboratory 

Techniques, A Microscale Approach, Part 5, 617-650. 

- 8 -

Experiment 

3 

Thin Layer and Column Chromatography 

by Pattara Sawasdee 

Objectives 

1) To perform separation techniques of column chromatography. 

2) To analyze the purity of the isolated compounds using TLC technique. 

Principles 

Chromatography is a laboratory method based on selective adsorption by which 

components in complex mixtures can be separated for identification or purification 

purposes. (Adsorption is the binding of molecules to the surface of another substance.) 

The utilization of a moving (or mobile) phase and a stationary phase is common to all 

chromatographic separation techniques. A mixture of compounds is introduced into the 

chromatographic system as a narrow zone and is partly retained by the stationary phase 

while the mobile phase carries the components through the system. Each component in 

the mixture will be distributed between the mobile phase and the stationary phase, but not 

to the same extent. Components that are more loosely interacted by the stationary phase 

will be swept through the column more rapidly, and a separation will be achieved. 

When the stationary phase is packed inside a column, the experimental 

procedure is called column chromatography. When the stationary phase is planar (e.g., 

a piece of filter paper or a coated glass), the mobile phase is usually liquid and the 

techniques are classified according to the stationary phase which is paper or thin-layer 

chromatography. 

Chromatography is a sophisticated separation method. Its versatility results from 

many adjustable factors which include: 

1. Types of adsorbent 

2. Types of solvent, which relates to the polarity of the mobile phase 

3. Column size, both length and diameter, relative to the amount of mixtures 

4. Rate of elution or solvent flow 

In this experiment, you will use an open column chromatography to separate a 

mixture of two compounds (benzil and benzoin). But first, you will have to determine a 

suitable solvent system using thin-layer chromatography (TLC). The ideal solvent system 

is the one that allows the faster-moving component to have an R f value of about 0.25- 

- 9 -

0.35 on the TLC plate while giving a good separation between the two spots (maximum 

R f difference). When you know the best solvent system, use this solvent as the eluent 

(mobile phase) for the column chromatography. After the separation of the two 

compounds, you will identify them by comparing the R f with that of the authentic 

compounds. 

*** 

R f stands for ratio of fronts or rate of flow*** 


Part A: Mobile phase determination by TLC method 

1) Obtain four TLC plates from the supply room. By using a pencil, not pen, lightly 

draw a line across the short side of each plate, on the silica gel side approximately 1 

cm from the bottom. Be careful not to scratch the silica gel as you are drawing the 

line. 

2) Use small capillary tubes to spot solutions of benzil and benzoin along the line. Keep 

a gap (0.8-1 cm) between the two spots. When spotting the solutions, gently and 

quickly touch the capillary to the surface of the plate so that the spots are not too 

large. Also, write a letter above or below to indicate what is spotted at each position 

(e.g. “A” for benzil and “B” for benzoin) as shown in Figure 1. 

A 

B 

Watch glass 

Beaker 

1 cm 

1 cm 

A = benzil solution 

B = benzoin solution 

Fig 1. TLC plate 

A B 

• • 

Filter paper 

TLC plate 

Solvent level below the 

pencil line on TLC plate 

Fig 2. TLC chamber 

- 10 -

3) Set up a TLC chamber as shown in Figure 2. Put a piece of filter paper in a 100 mL 

beaker. Place a small amount of n-hexane in this beaker. The liquid should cover the 

bottom of the beaker but the surface should be below the pencil line when the plate 

is placed in the beaker (that is, less than 1 cm in depth). The filter paper lining is 

used to saturate the atmosphere within the beaker with solvent fumes. 

4) Place one spotted TLC plate in the TLC chamber, cover with a watch glass and 

allow the solvent to move through the plate until it is approximately 0.5-1 cm from 

the top. Do not disturb the chamber while the plate is being developed!!! 

5) Remove the plate from the chamber and mark the solvent front with a pencil. Allow 

the plate to dry for a few minutes. Place it under short-wave ultraviolet light (254 nm) 

and circle dark spots appear on the plate under the UV light. 

6) Repeat step 2) to 5) for other three TLC plates but each time change the mobile 

phase from pure n-hexane to a mixture of ethyl acetate and n-hexane; 1:1, 1:2 and 

1:4, respectively. 

7) From the results, decide which solvent system would be appropriate for the 

separation of benzil and benzoin (PART B). Report the result to your instructor. 

PART B: Separation of a mixture by column chromatography 

1) Prepare a silica gel column as shown in Fig 3, first plugging the column with cotton 

and then affix to a clamp stand. Place a beaker under the outlet tap. 

Funnel 

Cotton 

Fig 3. Setting up a 

Chromatographic column 

Beaker 

2) In a clean and dry beaker, mix silica gel (~6 g) with your solvent of choice from 

PART A (~30 mL) (CAUTION: Silica gel dust is very harmful if inhaled). Then, slowly 

- 11 -

transfer the slurry into the column using a glass funnel until the silica gel level is 

about 10-12 cm (when settle). If necessary, gently tap the side of the column with a 

rubber tube during the packing process to compact the silica gel. 

3) Open the stopcock to allow liquid to drain into the beaker. Adjust the level of liquid 

around 2-3 cm above the level of silica gel and close the stopcock. 

4) Obtain 0.1 g of solid mixture (benzil+benzoin) from your instructor. Place it in a test 

tube and dissolve with a minimum amount of dichloromethane (~1-1.5 mL). 

5) Open the stopcock and drain the solvent in the column until it reaches the silica gel 

surface, and then close the stopcock. Slowly add the mixture solution into the 

column via a pipette. The flat surface of silica gel should be minimally disturbed. 

6) Open the stopcock to allow sample adsorption onto the silica gel, and then close the 

stopcock. 

7) Rinse the inside wall of the column with 1-2 mL of solvent. 

8) When the solvent reaches the top of the silica gel surface, carefully add 25 mL (or as 

much as your column can contain) of solvent for eluting. It is very important that the 

column never be dried out during the eluting process. 

9) Collect 10 fractions (2 mL each) in test tubes and label as 1, 2, 3,….. 

10) Analyze all of your collected fractions by TLC (Fig 4). 

1234 

Fig 4. TLC plate for 

checking each fraction 

5mm 

1cm 

11) Combine the fractions containing pure benzoin in a ceramic evaporating dish and 

place the dish on a steam bath until a solid or thick oil is obtained. 

12) Allow the dish to cool to room temperature and collect the solid into a pre-weighed 

plastic bag. 

13) Calculate the weight of isolated benzoin. 

14) Calculate recovery percentage. 

15) Repeat step 11) to 14) with fractions containing pure benzil. 

- 12 -


a. Wear safety goggles and lab coat at ALL times. 

b. n-hexane and ethyl acetate are flammable. Never use them near open flame or 

hot plate. 

c. Cover all vessels containing silica dust. It is dangerous if inhaled. 

Waste Disposal 

Place all organic solvents and chemicals into container marked “Organic Waste”. 

Reference 

1) D. L. Pavia, G. M. Lampman, G. S. Kriz, R. G. Engel, Introduction to Organic 

Laboratory Techniques, 3 rd edition, Part 2: technique 10 column chromatography 

and technique 11 thin-layer chromatography, pp.593-629. 

- 13 -

Experiment 

4 

Preparation of Cyclohexene from Cyclohexanol 

by Pattara Sawasdee 

Objectives 

1) To synthesize cyclohexene from cyclohexanol via dehydration reaction of an alcohol. 

2) To classify alkanes and alkenes using chemical reaction test 

Principles 

In this experiment some important properties of hydrocarbons will be studied. You’ll 

perform tests which to distinguishing between saturated hydrocarbons (alkanes) and 

unsaturated hydrocarbons (alkenes). An alkene (cyclohexene) will be prepared by 

dehydration of an alcohol (cyclohexanol). 

Alcohol dehydration is an acid-catalyzed elimination reaction, which can be 

performed by strong, concentrated mineral acids such as phosphoric acid. 

OH 

H 3 PO 4 

+ H 2 O 

cyclohexanol 

b.p. 161 o C 

cyclohexene 

b.p. 83 o C 

In the reaction above, cyclohexene is the only alkene that can be formed under 

these conditions. Cyclohexene and water are removed via azeotropic distillation to drive 

the equilibrium to product (Le Chatelier’s Principle). Traces of acid in the crude product 

are removed by treatment with sodium carbonate solution. A final wash with water 

removes any remaining carbonate. 

Cyclohexene is an unsaturated hydrocarbon. In chemistry, a hydrocarbon is any 

chemical compound that consists only of the elements carbon (C) and hydrogen (H). The 

major classes of hydrocarbons are alkanes, alkenes, alkynes and aromatic hydrocarbons. 

The alkanes are the least reactive class, because they contain only carbon and hydrogen 

and they have no reactive functional groups. There are a number of useful chemical tests 

that can differentiate alkanes from alkenes. These tests are based upon the reactivity of 

alkenes with a variety of reagents to which the alkanes are insensitive. In this experiment, 

you will distinguish cyclohexene (alkene) from cyclohexane (alkane) using bromine and 

permanganate tests. 

- 14 -

1) Bromine test using bromine in chloroform (Br 2 /CHCl 3 ) 

R C C 

H H 

R 

Br H 

+ Br 2 R C C R 

H Br 

A solutions of bromine in CHCl 3 has an intense red-orange color. When Br 2 in 

CHCl 3 is mixed with an alkane, no change is initially observed. When it is mixed with 

an alkene or alkyne, the color of Br 2 rapidly disappears as an addition reaction takes 

place. 

2) Permanganate Test (Baeyer’s test) 

OH OH 

3 R C C R + 2KMnO 4 + 4H 2 O 3 R C C R + 2MnO 2 + 2KOH 

H H 

H H 

The disappearance of the purple color of potassium permanganate and the 

formation of brown precipitate (MnO 2 ) is a positive test. The alkenes are readily 

oxidized by potassium permanganate to form diols. The alkanes are not reacted with 

the potassium permanganate as the purple color remains. 


Part A: Preparation of cyclohexene by dehydration of cyclohexanol 

OH 

H3 PO 4 

heat 

1) Transfer 10 mL (9.4 g.) of cyclohexanol to a 100 mL round-bottomed flask. Add 5 mL 

of 85% H 3 PO 4 . Thoroughly mix the solution by swirling. 

CAUTION: Phosphoric acid is strongly corrosive. If it is in contact with your skin, 

rinse with tap water immediately and report the incident to your instructor.) 

2) Add a few pieces of boiling chips, and assemble the flask for fractional distillation 

(Figure 1) using a 25 mL graduated cylinder in an ice-water bath as a receiver. 

(cyclohexene is very volatile and it will evaporate quite rapidly) 

3) Start circulating the cooling water in the condenser and heat the reaction flask using 

a heating mantle (avoid overheating). The temperature of the distilling vapor should 

be regulated so that it does not exceed 100°C. 

- 15 -

4) When white fumes appear in the round bottom flask, and a few milliliters of liquid 

remains in the reaction flask, discontinue the distillation by turning off the heating 

mantle. 

5) Transfer the distillate* to a small separatory funnel and add about 10 mL of 10% 

aqueous Na 2 CO 3 . Swirl the solution slowly at first and then shake vigorously to 

neutralize the solution. Vent frequently to prevent CO 2 pressure build up. 

6) Allow the layers to separate, drain and test the pH of the aqueous layer (bottom 

layer). Repeat the neutralization until the aqueous layer is basic to litmus. The 

aqueous layer can then be discarded. 

7) Wash the organic layer with 10 mL of water. 

8) Transfer the organic layer to a dried 50 mL Erlenmeyer flask. Add anhydrous 

Na 2 SO 4 to the flask and swirl occasionally until the solution is dry and clear. 

9) Weigh the cyclohexene product and calculate the total yield. 

* distillate = Lliquid condensed from vapor in distillation 

distilland = The material in the distillation apparatus that is to be distilled. 

Thermometer 

Water out 

Water in 

Fractionating 

column 

Condensor 

Graduate cylinder 

ice-water bath 

Heating mental 

Figure 1: a fractional distillation apparatus 

Part B: Test of unsaturation 

Samples : 1. Cyclohexane (Alkanes): from the laboratory 

2. Cyclohexene (Alkenes): from your synthesis (Part A) 

1) Bromine test: Place 3 drops of samples in dry test tubes. Then, add about 3-4 

drops of a bromine solution in chloroform. Stopper each tube, shake, and record 

the observation. If decolorization occurs, test for hydrogen bromide with wet 

litmus. 

- 16 -

2) Permanganate test: Place 3 drops of samples in clean test tubes. Add 3-4 

drops of 0.1% permanganate solution drop by drop while shaking. Watch for 

disappearance of the purple color and formation of a brown precipitate within 1 

minute. 


- Phosphoric acid is a strong acid capable of producing severe burns to skin or eyes. 

- Cyclohexanol can be irritating to the respiratory system and skin. 

- Cyclohexene is not particularly dangerous but is highly flammable and has an 

unpleasant smell. 

- Bromine is highly volatile, toxic, and causes severe skin burns. 


All contents of your test tubes from bromine test reactions go into “HALOGENATED 

organic waste”. 

Quiz 

Quiz will cover this material and the basic knowledge of hydrocarbons. 

- 17 -

Experiment 

5 

Alkyl Halides 

by Paitoon Rashatasakhon 

Objectives 

1) To classify alkyl halides according to their structures and reactivity. 

2) To understand the relationship between structures and reactivity of alkyl halides. 

3) To distinguish the differences in S N 1 and S N 2 reactions. 

Principles 

Alkyl halides are hydrocarbon compounds containing at least one atom of 

halogen directly bonded to an alkyl group. With a general formula R-X, the halogen 

atom could be F, Cl, Br, or I. If the halogen atom is attached to an aromatic ring, the 

compound will be referred to as an aryl halide. In terms of reactivity, aryl halides are 

usually less reactive than alkyl halides. 

Most of the reactions for alkyl halides are Nucleophilic Substitution reaction. 

Nucleophiles are molecules with high electron density or lone pairs of electrons, or 

ions with a negative charge. They can form bond by donating electrons to another 

molecule having a position of lower electron density (electrophiles). Examples of 

nucleophilic species are: water, amines, ammonia, cyanide ion, alkoxide ions, and 

hydroxide ion. 

Alkyl halides can react with a number of nucleophilic reagents, both organic and 

inorganic species. Therefore, alkyl halides are usually good starting materials in the 

synthesis of compounds with other functional groups. 

The reaction may occur by one of two mechanisms designated S N 

1 or S N 

2. 

Which mechanism operates depends on the structure of the R group, nucleophile, and 

the reaction conditions. 

General form of the S N 1 mechanism 

nuc: = nucleophile 

X = leaving group (usually halide) 

This mechanism involves the formation of a carbocation as the crucial 

intermediate in the rate-determining step. The reaction exhibits unimolecular (or "firstorder") 

kinetics, because only one molecule is involved in the rate-determining step. 

Since the mechanism goes through a carbocation, the leaving group must be attached 

- 18 -

to either a tertiary or secondary carbon to stabilize the intermediate. A methyl or primary 

leaving group will not form a carbocation. 

Because the intermediate carbocation, R + , is planar, the central carbon is not a 

stereocenter, even if it was a stereocenter in the original reactant, so the original 

configuration at that atom is lost. Nucleophilic attack can occur from either side of the 

plane, so the product may consist of a mixture of two stereoisomers. In fact, if the 

central carbon is the only stereocenter in the reaction, racemization may occur 

General form of the S N 2 mechanism 

nuc: = nucleophile 

X = leaving group (usually halide) 

The S N 2 reaction involves displacement of a leaving group by a nucleophile. The 

rate of an S N 2 reaction is second order, as the rate-determining step depends on the 

nucleophile concentration, as well as the concentration of alkyl halide. This reaction 

works best with methyl and primary halides because bulky alkyl groups block the 

backside attack of the nucleophile, but the reaction does work with secondary halides 

(although it is usually accompanied by elimination), and will not react at all with tertiary 

halides. In the following example, the hydroxide ion is acting as the nucleophile and 

bromide ion is the leaving group: Because of the backside attack of the nucleophile, 

inversion of configuration occurs. 

- 19 -


1. Reaction with NaI in acetone 

In this part of the experiment, you will test the reactivity of several alkyl halides in an 

S N 

2 reaction. Iodide ion (I - ) is an effective nucleophile in S N 

2 substitution. In acetone 

solution, other alkyl halides (alkyl chlorides or bromides) can be converted to alkyl 

iodides easily by this method. Although one might expect such a reaction to be 

reversible, it can be driven to formation of R-I by using anhydrous acetone as the 

solvent. Sodium iodide (NaI) is soluble in this solvent, but sodium chloride and sodium 

bromide are not. If a reaction occurs, a precipitate of sodium chloride or sodium 

bromide forms and thus the ion is not available in solution for the reverse reaction. 

The mechanism involves a one-step, concerted, S N 

2 reaction. Therefore, the reaction 

occurs most quickly when attack at the carbon that bears the halogen (X) is least 

sterically hindered. 

1. Place 2 drops of each of the following compounds in five separate clean and 

dry test tubes. Label them accordingly. 

n-butyl chloride, s-butyl chloride, t-butyl chloride, n-butyl bromide, bromobenzene. 

2. Add 1 mL of 18% NaI solution in acetone in each test tube. 

3. Stopper, and shake vigorously. 

4. Record the time required to observe precipitate. 

• If no precipitation takes place after 5 minutes, place the test tube in a 

steam bath (45-50°C). Do not allow complete evaporation by adding 

acetone to keep the solution at the same level. Record whether the 

precipitation take place and the time required. 

• If no precipitation takes place after 10 minutes in steam bath, record 

data as “no precipitation”. 

5. Present the final result to your instructor. 

2. Reaction with AgNO 3 in ethanol 

The silver nitrate test allows for the identification of alkyl halides by observing them in 

an alcoholic silver nitrate environment. The rate at which the silver halide salt 

precipitate forms is characteristic of different types of alkyl halides. You will test the 

reactivity of several alkyl halides in a S N 

1 reaction. Organic halides may react with 

ethanol to form ethyl ethers. When the ethanol contains silver ion, the rate of reaction 

increases because the silver ion acts as an electrophile toward the halogen and helps 

to break the carbon-halogen bond. Alkyl chlorides yield an observable silver chloride 

precipitate, which is insoluble in ethanol and thus provides an indicator that a reaction 

has occurred. In this case, the slow step being the breaking of the carbon-halogen 

bond. The carbocation then reacts rapidly with alcohol to form the ether. Organic 

halide reactivity parallels the stability of the corresponding carbocations. 

1. Place 2 drops of each of the following compounds in five separate clean and 

dry test tubes. Label the test tubes accordingly. 

- 20 -

n-butyl chloride, s-butyl chloride, t-butyl chloride, n-butyl bromide, bromobenzene. 

2. Add 1 mL of 1% AgNO 3 in ethanol. 

3. Stopper, and shake vigorously. 

4. Record the time required to observe precipitate. If no precipitation takes place 

after 5 minutes, place the test tube in a water bath (45-50°C). Do not allow 

complete evaporation by adding ethanol to keep the solution at the same 

level. Record whether the precipitation take place and the time required. If no 

precipitation takes place after 10 minutes in water bath, record data as “no 

precipitation”. 


3. Comparison of SN1 and SN2 

Blank test: 

1. Place 1 mL of ethanol, 5 drops of water, and 2 drops of bromophenol blue in a 

test tube. 

2. Stopper and shake the tube vigorously. 

Note: Bromophenol blue pH 3 (yellow) – pH 4.6 (blue) 

Test A: 

1. Place 1 mL of ethanol, 5 drops of water, 2 drops of bromophenol blue, and 5 

drops of n-butyl chloride in a test tube. 


3. Observe the result and compare with the blank test. 


Test B: 

1. Place 1 mL of ethanol, 5 drops of water, 2 drops of bromophenol blue, and 5 

drops of t-butyl chloride in a test tube. 


3. Observe the result and compare with the blank test. 


4. Preparation of alkyl halide 

Alkyl halides may be prepared in a variety of ways, the particular method to be 

employed depending largely upon the nature of the alkyl group and the halogen. 

Some of the general methods which may be used are illustrated in the following 

equations: 

R-H + Cl 2 R-Cl + HCl 

R – CH=CH 2 + HCl R-CHCl-CH 3 

R-OH + HCl R-Cl + H 2 O 

ROH + SOCl 2 RCl + SO 2 + HCl 

Replacement of the hydroxyl group of an alcohol is perhaps the most common 

method. When the reagent is a hydrogen halide, the ease with which this may be 

accomplished increases as one proceeds from a primary to a tertiary alcohol. Tertiary 

alcohols are readily converted to the corresponding chlorides simply by shaking for a 

few minutes, at room temperature, with concentrated hydrochloric acid. 

- 21 -

Synthesis of Tertiary Butyl Chloride 

1. Place 10 mL of cold concentrated hydrochloric acid in a 125 mL separatory 

funnel. 

2. Add 5 mL of tert-butyl alcohol 

3. Without closing the separatory funnel, mix the two components by swirling the 

funnel for a few minutes. 

4. Close the separatory funnel with its stopper, shake the separatory funnel at 

intervals for 10 minutes. From time to time, relieve any internal pressure by 

inverting the funnel and slowly opening the stopcock. 

5. Allow the mixture to stand until two distinct layers separate. 

6. Draw off and discard the aqueous layer (bottom layer) through the stopcock. 

7. Add 15 mL of distilled water, shake the separatory funnel for 30 seconds, put the 

funnel on the stand and relieve the pressure by opening the stopper. 



10. Add 15 mL of distilled water, shake the separatory funnel for 30 seconds, put the 

funnel on the stand and relieve the pressure by opening the stopper. 



13. Weigh and record the weight of an empty clean test tube. 

14. Draw off the organic layer (your product) into the test tube. Weigh the tube with 

your product. 

15. Calculate and record the weight of your product. 

16. Calculate the % yield of the product. 

17. In order to verify the identity of your product, test the product with both NaI and 

AgNO 3 (follow the procedures 1-2 above) 


- 22 -

Experiment 

6 

Alcohols and Phenols 

by Paitoon Rashatasakhon 

Objectives 

1) To classify alcohols according to their characteristic chemical reactions. 

2) To differentiate phenols and the three types of alcohols using simple chemical tests. 

Principles 

The alcohols, with a hydroxyl group attached to an alkyl chain, and the phenols, 

with the same group attached directly to the aromatic ring, have similar chemical 

properties in kind, but may differ considerably in the degree to which these properties are 

exhibited. Alcohols are classified as 1 0 , 2 0 and 3 0 , depending on the number of carbon 

atoms connected to the carbon bearing the OH group. The following experiments are 

designed to bring out these similarities and differences, as well as to demonstrate the 

properties of the hydroxyl group. 

Examples of primary alcohols 

OH 

OH 

OH 

OH 

Examples of secondary alcohols 

OH 

OH OH CH 2 Ph 

H OH 

Cl H 

CH 3 

Examples of tertiary alcohols 

OH 

OH 

OH 

OH 

CH 3 

Examples of phenolic compounds 

OH 

OH 

OH 

HO 

OCH 3 

HO 

- 23 -

Experimental Procedures 

1) Water solubility 

The more carbon atoms present in the molecule of alcohols, the less polar and 

more hydrophobic they become. However, alcohols with straight chain methylene (-CH 2 -) 

units are generally more hydrophobic than those with branch structure. 

1. Place 2 drops of each of the following compounds in six separate clean test tubes. 

Label them accordingly. 

ethanol, 1-butanol, 2-butanol, tert-butanol, cyclohexanol, phenol 

2. Add 10 drops of water into each test tube. Shake and let stand. 

3. Observe whether the compound is soluble in water. 

4. Record the data and present to your instructor. 

2) Alkali solubility 

Alcohols are weaker acids than water, but the aromatic ring makes phenols more 

acidic than water, which means that they may be neutralized by stronger bases such as 

sodium hydroxide. 

1. Place 2 drops of each of the following compounds in three separate clean test tubes. 

Label them accordingly. 

1-butanol, cyclohexanol, phenol 

2. Add 10 drops of 10% NaOH solution into each test tube. Shake and let stand. 

3. Observe whether the compound is soluble in NaOH solution. 


3) Reaction with metallic sodium 

The hydrogen atom of the hydroxyl group can be displaced by active metals such 

as sodium. This reaction can be used as an indication of the presence of an –OH group in 

an unknown compound. The reaction produces hydrogen gas and the corresponding 

sodium alkoxide as shown in the following equation. 

OH 

+ Na 

O - Na + + H 2 

tert-butanol 

sodium 

sodium tert-butoxide 

hydrogen 

1. Place 10 drops of each of the following compounds in three separate clean and dry 

test tubes. Label them accordingly. 

ethanol, 2-propanol, 

tert-butanol 

2. Add a very small piece of sodium into each test tube. 

3. Observe the result and note the relative rates of reaction (gas bubbling). 

4. Add a drop of phenolphthalein to the ethanol tube (only one tube). 

5. Observe the result. 


- 24 -

NOTE: Do not discard the waste from the experiment into the sink. Discard the 

solution into the bottle or beaker labeled “Sodium Waste”. 

4) Reaction with ceric nitrate 

Alcohols with less than 10 carbon atoms can form colorful complexes with Ce 3+ 

ion. This is a characteristic feature of alcohols as the positive test results will have 

different color compared to the blank test. 

1. Place 2 drops of each of the following compounds in four separate clean and dry test 

tubes. Label them accordingly. 

1-butanol, 2-butanol, tert-butanol, phenol 

2. Add 1 mL of ceric nitrate reagent into each test tube. 

3. Observe the result and compare the color with blank test (ceric nitrate reagent + a 

few drops of water) 

4. Present the data to your instructor 

5) Characteristic reactions of phenol 

a) Phenols and compounds with the hydroxyl group attached to an unsaturated 

carbon atom, give coloration (purple/violet) upon the addition of ferric chloride 

(FeCl 3 ) solution. This is due to the formation of complex between phenolic 

compounds and Fe 3+ ion. 

1. Place 1 mL of water into each of two clean test tubes. 

2. Add one drop of phenol in one tube and a few drops of ethanol in the other. 

3. Add 5 drops of 1% ferric chloride solution. Note the characteristic color developed in 

the phenol tube. This is a standard test for most phenol. 


b) The hydroxyl group of the phenols activates the benzene ring to further 

substitution. Bromination using bromine water can proceed smoothly under very 

mild conditions. 

1. Place 1 mL of water into each of two clean test tubes. 

2. Add one drop of phenol in one tube and a few drops of ethanol in the other. 

3. Add bromine water slowly into each tube. If the color disappears, continue adding 

until the color of the bromine just persists. 


6) Lucas test –Differentiation of primary, secondary and tertiary alcohols 

The Lucas test is a test for the ease of replacement of a hydroxyl group by a 

halogen atom, according to the reaction: 

R OH 

+ HCl 

ZnCl 2 

R Cl + H 2 O 

Since the product (alkyl halide) is insoluble in water, the solution becomes cloudy 

and may separate into two layers when the hydroxyl group is replaced with halogen. This 

cloudiness or appearance of a second layer (heterogeneous mixture) is evidence that a 

reaction has occurred. 

- 25 -

When Lucas reagent (ZnCl 2 + conc. HCl) is added to alcohols, H + from HCl will 

protonate the -OH group, so that the leaving group H 2 O, being a much weaker 

nucleophile than OH - , hence can be substituted by nucleophile Cl - . Lucas' reagent offers 

a polar medium in which S N 1 mechanism is favored. In unimolecular nucleophilic 

substitution, the reaction rate is faster when the carbocation intermediate is more stable. 

Therefore, tertiary alcohols react immediately with Lucas reagent to produce turbidity 

while secondary alcohols do so in about five minutes. Primary alcohols do not react 

appreciably with Lucas reagent at room temperature. Hence, the time taken for turbidity to 

appear is a measure of the reactivity of the class of alcohol with Lucas reagent, and this is 

used to differentiate between the three classes of alcohols 

1. Place 3 drops of each of the following compounds in four separate clean and dry test 

tubes. Label the test tubes accordingly. 

1-butanol, cyclohexanol, tert-butanol 

2. Add 1 mL of the Lucas reagent into each tube. 

3. Stopper and shake well. 

4. Observe the result immediately, after 5 min, and again after 30 min. 

5. Present the data to your instructor. 

7) The oxidation reaction 

The oxidation of the carbon atom is an important reaction for this class of 

compounds. When sodium dichromate is used as an oxidizing agent, the orange 

dichromate ion is reduced to the green chromic ion. In this reaction a chromate ester of 

the alcohol substrate is believed to be an intermediate, which undergoes an E2-like 

elimination to the carbonyl product. The oxidation state of carbon increases by 2, while 

the chromium decreases by 3 (it is reduced). The progress of these oxidations is easily 

observed. Indeed, this is the chemical transformation on which the Breathalyzer test is 

based. The secondary alcohols can be oxidized to ketones, while the oxidation of primary 

alcohols initially gives aldehydes which are oxidized further to carboxylic acids. Tertiary 

alcohols will not be oxidized under these conditions. 

1. Place 0.5 mL of a 10% solution of sodium dichromate in a clean test tube. 

2. Add 2 drops of concentrated sulfuric acid and stir with a glass rod. 

3. Add 5 drops of ethanol and warm gently (~40-50°C). 

4. Observe any change in color of the solution. 

5. Repeat step 1-4 with 2-propanol and tert-butanol. 

6. Present the data to your instructor. 

8) Identification of an unknown alcohol. 

After presenting the data from part 1-6 to your instructor, you will receive an 

unknown alcohol. Identify the possible structure of your unknown based on the data 

from ceric nitrate test, FeCl 3 test, Lucas test, and oxidation test. 

- 26 -

Experiment 

7 

Aromatic Chemistry – 

Nitration of Methyl Benzoate 

by Varawut Tangpasuthadol 

Objectives 

1) To perform nitration reaction of methyl benzoate. 

2) To purify the product by recrystallization. 

3) To analyze the purity of the product by TLC and melting point determination. 

Principles 

Nitration of methyl benzoate is a classic example of the electrophilic aromatic 

substitution reaction. In general, the electron-rich aromatic ring is attracted to 

electrophiles, which in this case is a ‘nitronium’ ion (NO 2 + ). The nitronium ion is 

generated from a mixture of concentrated nitric acid and concentrated sulfuric acid as 

shown below. 

HNO 3 + H 2 SO 4 NO + - 

2 + HSO 4 

nitronium iom 

O 

+ H 2 O 

O 

C 

OCH3 

+ NO 2 

+ 

H 2 SO 4 

15 o C 

C 

OCH3 

NO 2 

Methyl m-nitrobenzoate 

The –CO 2 CH 3 group on methyl benzoate is an electron withdrawing group and is 

therefore considered as a meta-directing group. If the reaction condition is very well 

controlled as directed, methyl m-nitrobenzoate should be the only product obtained from 

this reaction, otherwise by-products such as a di-nitro compound may occur. For this 

reason, students are also asked to check the purity of the product by performing TLC of 

their synthesized product. 

Chemicals 

Methyl benzoate 10 drops (~200 mg) irritate, flammable 

Conc. sulfuric acid 3 mL 

corrosive, reacts violently with water 

Conc. nitric acid 1 mL corrosive, oxidizing agent 

- 27 -


1) Withdraw 10 drops of methyl benzoate from a burette into a small conical flask 

with known weight. Record the exact weight of methyl benzoate used. 

2) Immerse the flask from step 1 in an ice-water bath. Slowly add 2 mL of 

concentrated sulfuric acid dropwise while swirling. 

3) Prepare the nitrating agent by mixing 1.0 mL of conc. sulfuric acid and 1.0 mL of 

conc. nitric acid in a test tube that is chilled in an ice-water bath. 

4) SLOWLY add, using a Pasteur pipet, the solution from step 3 into the mixture from 

step 2 while being chilled all the time. Stir the mixture regularly. The adding period 

should not be less than 15 minutes. If the mixture becomes cloudy, add a few 

drops of conc. sulfuric acid until the solution is clear. 

5) Let the reaction mixture stand in a water bath (no ice) at room temperature for 15 

minutes. 

6) Add about 10 g of crushed ice into the reaction mixture. Stir the mixture vigorously. 

7) After the ice has melted, isolate the solid product by vacuum filtration and wash 

with cool water, 5% NaHCO 3 , and cool water again until the filtrate is neutral. 

8) Dry the product in a watch glass over a steam bath, weigh and determine yield 

percentage of the product (before re-crystallization). 

9) Re-crystallize the product from methanol. (Use a minimum amount of hot methanol 

to dissolve the product.) 

10) Weigh the re-crystallized product. Calculate yield percentage after recrystalization 

and recovery percentage. 

11) Check the purity of the product by TLC comparing to the starting material. A 

mixture of hexane-ethyl acetate (3:1) is used as the eluent. 

12) Determine the melting point of the purified product and compare it with a 

reference. 

13) Submit the product to your instructor in a labeled plastic bag. 


All organic solvent wastes are discarded in a container marked, ‘Organic Waste’. 

- 28 -

Experiment 

8 

Aldehydes and Ketones 

by Varawut Tangpasuthadol 

Objective 

To carry out a series of chemical reactions for the classification of aldehydes and 

ketones. 

Principles 

Aldehydes and ketones are organic compounds that contain carbonyl functional groups 

connecting to either one hydrogen and one alkyl (or aryl) group or two alkyl (or aryl) 

groups, respectively. 

O 

O 

R C H R C R' R C CH R' 

O 

OH 

O 

R C CH 3 

Aldehyde ketone α-hydroxy ketone methyl ketone 

The chemistry of these compounds is primarily due to that of the carbonyl 

groups. Identifying methods for aldehydes and ketones can be performed by a number 

of chemical reagents listed in Table 1. 

Table 1 Classification tests for aldehydes and ketones 

Compound 

Aldehydes and ketones 

Aldehydes, α-hydroxy ketones 

Aliphetic aldehydes, α-hydroxy ketones 

Aldehydes 

Aldehydes, α-hydroxy ketones 

Methyl ketone, acetaldehyde 

Reagent 

2,4-Dinitrophenylhydrazine 

Tollen’s reagent 

Benedict’s reagent 

Schiff’s reagent 

Potassium permanganate 

Iodoform test 

Classification tests 

2,4-Dinitrophenylhydrazine (2,4-DNP) reagent 

Most aldehydes and ketones give a precipitate, 2,4-dinitrophenylhydrazone. The 

color of the precipitate is ranged from yellow-orange-red. The color of any precipitate 

must be judged cautiously, since the 2,4-DNP reagent is itself orange-red. Glucose and 

other carbohydrates also give positive results for this test but very slowly, therefore, 

boiling is also required. 

- 29 -

O 

R C R' + 

H 2 N 

H 

N 

NO 2 

NO 2 

2,4-dinitrophenylhydrazine 

H + 

R 

R' 

N 

NH 

NO 2 

NO 2 

2,4-dinitrophenylhydrazone 

+ H 2 O 

Tollen’s test 

Most aldehydes, α-hydroxy ketone and more carbohydrates can reduce silver 

nitrate (AgNO 3 ) solution in ammonia to give silver metal (Ag). The silver may precipitate 

as ‘silver mirror’ along the side of a test tube or as gray powder. The aldehyde is 

oxidized to a carboxylic acid: 

RCHO + 2 Ag(NH 3 ) 2 OH 2 Ag (s) + RCOO - NH 4 + + H 2 O + NH 3 

Benedict’s test 

Only aliphatic aldehydes and α-hydroxy ketone can be oxidized by copper (II) ion 

(Cu 2+ ) to give orange-red precipitate of cupric oxide (Cu 2 O). For some compounds, 

green, blue or brown precipitate can also be expected. Reducing sugars such as 

glucose, fructose, mannose, lactose and maltose can also be oxidized by the Benedict’s 

test. 

RCHO + 2 Cu 2+ + 5 OH - 

RCOO - + Cu 2 O (s) + 3 H 2 O 

Schiff’s test 

Schiff’s reagent is a solution of pararosaniline hydrochloride which is reacted with 

sulfuric acid to give a colorless solution. This solution will turn light purple or pink when 

reacting with an aldehyde. 

Potassium permanganate test 

Potassium permanganate solution contains manganese (VII) ion (Mn 7+ ) that can 

oxidize aldehydes and α-hydroxy ketone to carboxylic compounds and di-ketone, 

respectively. A brown precipitate of manganese dioxide (MnO 2 ) is observed. 

Iodoform test 

Methyl ketone and acetaldehyde form a precipitate of iodoform (CHI 3 ) when 

treated with a basic solution of iodine. The iodoform is yellow in color and may be 

produced in a very small amount. Therefore, the testing result should be carefully 

observed since the yellow precipitate usually sinks to the bottom of the test tubes. 

CH 3 

O 

C 

R' 

I 2 

NaOH 

CI 3 

O 

C 

R' 

OH - 

O 

- O C R' 

+ CHI 3 

iodoform 

Test samples 

Acetaldehyde, acetone, benzaldehyde, glucose (1% solution in water), benzoin (1% 

solution in ethanol), and one UNKNOWN sample 

- 30 -


Note- A blank test should be performed in every test. 

1) Reaction with 2,4-dinitrophenylhydrazine (2,4-DNP) 

Place one drop of a test sample in a test tube and add 1 mL of 2,4-DNP. Shake the 

solution well. If no precipitation is observed, gently heat the sample test tube about 15 

minutes. Compare the appearance of the precipitate and the time required for the 

precipitation of all test samples. 

2) Tollen’s reagent 

Place 2 drops of a test sample in a test tube. Add 1 mL Tollen’s reagent. Shake the 

solution well. If a mirror of silver is deposited on the inner walls of the test tube, the test 

is positive. Otherwise heat the test tube for 1 minute, let it cool down to room 

temperature. Record the change in appearance. 

3) Schiff’s reagent 

Place 1 mL Schiff’s reagent in a test tube. Add 1 drop of a test sample. Shake the 

solution. Record the result. 

4) Benedict’s test 

In a test tube, place 1 mL Benedict’s solution and add 1 drop of a test sample. Shake 

the solution well. If there is no change, warm up the test tube for about 5 minutes. 

Record the result. 

5) Oxidation with KMnO 4 

Place 4 drops of test sample in a test tube. Add 2 drops of 0.1% KMnO 4 solution. Shake 

the solution well. Record the result. 

6) Iodoform test 

Use only acetaldehyde, acetone and benzaldehyde for this test. 

Place 1 mL distilled water in each of 4 test tubes. Add one drop of the test sample in 

each tube. Add 1 mL of 5% NaOH solution. Add iodine-potassium iodide (I 2 -KI) solution 

one drop at a time until a permanent pale yellow solution is obtained. Shake the tube 

well after each addition. Allow the test tube to stand for 2-3 minutes. If the test is 

positive, the yellow solution will decolorize and a yellow precipitate of iodoform will form. 

If the color of the solution is discharged but no precipitation, add some more I 2 -KI 

solution until the pale yellow color returns. Allow the solution to stand for 2 more 

minutes. It may be necessary to warm the solution in a water bath (~60 °C) to aid in the 

discharge of the color. 

7) Classification test for an unknown compound 

Obtain an unknown sample from the instructor. Record the sample number. Perform the 

tests according to the procedure. From the results, try to identify the detailed structure 

of the unknown. Consult with your instructor and write down the structure in the report. 

- 31 -

Experiment 

9 

Synthesis of Esters and Reactions of 

Carboxylic Acids and their Derivatives 

by Warinthorn Chavasiri 

Objectives 

1) To synthesize an ester using acid-catalyzed esterification reaction. 

2) To study the reactions of carboxylic acids and their derivatives. 

Principles 

Carboxylic acid is an important class of organic compounds. Acid derivatives differ 

from their parent compound in that the hydroxyl (-OH) portion is replaced by another 

group. Important acid derivatives include acid halides, anhydrides, esters, and amides. 

O 

R 

C 

OH 

carboxylic acid 

O 

O 

O 

O 

O 

R 

C 

X 

R 

C 

O 

C 

R 

R 

C 

OR' 

R 

C 

NH 2 

acid halide 

anhydride 

ester 

amide 

Carboxylic acids can be converted into their salts by treatment with base. Because 

these salts are ionic and usually water-soluble, acids have low solubility in the presence 

of base in water can be extracted from a solution into an organic solvent. 

O 

O 

R 

C 

less soluble 

in water 

OH + NaOH R C 

O - Na + 

more soluble 

in water 

+ H 2 O 

Acid halides are usually prepared by reacting the corresponding acid with an 

inorganic halide such as PCl 3 , PCl 5 or thionyl chloride (SOCl 2 ). Acid anhydrides can be 

prepared by dehydration of the corresponding carboxylic acids. The acid-catalyzed 

reaction of an alcohol with a carboxylic acid is the most commonly used method for ester 

preparation. Amides can be prepared by heating ammonium salts of acids or by the 

reaction of ammonia, primary-, or secondary-amines with various acid derivatives. Most 

reactions of acid derivatives involve nucleophilic attack on the carbonyl carbon. 

- 32 -

Preparation of an ester 

Background 

Esters are found in many natural products, contributing to the scent of banana, 

orange, pineapple, and other fruits. The structure of ester determines its scent. By 

reacting different alcohols with carboxylic acids, you can produce esters of different 

scents. There are many different methods for synthesizing esters. However, the two 

most common ones are 1) acid-catalyzed Fischer esterification using a carboxylic acid 

and an alcohol; and 2) condensation of acid chloride with an alcohol. 

The Fischer esterification is an equilibrium reaction in which an acid and an alcohol 

combine to produce the ester and water. For example, the acid catalyzed reaction for 

the formation of ethyl acetate from acetic acid and ethanol. 

H 3 C 

O 

C 

OH 

+ CH 3 CH 2 OH 

H+ 

H 3 C 

O 

C 

OCH 2 CH 3 

+ H 2 O 

To drive the equilibrium towards completion, either the starting carboxylic acid or the 

alcohol is used in excess. Alternatively, if the ester has a significantly different boiling 

point than the alcohol or acid, it can be separated from the acid and alcohol by 

distillation. 

The second synthetic route to esters employs an acid chloride, which has to be 

prepared in an additional step. The reaction requires a base such as K 2 CO 3 or 

triethylamine to destroy the hydrochloric acid by-product. 

O 

O 

H 3 C 

C 

base 

Cl + CH 3 CH 2 OH H 3 C C 

OCH 2 CH 3 

Techniques 

Reflux 

Extraction using separatory funnel 

Simple distillation 

- 33 -


Part A Preparation of fruity ester (CARE: extremely corrosive) 

Each group will obtain an unknown carboxylic acid and an unknown alcohol. You will 

then synthesize an ester from the two starting materials, identify the scent of the ester 

product, purify the product by distillation, and predict the identity of the two starting 

materials based on the scent of the ester and its boiling point. 

Procedure 

1. Pour all of the unknown alcohol and carboxylic acid into a 50 mL round bottom 

flask. The alcohol was accurately measured to be 0.046 mol and the carboxylic 

acid was measured to be 0.12 mol. 

2. Carefully add 1.5 mL of concentrated sulfuric acid to the reaction flask. Add 

boiling stones to the mixture. 

3. Assemble the reflux apparatus. Bring the mixture to boil using a heating mantle 

and heat the mixture under reflux for 1 hour. During this time, you may proceed 

to Part B. 

4. Remove the heating source and allow the mixture to cool to room temperature. 

5. Pour the cooled mixture into a separatory funnel and carefully add 20 mL of 

distilled water. Rinse the reaction flask with 5 mL of distilled water and pour the 

rinsing into the separatory funnel. 

6. Stopper the funnel and shake it several times. Separate the lower aqueous layer 

from the upper organic layer. Label the unwanted aqueous layer and put it aside. 

7. The crude ester in the upper organic layer still contains some unreacted acid 

which can be removed by extraction with NaHCO 3 solution. Carefully add 10 mL 

of 5% NaHCO 3 to the organic layer in the separatory funnel. Without the stopper, 

swirl the funnel gently until carbon dioxide gas no longer evolved. 

8. Place the stopper, shake the funnel, relief the internal pressure, and settle the 

funnel to allow phase separation to occur, then remove and discard the lower 

layer. 

9. Add 10 mL of 5% NaHCO 3 to the organic layer in the separatory funnel. Place 

the stopper, shake the funnel, relief the internal pressure, and settle the funnel to 

allow phase separation to occur. 

10. Remove the lower layer and check whether it is basic to litmus paper. If it is not 

basic, repeat step 9 until the aqueous layer is basic. 

11. Add 10-mL portion of water. Place the stopper, shake the funnel, relief the 

internal pressure, and settle the funnel to allow phase separation. 

12. Add 10 mL of saturated sodium chloride to aid in layer separation. 

13. Carefully separate and discard the lower layer. 

- 34 -

14. Transfer the organic layer into an Erlenmeyer flask. Add about 2 g of anhydrous 

MgSO 4 to dry the solution. 

15. Assemble a simple distillation apparatus and carefully decant the ester solution 

into the distillating flask. 

16. Add boiling stones and distill the ester. Collect the fraction and observe the 

boiling range. 

17. Weigh the product. 

18. Report the scent and boiling point of your ester product to your instructor. Also, 

identify the unknown starting carboxylic acid and alcohol. 

19. Calculate the percentage yield based on the molecular weight of your product. 

- 35 -

Table of selected ester flavors and fragrances 

Complete this table before starting the experiments. 

Draw the structures of the carboxylic acids, alcohols and esters. 

Find out the boiling range of the esters from the references. 

Carboxylic Alcohol Ester Scent b.p. 

acid 

Acetic acid Isoamyl alcohol Isoamyl acetate banana 

Propionic acid Isobutyl alcohol Isobutyl propionate rum 

Anthranilic 

acid 

Methyl alcohol Methyl anthranilate grape 

Acetic acid Benzyl alcohol Benzyl acetate peach 

Butyric acid Methyl alcohol Methyl butyrate apple 

Butyric acid Ethyl alcohol Ethyl butyrate pineapple 

Acetic acid Octyl alcohol Octyl acetate orange 

Acetic acid n-Propyl alcohol n-Propyl acetate pear 

- 36 -

Part B 

Reactions of carboxylic acids and their derivatives 

Samples: 

O 

H OH 

Formic acid 

O 

H 3 C OH 

Acetic acid 

O 

O 

OH 

OH 

Oxalic acid 

O OH O OCH 3 

Benzoic acid 

Methyl benzoate 

O 

HN CH 3 

Acetanilide 

1. Solubility Experiments 

a) In water 

Add approximately 2 drops of liquid sample or 0.1 g of solid sample into water 

(3 mL). Shake well, observe the solubility, and record your observation. 

b) In 5% NaOH solution 

Use the same amount of sample as above, and use 3 mL of 5% NaOH solution. 

Observe the solubility in 5% NaOH solution and record your observation. 

c) In 5% NaHCO 3 solution 

Use the same amount of sample as above, and use 5% NaHCO 3 solution. 

Observe the solubility in 5% NaHCO 3 and look for evolution of gas. 

2. Ferric hydroxamate reaction 

Place a drop of methyl benzoate in a test tube and add 0.5 M NH 2 OH.HCl in 

ethanol (1 mL). Add 20% NaOH until the solution becomes basic to litmus. Warm 

up the mixture on the water bath for 5 min and then cool down to room 

temperature. Add 1M HCl until the solution becomes acidic or the brown 

precipitate dissolves. Add 5% FeCl 3 until a permanent color is observed. Record 

your observation and then repeat with benzoic acid. 

3. Reaction with Tollens’ reagent 

Place approximately 4 mL of Tollens’ reagent in each of 5 test tubes and add in 

the samples in the first four tubes as listed in the table below. The last tube is 

used as a blank test. 

Tube # Tollens’s reagent (mL) Sample 

1 

2 

3 

4 

5 

4 

4 

4 

4 

4 

Formic acid (5 drops) 

Acetic acid (5 drops) 

Oxalic acid (0.1 g) 

Benzoic acid (0.1 g) 

- = blank test 

Place the tubes in water bath for 5 min and compare with the blank test. Record 

your observation. 

- 37 -

4. Reaction with KMnO 4 

Place approximately 1 mL of distilled water in each of 5 test tubes and add in the 

reagents and samples as listed in the table below. The last tube is used as a 

blank test. 

Tube 

# 

Water (mL) 

Conc. 

H 2 SO 4 

(drop) 

0.3% 

KMnO 4 

(mL) 

Sample 

1 

2 

3 

4 

5 

4 

4 

4 

4 

4 

1 

1 

1 

1 

1 

1 

1 

1 

1 

1 

Formic acid (5 drops) 

Acetic acid (5 drops) 

Oxalic acid (0.1 g) 

Benzoic acid (0.1 g) 

None (= blank test) 

Gently shake and place the tubes in water bath for 1 min and compare with the 

blank test. Record your observation. 

Caution: Extreme care must be exercised to avoid contact with concentrated sulfuric 

acid (safely goggles must be worn at all times). If it comes in contact with the skin or 

clothes, it must be washed off immediately with excess water. In addition, sodium 

bicarbonate may be used to neutralize the acid. Clean up all spills immediately. 

Reference 

1. “Introduction to Organic Laboratory Techniques: A Small Scale Approach”, Pavia, 

Lampman, Kriz and Engel, Brooks/Cole, 2 nd Ed, 2005. 

2. “Theory and Practice in the Organic Laboratory” Landgrebe, Brooks/Cole, 5 th Ed, 

2005. 

3. “Microscale and Miniscale Organic Chemistry Laboratory Experiments” Schoffstall, 

Gaddis, Druelinger, McGraw Hill, 2 nd Ed, 2004. 

4. “Organic Chemistry”, Solomon and Fryhle, John Wiley & Sons, 8 th Ed, 2004. 

- 38 -

Experiment 

10 

Oxidation and Reduction of Benzoin 

by Sumrit Wacharasindhu 

Objectives 

1) To understand oxidation-reduction processes in organic reactions. 

2) To perform simple oxidation and reduction reactions of benzoin and related 

techniques in laboratory-scale synthesis. 

Principles 

Focusing on the functional groups in a molecule allows us to recognize patterns 

in the behavior of related compounds. Consider what we know about the reaction 

between sodium metal and water, for example; 

2 Na(s) + 2 H 2 O(l) H 2 (g) + 2 Na + (aq) + 2 OH - (aq) 

We can divide this reaction into two half-reactions. One involves the oxidation of sodium 

metal to form sodium ions. 

Oxidation: Na Na + + e - 

The other involves the reduction of an H + ion in water to form a neutral hydrogen atom 

that combines with another hydrogen atom to form an H 2 molecule. 

Reduction: 

2 

O 

H H H + OH 

2H 2e 2H H 2 

Once we recognize that water contains an -OH functional group, we can predict what 

might happen when sodium metal reacts with an alcohol that contains the same 

functional group. Sodium metal should react with methanol (CH 3 OH), for example, to 

give H 2 gas and a solution of the Na + and CH 3 O - ions dissolved in this alcohol. 

2 Na(s) + 2 CH 3 OH(l) H 2 (g) + 2 Na + (alc) + 2 CH 3 O - (alc) 

Because they involve the transfer of electrons, the reactions between sodium metal and 

either water or alcohol are examples of oxidation-reduction. But what about the 

following reaction, in which hydrogen gas reacts with an alkene in the presence of a 

transition metal catalyst to form an alkane 

- 39 -

H H H H 

Ni 

C C + H 2 H C C 

H H 

H H 

H 

There is no change in the number of valence electrons on any of the atoms in this 

reaction. Both before and after the reaction, each carbon atom shares a total of eight 

valence electrons and each hydrogen atom shares two electrons. Instead of electrons, 

this reaction involves the transfer of atoms in this case, hydrogen atoms. There are 

so many atom-transfer reactions that chemists developed the concept of oxidation 

number to extend the idea of oxidation and reduction to reactions in which electrons 

aren't necessarily gained or lost. 

Oxidation involves an increase in the oxidation number of an atom. 

Reduction occurs when the oxidation number of an atom decreases. 

During the transformation of ethene into ethane, there is a decrease in the oxidation 

number of the carbon atom. This reaction therefore involves the reduction of ethene to 

ethane. 

-2 

H H H H 

Ni 

C C + H 2 H C C 

H H 

H H 

-3 

H 

Reactions in which none of the atoms undergo a change in oxidation number are called 

metathesis reactions. Consider the reaction between a carboxylic acid and an amine, 

for example. 

CH 3 CO 2 H + CH 3 NH 2 CH 3 CO 2 + CH 3 NH 3 

Or the reaction between an alcohol and hydrogen bromide. 

CH 3 CH 2 OH + HBr CH 3 CH 2 Br + H 2 O 

These are metathesis reactions because there is no change in the oxidation number of 

any atom in either reaction. The oxidation numbers of the carbon atoms in a variety of 

compounds are given in the following table. 

- 40 -

Typical Oxidation Numbers of Carbon 

Functional Group Example 

Oxidation Number of 

Carbon in the Example 

Alkane CH 4 -4 

Alkyllithium CH 3 Li -4 

Alkene H 2 C=CH 2 -2 

Alcohol CH 3 OH -2 

Ether CH 3 OCH 3 -2 

Alkyl halide CH 3 Cl -2 

Amine CH 3 NH 2 -2 

Alkyne HC CH -1 

Formaldehyde H 2 CO 0 

Aldehyde RCHO +1 

Carboxylic acid RCO 2 H +3 

Carbon dioxide CO 2 +4 

In this experiment, benzoin having both a secondary alcohol and a ketone 

functional group can be oxidized to a diketone, benzil, or reduced to a diol, hydrobenzoin. 

In this reaction, the commonly used reducing agent, sodium borohydride, is used for the 

reduction. The oxidation can be accomplished with any of several oxidizing agents, such 

as nitric acid. 

O 

OH 

HNO 3 

Benzoin 

NaBH 4 

MW 212.24 

O 

OH 

O 

Benzil 

MW 210.23 

OH 

Hydrobenzoin 

MW 214.26 

- 41 -


Part I Oxidation of benzoin to benzil (step 1-3 must be carried out in fume hood) 

1. Place 1.0 g of benzoin into a 50 mL Erlenmeyer flask and carefully add 7 mL of 

concentrated nitric acid. 

2. Heat the mixture on a steam bath with occasional slow swirling for 30 min or until the 

brown-red nitric oxide gas longer evolves. Make sure that the fume hood safety 

shield is pulled down. 

Caution: Concentrated nitric acid is highly corrosive and can cause severe burns. 

Nitric oxide (NO 2 ) fume is highly toxic and can damage the lungs. 

3. Pour the reaction mixture into 25 mL of cool water and stir to coagulate the 

precipitated product. 

4. Collect the yellow solid by suction filtration and wash the precipitate twice with 5 mL 

of cool water to remove trace amount of nitric acid. 

5. Air dry the product on the filtration set for 1 min and transfer it onto a watch glass. 

6. Remove trace amount of water in the product by placing another piece of dry filter 

paper over and pressing with the bottom of a small beaker or round-bottom flask. 

7. Recrystallize the product by adding 5 mL of 95% ethanol, heating the mixture in the 

steam bath to make a clear solution, and adding water dropwise until cloudiness 

occurs. Allow the solution to cool to room temperature and then place in an ice bath. 

8. Collect the yellow solid by suction filtration. 

9. Air dry the product on the filtration set for 1 min and transfer onto a watch glass. 

10. Weigh the product, record the yield, and determine the melting point. 

Part II Reduction of benzoin to hydrobenzoin 

1. Place 1.0 g of benzoin into a 50 mL Erlenmeyer flask and dissolve it with 10 mL of 

95% ethanol. 

2. Carefully add, in small portions, over 5 minutes, 0.20 g of sodium borohydride while 

swirling. This reaction is exothermic. Do not add sodium borohydride too rapidly. 

3. Allow the reaction to proceed at room temperature for 20 min with frequent swirling. 

4. Cool the reaction in an ice bath, add 15 mL of water, then 0.5 mL of 6 M HCl. 

5. Add another 5 mL of water and allow the mixture to stand for 20 min at room 

temperature with frequent swirling. 

6. Collect the product by suction filtration and rinse with 50 mL of water. 

7. Recrystallize the product by adding 5 mL of 95% ethanol, heating the mixture in the 

steam bath to make a clear solution, and adding water dropwise until cloudiness 

occurs. Allow the solution to cool to room temperature and then place in an ice bath. 

8. Collect the yellow solid by suction filtration. 

9. Air dry the product on the filtration set for 1 min and transfer onto a watch glass. 

10. Weigh the product, record the yield and determine the melting point. 

- 42 -

Experiment 

11 

Amines 

by Duangamol Nuntasri 

Objectives 

1. To classify amines according to their characteristic chemical reactions 

2. To use the chemical characteristics to identify amine sample 

Principles 

Amines are organic compounds that resemble ammonia but at least one hydrogen 

atom is replaced by organic substituents like alkyl (alkane chain) and aryl (aromatic 

ring) groups. 

Types of Amines 

Amines can be classified as primary, secondary or tertiary. If there is only one carboncontaining 

group (such as in CH 3 NH 2 ) then it is considered primary. Two carboncontaining 

groups make it secondary, and three groups make it tertiary. The lone pair of 

electrons on the nitrogen is sometimes donated to form a fourth carbon-containing 

group to the amine. In this case, quaternary ammonium salt (R 4 N + X - ) is obtained. 

Primary amine Secondary amine Tertiary amine 

H 

N 

R 1 

H 

N 

R 1 

R 1 

R 3 N 

H 

R 2 

R 2 

An organic compound with multiple amine groups is called a diamine, triamine, 

tetraamine and so forth, based on the number of amino groups on the molecule. The 

formula for the simplest diamine, for example, is 

Aromatic amines 

H 2 N-CH 2 -NH 2 

Aromatic amines have the nitrogen atom directly bonded to an aromatic ring. Due to its 

delocalization properties, the aromatic ring greatly decreases the basicity of the amine - 

this effect can be either enhanced or offset depending on the substituents on the ring 

and on nitrogen. The presence of the lone pair electrons from the nitrogen has an 

opposite effect on the electron density of the ring. It causes the ring to become much 

more reactive, particularly towards electrophiles. 

- 43 -

Naming conventions 

Generally amines are named for their carbon structures with the amine functionality 

included as either a prefix (amino-) or a suffix (-amine"). Generally, smaller molecules 

will use the suffix form, while larger chains will list amine functionality as if it were any 

other type of functional group. Examples include methlyamine (CH 3 NH 2 ), and 2- 

aminopentane (CH 3 NH 2 CH(CH 2 ) 2 CH 3 ). 

• as prefix: "amino-" 

• as suffix: "-amine" 

• the prefix "N-" shows substitution on the nitrogen atom (in the case of secondary, 

tertiary and quaternary amines) 

Systematic names for some common amines: 

small amines are named 

with the suffix -amine. 

large amines have the prefix 

amino as a functional group. 

H 3 C CH 3 

H 2 N CH 3 NH 2 

Methylamine 

2-Aminopentane 

Physical properties 

1. Basicity 

1.1 Solubility and 1.2 Indicator testing 

All three classes of amines form hydrogen bonds with water. 

N + H OH 

N H OH 

For this reason, low-molecular weight amines are water soluble. Borderline water 

solubility is observed when the amine has about six carbon atoms. Aqueous solutions of 

amines are alkaline (basic). 

N 

H OH 

N H + OH 

Like ammonia, amines act as reasonably strong bases (see the provided table for some 

examples of conjugate acid K a values). The basicity of amines varies by molecule, and 

it largely depends on: 

• The availability of the lone pair electrons on the nitrogen 

- 44 -

Ions of compound 

K b 

ammonia NH 3 1.8 × 10 -5 

methylamine CH 3 NH 2 4.4 × 10 -4 

propylamine CH 3 CH 2 CH 2 NH 2 4.7 × 10 -4 

2-propylamine (CH 3 ) 2 CHNH 2 5.3 × 10 -4 

diethylamine (CH 3 ) 2 NH 2 9.6 × 10 -4 

+I inductive effect of alkyl groups raises the energy of the lone pair of electrons, thus 

elevating the basicity. 

• The electronic properties of the attached substituent groups (e.g., alkyl groups 

enhance the basicity, aryl groups diminish it, etc.) 


K b 

ammonia NH 3 1.8 × 10 -5 

aniline C 6 H 5 NH 2 3.8 × 10 -10 

4-methylphenylamine 4- 

CH 3 C 6 H 4 NH 2 

1.2 × 10 -9 

+M mesomeric effect of aromatic ring delocalizes the lone pair electron into the ring, 

resulting in decreased basicity. 

• The degree of solvation of the protonated amine, which depends mostly on the 

solvent used in the reaction 

The nitrogen atom of a typical amine features a lone electron pair which can bind a 

hydrogen ion (H + ) in order to form an ammonium ion -- R 3 NH + . The water solubility of 

simple amines is largely due to the extent of hydrogen bonding that occurs between 

protons of water and N lone pairs. 

Amine protonation VS water solubility: 


NH 4 

+ 

+ 

RNH 3 

+ 

R 2 NH 2 

R 3 NH + 

Maximum number of H-bond 

4 Very Soluble in H 2 O 

3 

2 

1 Least Soluble in H 2 O 

- 45 -

1.3 Salt Formation 

Amines are weak organic bases that form salts with strong acids. 

N + H + Cl - N H Cl 

Like ordinary salts, ammonium salts are readily dissociate into ions and are therefore 

water soluble. The salts can be reconverted to amines by making their solutions 

alkaline. 

N H Cl + Na + OH - N + Na + Cl - + H 2 O 

2. Reaction with bromine in water 

The amino functional group in aniline works as electron donating group to the benzene 

ring via resonance effect, making electrophilic aromatic substitution reaction more 

favorable than for non-substituted aromatics. For the bromination of aniline, bromine 

acts as electrophile which gets directed to substitute hydrogen atoms at ortho- and 

para- positions, producing 2-4-6-tribromoaniline. With, aliphatic amines this type of 

reaction cannot occur. 

NH 2 

+ 3 Br 2 

NH 2 

Br 

Br 

+ 3 HBr 

3. Reaction with nitrous acid 

Nitrous acid (HNO 2 ) is unstable, and so a mixture of NaNO 2 and dilute acid is usually 

used to produce nitrous acid in situ. Primary aliphatic amines with nitrous acid give very 

unstable diazonium salts which spontaneously decompose by losing N 2 gas, forming a 

carbonium ion. The carbonium ion goes on to produce a mixture of alkenes, alcohols or 

alkyl halides, with alcohols as the major product. This reaction is of little synthetic 

importance because the diazonium salt formed is too unstable, even under quite cold 

conditions. 

NaNO 2 + HCl → HNO 2 + NaCl 

Br 

R 

H 2 C NH 2 

HCl 

NaNO 2 

R 

C 

H 2 

N 

N 

R CH 2 + N N H 2 C OH 

R 

Primary aromatic amines, such as aniline (phenylamine) form a more stable diazonium 

ion at 0–5°C. Above 5°C, it decomposes to give phenol and N 2 . Diazonium salts can be 

isolated in the crystalline form but are usually used in solution and immediately after 

preparation, due to rapid decomposition on standing even with little ambient heat. Solid 

diazonium salts can be explosive on mild warming. 

- 46 -

NH 2 

HCl / NaNO 2 

5 o C 

N 

N 

If the primary amine is readily converted to a diazonium salt and loses nitrogen, bubbles 

will be visible in a very short period of time. If no bubbles of nitrogen appear, you can 

assume that the amine is secondary or tertiary rather than primary. 

At low temperatures, diazonium salts are stable and will couple with the alpha position 

with the sodium salt of β-naphthol to form a red colored precipitates. 

Cl 

N 

N 

+ 

OH 

OH 

N 

N 

β-naphthol 

(colorless) 

Azo compound 

(red) 

4. Hinsberg test 

The reaction between primary or secondary amines and benzenesulfonyl chloride in the 

presence of sodium hydroxide solution yields the corresponding substituted 

benzenesulfonamide, whereas tertiary amines do not react with this combination of 

reagents. 

With primary amines the sulfonamide formed is acidic and dissolves in the excess base 

used to yield a solution of the corresponding anion. Addition of excess hydrochloric acid 

converts the anion into the water-insoluble sulfonamide. 

NaOH 

RNH 2 + SO 2 Cl SO 2 NHR 

+ NaCl + H 2 O 

Insoluble in water 

excess 

HCl 

excess 

NaOH 

SO 2 NRNa + H 2 O 

Solubility in water 

Secondary amines react with benzenesulfonyl chloride but the sulfonamide lacks an 

amide hydrogen and is insoluble in the both aqueous basic and acid reagents. 

- 47 -

R 2 NH 

NaOH 

+ SO 2 Cl SO 2 NR 2 

+ NaCl + H 2 O 

Insoluble in water 

excess 

NaOH 

No reaction 

Tertiary amines react differently with benzenesulfonyl chloride; the intermediate 

ammonium ion does not have a proton to lose and reacts rapidly with hydroxide ion to 

displace the benzenesulfonate anion and regenerate the tertiary amine. 

NaOH 

R 3 N + SO 2 Cl SO 2 NR 3 Cl 

water soluble 

excess 

NaOH 

SO 3 + NR 3 + Cl 

water soluble 

The overall reaction amounts to an amine-catalyzed hydrolysis of the benzenesulfonyl 

chloride. With tertiary amines there can be a side reaction between the amine and the 

intermediate ammonium ion to produce a complex mixture of water-insoluble products, 

which could lead to confusion with the results for a secondary amine. This complication 

can be minimized by keeping the concentration of the amine low. If there is any doubt 

regarding the interpretation of the results, carry out the test on known compounds and 

compare the results with those of the unknown. 

5. Rimini’s test, Simon’s test, modified Rimini’s test and modified Simon’s test 

Rimini’s reagent (sodium nitroprusside and acetone) and Simon’s reagent (sodium 

nitroprusside and acetaldehyde) can be used as to distinguish primary amines from 

secondary aliphatic ones. When amines react with Rimini’s reagent, different colors are 

produced, according to Table 1. Modified versions of Rimini’s reagent and Simon’s 

reagent, however work well for identifying aryl amines. i.e. primary, secondary and 

tertiary aromatic amines. 

- 48 -

Table 1 The developed color of Rimini’s test, Simon’s test, modified Rimini’s test and 

modified Simon’s test 

Reagent 1 o aliphatic 2 o aliphatic 1 o aromatic 2 o aromatic 3 o aromatic 

Rimini purple deep red - - - 

Simon 

Modified 

Rimini 

Modified 

Simon 

pale yellow 

to 

red brown 

- - 

- - 

deep blue - - - 

orange red 

to 

red brown 

orange red 

to 

red brown 

Blue green 

purple 

blue green 

blue green 

Amine samples for chemical reaction testing 

Aliphatic amine 

C 4 H 7 

C 4 H 7 NH 2 

C 4 H 7 NH 

n-butylamine 

di-n-butylamine 

Aromatic amine 

NH 2 

CH 3 

NH 

CH 3 

N CH 3 

Aniline N-methylaniline N,N-dimethylaniline 

- 49 -


1. Basicity testing 

1.1 Water solubility 

Test with n-butylamine, di-n-butylamine and N,N-dimethylaniline 

To 2 mL of distilled water in a test tube, add 5 drops of amine and note its solubility. 

Test the resulting solution with pink litmus paper. 

1.2 Indicator test 

Test with n-butylamine and aniline 

Add 2 drops of amine to each of two test tubes containing 1 ml of distilled water. To one 

tube add a drop of phenolphthalein indicator and to the other add a drop of 

bromothymol blue solution. Interpret the results observed on the basis of the following 

information: bromothymol blue; below pH 6 the color is yellow, at pH 7 it is yellow-green, 

and above pH 7.5 the color is blue. Phenolphthalein is colorless at pH 8.4 and red 

above pH 8.6. 

1.3 Salt formation 


To 10 drops of 6M HCl, add 2 drops of amine and shake. Note whether the amine is 

soluble in aquous acid. Add 6M NaOH dropwise until the solution becomes alkaline, and 

observe. 

2. Reaction with bromine in water 


To 1 ml of distilled water in test tubes, add 1 drop of amine then bromine water 

dropwise until excess of bromine and shake. 

3. Reaction with nitrous acid (Azo dye) 


To 1 mL of conc. HCl and 1mL of water, add amine 5 drops. Stir the solution while 

cooling in ice bath (ensure temperature does not exceed 5 o C). Dissolve 0.2 g NaNO 2 in 

1 mL distilled water and cool in ice bath at the same temperature. Add NaNO 2 solution 

dropwise with stirring or shaking to the cold amine hydrochloride. The endpoint can be 

determined by putting a drop of the solution on starch-KI paper. A blue color is observed 

when excess nitrite is present. Observe whether bubbles appear or not. 

Compare the result between n-butylamine and aniline. 

Remove about 1 mL of the aniline diazonium chloride solution into test tube and heat on 

water bath at 50 o C. Observe the change and odour. 

To another portion of the cold aniline diazonium solution, add a pre-chilled solution of 

0.1 g of β-naphthol in 1 mL of 5% NaOH. Observe the result. 

- 50 -

4. Hinsberg Test 

Test with aniline, N-methylaniline and N,N-dimethylaniline 

To 2 drops of amine in test tube with 1 mL of water, add 1 mL of 10% aqueous NaOH 

and 2 drops of benzenesulfonyl chloride (in hood). Close test tube tightly with stopcock 

and vigorously shake the tube for 5 mins and note any reaction. If unreacted 

benzenesulfonyl chloride is left as an oil drop at bottom of tube. In this case, warm the 

mixture do not boil in hood with shaking for 10 min. The reaction mixture should still be 

strongly alkaline at this point. Cool the test tube to room temperature, shake well, and 

note whether any solid or liquid separates. Do not confuse any separated material with 

unreacted benzenesulfonyl chloride. Observe the result. 

Separate half amount of reaction mixture (both liquid and solid) into new test tube. Add 

2M of HCl until solution turns acidic. Observe if a precipitate forms. If so, this indicates 

that the unknown was a primary amine. However, if the organic material does not 

dissolve in the hydrochloric acid, it indicates a secondary amine (acid- and baseinsoluble 

secondary sulfonamide). If the organic material is soluble, it indicates a tertiary 

amine (acid soluble, unreactive toward benzenesufonyl chloride). 

5. Rimini test, Simon test, modified Rimini test and modified Simon test 

5.1 and 5.2 test with aliphatic amine (n-butylamine and di-n-butylamine) 

5.3 and 5.4 test with aromatic amine (aniline, N-methylaniline and N,N-dimethylaniline) 

5.1 Rimini test 

To 1 mL of sodium nitroprusside reagent A (0.13 M in 50% aq. methanol) in test tube 

with 1 mL of water, add 5 drops of acetone and 3 drops of amine. Shake the tube and 

observe result within 2 min. 

5.2 Simon test 

To 1 mL of sodium nitroprusside reagent A (0.13 M in 50% aq. methanol) in test tube 

with 1 mL of water, add 5 drops of 2.5 M acetaldehyde solution and 3 drops of amine. 

Shake the tube and observe result within 2 min. 

5.3 Modified Rimini test 

To 1 mL of sodium nitroprusside reagent B (0.13 M in 80% aq. dimethyl sulfoxide) in 

test tube with 1 mL of saturated ZnCl 2 aqueous solution, add 5 drops of acetone and 3 

drops of amine. Shake the tube and observe result within 2 min. 

5.4 Modified Simon test 

To 1 mL of sodium nitroprusside reagent B (0.13 M in 80% aq. dimethyl sulfoxide) in 

test tube with 1 mL of saturated ZnCl 2 aqueous solution, add 5 drops of 2.5 M 

acetaldehyde solution and 3 drops of amine. Shake the tube and observe result within 2 

min. 

6. Classification test for an unknown compound 

Obtain an unknown sample from your instructor. Record the sample number. First, 

determine if the unknown is an alkyl or aryl amine and then determine whether the 

amine is primary, secondary, or tertiary. Consult with your instructor and write down the 

answer in your report sheet. 

Caution 

- Carefully dispose the red azo compound and reaction mixture from reaction with 

nitrous acid by washing down the drain without contact to your skin. 

- 51 -

Experiment 

12 

Identification of Organic Compounds 

by Duangamol Nuntasri 

and Paitoon Rashatasakhon 

Objectives 

1. To classify organic compounds to the corresponding solubility classes. 

2. To determine functional groups of organic compounds by chemical tests 

Principles 

Qualitative organic analysis, the identification and characterization of organic 

compounds, is an important part of organic chemistry. In this experiment you will be 

issued an unknown compound and asked to identify it through chemical methods. Your 

instructor will give you an unknown, you must first determine the class of compound to 

which the unknown belongs. Then identify its main functional group and determine the 

specific compound in the class that corresponds to the unknown. Unknown compounds 

are restricted to include only eight possible important functional groups: aldehydeketone, 

carboxylic acid, phenol, amine, alcohol, alkyl halide, unsaturated, and 

ester. The experiment presents all of the chemical methods of determining the main 

functional groups, and it includes methods for verifying the presence of the subsidiary 

functional groups as well. 

How to proceed 

1. Preliminary classification by physical state. 

2. Determination of solubility in different solvents 

3. Application of relevant chemical classification tests 

Preliminary classification 

Note the physical characteristics of the unknown. These include its color, its 

odor, and its physical state (liquid, solid, crystalline form). Compounds with a high 

extent of conjugation are frequently yellow to red. Amines often have a fishlike odor. 

Esters have a pleasant fruity or floral odor. Acids have a sharp and pungent odor. As a 

note of caution, many compounds have distinctly unpleasant or nauseating odors. 

Some have corrosive vapors. Any unknown substance should be smelled with the 

greatest caution. Initially, open the unknown container, holding it away from you. Then, 

using your hand, carefully waft the vapors toward your nose. 

Solubility behavior 

Solubility tests are important in determining the nature of the main functional 

group of the unknown compound. These a tests are simple, require only small amounts 

of unknown, and can reveal whether the compound is a base (amine), an acid (phenol), 

a stronger acid (carboxylic acid), or a neutral substance (aldehyde, ketone, alcohol, 

ester). Common solvents used to determine solubility include water, 5% NaOH, 5% 

NaHCO 3 , 5% HCl, and conc. H 2 SO 4 . 

- 52 -

The solubility tests and solubility classes (Table 1) provides a useful and logical 

approach to determining compound class from solubility observations. The possible 

functional groups for each solubility class are listed in Table 2. 

Chemical classification tests 

The solubility tests usually suggest or eliminate several possible functional 

groups. The chemical classification tests listed in Table 3 allows one to distinguish 

among these possible choices. You should choose to perform only those meaningful 

tests; time is wasted doing unnecessary tests. The solubility and the main chemical 

tests will be possible to identify the class of compound. Solubility tests automatically 

eliminate the need for some of the chemical tests. Each successive test will either 

eliminate the need for another test or dictate its use. Many possibilities may be 

eliminated on the basis of logic alone. 


Part 1 Solubility tests 

Five unknowns (A to E) will be used for this experiment. The order of solvents for 

this test should be as follows: 

Water 5% NaOH 5% NaHCO 3 5% HCl concentrated H 2 SO 4 

Place about 1 mL of the solvent in a test tube. Add two drop or 50 mg (about the 

size of one grain of rice) of an unknown directly into solvent. Shake the test tube to 

ensure good mixing, and then observe whether the unknown is soluble. The 

disappearance of the liquid or solid, or the appearance of the mixing lines, indicates that 

dissolution is taking place. Add several more drops of the liquid, or a few more crystals 

of the solid, to determine the extent of the compound’s solubility. A common mistake in 

determining the solubility of a compound is testing with a quantity of the unknown too 

large to dissolve in the chosen solvent. Use small amounts. It may take several minutes 

to dissolve solids. Compounds in the form of large crystals will need more time to 

dissolve than powders or very small crystals. In some cases it is helpful to first pulverize 

a compound composed of large crystals. Sometime gentle heating helps, but strong 

heating is discouraged, as it often leads to reaction. When colored compounds dissolve, 

the solution often assumes the same color. 

By the above procedure, the solubility of the unknown should be determined in 

each of the following solvents: water, 5% NaOH, 5% NaHCO 3 , 5% HCl, and 

concentrated H 2 SO 4 . With sulfuric acid, a color change may be observed rather than 

dissolution. A color change should be regarded as a positive solubility test. 

If compound is found to dissolve in water, the pH of the aqueous solution should 

be estimated with pH paper or litmus. Compounds soluble in water are usually soluble 

in all the aqueous solvents. If a compound is only slightly soluble in water, it may be 

more soluble in acid or alkali. For instance, a carboxylic acid may be only slightly 

soluble in water but very soluble in dilute base. It will often not be necessary to 

determine the solubility of the unknown in every solvent. 

- 53 -

Part 2 Solubility test and determine functional group of specific unknown 

- Obtain your specific unknown from the instructor 

- Record the unknown number 

- Report the physical character of the unknown 

- Classify unknown into the solubility class, and check your answer before 

proceed to the next step 

- Examine unknown functional group using appropriate chemical tests as shown in 

Table 3 

- Report the answer of unknown in your report sheet 

Ref. Pavia, D.L.; Lampman, G.L.; Kriz, G.S. Introduction to Organic Laboratory 

Techniques, 3 rd ed., W.B. Saunders Company, New York. 

- 54 -

Table 1 

Solubility tests and Solubility classes 

Water 

Soluble and 

change blue 

litmus into red 

Soluble but does 

not change blue 

litmus into red 

Soluble and 

change red 

litmus into blue 

Soluble but does 

not change red 

litmus into blue 

Insoluble 

Insoluble 

Insoluble 

Solvent (1 ml) : unknown (2 drops or 1 grain of rice) 

5% 

NaOH 

- 

- 

5% 

NaHCO 3 

5% 

HCl 

Produces 

CO 2 gas 

No gas 

evolves 

conc. 

H 2 SO 4 

5% 

FeCl 3 

- - - 

- - 

- - - - 

- - - - 

Soluble 

Slightly 

soluble 

Insoluble 

Insoluble Insoluble - 

Insoluble Insoluble - 

Note 1. “ - ” = unnecessary. 

Change color of 

FeCl 3 solution 

Produce redbrown 

precipitate 

Does not change 

color of FeCl 3 

solution 

Solubility class 

Water soluble strong 

acid 

Water soluble weak acid 

Water soluble base 

Water soluble neutral 

compound 

Slightly 

soluble and 

produces 

CO 2 gas 

- - - Strong acid 

Insoluble - 

- - Weak acid 

- 

Soluble 

(contains N 

atoms) 

Insoluble 

(contains N or 

S atoms) 

Insoluble 

(no N or S 

atom) 

- - Basic compound 

- - 

Miscellaneous neutral 

compound 

(this class of compounds is not 

available in this course) 

Soluble - Neutral compound 

- 55 -

Table 2 

Functional group tests for compounds in each solubility class 

Functional groups 

Solubility class 

Neutral 

Carboxylic 

Phenol 

(see note 1) 

acid 

Water soluble strong acid √ √ 

Water soluble weak acid 

√ 

Water soluble base 

Water soluble neutral compound √ 

Strong acid √ √ 

Weak acid 

√ 

Basic compound 

Neutral compound 

√ 

Amine 

Note : 

1. The following functional groups are possible for this class of compounds. 

1.1. carbonyl compounds (aromatic or aliphatic aldehyde, α-hydroxy ketone, methyl ketone, 

or other ketone) 

1.2. alcohol (primary, secondary, or tertiary alcohol) 

1.3. ester 

1.4. unsaturated hydrocarbon (alkene or alkyne) 

1.5. alkyl halide (primary, secondary, or tertiary alkyl halides) 

1.6. nitro compound (not available in this course) 

1.7. thiol (not available in this course) 

1.8. amide (not available in this course) 

1.9. ether (not available in this course) 

√ 

√ 

The solubility flow-chart below provides a summary of information from Table 1 and Table 2. 

- 56 -

turn red litmus to blue 

turn blue litmus to red 

low MW 

amines 

low MW 

carboxylic acids 

soluble 

No change with 

litmus 

neutral, hydrogen 

bonding or neutral but 

polar 

Unknown 

H 2 O 

insoluble 

5%NaHCO 3 

soluble 

soluble 

insoluble 

strong acids 

weak acids 

carboxylic acids 

some phenols 

phenols 

5%NaOH 

insoluble 

soluble 

bases 

amines 

5%HCl 

insoluble 

conc.H 2 SO 4 

soluble 

insoluble 

neutral 

compounds 

alkenes esters 

alkynes ethers 

alcohols amides 

ketones thiols 

aldehydes 

alkyl halides 

nitro compounds 

inert 

compounds 

alkanes 

aromatic 

- 57 -

Table 3 Functional group tests and required reagents 

Functional group Sub-functional group Reagents 

(see note 1) 

Solvent for solid 

unknown 

Aldehyde, ketone Aldehyde: aliphatic,aromatic 1. 2,4-DNP Ethanol 

(see exp. 8) 

2. Schiff’s reagent Ethanol 

Alcohol 

(see exp. 6) 

Ester 

(see exp. 9) 

Unsaturation 

(see exp. 4) 

Carboxylic acid 

(see exp. 9) 

Phenol 

(see exp. 6) 

Amine 

(see exp. 11) 

Alkyl halides 

(see exp. 5) 

Ketone: α-hydroxy ketone, 

methyl ketone, other ketones 

3. Tollens’ reagent Ethanol 

4. Benedict’s reagent Ethanol 

5. Iodoform test Ethanol 

1°, 2°, 3° alcohol 1. Ceric nitrate reagent 

(positive results= red or orange) 

2. Lucas reagent 

3. Oxidation test 

Aliphatic, aromatic amine 

1°, 2°, 3° amine 

- Ferric hydroxamate test - 

- 1. KMnO 4 

2. Br 2 /CCl 4 

- 5% NaHCO 3 - 

- 

1. 5% FeCl 3 Ethanol 

2. Ceric nitrate reagent Dioxane or acetone 

3. Br 2 /H 2 O Ethanol 

1. 2 M HCl 

2. Sodium Nitroprusside Test 

(Ramini, Simon, Modified 

Ramini and Simon) 

3. Diazotization & Coupling 

4. Br 2 /H 2 O 

1°, 2°, 3° alkyl halide 1. AgNO 3 / EtOH 

2. NaI / actone 

Note: 

1. For liquid unknown, perform the tests without any solvent. 

2. For solid unknown which is soluble in water, perform the tests using a solution of the 

unknown in water. 

3. For solid unknown which is insoluble in water, perform the tests using a solution of the 

unknown in solvent listed in Table 3. 

- 

- 

- 

- 

- 58 -

Crystallization and Determination of Melting and Boiling Points

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