Grignard Synthesis of Triphenylmethanol

Objective:

1. To synthesis triphenylmethanol from Grignard reaction

2. To study the method to produce Grignard reagent

Introduction:

Grignard reagents are organomagnesium halides (RMgX), and are one of the most synthetically useful and versatile classes of reagents available to the organic chemist. An alkyl, benzyl, or aromatic halide is reacted with a magnesium metal by using an anhydrous solvent in order to produce Grignard reagent. Ether or tetrahydrofuran are usually can be used as the anhydrous solvent in producing the particular reagent. Tetrahydrofuran is a strong base and it has a better solvating ability, it may used when Grignard reagent does not readily form in diethyl ether. This is considered as an organometallic compound which consists of the combination of a metal and organic molecule. Figure 1 in below shows the general reaction mechanism for the formation of Grignard reagent.

The chemical reaction between an organic halide and a magnesium metal can produce an alkyl or aryl free radical and magnesium free radical. The formation of Grignard reagent has been occurs. The bonding between carbon and magnesium is a covalent bond but it is highly polarized because the magnesium is bonded to halide which is an electron withdrawing group. This causes the formation of partial positively charge and partial negatively charge on the magnesium atom and alkyl or aryl group respectively. Hence, the carbanion has both characteristics of a good nucleophile and a strong base. Its basicity allows it to react with the electrophile carbon in a carbonyl group. Besides, Grignard reagent also works with acidic compound such as carboxylic acid, phenol, thiol, alcohol, and even water. One of the most important reactions is the addition of Grignard reagent to the carbonyl compound like aldehyde, ketone, and ester in order to produce the corresponding secondary alcohol and tertiary alcohol.

Figure 2

The most challenging part of this experiment is to avoid the Grignard reagent contact with water. The partial negative charge on the carbon atom that bonded to magnesium exhibits a very basic property. The water molecule will destroy the nucleophilic property of Grignard reagent. So, several precaution steps must be taken in the procedures to avoid the Grignard reagent reacts with water: the reaction flask is dried in the oven before use; iodine is vaporized in the flask tie up traces of water and to activate the surface of magnesium; the anhydrous diethyl ether should be used.

Figure 3

The figure 3 above shows the Grignard reagent and water reaction. The metal hydroxide (alkoxide) formed in the above reaction is appears as insoluble white solid (HO-MgBr or RO-MgBr) in the diethyl ether solvent. Thus, the process of producing Grignard reagent must be start over again in a dry glass if the insoluble white solid is observable during the formation of Grignard reagent.

For a variety of reasons, anhydrous diethyl ether is the solvent choice for carrying out a Grignard synthesis. One of the reasons is the vapors of the highly volatile diethyl ether helps to prevent the oxygen in the atmosphere to reach the reaction solution. Grignard reagent will reacts with oxygen which hydroperoxides is produced. This compound is highly unstable if exposed to the air so that the compound is usually not isolated from the solvent. Other than that, the basic oxygen atoms in ether molecules are actually coordinate with and help to stabilize the Grignard reagent. The figure 4 below shows how the anhydrous diethyl ether protects Grignard reagent from oxidation:

Figure 4

There is a layer of oxide coated on the surface of magnesium which used to synthesis Grignard reagent. The oxide layer works to prevent it to react with alkyl bromide. The formation of Grignard reagent is highly exothermic which will produce a lot of heat energy from the system. Once the reaction has been initiated, the system will reflux itself in the bottom flask without any external heat source. The adding of a drying tube that contains calcium chloride to the reflux apparatus is used to protect the reaction from atmospheric moisture.

In this experiment, bromobenzene is the alkyl halide used to generate Grignard reagent.

Figure 5

Once the Grignard reagent is readily formed, the carbonyl compound has been introduced into the reagent in order to synthesis the expected product. The methyl benzoate (ester) acts as the carbonyl containing compound in the experiment. The reaction between methyl benzoate and Grignard reagent is showing in the following figure 6:

Figure 6

Dissociation of magnesium alkoxide produces a ketone which tends to react further with more Grignard reagent. The final step of the synthesis is involving hydrolysis of the magnesium alkoxide by using a mineral acid. As the result, the reaction synthesizes an alcohol, triphenylmethanol and magnesium salt (water soluble).

A side reaction may take place in the reaction between phenylmagnesium bromide and bromobenzene. The side product has been produced is biphenyl which consists of two phenyl rings. The biphenyl is known as impurity in the experiment. The impurity can be removed from the product through a method of recrystallization since biphenyl is much more soluble in ligroin compared to triphenylmethanol.

Apparatus: dropping funnel, two neck round bottomed flask, drying tube, condenser, sonicator, magnetic stirrer, hot plate, separating funnel, beaker, Buchner funnel, glass funnel, melting point apparatus

Materials: magnesium turning, anhydrous diethyl ether, bromobenzene, calcium chloride, methyl benzoate, iodine crystal, 10% H₂SO₄, ice, sodium bisulfate, sodium sulfate anhydrous, petroleum ether, methyl spirit

Procedure:

Part A – Preparation of phenylmagnesium bromide (phenyl Grignard reagent)

1. A condenser and a 50ml dropping funnel were set up to a 250ml two neck round bottom flask.

2. A calcium chloride drying tube was inserted into the top of the condenser.

3. 1.4g of magnesium turning was weighed and placed into the two neck round bottom flask with a stir bar. 10ml of anhydrous diethyl ether was added immediately into the two neck bottom flask.

4. 6.2ml of bromobenzene and 30ml of anhydrous diethyl ether were added into the 50ml dropping funnel.

5. 5ml of the mixture in dropping funnel was added to the ether/magnesium mixture. The mixture is stirred with a magnetic stirrer.

6. If the reaction does not begin immediately, both palms were placed around the bottom of the flask to keep it warm.

7. A small amount of iodine crystal was added directly into the magnesium surface if the reaction does not take place after 5-10minutes.

8. Once the reaction was initiated and the formation of Grignard reagent became steady, the ether refluxed itself. The remaining mixture in dropping funnel was added dropwise into the round bottom flask.

9. The solution was allowed to reflux for 10 minutes.

Part B – Preparation of triphenylmethanol from Grignard reagent

1. After reflux, the round bottom flask was cooled down in ice bath and a solution of 3.2 ml of benzoate dissolved in 15ml anhydrous diethyl ether was placed in the dropping funnel.

2. The solution was added over 5 minutes to avoid the solution to get too exothermic.

3. Once the reaction was completed, the solution was heated to reflux for 10 minutes to complete the reaction.

4. After reflux, the round bottom flask was cooled with ice bath and the mixture was poured into a 600ml beaker with contains 75g of ice and 30ml of 10% H₂SO₄.

5. The mixture was stirred until all white solid was dissolved.

6. The mixture was poured into the separating funnel and separated the two layers. The ether layer was washed with 15ml of H₂SO₄, followed by 15ml of water and then with a solution of 1g of sodium bisulfite dissolved in 12ml of water.

7. The ether layer was dried over sodium sulfate anhydrous and filtered with cotton wool. 15ml of petroleum ether was added to the ether layer. The ether layer was concentrated over a steam bath at 60-80 °C until the triphenylmethanol was formed. The product was cooled in an ice bath.

8. The product was recrystallized from methyl spirit.

9. The weight, yield and melting point of triphenylmethanol were determined.

Results and calculation

Weight of magnesium = 1.4009g

Volume of bromobenzene = 6.2ml

Volume of methyl benzoate = 3.2ml

Weight of watch glass = 14.8091g

Weight of watch glass + weight of product = 13.7410

Weight of product = 1.0681g

Atomic weight of magnesium = 24.31g/mol

Number of mole of magnesium = 1.4009 g / 24.31 g mol^-1

= 0.0576mol

Molecular weight of bromobenzene = 157g/mol

Density of bromobenzene = 1.495 g/cm³

Weight of bromobenzene = density x volume

= 1.495 g/cm³ x 6.2cm³

= 9.269g

Number of mole of bromobenzene = 9.269g/ 157 g mol^-1

= 0.0590mol

Since the magnesium is limiting reagent, so the number of mole of Grignard reagent produced is limited by number of mole of magnesium.

Number of mole of Grignard reagent produced = 0.0576mol

Density of methyl benzoate = 1.0837 g/cm³

Weight of methyl benzoate = density x volume

= 1.0837 g/cm³ x 3.2cm³

= 3.468 g

Molecular weight of methyl benzoate = 136.144g/mol

Number of mole of methyl benzoate = 3.468 g / 136.144g mol^-1

= 0.0255 mol

Since the Grignard reagent is excess, the product is limited by methyl benzoate so that it is limiting reagent in the reaction.

Thus, the number of mole of triphenylmethanol = 0.0255 mol.

Molecular weight of triphenylmethanol = 260.318g/mol

Theoretical weight of triphenylmethanol = 0.0255mol x 260.318g/mol

= 6.6381 g

Actual weight of triphenylmethanol =

Percentage yield = actual yield / theoretical yield x 100%

= 1.0681g / 6.6381g x 100%

= 16.09%

Melting point of triphenylmethanol = 153°C ~ 156°C

Discussion:

The purpose of this experiment is to synthesis triphenylmethanol by using Grignard reagent. In order to synthesis triphenylmethanol, Grignard reagent is playing an important role because Grignard reagent is the key reagent in this experiment. The presence of water in the process of generating Grignard reagent will causes the particular reagent to be decomposed. So, the solvent used in the experiment must not contain any water such as diethyl ether since it is a water free solvent. In order to make sure the water in air can be eliminated, a small amount of calcium chloride has been placed with the drying tube on the top the condenser. The calcium chloride acts as a drying agent which to absorb all the water from the air in the reflux apparatus and it prevent the atmospheric moisture. The purpose of using magnetic bar is to increase the rate of reaction for Grignard reagent.

In order to produce Grignard reagent, the magnesium turning was added with anhydrous diethyl ether. Magnesium turning (thin shaving with high surface area) is usually used in preparation of Grignard reagent due to its large surface area that can increase the reaction rate. The diethyl ether functions as the medium for the Grignard reaction to take place and stabilize the reagent. This is because the solvent (diethyl ether) is highly volatile solvent which can prevent the water from atmosphere approaching to the Grignard reagent after the Grignard reagent is formed. Besides, diethyl ether is easily removed from the reaction mixture since it has a low boiling point of 36°C. A mixture of bromobenzene and diethyl ether was prepared in the dropping funnel. The adding of diethyl ether in the mixture is works for the similar function which make sure the solvent is free from water. The addition of ether and bromobenzene mixture into the magnesium surface might not allow the reaction to occur due to lack of heat energy.

The iodine crystal was added into the magnesium surface because the heats from water bath or palm were not enough to initiate the reaction. Alternatively, the iodine crystal was added instead of increasing the temperature that supplied to the system in order to prevent the explosion since diethyl ether is highly flammable. The iodine crystal facilitate the reaction either activating the magnesium through removal of its oxide coating or by oxidizing the bromide in organic compound to form negatively charged bromide which is more reactive towards magnesium. A second alternative is place the flask containing reaction mixture over a sonicator to start the Grignard reaction. The sonicator is used to produce ultrasonic wave in which helps to remove the oxide coating physically.

When the reaction has been initiated, the appearance of bubbles on the solvent surface indicated that the formation of phenylmagnesium bromide start to occurr. The bromobenzene is reacted with magnesium metal to form phenylmagnesium bromide which is known as Grignard reagent. The chemical reaction between magnesium and bromobenzene is shows in below:

The formation of Grignard reagent is solvated by diethyl ether which protects the reagent from attacked by water molecules. If the water reacts with Grignard reagent, the decomposition of the particular reagent will occur. The mixture of ether and bromobenzene was added slowly to make sure that the Grignard reagent form steadily in the reaction. The side product may be generated in high yield if the mixture is added in a large volume at the same time. The formation of Grignard reagent is an exothermic process. Thus, the system can refluxed itself without any heat supply to it.

After reflux, the Grignard reagent produced was cooled down in an ice bath in order to reduce its temperature. This is to prevent the immediate addition of solvent from evaporating quickly due to high temperature of Grignard reagent if cooling down process is not taken. The methyl benzoate is the subsequent reactant which was used to react with Grignard reagent in this experiment. In order to avoid the reaction between Grignard reagent and methyl benzoate get too exothermic, the methyl benzoate in anhydrous diethyl ether must be added in a small amount. The system was refluxed itself to produce (Ph)₃CO-Mg-Br as product.

The Grignard reagent can be dissociated to form negatively charged carbanion which attacked the carbonyl carbon with partial positively charged. The carbonyl carbon of methyl benzoate was attacked by the nucleophilic carbanion during reflux. The nucleophilic addition of Grignard reagent to methyl benzoate caused the methoxide became the leaving group from the intermediate and the formation of benzophenone. Since the benzophenone consists of a carbonyl carbon as functional group, this favored the second nucleophilic attack of Grignard reagent and (Ph)₃CO^- anion with three benzene ring has been produced in the solution through reflux. The solution was treated by using sulphuric acid to protonate the (Ph)₃CO^- anion to generate the triphenylmethanol, (Ph)₃COH as product. Triphenylmethanol has a synonym which is known as triphenylcarbinol. The formation of triphenylmethanol is highly exothermic, so ice bath was used to reduce the temperature and heat energy produced from the system. In this stage, some white solid were precipitated out in the cold solution, the white solid is the desired product. The mechanism of formation of triphenylmethanol by using Grignard reagent via nucleophilic addition is shown in the following Diagram 1:

Diagram 1

In order to remove the impurities and side product, the washing process is necessary. The ether layer was washed with sulphuric acid and followed by water. The aqueous solution was used to remove the water-soluble impurities in the mixture. Then, sodium bisulfide was a base which was used to neutralize the acid added before. Sodium sulfate anhydrous has been introduced to remove all the water in the mixture since it is a drying agent and the clump of solid sodium sulfate was filtered. In the process of producing triphenylmethanol, some side products have been produced at the same time such as biphenyl.

Therefore, petroleum ether was used in the experiment in order to let the biphenyl dissolved in it so that this side product can be removed via recrystallization. This might be due to the higher polarity of the triphenylmethanol compared to biphenyl which enables the triphenylmethanol to dissolve easily in the polar methyl spirit.

Recrystallization of triphenylmethanol has been carried out to purify the product. Methyl spirit (denatured alcohol) is a mixture of methanol and ethanol which was used as the dissolve medium in recrystallization. After recrystallization, the product is still not pure enough since its melting point of 153°C ~ 156°C is lower than the theoretical melting point which is 162°C. This deviation may be due to the product is not completely dry so that it affects the melting point of the product. The yield of the product in the experiment is 1.0681g which contributes to the percentage yield of 16.09%. The percentage yield is very low may be due to there are many impurities were formed in the reaction since the impurities compete the material which required for the formation of desired product.

Precaution steps:

1. Avoid the diethyl ether from any heat source since it is extremely flammable.

2. Carry out the reaction away from any heat source.

3. Place all the glassware in a 110°C in order to make sure all the glassware is totally dry.

4. Use solvent to wash the glassware instead of distilled water.

Synthesis of Dibenzalacetone by Aldol Condensation

Objective:

1. To carry out a mixed aldol condensation reaction

2. To study the mechanism of aldol condensation reaction

Introduction:

The reaction of an aldehyde with a ketone employing sodium hydroxide as the base is an example of a mixed aldol condensation reaction, the Claisen-Schmidt reaction. The double mixed-aldol condensation reaction between acetone and benzaldehyde was carried out. Acetone has α-hydrogens (on both sides) and thus can be deprotonated to give a nucleophilic enolate anion. The alkoxide produced is protonated by solvent, giving a β-hydroxyketone, which undergoes base-catalyzed dehydration. The elimination process is particularly fast in this case because the alkene is stabilized by conjugation to not only the carbonyl but also the benzene. In this experiment, excess benzaldehyde such that the aldol condensation can occur on both sides of the ketone.

Dibenzalacetone is readily prepared by condensation of acetone with two equivalent of benzaldehyde. The aldehyde carbonyl is more reactive than that of the ketone and therefore reacts rapidly with the anion of the the ketone to give a β-hydroxyketone, which easily undergoes base catalyzed dehydration. Depending on the relative quantities of the reactants, the reaction can give either mono- or dibenzalacetone.

Dibenzalacetone is a fairly innocuous substance in which its spectral properties indicate why it is used in sun-protection preparations. In the present experiment, sufficient ethanol is present as solvent to readily dissolve the starting material, benzaldehyde and also the intermediate, benzalacetone. The benzalacetone once formed, can then easily to react with another mole of benzaldehyde to give the desired product in this experiment, dibenzalacetone.

Apparatus: Erlenmeyer flask, Buchner funnel, glass funnel, melting point apparatus, UV/Vis spectrometer, FTIR spectrometer

Materials: Benzaladehyde, acetone, sodium hydroxide, 95% ethanol, ethyl acetate, ice

Procedure:

1. 5g of NaOH was added to 25ml of H­₂O in an Erlenmeyer flask and the solution was swirled.

2. 25ml of 95% ethanol was added and the solution was allowed to come nearly to room temperature.

3. 2.9g of acetone and 10.5ml of benzaladehyde were added.

4. After 15 minutes of occasional swirling, the products was filtered on a Buchner funnel.

5. The product was washed with cold ethanol and was allowed to suck dry.

6. The yellowish product was recrystallized from ethyl acetate.

7. After recrystallization, a yellow crystalline was obtained.

8. The weight, yield, and melting point of the product were determined.

9. The UV and IR spectra of dibenzalacetone were ontained.

Results:

Weight of watch glass = 36.1291 g

Weight of watch glass + products = 45.6878 g

Weight of products = 9.5587

Melting point of products = 109 °C

Moles of benzaldehyde used = 0.1 mol

Moles of acetone used = 0.05 mol

Moles of dibenzalacetone produced = Moles of acetone used

= 0.05 mol

Theoretical weight of dibenzalacetone produced = 0.05 mol * 234.29 g mol^-1

= 11.7145 g

Percentage yield of dibenzalacetone = 9.5587 g / 11.7145 g × 100 %

= 81.60 %

Discussion:

Condensation is a process which joins two or more molecules usually with the loss of a small molecule such as water or an alcohol. Aldol condensation (Claisen-Schmidt reaction) definitely is a process which join two carbonyl groups with a loss of water molecule in order to form β-hydroxyketone. The product is also known as adol because it containing two functional groups which includes aldehyde (or ketone) group and alcohol group. The product dibenzalacetone was formed from the reaction between an acetone molecule and two benzaldehyde molecules. Generally, the aldol condensation is carried out under a base condition.

Sodium hydroxide was mixed with distilled water then was used to react with sufficient ethanol as the first step. The particular reaction is an exothermic reaction which released the heat energy to the surrounding from the reaction. The sodium hydroxide was functioned as a catalyst in the reaction. The ethanol acts as a solvent which allows the acetone and benzaldehyde to dissolve and react with each other. After that, acetone and benzaldehyde were mixed in the solvent which turns to yellow colour quickly. Eventually, the product was formed with a yellow precipitate appear in the reaction after a few seconds. However, there are some impurities and side products were formed in the yellow precipitate. So, recrystallization was carried out by using ethyl acetate as solvent in order to purify the product and hence a pure product could be obtained for the ultraviolet (UV) and IR spectra analysis. In the recrystallization process, the yellow precipitate in ethyl acetate was immersed into an ice-bath in order to obtain a higher yield of product. This is because the heat energy in the precipitate easily to be released since the precipitation formation is an exothermic reaction and hence it maximizes the formation rate of the product.

Acetone is considered as a stable and unreactive compound, so it should be converted into anionic form to increase its nucleophile properties to initiate the reaction. The sodium hydroxide dissolves in water to produce hydroxide ion and it tends to attack the α-hydrogen in acetone and to form water molecule. The deprotonation of acetone caused the enolate ion was produced as nucleophile which will be used in the synthesis of dibenzalacetone. An enolate ion was formed which it exists as resonance-stabilized structure which shown in the following diagram:

Diagram 1

The acetaldehyde enolate ion attack to the benzylic carbon of benzaldehyde via nucleophilic addition to form the intermediate as shown in below:

Diagram 2

The oxygen attached to the benzylic position of carbon tends to attract one proton from water molecule to form hydroxide group in the intermediate. This is the formation of an aldol since the molecule consists of a carbonyl group and an alcohol group. In the basic condition, the hydroxide ion tends to remove one proton from the α-carbon resulting the formation of C=C double bond at the α and β carbon. At the same time, the hydroxide group attached to the β carbon forms a leaving group. After the condensation, benzalacetone was formed after two water molecules leaved as shown:

Diagram 3

The benzalacetone tends to form benzalacetone enolate ion after the hydroxide group from the surrounding attack the proton which attached to the carbon at benzylic position.

Diagram 4

The same process has been take place as in the Diagram 2 but with the more bulky benzalacetone enolate ion as the material. The benzalacetone enolate ion acts as a nucleophile which attacks another benzaldehyde. The protonation of the aldol took place followed by the hydroxide groups have been eliminated as leaving groups. As a result, the nucleophilic addition and base-catalyzed dehydration lead to the formation of the desired product which is dibenzalacetone. The mechanism of dibenzalacetone formation was shown in the Diagram 5:

 

 

 

 

 

 

 

 

Diagram 5

The overall mechanism of the dibenzalacetone was summarized in the Diagram 7 as shown in below:

Diagram 7

The percentage yield of dibenzalacetone in this experiment is 83.27%. Some of the product has been lost during the process of recrystallization. In recrystallization, some of the product dissolved in the ethyl acetate. The melting point of the product is lower than the actual melting point (110 °C ~ 111 °C). This is because there is some impurities exist in the particular compound which will tend to lower the melting point of the dibenzalacetone.

Precaution steps:

1. Avoid to carry out the experiment near the fire since the organic solvent are mostly flammable.

2. Avoid to smell benzaldehyde directly.

3. Handle carefully with sodium hydroxide since it is corrosive.

Thin Layer Chromatography and Column Chromatography

Objective:

Part I:

1. To learn the technique of TLC and the visualization of colourless components.

2. To identify an unknown drug by a TLC comparison with standard compounds.

Part II:

1. To learn the technique of column chromatography.

2. To separate the mixture of pyrene and p-nitoraniline by column chromatography.

Introduction:

Chromatography is a common laboratory technique to separate and analyze two or more analytes in the mixture by distribution of two phases: a stationary phase and a mobile phase. The stationary phase is a phase which allows the mobile phase to travel along. These two phases can be solid-liquid, liquid-liquid or gas liquid. This method works on the principle that different compounds with different solubilities and adsorptions to the two phases which they are to be partitioned. The compounds to be retained on the stationary phase are more interacted with it while the compounds to be moving carried along by the mobile phase. The rates of migration for each component on the system is depends on the degree of the compounds of mixture are adsorbed by the stationary phase and their degree of solubilities on the mobile phase. The stronger the adsorption by stationary phase, the slower the compounds travels along the mobile phase. As a result, the distance of separation for each compound in the mixture will be different. The types of chromatography is divided into few types which include gas chromatography(GC), high performance liquid chromatography(HPLC), thin layer chromatography(TLC), and column chromatography(CC). However, only TLC and CC are applied in this experiment.

Thin layer chromatography (TLC) is a solid-liquid form of chromatography where the stationary phase is usually a polar adsorbent while the mobile phase can be one single solvent or a combination of solvents. TLC is a quick and inexpensive technique that can be used to 1) determine the number of compounds in a mixture, 2) identify the compounds, 3) monitor the progress of a reaction, 4) determine the effectiveness of a purification, 5) determine the appropriate conditions for column chromatography separation, and 6) analyze the fractions obtained from column chromatography. In thin layer chromatography, the stationary phase is refers to polar adsorbent, usually is silica gel or aluminium oxide which is coated on an aluminium plate. Generally, polar solvent is used as the mobile phase in the system to carry the analytes by passing through the stationary phase. The stationary phase in TLC chromatography is typically silica gel, (SiO₂.xH₂O)_n which has been shown in the diagram 1 below:

Diagram 1

Different compounds are spotted on the silica gel plate (stationary phase), the prepared solvent (mobile phase) will carry the compounds goes up along the plate through capillary action which the solvent travels from the bottom to the solvent front. Once the dilute solutions of compounds are spotted on the plate, then development of solvent is the next. After that, the position of each compounds can be visualized under the presence of ultraviolet (UV) light. The compounds in the mobile phase will have different interaction with the polar stationary phase. The factors are mainly depends on the polarity of adsorbent (silica gel in this experiment), solvent polarities, and functional groups of the compounds. The polar adsorbent will more strongly attract the polar molecules of compounds and it will have lower affinity to the non-polar compounds. Hence, the movement of compounds with different polarities could be different. In addition, the polarity of solvent is very important to the compound separations, a solvent system may increase in its polarity by changing the composition of the solvent mixture. The more polar the solvent, the faster the compounds can be drawn up, which means the further the compounds move. The comparison of the polarities of solvent are listed down in the diagram 2.

Diagram 2

The second factor that affects the interaction between stationary phase and compounds is functional group of each compound. The highly polar groups in compound will cause the stronger adsorption and eluted less readily to the stationary phase compared to less polar compounds. Hence, the highly polar compounds will tend to interact strongly with the polar adsorbents and absorb onto the fine particles of the absorbent, hence it cannot travel further. The adsorption strengths of each compound having the following types of functional groups in the order of increasing group polarities.

However, the variation may take place which depends on the overall structure of each compound.

Column chromatography is one of the most useful methods for the separation and purification of both solids and liquids when carrying out small-scale experiments. Like TLC, the silica gel is used as a stationary phase while an organic solvent is used as the mobile phase which its polarity should be lower than silica gel. Column chromatography is carried out in a glass tube that is clamped vertically with the mixture of samples at the top. The samples are dissolved in a small quantity of solvent which is used to apply on the top of the vertical column. In this case, the solvent (mobile phase) will tend to flow down through stationary phase (silica gel) instead of the capillary action. The compound with less polar characteristics will elute faster through the column due to the silica gel has the strong affinity towards the more polar compounds. Eventually, the compounds start to be separated as the solvent is allowed to flow through the stationary phase. Due to the difference in their polarities, the solvent acts as the mobile phase will carry the less polar compounds further down from the top in the system. Below the column, several flasks are used to collect the solvent with compound in various fractions. The solvent is continually added to the top of the column until each band resolves and is carefully collected. As a result, the experiment is end up with the separation of two compounds from the mixture into two different portions. With coloured substances, the bands might be visible and easily to be collected as they run off the column. However, colourless compounds can be observed directly. So, the particular compound in eluting solvent is collected in many small fractions and testing each of them by using TLC. A fresh solvent (mixture of solvent similar to the eluent in column chromatography) is being used in TLC for the next step of identifying the compounds.

The total distance traveled by the compounds on silica gel plate are measured and is being compared to each other. The migration rate of each compound is compared by using the retardation factor, R_f. Retardation factor is the ratio of distance traveled of the compound to the distance of the solvent traveled. Retardation factor, R_f is the distance of compound traveled divided by the total distance of solvent travelled in TLC plate. If two spots in the TLC plate travel the same distance or have the same value of R_f, then both compounds might be concluded as the same compounds.

Part I: TLC analysis on analgesics drugs

Apparatus: UV lamp, capillary tube, 250ml beaker

Materials: aspirin, acetaminophen, caffeine, unknown A, unknown B, TLC plates, ethyl acetate, hexane, iodine

Procedure:

PartA: Spotting the TLC plate

1. A TLC plate was obtained from instructor. Holding the edges of the plate carefully.

2. Set the plate down on a clean and dry surface, then a line was drawn across the plate about 1.0cm from the bottom of the plate by using a 2B pencil.

3. Five lines of 2-3mm were drawn, spaced about 0.6cm apart and running perpendicularly through the lines across the bottom of the TLC plate. 0.5cm must be spaced from each side of the edges.

4. 5 different analgesics were spotted on each 2-3mm lines. Firstly, acetaminophen was spotted on the plate, followed by caffeine, unknown A, aspirin and lastly the unknown B. The plate was examined under the UV light to check whether enough each solution has been applied.

Part B: Developing the TLC plate

1. A developing chamber was prepared by using a 250ml beaker, a half-piece of filter paper inside and aluminium foil to cover.

2. Mixture of 1:3 of ethyl acetate : hexane was poured into the beaker to the depth of about 1cm. The TLC plate was placed in the developing chamber.

3. After the solvent has risen to near the top of the plate, the plate was removed.

Part C: Visualization

1. The colourless compounds were visualized by illumination of the plate with UV lamp.

2. The spots were outlined by using a 2B pencil. The spots may be visualized by putting the plate in an iodine chamber for a couples of minutes.

Part D: Comparison of the unknown with reference standards

1. The plate was sketched in notebook and the R_f value was calculated for each spot.

2. The unknown drug was determined based on R_f value.

Part II: The separation of pyrene and p-nitroaniline by column chromatography

Apparatus: glass column, UV lamp, capillary tube, 250ml beaker, test tubes, glass funnel

Materials: pyrene, p-nitroaniline, TLC plates, ethyl acetate, hexane, iodine

Procedure:

Part A: Column preparation

1. A 49cm chromatography column, 15g of deactivated Silica gel and 110ml of developing solvent mixture (ethyl acetate:hexane; 1:3) were obtained.

2. A slurry of the adsorbent( silica gel) was prepared with a solvent in a 250ml Erlenmeyer flask.

3. A small plug of cotton was pushed into the constriction at the bottom of the column. The column was clamped in a vertical position and 0.5cm layer of the sodium sulfate anhydrous was added on top of the cotton.

4. Ensuring the stopcock of the column is closed, 15ml of solvent was poured in. After setting, all the slurry was quickly decanted through a funnel into the column.

5. The stopcock was opened and allowed the solvent to drain while tapping the walls of the column with the ends of a folded price of rubber tubing.

6. Once the solvent level is within 6cm of the top of the adsorbent, the packing should be essentially complete. 0.5cm level layer of sodium sulfate anhydrous was added on the adsorbent.

7. Excess solvent was drain off until its level is precisely on top of the sodium sulfate anhydrous and the stopcock was closed.

Part B: Separation and collection of pyrene and p-nitroaniline

1. The mixture of pyrene and p-nitroaniline was took and a few drops of ethyl acetate was added to dissolve as much as possible.

2. The solution was transferred directly to the top of the sulfate anhydrous layer with a dropper.

3. The solvent was drain off until the mixture solution is just below the top of the sodium sulfate anhydrous.

4. The wall was rinse with 1ml of fresh solvent(ethyl acetate/hexane 1:3) and was drain until the level was once again below the top of sodium sulfate anhydrous. The rinsing of the walls was repeated until the solvent above the silica gel is virtually colourless.

5. The column was filled carefully with the fresh solvent(ethyl acetate/hexane 1:3) and allowed solvent to drain.

6. The separation of bands was observed as the column develops. The colourless band of pyrene was collected into 3 test tubes.

7. When the edge of the yellow band (p-nitroaniline) reached the lower part of column, a new test tube was replaced and the yellow band was collected into three fractions.

8. Each fraction was concentrated to a small volume by evaporation for analysis by TLC.

Part C: Analysis of the fraction

1. The fractions were spotted on a TLC silica gel plate along with the reference pyrene and p-nitroaniline.

2. The TLC plate was developed in a developing chamber containing a mixture of ethyl acetate/hexane 1:3.

3. The TLC plate was visualized with the UV lamp to determine the fraction of pure pyrene and pure p-nitroaniline.

4. The chromatogram was drawn in the notebook.

5. The R_f value was calculated for pyrene and p-nitroaniline.

6. The pure fraction of pyrene was combined in a pre-weigh test tube and the pure fraction of p-nitroaniline in another test tube. Both solvents were evaporated on a stem bath.

7. Once the solvent has evaporated, the weight of the pure pyrene and p-nitroaniline were calculated.

Results:

Part I:

Total distance of solvent travelled from bottom line in TLC plate = 8.0cm

Retardation factor, R_f

= Distance of sample travelled from the bottom line / Total distance of solvent travelled from bottom line in TLC plate

 

Samples	Distance travelled from the bottom line (cm)	Retardation factor, Rf
Acetaminophen	0.50cm	0.0625
Caffeine	0.35cm	0.0438
Unknown A	0.50cm	0.0625
Aspirin	2.40cm	0.3344
Unknown B	2.20cm	0.2750

Diagram of acetaminophen, caffeine, unknown A, aspirin, and unknown B travelled on the TLC plate

Inference: unknown A is acetaminophen whereas the unknown B is aspirin.

Part II:

Total distance of solvent travelled from bottom line in TLC plate = 8.0cm

Retardation factor, R_f

= Distance of sample travelled from the bottom line / Total distance of solvent travelled from bottom line in TLC plate

 

Samples	Distance travelled from the bottom line (cm)	Retardation factor, Rf
1st fraction of pyrene	6.1cm	0.7625
2nd fraction of pyrene	6.1cm	0.7625
3rd fraction of pyrene	6.1cm	0.7625
4th fraction of pyrene	6.1cm	0.7625
1st fraction of p-notroaniline	1.6cm	0.2000
2nd fraction of p-notroaniline	1.7cm	0.2175
3rd fraction of p-notroaniline	1.7cm	0.2175
4th fraction of p-notroaniline	1.7cm	0.2175

Diagram of pyrene and p-nitoraniline travelled on the TLC plate

Inference: pyrene is present in the 1^st to 4^th spots while 5^th to 8^th spots containing p-nitroaniline.

Discussion:

In the thin layer chromatography, the eluent (solvent) is prepared by using a mixture of hexane and ethyl acetate in the ratio of 3:1. The polarity of the particular solvent cannot be too low because the polar compounds will not be able to carry by the eluent and will not be separated, so that the separation might not be observable. If the solvent of too high polarity is used, the polar compound will travel so fast that the separation between non-polar compound and polar compound to become so small and poor separation will be observed. The solvent mixture of ethyl acetate and hexane (1:3) is believed that it has the optimized solubility for the organic compounds to dissolve in the solvent. In another word, the compounds can be easily to be carried by the solvent in the TLC plate. A few drops of acetic acid were added into the particular solvent in order to protonate the organic compound on the TLC plate and prevent them from ionization. This is due to the deprotonation of organic compounds will cause the compound to form ions. So, the adding of acetic acid is used to maintain the structure of organic compound as they can travel up through the TLC plate. Before the TLC plate is placed into the solvent. A filter paper was dipped inside the solvent in a beaker which is covered by using an aluminium foil. This is to create a system that prevents the vapourization of organic solvent and hence the solvent is allowed to travel up along the plate faster. After the TLC plate was introduced into the solvent, the solvent is starting to migrate itself and the compounds on the TLC plate until the solvent front has been reached.

There are three components in the TLC which include the TLC plate with adsorbent, the development solvent and the organic compounds that to be analyzed. The adsorbent, silica gel consists of a three dimensional network of thousands of alternating silicon and oxygen bonds. It is a very polar and is capable of hydrogen bonding due to its partial positive charge in silicon and partial negative in oxygen. The silica gel with compete with the development solvent for the organic compounds as the solvent is traveling up through the TLC plate. The silica gel tends to bind the compounds (on stationary phase) while the development solvent tried to dissolve the compounds (on mobile phase) in order to carry the compounds along the plate as the solvent travels up. All the compounds are possible to be adsorbed into the stationary phase however the time of adsorption of compounds in the particular phase is depends on the polarity of each compound. The more polar the compound is, the longer the time taken that the compound adsorbed into the stationary phase so it eluting speed is slower (more time on stationary phase). Less polar compounds are weakly adsorbed, so the time taken for less polar compounds to be adsorbed on stationary phase is shorter. As a result, the less polar compounds can travel further along the plate compared to the more polar compounds.

In this experiment, the analgesics drugs have been analyzed by using TLC are acetaminophen, caffeine, and aspirin. The structure of each compounds are shown as below:

The polarity of the compounds could be compared by looking at the structure of these compounds. The polarity of the compound is due to the effect of electronegativity in atoms and the asymmetrical structure of compound. The unequal sharing of electron within the bond causes the formation of an electric dipole which leads to partial positive and partial negative exist in the several atoms. The highly electronegative atoms present in these compounds are O, N, and F. These highly electronegative atoms tend to withdraw the electrons towards themselves from the aromatic ring and hence it polarizes the compounds. Nitrogen and oxygen atom present in these compounds have higher electronegativity compared to the carbon atom and hydrogen atom. Thus, nitrogen and oxygen atom acquire partial negative charge while the carbon and hydrogen atom acquire partial positive charge. Hence, caffeine has the highest polarity, followed by acetaminophen while aspirin possesses the lowest polarity. The sequence of increasing in polarity is arranged in the order below: 

Based on the retardation factor, R_f, the polarities of these compounds can be compared. The higher the R_f, the lower the polarity and hence the distance traveled by the compound would be longer. Aspirin has the highest R_f value. The interaction between aspirin and stationary phase (silica gel) is the weakest which allows the aspirin can travels up along the plate fastest. A stronger interaction in between the silica gel and acetaminophen causes the compound move slower when travels up the TLC plate. The polarity of the acetaminophen is considered lower than caffeine but the difference is quite small. This can be shown in the distance traveled by both compounds which corresponding to their R_f value. In this part, TLC method can be used to identify the two unknowns spotted on the TLC plate. The compounds with same polarities usually travel up through the plate in the same distance and possess the same R_f value provided the stationary phase and mobile phase are identical. Unknown A is predicted as acetaminophen while unknown B is forecasted as the aspirin. This is because the R_f value of unknown A and unknown B are same with the R_f value of acetaminophen and aspirin respectively.

Column chromatography is used to purify the individual organic compound from the mixture of compounds. In column chromatography, the stationary phase and mobile phase used are same as used in thin layer chromatography. The adsorbent (stationary phase) used is a solid which silica gel is usually being used. The eluent (mobile phase) used is the mixture of hexane and ethyl acetate in the ratio of 1:2. The compounds used were the pyrene and p-nitroaniline. The structure of each compound are shown as below: Based on the structure of the compound, the pyrene has the lower polarity due to its delocalized electron in the aromatic ring. The delocalizing π electrons are distributed evenly in the whole structure and hence it stabilize the pyrene. In addition, there is no other electronegative atom attach to the pyrene as substituent, so the electron are just delocalize within the four aromatic ring of pyrene. On the other hand, the p-nitroaniline has higher polarity due to its electronegative substituent in the structure. N and O atoms are the highly electronegative atoms which tend to withdraw the electrons from the benzene ring towards the atoms. This causes the arisen of partial positive charge and partial negative charge in the substituent and within the benzene ring. Consequently, the structure of p-nitroaniline induces the polar property of the compound and hence the polarity of p-nitroaniline is much more higher than pyrene. Due to the difference in polarities in each compound, the interaction with the silica gel would be different. The polar compound would have the stronger interaction with the silica gel. This is because the polar silica gel tends to pull the polar p-nitroaniline toward the stationary site and causes the compound becomes more difficult carried by the solvent through the system. As a comparison, the less polar pyrene was weakly adsorbed into the stationary phase. The silica gel would not tend to attract the less polar compound into the stationary site and hence the particular compound can be carried by the solvent to reach the bottom of the column. The pyrene is said that it elute faster than the more polar p-nitroaniline.

The total number of fraction have been collected was 19 fractions after all the yellow bands (p-nitroaniline) transferred out from the column. After evaporated most of the solvent, the 19 fractions were concentrated into eight fractions which each of the fraction have been used to apply on a TLC plate by using TLC. The separation of compounds is considered as successful since the 1^st to 4^th spots are containing pyrene only. The 5^th to 8^th spots are only contain the compound of p-nitroaniline. These can be proved by checking the distance of compounds traveled on TLC plate of each fraction and the corresponding to the R_f­ value are the same. However, the colours of the 1^st and 4^th fractions of pyreme on the TLC plate were faded compared to the 2^nd and 3^rd fractions. This might be due to the concentration of pyrene in the former fractions were not the same with the latter fractions. The amount of pyrene collected in the 1^st fraction is less because it collected more solvent. The 4^th fraction also has less concentration of compound because the concentration left in the column after 2^nd and 3^rd fractions of compound have been collected. The four spotted compounds were noticed that they were connected to each other on the TLC plate. This might be due to the diameter of the spotted place was too big. The 1^st and 4^th fractions of p-nitroaniline also have the same condition which their colours on the TLC plate were faded. It was believed that the 1^st fraction and 4^th fraction of p-nitroaniline have the similar condition with the 1^st and 4^th fractions of pyrene. The concentrations of the p-nitronaniline were not enough so they appear lighter colours on the plate under the sources of ultraviolet light.

By using the iodine as the development solvent in TLC, the similar observation for the both pyrene and p-nitroaniline were observed. The distance traveled of compounds were almost similar compared to previous solvent. But, the difference was just the colour of the spots on the TLC plate.

Precaution steps:

1. Do not move the beaker after the TLC plate had introduced into the beaker.

2. Do not look directly to the ultraviolet lamp.

3. Avoid to use pencil on the TLC plate

4. Do not touch on the surface of silica gel of TLC plate by using finger.

Determination of the Molar Entropy of Camphor

Objectives:

1. To study the molar entropy of fusion of camphor

2. To determine the molar entropy of fusion ( ) of camphor

Introduction:

Camphor, C₁₀H₁₆O is an aromatic crystalline compound which is obtained from naturally from the wood or leaves of the camphor tree or synthesized. The camphor is used in this experiment to determine its molar entropy of fusion. Molar entropy of fusion is denoted as ( ), which represents the entropy increases when a particular substance is melting. The entropy is always depends on the state of the substance. The disorder in entropy of a substance is the lowest in solid state, whereas liquid is more disorder and followed by gas state with highest disorder in entropy. In another words, the degree of disorder increases in the transition from a closely packed molecule arrangement solid to the disorganized molecules arrangement of liquid state. But, in this experiment, the entropy of camphor is the energy absorbed by the camphor is disturbed by the adding of naphthalene in the system.

The molar entropy of fusion ( ) of camphor can be determined cryoscopically. Entropy, S is a thermodynamic quantity that is a measure of how dispersed the energy of a system is among the different possible ways that system can contain energy. When the energy of a thermodynamic system is concentrated in a relatively few energy states, the entropy of the system is low. When the same energy is dispersed or spread out over a great many energy states, the entropy of the system is high. Cryoscopy is a technique used for determining the molecular weight of a solute by dissolving a known quantity of the solute into a solvent and records the drops of freezing point of the solvent. For dilute solutions of naphthalene in camphor is described as below

Colligative properties are defined as those properties of solutions that depend on the number of dissolved particles in solution, but not on the natures of the solutes. The freezing point and boiling point of the particular solutions will be affected. This is because a liquid solution containing a solvent with an impurity is more disordered than a pure solvent. Therefore more energy must be removed to take the solution from its highly disordered liquid state to its ordered solid state than to do the same for a pure solvent. Hence, solutions of solvents with impurities freeze at a lower temperature than pure solvents due to colligative effect. In this experiment, colligative property is related to the freezing point depression and molality of a liquid.

The molality of a liquid is calculated by using the number of mole of solut is divided by the mass of solvent used in kg

The freezing point depression is calculated by

Apparatus: test tube, beaker (250cm³), Bunsen burner, thermometer (0-200 )

Materials: camphor (M=152g/mol), naphthalene (M=128g/mol)

Procedure:

1. 2g of camphor is being melted and stirred in a test tube.

2. The test tube in an air jacket and is allowed it to solidify.

3. The air jacket is removed and the test tube is heated gently until melting commerce.

4. The content is stirred with the thermometer and the temperature at the last crystal just disappears.

5. 0.05g of naphthalene is added into this test tube and the process is repeated.

6. The procedure is repeated for further 0.05g of naphthalene up to a total weight of 0.20g of naphthalene.

Results and calculation:

Table 1 Weight of naphthalene and camphor, no of mole for naphthalene and camphor, mole fraction, freezing point and T

Weight of naphthalene (g)	Number of moles of naphthalene (mol×10-4)	Weight of camphor (g)	Number of moles of camphor (mol)	X	Freezing point (K)	∆ T	Kf
-	-	2.00	0.0132	-	445	-	-
0.05	3.91	2.00	0.0132	0.0288	433	12	417
0.10	7.81	2.00	0.0132	0.0559	429	16	286
0.15	11.72	2.00	0.0132	0.0815	421	24	294
0.20	15.63	2.00	0.0132	0.1059	409	36	340

*Molecular mass of naphthalene and camphor are 128.06 g/mol and 152.13g/mol

ΔT= depression of freezing point of the mixture in K

= K_fp x m

K_{f p}= freezing point depression constant (K_fp of camphor= 37.7 /m)

m = molality of the solute= No. of solute mole (mol)/ mass of solvent (kg)

 

A graph of K_f versus X was plotted.

∆T = K_f X

= 12 / 0.0288

= 417

= 16 / 0.0559

= 286

= 24 / 0.0815

= 294

= 36 / 0.1059

= 340

 

Since

_f = RT_o²/ H_f Equation 1

_f = H_f / T_o Equation 2

thus substitute Equation 2 Equation 1

_f = RT_o/ S_f

S_f = RT₀/ _f

From the graph plotted,

Lim K_f = RT₀ / ∆S_f

X 0

RT₀ / ∆S_f = 222.5

222.5 = 8.314 * (172 + 273) / ∆S_f

∆S_f = 16.628 J K­^-1 mol^-1

∆S_f = ∆H_f / T₀

∆H_f = 16.628 J K­^-1 mol^-1 * (172+273) K

= 7.399 kJ mol^-1

 

Discussion:

In this experiment, the molar entropy, S_f of the camphor is obtained which has the value of 15 J/mol. Before obtain the molar entropy, we have to measure the temperatures change in each of the different amount of naphthalene added into the camphor. After the temperature changes are obtained, the cryoscopic constant for each amount of naphthalene added are calculated, the value of cryoscopic costant for each are shown in table 2. By using the value obtained, a graph of cryoscopic constant against mole fraction of naphthalene in the mixture is plotted and the y-intercept represents the cryoscopic constant of pure camphor which is 246.5.

According to the Thermodynamics Second Law, the entropy of a substance is the measure of the degree of disorder in a particular system. The value of the entropy of a distribution of atoms and molecules in a thermodynamic system is a measure of the disorder in the arrangements of its particles. State of the substance can determine the entropy of its substance. The degree of disorder is increase from solid state to liquid state and finally to gaseous state. In another word, the entropy of the same substance has higher entropy when it is in gaseous state compared with solid or liquid state. Molar entropy of fusion of camphor is denoted as ( ), which represents the entropy increases when a pure camphor is melting.

The different amounts of naphthalene are added into camphor as the pure camphor act as the control experiment. The pure freezing point of camphor is 172 in this experiment. After added the naphthalene, the freezing temperature are decrease as more amount of naphthalene is added into the same amount of camphor. The more the naphthalene added, the lower the freezing temperature of the particular mixture. This is showed that the system of camphor is being disturbed by naphthalene. The naphthalene molecules are rearranged with the camphor naphthalene in order to form bond in between each other. For a substance in liquid state to freeze, the naphthalene and camphor molecules begin to form cluster of molecules to convert to solid state. The molecule moves in high speed are not able to form a cluster of molecule with each other, so the molecules prefer move slower by reducing their kinetic energy. The kinetic energy is converted into heat energy that is being released to the surrounding as the temperature of the mixture decrease. Thus, the freezing point of the solvent with different solute is lower than the freezing temperature of pure solvent.

Precaution steps:

In this experiment, there is some precaution steps should be taken. Firstly, the zero error should be avoided when measuring the weight of naphthalene and camphor. This can be avoided by press the tare button when use it. Secondly, the reading of temperature on thermometer should be taken parallel to the eyes to avoid parallax error.

Reactions of Aldehydes, Ketones And Phenols

Objective:

1. To carry out some simple chemicals test in order to distinguish between aldehydes, ketones and phenols

2. To study the properties of aldehydes, ketones and phenols.

3. To identify the unknowns A, B, C, D and E.

Introduction:

Part I : Reaction of Aldehydes and Ketones

The carbonyl group is C=O and any compound containing this group that can be described as a carbonyl compound. Carbonyl compounds fall into two main classes: aldehydes and ketones on the one hand and carboxylic acids and their derivatives on the other hand. The characteristic reactions of the aldehydes and ketones are addition and oxidation reactions occurring at the unsaturated carbonyl group. With the same reagent, aldehydes usually react faster than ketones, mainly because there is lees crowding at the carbonyl carbon and the steric effect. Aldehydes are also more easily oxidized than ketones. The carbonyl and other compounds investigated in this experiment are tested in each of the following ways:

A) Chromic Acid (H₂CrO₄)

Chromic acid is a strong oxidant. Aldehydes are oxidized to carboxylic acids by chromic acid. The Cr⁶⁺ in the chromic acid which is orange, then is reduced to Cr³⁺ which is green/blue. Ketones are not oxidized by chromic acid. B) Tollen’s Test (Ag(NH₃)₂⁺ / OH^-

Tollen’s reagent (Ag(NH₃)₂⁺ / OH^- is a weak oxidant. Aldehydes are readily oxidized to carboxylic acids by Tollen’s reagent to produce a silver mirror on the inside of a clean test tube. Ketones are not oxidized by Tollen’s reagent.

C) Fehling’s solution

Fehling’s solution is an oxidizing agent. It is prepared by mixing equal part of Fehling’s solution I (copper(II) sulfate) and Fehling’s solution II (sodium potassium tartate and sodium hydroxide). Aldehydes are easily oxidized to carboxylic acid by Fehling’s solution and will reduce the cupric ion which complexed with tartate ion to cuprous oxide. A positive result is indicated by the formation of a brick red precipitate. Ketones are not oxidized by Fehling’s solution.

D) 2,4-dinitrophenylhydrazine (DNP Test or Brady’s Reagent)

2,4-dinitrophenylhydrazine (Brady’s reagent) is an important reagent related to hydrazine. Most aldehydes and ketones very readily with this reagent to give the yellow orange and red precipitates of 2,4-dinitrophenylhydrazones. Unconjugated aldehydes and ketones give precipitates toward the yellow while conjugated compound tend to be deeper colour of red. The conversion of aldehydes and ketones into hydrazone is an example of the addition-elimination reaction occurring at the unsaturated carbonyl group.

E) Iodoform Test

Iodoform test can be used for the detection of acetalaldehyde and all methyl ketone which have the formula:

Iodoform, CHI₃ is a yellow solid with a strong medicinal smell. Iodoform will precipitate out of a mixture of methyl ketone, iodine and base. For acetaldehyde, the following reaction shows the formation of iodoform:

Compounds that are easily oxidized to acetaldehyde and methyl ketones also give a positive iodoform test. Only ethanol can be oxidized to acetaldehyde and secondary alcohol that have the general formula CH₃CH(OH)R can be oxidized to methyl ketones.

Part II: Reactions of Phenols

Compounds in which a hydroxyl group is bonded to an aromatic ring are called phenols. Alcohols and phenols are similar in some ways, but there are enough differences so that they are considered different functional groups. One major difference is that phenols are typically about a million time more acidic than alcohols. We shall focus on chemical reactions that can help to distinguish phenols from alcohols.

A) Solubility of Phenols

The presence of a hydroxyl group in phenols permits hydrogen bonding between them and the similar substance water H-OH. This leads to appreciable water solubility of phenols. If non polar groups like alkyl groups are attached to the aromatic ring, the water solubility of the phenols decreases.

B) Acidity of Phenols

Most phenols are weaker acids than carboxylic acids and stronger than alcohols. When phenols react with a base, the phenol is converted phenoxide anion. The phenoxide anion is more soluble in water than the corresponding phenol. Consequently, if a water-insoluble phenol is treated with an aqueous solution of a base that is strong enough to convert most of the phenol to phenoloxide anion, that phenol will dissolve in the aqueous base as the phenoxide salt. None of the above-mentioned bases is strong enough to convert a substantial amount of a typical alcohol into an alkoxide anion which would cause a water-soluble alcohol to dissolve as its alkoxide anion.

C) Reaction with Bromine Water

The hydroxyl groups of phenols activate the ring to electrophilic substitution, so that reaction occurs under very mild conditions. With bromine water and phenol, the product is 2,4,6-tribromophenol, which has such a low solubility in water and appears as a white precipitate.

D) Ferric Chloride Test

Phenols give a colouration (pink, green or violet depending on the structure of the phenol) with ferric chloride. This is due to the formation of certain coordination complexes with the iron. Ordinary alcohols do not react. This test may be used to distinguish most phenols from alcohols.

Apparatus: test tubes

Materials: chromic acid, Tollen’s reagent, Fehling’s solution I, Fehling’s solution II, 2,4-dinitrophenylhydrazine, iodine solution, 1,4-dioxane, acetophenone, 3-pentanone, acetaldehyde, benzaldehyde, isopropyl alcohol, 1-propanol, unknown A, unknown B, unknown C, 5% sodium hydroxide, 5% sodium carbonate, 5% sodium bicarbonate, bromine water, 1% ferric chloride, phenol, 2-naphthol, unknown D, unknown E

Procedure:

Part I: Reactions of aldehydes and ketones

A) Chromic acid test

1. To 1ml of acetaldehyde, benzaldehyde, acetophenone, 3-pentanone, isopropyl alcohol and 1-propanol in separate dry test tube, 1ml of chromic acid was added. Any changes in colour was observed.

2. The above procedurs was repeated with unknown A,B and C.

B) Tollen’s test (Ag(NH₃)₂⁺/OH^-)

1. Preparation: Tollen’s reagent is prepared freshly before it is needed. 2 drops of 3M of NaOH was added to 10ml of 0.2M AgNO₃. Then 2.8% ammonium hydroxide was added dropwise with stirring until the precipitate of AgNO₃ just dissolves. Allow time for the solid to dissolve.

2. Cleaning test tube: A set of test tubes were cleaned by adding 5-10ml of 3M NaOH to each and heating them in water bath while preparing the Tollen’s reagent.

3. Carrying out the test: The NaOH was emptied and rinse them with distilled water and 2ml of Tollen’s reagent was added to 1ml of acetaldehyde, benzaldehyde, acetophenone, 3-pentanone, isopropyl alcohol and 1-propanol in separate test tube. Set the tubes aside with agitating the contents for few minutes. Warm the mixture briefly on water bath if no reaction.

4. Step 3 was repeated with unknown A, B and C and observations were made.

5. Destroying the reagent: Add a few drops of HNO₃ to Tollen’s reagent after done the test.

C) Fehling’s test

1. Preparation: Equal volume of Fehling’s solution I (copper(II) sulfate) and Fehling’s solution II (sodium potassium tartrate and sodium hydroxide) were mixed.

2. To 1ml of acetaldehyde, benzaldehyde, acetophenone, 3-pentanone, isopropyl alcohol and 1-propanol in separate dry test tube, 6 drops of Fehling’s solution were added. The test tubes were heated in a hot water bath for several minutes. Reddish precipitate of Cu₂O will indicate a positive test for aldehyde.

3. Step 2 was repeated with the unknown A, B and C and observations were made.

D) 2,4-dinitrophenylhydrazine

1. Preparation: 4g of 2,4-dinitrophenylhydrazine was dissolved in 8ml of concentrated H₂SO₄. The mixture was cooled and added with 90ml of methanol and 10 ml of water.

2. To 1ml of acetaldehyde, benzaldehyde, acetophenone, 3-pentanone, isopropyl alcohol and 1-propanol in separate dry test tube, 2ml of Brady’s reagent was added and were shaked vigorously. Warm and allow to stand for 5-10 minutes if no precipitation form immediately. A crystalline precipitate indicates the presence of a carbonyl compound.

3. Step 2 was repeated with the unknown A, B and C and observations were made.

E) Iodoform test

1. Preparation: 20g of KI and 10g of iodine crystals were dissolved in 100ml of water.

2. To 1ml of acetaldehyde, benzaldehyde, acetophenone, 3-pentanone, isopropyl alcohol and 1-propanol in separate dry test tube, 2ml of water was added. For insoluble compound, 1,4-dioxane was added one drop at a time with swirling until a homogenous solution is formed. Then, 2ml of 5% NaOH solution was added with swirling and potassium iodide-iodine reagent was added dropwise until a deep red-brown colour persists. The mixture were heated in water bath of 60 °C at least two minutes. The test tubes were removed from water bath and sufficient 5% NaOH was added dropwise to cause the disappearance of red brown colour. Cool down to room temperature and record whether a yellow precipitate forms within 15 minutes.

3. Step 2 was repeated with the unknown A, B and C and observations were made.

Part II: Reactions of phenols

A) Solubility of phenols

1. To 0.05g of phenol and 2-naphthol in separate dry test tube, 2ml of water was added. Mix and observe whether the compound is completely soluble, partially soluble and insoluble.

2. The above procedure was repeated with unknown D and E. Observation were made.

B) Acidity of phenols

1. To 0.05g of phenol and 2-naphthol in separate dry test tube, 2ml of NaOH, Na₂CO₃, NaHCO₃ were added. Mix and observe whether the compound is completely soluble, partially soluble and insoluble.

2. The above procedure was repeated with unknown D and E. Observation were made.

C) Reaction with bromine water

1. 0.05g of phenol and 2-napthol were dissolved in 2ml of dilute HCl in separate dry test tube and bromine water was added with shaking until the yellow colour persists. Observation was made.

2. The above procedure was repeated with unknown D and E. Observation were made.

D) Ferric chloride test

1. 0.05g of phenol and 2-napthol were dissolved in water (or mixture of water and ethanol if the compound is not water soluble) in separate dry test tube and 1% of ferric chloride was added dropwise. Shake and observe any colour changes.

2. The above procedure was repeated with unknown D and E. Observation were made.

Results:

Part I: Reaction of Aldehydes and Ketones

Table 1.1 Chemical tests on unknown A, B and C.

Table 1.2 Inferences based on observation for unknown A,B and C.Part II : Reactions of phenols

Table 2.1 Chemical tests on Unknown D and E

Table 2.2 Inferences based on observation for unknown D and E.

Discussion:

Chromic acid test can be used to differentiate aldehyde and ketone. Since chromic acid is a strong oxidizing agent. It can oxidize alcohols and aldehydes to form carboxylic acid but it will not oxidize ketone. The change in colour from orange solution to green solution shows a positive test which due to the change in oxidation state of the chromic metal. The chromic acid will be reduced from Cr⁶⁺ (orange) to Cr³⁺(green) which undergoes reduction. Tollen’s reagent is a test used to identify a carbonyl compound whether the compound is aldehyde or ketone. Aldehyde is readily oxidized to form carbocylic acid but ketone is not. The positive result in Tollen’s reagent is the formation of silver mirror or black precipitate in the solution. For Fehling’s test, the functional group of aldehyde can be tested via reduction of deep blue solution of copper(II) to a formation of brick red precipitate. In DNP test, Brady’s reagent is usually used to test on aldehydes and ketones. The formation of yellow, orange or red precipitate give the positive result to the compounds with carbonyl groups. Besides, iodoform test is functioned as the test for α-methyl group present in aldehydes, ketones and secondary alcohols. The positive result can be seen through the formation of yellow precipitate in the particular test.

For unknown A, it gave negative results in the chromic acid test and Tollen’s test. The unknown A cannot be oxidized by the chromic acid and Tollen’s reagent. In Fehling’s solution, there is a formation of greenish brown precipitate in this test. The unknown A in this test gave negative result because unknown A cannot shows positive test in the first two tests respectively. So, it is believed that the formation of the precipitate is due to the reaction between the unknown A and the metal in Fehling’s reagent. Unknown A shown a positive result in DNP test which the reddish orange precipitate was formed. This is because the C=O functional group is exists in the unknown A. In iodoform test, the present of yellow precipitate indicated that the positive test for iodoform test. This is shows that unknown A consists of α-methyl group in its structure. The unknown A is predicted has ketone group, with more specified is predicted as acetophenone.

In chromic acid test, unknown B gave a positive result which orange solution is turned to green solution. This is shows that unknown B is oxidized by the chromic acid. So, the following test Tollen’s test and Fehling’s test also shows the positive results. The formation of silver mirror and brick red precipitate are formed in the Tollen’s test and Fehling’s test respectively. In DNP test, unknown B also gave the positive result which the bright yellow precipitate is formed. This is indicates that the unknown B has the functional group of C=O in its compound. However, the unknown B did not give a positive result in the iodoform test. So, we can predict that the α-methyl group is absent in the compound. Unknown B is predicted as benzaldehyde which is categorized into aldehyde group.

The positive result in chromic acid test for unknown C shows that the reduction of Cr⁶⁺ to Cr³⁺ take places in the reaction. Unknown C is considered as undergoes oxidation via chromic acid test since chromic acid is a strong oxidizing agent. The unknown C cannot be oxidized by Tollen’s reagent and Fehling’s solution. Unknown C gave negative result in both tests which no formation of silver mirror in Tollen’s test and no brick red precipitate in Fehling’s solution. It is also show a negative result in DNP test due to no precipitate form. This is shows that unknown C does not consist of α-methyl group in its compound. Unknown C is predicted as isopropyl alchol which is under the category of alcohol.

There are four tests to differentiate phenolic compound which include solubility, acidity, bromine water test, and ferric chloride test. Solubility is used to differentiate the compound is completely soluble, partially soluble or non soluble in water. Solubility of compounds is the tendency of the compound to form the hydrogen bonding with the water molecules. The formation of homogeneous solution after the compound dissolved completely in water indicates the positive result. In the acidity of phenolic compound test, the strong base will tend to react with the compound by deprotonating the phenolic compound. Normally, sodium hydroxide and sodium carbonate are strong enough bases to dissolve most water insoluble phenols instead of sodium bicarbonate. The degree of dissociation can be shown in the solubility of the phenolic compound in the strong base solution. Positive result is indicated by the tendency of phenolic compound dissolve in all the base solution. In bromine water test, the decolouration of yellow solution shows a positive test which is followed by the formation of white precipitate. In ferric chloride test, phenol tends to react with ferric chloride to give a coloured complex. The change of colour in the solution shows the positive result.

Unknown D is soluble in the water and tends to form a homogeneous solution. Similarly, it is also dissolved completely in sodium hydroxide, sodium carbonate, and sodium bicarbonate. The acidity of unknown D can be shown in the solubility in sodium hydroxide, sodium carbonate, and sodium bicarbonate. Unknown D is soluble in all base solution although it is a weak acid. However, it is a water soluble phenol. So, it can soluble in the three base solutions. In the bromine water test, the unknown D had decolourized the yellow bromine water. This is shows that unknown D is undergoes bromination to form a brominated phenolic compound. In the ferric chloride test, the solution turns to purple solution after ferric chloride was added. This is shows that unknown D is not an ordinary alcohol. Thus, unknown D is predicted as phenol.

For unknown E, it is soluble in water and form homogenous solution. However, unknown E is only soluble in sodium hydroxide but not in sodium carbonate and sodium bicarbonate either. This is shows that unknown E is a water-insoluble phenol and is categorized as a weak acid. In bromine water test and ferric chloride test, the colours of bromine water and ferric chloride were remained unchanged in each test. So, this is means that unknown E does not react with bromine and ferric chloride. As a result, the unknown E is considered as ordinary alcohol.

Precaution steps:

1. Phenolic compounds are corrosive and cause severe chemical burns on contact.

2. When phenol is applied directly to the skin, a white covering of precipitated protein forms. This soon turns red and eventually sloughs, leaving the surface stained slightly brown. If phenol is left on the skin, it will precipitate rapidly and lead to cell death and gangrene. Phenol appears to have local anesthetic properties and can cause extensive damage before pain is felt.

3. When skin comes in contact with phenol, irrigate exposed area with large volume of water.