Determination of the critical micelle concentration (CMC) of an amphiphile by conductivity method

Title: Determination of the critical micelle concentration (CMC) of an amphiphile by conductivity method

Objective:

1. To learn the process involved in CMC determination of a surfactant

2. To determine the critical micelle concentration of the amphiphile sodium dodecyl sulphate (SDS)

Introduction:

Surfactants are amphiphilic molecules that possess both hydrophobic and hydrophilic properties. A typical surfactant molecule consists of a long hydrocarbon ‘tail’ that dissolves in hydrocarbon and other non-polar solvents, and a hydrophilic ‘headgroup’ that dissolves in polar solvents (typically water). One example of a dualcharacter molecule having a head-group and a non-polar tail is sodium dodecyl sulphate (SDS), Na^{+ -}OSO₃Cl₂H₂₅. When a sufficient amount of SDS is dissolved in water, several bulk solution properties are significantly changed, particularly the surface tension (which decreases) and the ability of the solution to solubilise hydrocarbons, (which increases). These changes do not occur until a minimum bulk SDS concentration is reached. This concentration is called the critical micelle concentration (CMC). Several experiments, including light scattering and NMR, show that below the CMC, the surfactant exists mainly as solvated monomeric species, whereas above the CMC these monomers undergo self-assembly to form roughly spherical structures (having an overall diameter of ~5 nm) known as micelles (see Fig 1). Micelles are the simplest of all self-assembly structures.

Technically, a micellar solution is a colloidal dispersion of organised surfactant molecules. Non-ionic surfactant molecules can cluster together in micelles of 1000 molecules or more, but ionic species tend to form micelles of between 10 and about 100 molecules because of electrostatic repulsions between head-groups. One of the key aspects of micelle structure is that the interior of the micelle consists of an associated arrangement of hydrocarbon chains (an ‘oil droplet’). The exterior coat is constructed of the polar, ionic moieties (the OSO₃^- groups in the case of SDS). This ionic surface (which also contains associated water of hydration) is called the Stern layer. Surrounding this ionic mantle is a region that contains both counterions and oriented water molecules – the Gouy-Chapman layer. Together the Stern and Gouy-Chapman layers are known as the electrical double layer. But it is the oil-like interior of the micelle that gives it its many diverse and interesting properties. The hydrocarbon core (~3 nm in diameter) has the capacity to accommodate guest molecules. The most common application of micelles is as detergents but they can also act as micro-reaction vessels for organic syntheses and drug delivery agents

In this experiment you will determine some fundamental properties of the SDS micelle: the CMC and the free energy, enthalpy and entropy of micellisation. You will measure the CMC by measuring the conductivity of the system as a function of SDS concentration. The thermodynamic properties are obtained by determining the CMC at a variety of temperatures. You will need to pool your data – each member of the team will determine the CMC at a different temperature.

Conductometric Determination of the CMC

Below the CMC, the addition of surfactant to an aqueous solution causes an increase in the number of charge carriers ( (aq) Na⁺ and (aq) ^-OSO₃Cl₂H₂₅ ) and consequently, an increase in the conductivity. Above the CMC, further addition of surfactant increases the micelle concentration while the monomer concentration remains approximately constant (at the CMC level). Since a micelle is much larger than a SDS monomer it diffuses more slowly through solution and so is a less efficient charge carrier. A plot of conductivity against surfactant concentration is, thus expected to show a break at the CMC (Figure 1). Figure 1

Apparatus: beaker, pipette, conductivity meter, glass rod

Materials: SDS, deionised water

Procedure:

1. 50ml of an approximately 0.04M aqueous stock solution SDS was prepared.

2. 25ml of deionised water was pipetted into a 200ml beaker.

3. 0.5ml of SDS stock solution was pipetted into water and stir.

4. The conductivity was recorded.

5. Repeat steps 3 and 4 until all the SDS have been added into the beaker.

6. A conductivity as a function of the SDS concentration was plotted and CMC was estimated.

7. The standard change in Gibbs free energy was calculated.

Results and calculation:

Table 1 Total volume of SDS added, concentration of SDS in solution and conductivity of solution

Volume of Stock Solution of SDS added, V1 (ml)	Concentration of SDS in solution, M2 (M)	Conductivity, (mS)
0.0	0.0000	9.17
0.5	0.0007	27.2
1.0	0.0015	56.1
1.5	0.0023	67.3
2.0	0.0030	102.8
2.5	0.0036	108.4
3.0	0.0043	96.5
3.5	0.0049	137.8
4.0	0.0055	99.6
4.5	0.0061	160.4
5.0	0.0067	207.0
5.5	0.0072	204.0
6.0	0.0077	238.0
6.5	0.0083	247.0
7.0	0.0088	257.0
7.5	0.0092	245.0
8.0	0.0097	253.0
8.5	0.0101	272.0
9.0	0.0105	286.0
9.5	0.0011	301.0
10.0	0.0114	307.0
10.5	0.0118	313.0
11.0	0.0122	322.0
11.5	0.0126	331.0
12.0	0.0130	337.0
12.5	0.0133	344.0
13.0	0.0137	348.0
13.5	0.0140	354.0
14.0	0.0144	357.0
14.5	0.0147	365.0
15.0	0.0150	371.0
15.5	0.0153	375.0
16.0	0.0156	380.0
16.5	0.0159	384.0
17.0	0.0162	389.0
17.5	0.0165	393.0
18.5	0.0170	408.0
19.5	0.0175	418.0
20.5	0.0180	422.0
21.5	0.0185	425.0
22.5	0.0190	434.0
23.5	0.0194	440.0
24.5	0.0198	445.0
26	0.0204	454.0
27	0.0208	459.0
28	0.0211	464.0
29	0.0215	473.0
30	0.0218	475.0
31	0.0221	480.0
32	0.0225	483.0
33	0.0228	493.0
34	0.0231	714.0
35	0.0233	879.0
36	0.0236	907.0
37	0.0239	946.0
38	0.0241	964.0
39	0.0244	975.0
40	0.0246	988.0
41	0.0248	996.0
42	0.0251	1004.0
43	0.0253	1014.0
44	0.0255	1018.0
45	0.0257	1021.0
46	0.0259	1025.0
48	0.0263	1035.0
49	0.0265	1041.0
50	0.0267	1055.0

Graph 1 Conductivity against SDS concentration in the solution

At the critical micelle concentration (CMC), the conductivity of the solution is approximately 100, hence the concentration of the SDS solution is approximately 0.003M. and provided the value of p/n = 0.3.

ΔG _M, _m° ≈ RT (2 – p/n) ln [CMC]

= 8.3145 J mol^-1 K^-1 x (25 + 273) K x (2 – 0.3) ln 0.092

= - 10.05 kJ mol^-1

Discussion:

In this experiment, the critical micelle concentration (CMC) of sodium dodecyl sulfate was determined by using the method of conductivity. Sodium dodecyl sulfate (SDS), NaOSO­₃C₁₂H₂₅ is known as amphiphilic surfactant which possesses both hydrophobic and hydrophilic properties. SDS was ionized in the aqueous solution to form Na ⁺ and ^-OSO­₃C₁₂H₂₅ ions in the solution. Self-dissociation of SDS into micelle is strongly cooperative and occurs at the defined concentration called critical micelle concentration. Below CMC, the amphiphile dissolves as monomers. Once the concentration beyonds CMC, the monomers concentration remains unchanged while the micelle concentration increases. The CMC can be determined by the conductivity method of the SDS solution. Na ⁺ and ^-OSO­₃C₁₂H₂₅ ions are known as charge carriers which will increase the conductivity of the solution when ionization takes place.

At the beginning of the experiment, a small amount of SDS is added into the distilled water. In a SDS dilute solution, the concentration of SDS is below its CMC, hence it behaves as normal electrolyte and ionizes to give out Na ⁺ which soluble in the aqueous phase while ^-OSO­₃C₁₂H₂₅ ions solubilize its hydrophilic head in the water and hydrophobic tail extent out the water surface. The ions exist as solvated monomer instead of micelle due to low SDS concentration. The number of monomers was increased as the amount of the SDS solution was added into the solution. At the same time, the increase of conductivity that had been detected due to the increase of SDS ions carried more charges within the solution. Once the amount of SDS solution added into the aqueous solution is equals to the CMC, the first micelle start to form spontaneously in the solution.

The micelle formation occurs at the above of CMC which the monomers undergo self-assembly to form aggregate in the solution. This caused the solution converted from true solution to become a colloidal system. The micellar solution is known as a colloidal dispersion (association colloid) of organized surfactant molecules. The micelle formed in the solution is a spherical structure which the hydrophilic head groups were exposed to the solution while the hydrophobic tails were faced toward the interior of the micelle structure. The exterior of the micelle is built up from the ionic ^–OSO₃ groups which form the Stern layer which associated by water molecules. The further layer that surrounding the Stern layer is composed of the positive counter ions and oriented water molecule called Gouy-Chapman layer. Both Stern layer and Gouy-Chapman layer are known as electric double layer. This double layer will maintain the stability of the colloidal system.

The higher concentration of SDS caused nucleation for the micelle to form increased and hence more micelle was formed in the solution. Above the CMC, the concentration of micelle definitely increases. However, the concentration of monomers almost remained unchanged in the solution. Monomers tend to form the micelle at the same time the added SDS solution ionized in the solution to replace the monomers that used to build micelle. But, the charge carriers could be increased slowly because the rate of micellisation is slower than the rate of monomers were used in the building of micelle and hence the conductivity of the solution increased at a slower rate in an ideal condition. This can be noticed in the graph 1 which shows the increasing rate of conductivity had became slower obviously. This is because the formation of micelle required the ionic monomers and some of the ions had been attracted towards the micelle surrounding to form the electric double layer. As a result, some monomers are no longer free in the solution but for those ions are not strongly attracted still can carry charge in the solution. Hence, the conductivity of the solution increased slower. However, at the final part in graph shows a sudden increase in the conductivity of the may be due to the formation of bubbles inside the solution. Above the CMC, when bubbles start forming, micelles will be broken down to form monomers to expand the bubbles. As more SDS monomers being formed back, the conductivity shoot up because SDS monomers is a more effective charge carrier than micelles.

Precaution Steps

1. The stirring is controlled not to be too fast during the experiment to avoid the formation of bubbles as bubbles can affect the conductivity.

2. SDS solution is added slowly to the water to prevent the formation of bubbles.

Recrystallization

Objectives

1. To separate benzoic acid from impurities by recrystallization.

2. To learn the technique of recrystallization.

3. To determine the percent recovery of benzoic acid from recrystallization.

Introduction

         A pure compound is a homogeneous sample that consisting only of molecules having the same structure. However, each substance believed to be pure may actually contain small amounts of contaminants. This includes the formation of side products during reaction, unreacted starting materials, inorganic materials, and solvents. However, recrystallization technique can be used to purify a solid and remove the impurities.

        Recrystallization is a method of purifying a solid which takes the advantage of differences in the solubility of the desired products and impurities to obtain the pure desired products. Almost all solute are more soluble in hot solvent than in a cold solvent. Thus, if a solid is dissolved in a hot solvent but is insufficient to dissolve it in the cold solvent, the crystals should form when the hot solution is allowed to cool. In the simplest case all of the impurities present in a solid sample will be so much more insoluble in the chosen solvent that all that remains in solution is the pure dissolved product (the solute).

Step 1: Choosing the solvent

An essential characteristic of a successful solvent is that the compound be soluble in the hot solvent but insoluble when the solvent is cold. Tests can be performed with small amounts of material in test tubes: a few drops of a solvent are added and if the material proves insoluble then the tube is heated to see if the material will dissolve at a higher temperature-if so, then a good solvent for recrystallization of that material may have been identified. A solvent should be rejected if the material appears readily soluble in cold solvent, is not soluble to any appreciable extent in the hot solvent even when the volume of solvent is increased, or requires an impractically large volume in order to fully dissolve the crystals.

Step 2: Dissolving the sample

An Erlenmeyer flask should be used of such a size that it will only be filled to around half-way when all the solvent has been added. The solid sample is introduced together with around 75% of the amount of solvent thought to be required. It is always advisable to use less solvent at this stage. The flask is heated on a hotplate until dissolution of the solute is complete, additional solvent can be added to the hot solution as necessary to ensure complete dissolution. A boiling wooden stick should be added to provide a nucleation side for bubbles to form and facilitate an even boiling process. A process of gradual addition of solvent to the flask will ensure that the sample has dissolved to form saturea and will deposit crystalline material once it is cooled. Using excessive amount of solvents will only decrease the percent recovery of the products.

Step 3: Hot filtration

Once the solute is fully dissolved, the remaining impurities can be removed by filtering the hot solution through a filter paper folded into a cone and placed inside a glass filter. A problem here is that the solution will cool rapidly as soon as the Erlenmeyer flask is removed from the hotplate. In most cases, this problem can be minimized or avoided entirely by using a stemless funnel placed on the top of beaker containing a few millimeters of the recrystallization solvent. The beaker is placed on the hotplate and the boiling solvent serves to heat the funnel and prevent the solute from crystalling during the filtration process.

Step 4: Cooling

Cooling the filtered solution will allow crystals to form and rate of cooling can determine the size of the crystals formed. Fast cooling generally produced more crystals of relatively small dimensions, but slow cooling might allow larger crystals to form. The solution usually is left to cool to room temperature before cooled in the ice-bath to ensure maximum recovery.

Step 5: Cool filtration

When the crystallization process is judged to be completed the crystals need to be collected by suction filtration. Both the funnel and suction flask should be chosen so that neither will become more than half full during the filtration process. It is preferable that all of the crystalline material is being transferred to the funnel as a suspension in the crystallization solvent, however it is sometimes hard to get all of the crystals moving freely by swirling the flask and occasionally it will be necessary to add more ice-cold solvent in order to transfer the last of the crystalline material. It may also be necessary to dislodge crystalline material from the sides of the flask with a spatula prior to filtration.

Step 6: Washing the crystals

Once the suction filtration process is completed, the collected crystals should be washed with a little more ice-cold solvent to remove final soluble impurities which would otherwise be left on the surface of the crystals. The solvent used for this final washing should be as cold as possible to minimize losses from the crystals re-dissolving.

Step 7: Drying the crystals

Once the crystals have been collected on the suction funnel they can usually be satisfactorily dried by continuing to draw air over them for a few minutes. The almost dry crystals should then be spread on a filter paper to allow the last traces of volatiles solvent to evaporate.

Apparatus and Materials

Erlenmeyer flask (125 mL), short-stemmed funnel, hot plate, boiling chips, benzoic acid and charcoal.

Procedures

1. 2.0 g of crude benzoic acid were weighed into a 125-mL Erlenmeyer flask.

2. 200 mL of water was heated to boiling in a beaker on a hot plate with boiling chips.

3. An Erlenmeyer flask with a little water in it with boiling chips also was heated on a hot plate, with a short-stemmed funnel resting in its neck.

4. A filter paper was fluted to fit the funnel.

5. A few boiling chips was added to the benzoic acid and the adding of hot water to the benzoic acid was started until the benzoic acid has dissolved.

6. About 0.2 g of decolorizing charcoal was added.

7. The hot solution was filtered through the fluted filter paper into the heated flask.

8. The original flask and the filter paper were rinsed with a little hot water.

9. The solution of benzoic acid was removed from the hot plate and allowed to cool to the room temperature.

10. The solution is then cooled in an ice bath after 15 minutes for 10 minutes.

11. The crystals were collected by suction filtration using Buchner funnel.

12. Vacuum is continued to pull on the funnel for 5 minutes.

13. The filter paper with crystals was transferred onto a fresh piece of filter paper, and the crystals are allowed to air-dry.

14. The percent recovery and the melting point of benzoic acid were determined.

Results & Calculation

Weight of crude benzoic acid = 2.0007 g

Weight of filter paper = 0.8459 g

Weight of benzoic acid crystals + filter paper = 1.8776 g

Weight of benzoic acid crystals = 1.0317 g

Melting point of benzoic acid crystals = 120 °C

Percent Recovery

= weight of compound recovered / weight of compound started with x100%

= (1.0317 g / 2.0007 g) × 100%

= 51.57%

Relative Accuracy of Melting Point

= melting point of benzoic acid/ melting point of recovered benzoic acid × 100%

= (120 °C / 122 °C) × 100%

= 98.36%

Discussion

The percentage recovery of benzoic acid is only 51.57% may be due to several factors that caused the loss of products. One of the factors is that the volume of water added to the solution is too much which making the solution not saturated enough to produce maximum yield of benzoic acid after cooling. Normally, larger volume of water used will tend to the products to dissolve more easily. The benzoic acid crystallized on the filter paper during the hot filtration. The additional hot water need to be added to dissolve the benzoic acid crystals on the filter paper which causes the solution to be more dilute. So that, the products is lost in the solution.

Besides, too much decolorizing charcoal is added to the solution is considered as one of the factors. Decolorizing charcoal functions to provide vacant sites to the organic compounds to accommodate to it in which it removes the unwanted colored impurities. However, this also caused the loss of products in this process because some of the benzoic acid also will be adsorbed onto the surface of charcoal. Generally, the charcoal added should be only about 1-5% of the weight of the sample being recrystallized. A little amount of charcoal is sufficiently to remove the colored impurities. Otherwise, excessive use of charcoal will only caused the products to be removed together with the colored impurities.

In this experiment, the benzoic acid is dissolved in hot water while only a little amount of benzoic acid are able to dissolve in cold water. The benzoic acid cannot dissolve well in cold solution because of its hydrophobic benzene ring. However, the carboxyl group, -COOH that attached to the benzene ring allows some of the benzoic acid solubilise in water. In hot solution, the increase in temperature causes the water molecules has more kinetic energy and move faster. As a result, it allows the water molecules to penetrate through the benzoic acid solid and hence solubilization of benzoic acid occurs. In addition, the charcoal and other impurities present in mixture cannot dissolve in water. So, water is a good solvent to be chosen in the experiment.

The boiling chips were added in the experiment. Boiling chips are small, insoluble, and porous stones made of calcium carbonate or silicon carbide. There is a lot of pores inside the boiling chips in which it provides nucleation site to trap air and creates space to allow the bubble of solvent to form. When the boiling chips are heated, it will release tiny bubbles which can prevent bumping and boiling over of the mixture so that the loss of solution can be avoided even it is boiled. The adding of boiling chips must be added before boiling of solution instead of after boiling. This is because adding boiling chips to a solution near its boiling point will induce flash boiling as well. The boiling chips are not soluble in the solvent and hence they can be filtered out by using filter paper, but they are not reusable.

During the hot filtration, most of the charcoal powders were removed and stay on the filter paper while the benzoic acid solution pass through the filter paper and goes into the conical flask. However, some of the charcoal powder was noticed in the conical flask as well, although in a small amount. This might be due to the size of charcoal powder is too small that can pass through the pores of filter paper. During the cooling process, the hot solution was allowed to cool slowly to the room temperature, and then only immersed in an ice-bath. The solution should be protected from contaminants by covering with a piece of filter paper. Fast cooling always produces relatively small crystals because the particles do not have sufficient time to arrange themselves in proper conformation, so it is not advisable to cool down the hot solution immediately in ice-bath. The small size of crystal form may trap impurities easily. Oppositely, slow cooling allowed the molecules to interact and arrange themselves properly and hence they form larger size of crystals. But, large particles may causes some solvent being trapped inside the crystals.

During the cold filtration, the water soluble impurities that might dissolve in water which was filtered out through the suction filtration. However, some of the impurities might be trapped on the surface of the benzoic acid crystals, so a small volume of ice-cold water should be used to wash the benzoic acid crystals to dissolve the particular impurities. The crystal was dried in the oven at 100 °C. A fresh piece of filter paper can be used to place under the filter paper with benzoic acid crystals.

The purity of a crystal can be determined by its melting point. A narrow range of melting point indicates high purity of the sample, otherwise broad range of melting point indicates the presence of impurities in the crystal. The melting point of the recovered benzoic acid obtained experimentally is 120 °C. Compared to the pure benzoic acid with 122°C of melting point, the purity of the recovered benzoic acid is very high which is98.36%. Although the accuracy is high enough, but it also means that the compound is slightly contaminated with impurities which included the charcoal powder or the water molecules that trapped inside the benzoic acid crystals. The melting point of recovered benzoic acid is lower because the crystals cannot arrange properly due to impurities.

Structure of benzoic acid

Suction filtration

Surface tension of liquids

Objective:

1. To understand the basic concept of surface tension in liquids and how it affects the properties of liquids

2. To study the effects of detergent on the surface tension of liquids.

Introduction:

Water molecule consists on a big oxygen atom and two smaller hydrogen atoms. The hydrogen atoms hold the slightly negative charges which making the entire water molecule becomes polar. Eventually, the hydrogen bond is exists in between two neighboring water molecules. Each water molecule experiences a pull from other water molecules from every direction, but water molecules at the surface do not have molecules above the surface of the water to pull at them. These water molecules experience an inward pulling from the water molecules below than them. The difference in force draws the water molecules at the surface creating the surface tension.

Surface tension is the force acting at the angles of 90º to any line on the liquid surface. Surface tension is a property of a liquid surface that caused by the cohesive force between the same liquid molecules. In a liquid, each molecule within the body of the liquid tends to attracted equally at all directions by the cohesive force, so that it experiences no net force. However, those molecules on the surface of liquid have no neighboring molecules above which exhibits stronger attractive force upon on their nearest neighboring molecules on the surface. This results an inward force which pulling the molecules towards the interior of the liquid.

Due to the effect of surface tension, the surface of liquid will form a thin “film” which makes it more difficult to move an object floating on the surface than to move it when it is completely submersed in the liquid. Surface tension is typically measured in the unit of dyne/cm which is the force in the unit of dyne that required to breaks the surface film of water with the length of 1cm. The surface tension in room temperature is 72dyness/cm. It means 72 dynes of forces would be taken to break down a surface film of water 1cm long. The increase in temperature will affects the surface tension of the water dramatically. When the kinetic energy of water molecules increases, it will tend to break down the surface tension of water. Besides, the addition of some solute can influence the surface tension of the liquid but it is depends on the nature of the solutes added. However, some of the solute concentration may not have effect to the surface tension of liquid once the minimum is reached which is surfactant.

Surfactant (surface-active-agents) is a compound in which can lower the surface tension of a liquid or interface tension between two different liquids (or a liquid and a solid). Examples of surfactants are detergent, wetting agent, emulsifiers foaming agents and dispersants. Surfactants usually are amphiphilic, which means they contain both the hydrophobic groups and hydrophilic groups. Due to these properties, the surfactant molecules will migrate to the water surface, where the insoluble component will project out of the water surface and the hydrophilic water soluble component will remain in the water phase. Detergents are the chemicals which consist of hydrophobic (non-polar) hydrocarbon "tails" and a hydrophilic (polar) "head" group. Surfactants can interact with water in a variety of ways which able to disrupt the hydrogen bonding network between the water molecules. Since this will reduce the cohesive force between the molecules, so the surface tension of the water will be altered.

Surface tension is a term that used to describe the cause of the phenomena in which the difference between meniscus of water and mercury. The influence of intermolecular force and surface tension on the interfacial properties of liquid causes the meniscus formation. Stronger intermolecular force between water molecules and glass surface (adhesive force) compared to those between water molecules (cohesive force) cause the water to be drawn up onto the glass wall forming a concave meniscus. For the formation of mercury meniscus, a convex meniscus is formed due to the stronger intermolecular force between mercury atoms (cohesive force) compared to those between mercury atoms and glass surface which prevent mercury from wetting the glass surface.

Apparatus: Petri dish, Toothpick

Materials: Sulfur powder, liquid dish detergent, low fat milk, full cream milk, food colouring

Procedure:

Part A

1. Clean water was poured into a petri dish with depth of 1cm.

2. Powdered sulfur was dusted on the surface of clean water and observation was recorded.

3. The surface of water was touched by a toothpick and observation was recorded.

4. The tip of toothpick was dipped in detergent. The surface of water was touched and toothpick was hold in a place for a while. Observation was recorded.

Part B

1. Low fat milk was poured into a petri dish.

2. Four different coloured food colouring were added. The drops were kept close together in the centre of the petri dish and observation was recorded.

3. The surface of milk was touched with a toothpick and observation was recorded.

4. The tip of a toothpick was dipped in detergent. The surface of milk was touched and toothpick was hold in a place for a while. Observation was recorded.

5. The process was repeated by replacing low fat milk with full cream milk and water.

Results: Observation

Part A: Water with sulphur powder on surface

Before Touched by toothpick                 Touched by toothpick

Touched by toothpick with detergent             After 3 minutes

             After 5 minutes

Observation: The sulphur powder on the water surface where touched by toothpick with detergent was sank to the bottom.

Part B: Low fat milk with colouring

Before                                                    Touched by toothpick            

        

Touched by toothpick with detergent                   After 3 minutes

                After 5 minutes

Observation: The food colouring are being pushed away rapidly from the region where touched by the toothpick with detergent.

Part B: Full cream milk with colouring

Touched by toothpick                     Touched by toothpick with detergent

                After 1 minute                                After 3 minutes

               After 5 minutes

Observation: The food colouring are being pushed away slowly from the region where touched by the toothpick with detergent.

Part B: Water with food colouring

                Before                                    Touched by toothpick

Touched by toothpick with detergent               After 3 minutes

               After 5 minutes

Observation: The food colouring are not changed significantly from the region where touched by the toothpick with detergent.

Discussion:

In Part A of the experiment, the sulphur powder was sprinkled evenly on the water surface. The sulphur powder was floated on the surface because the size of sulphur is light and small which the surface tension of water can withstand and maintain them on the surface. When the surface was touched by the toothpick, the sulphur powder starts to float further from the centre where the place touched by the particular toothpick. However, the sulphur powder was started to sink into the bottom of water very quickly once it was touched by the toothpick with detergent. This is because the surface tension of water was disrupted by the detergent since detergent is a surfactant which acts to lower the surface tension of liquid. As a result, the weak surface tension of water no longer able to withstand the sulphur powder on its surface, so the sulphur sank into the bottom of water.

In part B, low fat milk is used as the liquid to show the property of surface tension when exposed to detergent. Low fat milk has a small amount of fat and the water is predominately in it. Due to the densities of colourigs are lower than milk, so they are floated on the surface when the food colourings were added on the surface of milk. When the toothpick was used to touch on the surface of milk, the food colourings were disrupted insignificantly due to the wave generated by the touch. However, the food colourings were dispersed rapidly and became faint in colours when the toothpick with detergent touched on the surface of milk. This is because the surface tension of the milk in the centre was lowered by the surfactant. The stronger surface tensions of the surrounding milk molecules pull the surface of milk away from the weak region where towards the edge of the plate. The detergent decreased the surface tension of milk by dissolving the fat molecules which caused the turbulence. This moment caused the food colourings to swirl, but the swirling of the colors continues for some times before stopping.

Full cream milk was used in the part C of the experiment to identify the difference in surface tension for two different milks. Full cream milk has a larger amount of fat compared to low fat milk. The toothpick without detergent did not affect the distribution of food colourings on the surface of milk. The fat globules in milk were steady and undisturbed. When the surface was touched by the toothpick with detergent, the food colourings started to disperse with a slower speed compared to the dispersion in low fat milk. This may be due to the full cream milk contains less water which limiting the movement of the milk so the colourings spread slower in full cream milk compared to low fat milk. The milk molecules with lower surface tension in the spot were pulled by the milk molecules with higher surface tension in the surrounding. Hence this caused the food colourings moved with the milk molecules streaming away from the detergent dropped. Comparing to the low fat milk, the colourings were scattered more in full cream milk. This may be due to the detergent weakens the milk's bonds more in the full cream milk because it had more fat. Hence, food coloring scattered more in the full cream milk.

In part D of the experiment, food colourings were added on the surface of water. The colourings were sunk to the bottom of water because they have higher density compared to water while there is some amount of colourings were floated on water surface. When the surface was touched by the toothpick, the distribution of the food colourings were not affected much. When the toothpick with detergent was used to touch on the water surface, the floated food colourings were pushed away from the centre where the toothpick had touched. The detergent reduced the surface tension of the water at middle and hence it caused food colourings to spread. This phenomenon was due to the water molecules in the edge with higher surface tension pulled the water molecules with lower surface tension from the centre.

Precaution steps:

1. Do not shake and stir the surface of liquid because it will influence its surface tension.

2. Make sure that all the fans have been switched off before carry out the experiment.

3. Make sure the milk is not expired because it will affect the property of the milk.

4. Do not exhale deeply which may affect the surface tension of liquid when carrying out the experiment.

Relationship between osmosis process and hypertonic, hypotonic and isotonic effects

Objective:

1. To understand the phenomenon in which molecules flow from a location of higher chemical potential to a region of lower potential in which both regions are separated by a semi permeable membrane.

2. To investigate the hypertonic, hypotonic and isotonic effect.

Introduction:

Osmosis is the net diffusion of water molecules from a region of lower solute concentration to a region of higher concentration by passing through a semi permeable membrane. A semi permeable membrane is otherwise known as selectively membrane which only allows some of the molecules to pass through it but not others. However, the membrane is permeable to the solvent. The permeability is depends on the size and property of molecule. In osmosis, the net movement of water molecules will only move in one direction. The water molecules move from the side where water concentration is higher to the other side of the membrane which down the concentration gradient.

Osmosis is a passive process which does not involve any energy input into the system. The osmosis process is spontaneously and naturally. The difference in the concentration of solutions creates an osmotic pressure. Osmotic pressure is the pressure that exists across a semi permeable membrane between two solutions of different concentration. Osmotic pressure of a solution is a colligative property and at given temperature its magnitude depends only on the concentration of the solute but not its identity. Osmotic pressure is calculated by using the van't Hoff's formula below:

Π = cRT

where π = osmotic pressure (atm)

c = concentration of solution (mol/L)

R = gas constant

T = Temperature(K)

The concentration of solution can be categorized into three basic categorizes which have different concentration with respect to a particular concentration of solution. The types of solution are divided into hypertonic, hypotonic, and isotonic. Hypertonic solution is the solution containing a higher solute concentration when compared to a particular solution. The higher the solute concentration, the lower the water potential and osmotic pressure. This is because some water molecules will be attracted to the solute molecule and will no longer be free. In osmosis, the water molecules will flow into the hypertonic solution due to lower water potential. For hypotonic solution, is defined as a solution with lower solute concentration with respect to a solution. Its water potential and osmotic pressure are higher. In hypotonic solution, there are more free water molecules which always diffuse into a region of lower water potential. The solution has same concentration respect to another solution is known as isotonic solution. It has the same water potential and osmotic pressure which the net movements of water molecules are the same for both solutions.

Net movement of the water molecules are from hypotonic (low-concentrated solute) to hypertonic (high-concentrated solute) due to difference of osmotic pressure in two solutions. Due to the thermal agitation, the water molecules will diffuse spontaneously (Δ G < 0) from the region of lower solute concentration towards higher solute concentration. The addition of solute will increase the entropy of water molecules since the solute is being dispersed overall in the solution. The direct osmosis (follow concentration gradient) is entropy generating process (ΔS > 0) which having tendency to equalize the chemical potentials and concentrations of two different solutions which are separated by the semi permeable membrane. The water is acting to dilute the solution with higher solute concentration until the concentration is the same for both sides. However, the water will continue to flow with the similar rate in the both directions.

Apparatus: 50ml beaker, analytical balance

Materials: dialysis tubing, tap water, 30% sucrose solution, 60% sucrose solution, string

Procedure:

1. Three pieces of equal length of dialysis tubing and several lengths of string were obtained.

2. One end of the tubing was folded over and tied closed with the string.

3. To each tube, 5 ml of 30% sucrose solution was added. Then, the bag was squeezed gently to remove excess air and was tied off with string.

4. Some slack was left in the bag as room for expansion, but the air trapped inside was removed. The bags were briefly rinsed in running tap water and then were dried on paper towels.

5. Each bag was weighed and the results were recorded.

6. After weighing, three bags were placed into tap water, 30% sucrose solution and 60% sucrose solution each.

7. The bags were remained undisturbed for 30 minutes.

8. The bags were removed, rinsed, and reweighed. All results were recorded into table.

Results and calculations:

	Tap water	30% Sucrose	60% sucrose
Weight at 0 min	5.7739g	5.9171g	5.9795g
30 min	7.1150g	5.8309g	5.3015g
Change	1.3411g	-0.0862g	-0.6780g

Percentage of weight of dialysis tubing (in tap water) increase

= 1.3411g/5.7739g x 100%

= 23.22%

Percentage of weight of dialysis tubing (in 30% sucrose solution) decrease

= 0.0862g/ 5.8309g x 100%

= 1.48%

Percentage of weight of dialysis tubing (in 60% sucrose solution) decrease

= 0.6780g/5.9795g x 100%

= 11.34%

Discussion:

In this experiment, three dialysis tubing were filled with 5 ml of 30% sucrose solution. These dialysis tubing were separately put into the three beaker which containing tap water, 30% sucrose solution and 60% sucrose solution respectively. The dialysis tubing is semi permeable which allows some molecules that are small enough to pass through it. In this case, the small molecules are referred to water molecules. The contents in the dialysis tubing are not flow out through the small pores before place them into beakers with 30% sucrose solution. This may be due to the cohesion force between the water molecules in which try to pull the outermost water molecules in dialysis tubing inwardly.

When one of the dialysis tubing containing 30% sucrose is placed into the beaker with tap water, this tap water is said to be hypotonic with respect to 30% sucrose solution. Due to the high water potential in tap water, the water molecules will tend to move from the tap water into the dialysis tubing. The osmotic pressure of tap water is higher than the osmotic pressure of sucrose solution. The water molecules are forced to move out from the tap water to 30% sucrose solution and move to 30% sucrose solution by diffusing the membrane of dialysis tubing. The reason is because the higher osmotic pressure in the tap water forced the water molecules to diffuse into sucrose solution. Eventually, the rate of water diffuse into the tubing is higher than the rate of water diffuse out from the tubing until equilibrium is reached. The increase in the water caused the weight of dialysis increases as well.

The 60% sucrose solution is considered as hypertonic solution when compared to the 30% sucrose solution in it. The 60% sucrose solution has higher solute concentration and lower water potential due to lack of free water molecules. This is because more water molecules are attracted to the sucrose molecules in the solution and hence the particular water molecules are not free to move. The low osmotic pressure and low water potential in 60% sucrose solution causes the water molecules in 30% sucrose solution to move into it by osmosis. Spontaneously, a net movement of water tends to minimize the difference in concentration between two solutions. The diffusing rate of water molecules move out from the dialysis tubing is higher than the water molecules move into the particular tubing. As a result, the volume of tubing has been decreased as well as its weight. The net movement will become zero after the same concentration of solution is reached for both sides of dialysis membrane.

Isotonic solution is a solution has the same concentration with respect to another solution. In this experiment, one of the 30% sucrose solutions was placed into a beaker containing 30% sucrose solution. So, the solution in the beaker is called isotonic since it has the same concentration with the solution in dialysis tubing. The water potentials of both solutions are the same as the two solutions have the same amount of water molecules free to move in the solutions. Thus, the rates of water molecules diffuse in and out of dialysis tubing are the same since there is no difference in osmotic pressure for both solutions. The net movement of water molecules is zero as the water potentials between two solutions has reached equilibrium. Theoretically, the weight and volume of dialysis has no change. However, the weight of dialysis tubing was reduced in practical. This might be due to the content in beaker was contaminated by impurity which caused the water molecules to move out. The second reason might be due to there is some water left inside the tubing which diluted its content, so the concentration has been reduced and the solution is beaker is considered as hypertonic with respect to the sucrose solution in dialysis tubing. The decrease in concentration might be small but it is enough to dilute the contents which caused the water potential increased and hence diffuse out to the solution (hypertonic) in the beaker.

Precaution steps:

1. The dialysis tubing should not be too tight and turgid in order to leave some slack in the tubing as space for expansion when water flow into it.

2. The exterior of dialysis tubing should be roughly dried before weighing to prevent the weight of water is being counted.

3. The interior of dialysis tubing should be dried in order to prevent dilution of sucrose solution.

Determination of the activation energy for the reaction of bromide and bromate ions in acid solution.

Objectives

1. To understand the chemistry of activation energy.

2. To determine the activation energy for the reaction of bromide and bromate ions in acid solution.

Introduction

Activation energy is defined as the minimum energy barrier that must be overcome for a chemical reaction to take place. It is usually denoted as E_a, and given in unit of kiloJoule, kJ/mol. For a chemical reaction, an appreciable number of molecules with the energy equal to or greater than activation energy should be exist in the system. In order for a reaction to occur, the reactant particles must collide according to the collision theory. However, not all collision are able cause the reaction to happen, only a certain collisions in the system can cause chemical reaction, which is called effective collision. The effective collisions of molecules must collide with the correct orientation and sufficient energy to overcome the activation energy barrier. The energy is needed to break the existing bonds and form the new bonds of the molecules which resulting in the formation of products.

The activation energy of a reaction can be measured by using Arrhenius equation as shown in the equation below:

k = Ae^-Ea/RT

where k = rate constant

T = absolute temperature

E_{a =} energy of activation

R = gas constant

The pre-exponential term, A is the property of particular reaction related to the collision frequency of the reactive species and thus is temperature dependent. However, according to the equation, the dependence of k on temperature is dominated by the strong exponential term, so the dependence of A on temperature is usually ignored as a first approximation. By taking logarithms of both sides,

Log₁₀ k = -E_a / 2.303RT + log ₁₀ A

= -E_a /2.303RT + constant

So, when a reaction has a rate constant that obeys Arrhenius equation, a plot of log₁₀ k versus 1/T gives a straight line. The gradient of the straight line is –E_a / 2.303R while the interception of the straight line on the y-axis of the graph can be used to determine the values of log₁₀ A.

Now, the rate of reaction is higher when the time taken for a fixed amount of reaction to complete is shorter. This makes the time taken, t to complete a fixed amount of reaction is inversely proportional to the rate constant, k.

T α 1/k

or t = constant/k

By taking logarithms of both sides,

Log₁₀ t = - log₁₀ k + constant

= E_a / 2.303RT + constant

A plot of log₁₀ t versus 1/T gives a straight line as well and the slope of the graph is E_a / 2.303R. Thus, if t is measured at several temperatures then the energy of activation can be found.

In this experiment, the above method is applied to the reaction of bromide and bromate ions in an acid solution which occurs slowly at room temperature.

KBrO₃ + 5 KBr + 3 H₂SO₄ --> 3 K₂SO₄ + 3 Br₂ + 3 H₂O

or BrO₃^- + 5 Br^- + 6H⁺ --> 3 Br₂ + 3 H₂O

The time required for a fixed amount of the reaction to be completed, t is found by adding a fixed amount of phenol and some methyl red indicator to the reaction mixture. The bromine produced in the first reaction reacts very rapidly with the phenol to form tribromophenol.

C₆H₅OH + 3 Br₂ -->  C₆H₂Br₃OH + 3 HBr

When all the phenol has reacted, the bromine continuously produced in the first reaction will then react with the methyl red indicator and bleaches its colour.

Methyl red + Br₂ --> colourless compound

Apparatus : 1 dm^-3 beaker, 3 100cm³ beakers, 2 boiling tubes, 1 5cm³ pipette, 1 10-cm³ pipette, thermometer (0 - 110°C), stopwatch

Material : 0.01 mole dm^-3 aqueous phenol solution, bromide/bromate solution (0.0833 mole dm^-3 potassium bromide and 0.0167 mole dm^-3 potassium bromate, equivalent to 0.05 mole dm^-3 bromine), 0.3 mole dm^-3 sulphuric acid, methyl red indicator.

Procedures

1. 10 cm³ of phenol solution and 10 cm³ of bromide/bromate solution were pipette into one boiling tube.

2. Four drops of methyl red indicator were added to the mixture.

3. 5 cm³ of sulphuric acid was pipette into another boiling tube.

4. The two boiling tubes were immersed in the water bath of (75 ± 1) °C.

5. The contents of the two tubes were mixed by pouring rapidly from one tube to the other twice and the stopwatch was started at the same time.

6. The boiling tube containing the reaction mixture was kept immersed in the water.

7. The time required for the red colour of the methyl red indicator to disappear was determined.

8. The whole experiment was repeated at 65, 55, 45, 35, 25 and 15 °C.

9. Ice was used to achieve the lowest temperature.

Results & Calculations

(Assume R = 8.314 J K^-1 mol^-1)

From Graph 1,

Slope of the graph = E_a / 2.303R

E_a / 2.303R = (2.94 – 2.20) / (3.35×10^-3 – 3.10×10^-3)

E_a /2.303R = 0.74 / (2.50×10^-3)

E_a = 2960 x 2.303R

E_a = 2960 x 2.303 x 8.314

E_a = 56676 J/ mol

Energy of activation, E_a = 56676 J /mol

Discussion

The reaction between bromide and bromate ions in acid solution is a slow chemical reaction at room temperature. This may be due to the high activation energy of the reaction, which required 56.676 kJ of energy in order for a reaction to take place. According to collision theory, high activation energy will cause the product more difficult to form since it is not sufficient energy for that molecule collide without enough energy. Any reaction cannot occur if the colliding molecules do not have the energy equal or higher than its activation energy. There are also other factors that can reduce the effectiveness of collisions of molecules such as present or absent of catalyst.

According to Table 1, we can observed that the higher the temperature of the reactant species, the shorter the time taken for the disappearance of red colour of methyl red indicator. This is shows that shorter time taken in the reaction once the reaction is faster and higher rate of reaction. Although the activation energy for the reaction to occur remains unchanged at all the temperature, but the rate of reaction increased as the temperature increased. This means that the rate of reaction is depends on the temperature of the reactant. This has also been proven in Arrhenius equation where rate constant, log₁₀ k is proportionally to 1/T with the slope of the graph, -E_a/R and a constant of log₁₀ A.

Log₁₀ k = -E_a/2.303RT + log₁₀ A

The higher temperature caused the value of the -E_a/2.303RT closer to the value of 0. The constant log₁₀ A will then minus off the value of -E_a/2.303RT and results in a larger value. The larger value will caused the value of the rate constant, k to become larger as well. Larger value of k will then results in faster reaction. Hence, this is proven that the higher the temperature of reactants, the reaction will proceed faster with the higher rate of reaction.

Besides, the rate of reaction roughly doubles for every 10 °C increase in temperature. This is because increase in temperature increase the kinetic energy of the molecules, the molecules with higher kinetic energy can move faster. With the higher speed, the molecules will collide more frequently results the larger amount of successful collision. With the higher kinetic energy of the molecules, the molecules can overcome the activation energy barrier during the collision and hence the reaction can be take places. As a result, the rate of the reaction could be increased if more heat energy is provided to the reactants.

Fixed amount of phenol and methyl red indicator were added to the mixture contents for the different temperature. This is because phenol can provides an intermediate state before the bromine molecules produced in the reaction between bromate and bromide ions in acid solution which is able to bleach the methyl red immediately. In other words, phenol is used to observe the time taken for the bromine molecules to react completely with phenol before bleach the methyl red indicator. The purpose of adding methyl red into the solution is to provide a colour which can easily to be observed. When the sulphuric acid was poured to the bromate/bromide ions solution, the methyl red indicator turns to pink colour. The bleaching effect from the bromine molecules caused the methyl red indicator to turn colourless after all the phenol is used up. This shows a colour changes which the time taken should be stopped. In order to compare the time taken for the bleaching of methyl red colour, the amount of phenol used in the repeated experiment must be equal.

The reason of choosing phenol as the reactant is due to it can form an intermediate state because phenol can react with bromine molecules quickly to produce tribromophenol and hydrogen bromide. This is happens before the bromine molecules react with the methyl red.

The –OH group in the phenol is an activating group in the benzene ring which can donates electron into the benzene ring to stabilize it. So, the product can be easily to form in the reaction. –OH group is ortho-para activating group in the benzene ring. This means that the incoming substituent will go into the ortho position or the para position, but hardly go into the meta position. This is due to the high energy is required for the product with a substituent at meta position to form in the reaction. The bromine molecules undergo substitution reaction in this reaction by substituting three hydrogen atoms from the benzene ring with three bromine atoms to maintain the aromaticity of the ring in phenol.

The reaction between bromate and bromide ions in acid solution is a redox reaction.

BrO₃^- + 5 Br^- + 6H⁺ --> Br₂ + 3H₂O

The potassium and sulphate ions act as spectator ions in this experiment and they are not participate in any of the redox reaction or changing of their state. The bromide ions undergo oxidation by donating one of its electrons to the bromine atom in the bromate ions. The bromine atom in the bromate ions then undergoes reduction by receiving electron from the bromide ions. The hydrogen ions and oxygen atom in bromate ions does not involve in increase or decrease in oxidation number but they were involved in changing the state from the ions in aqueous solution to the water molecule in liquid state.

Synthesis of Tert-Butyl Chloride

Objectives:

1. To produce tert-butyl chloride from tert-butyl alcohol

2. To understand the S_N1 and S_N2 mechanism involved in the reaction

3. To determine the yield of percentage of t-butyl chloride

Introduction:

Alkyl halide is also known as haloalkane or halogenalkane. Alkyl halide is a hydrocarbon group which attached with at least one halide atom in the molecule. Alkyl halides always resemble the parent alkanes in being colourless, relatively odorless and hydrophopic. Their boiling point is always increase as the parent chain increase, the longer the parent chain, the higher the melting point. This is due to the increased strength of the intermolecular forces—from London dispersion to dipole-dipole interaction because of the increased polarity. The molecules in the following show some of the example of alkyl halides:

Alkyl halide can be prepared from alcohol by reacting them with a hydrogen halide, HX (X=Cl,Br, or I). The mechanism of acid catalyzed substitution of alcohols are termed S_N1 and S_N2, where “S” stands for substitution while sub-“N” stands for nucleophilic, and the number “1” and “2” is described as first order and second order respectively. The “1” or “2” is also represent the reaction is unimolecular or bimolecular reaction. The secondary alcohols are more favor to react with hydrogen halides by both S_N1 and S_N2 mechanisms. For primary or methyl alcohol, both molecules undergo S_N2 mechanism while tertiary alcohol undergoes S_N1 mechanism.

R₃COH > R₂CHOH > RCH₂OH > CH₃OH

Tertiary alcohols react readily with HX alone to form alkyl halide, while secondary and primary require catalyze in the halohydrogenation reaction. Zinc chloride acts as the catalyze in the reaction. In some condition, heat supply is needed in the reaction. The mechanism for S_N1 and S_N2 are shown in the diagram 1.

Diagram 1

In an S_N1 reaction, the protonated alcohol, or oxonium ion losses a water molecule to form a carbocation intermediate in the rate-determining step. The carbocation is then rapidly attacked by halide ion (X^-) to form alkyl halide. Since tertiary alcohols form more stable carbocation intermediates than do primary and secondary alcohols, tertiary alcohols are the most likely follow the S_N1 pathway.

In S_N2 reaction, the nucleophile (X^-) assists in the explusion of H₂O from the oxonium ion via a bimolecular transition state. The S_N2 process is expected to be especially slow and even is not be observed for tertiary alcohols since the transition state will be particularly crowded; as the degree of substitution decreases at the reacting center the rate of the S_N2 process becomes greater and the rate of the S_N1 process decreases (vide supra). Consequently, the S_N2 process is the predominant one for primary alcohols.

In this experiment, t-butyl chlorride is synthesized from 2-methyl-2-propanol (t-butyl alcohol) by using HCl as the hydrogen halide. The chemical equation below show the formation of t-butyl chloride:

Diagram 2

The presence of a tertiary alkyl halide can be determined by reacting a small amount of the product with a silver nitrate (AgNO₃) in ethanol. Tertiary alkyl halides will react rapidly via S_N1 mechanism with the AgNO₃ to form a precipitate of AgCl:

Diagram 3

To promote the above S_N1 reaction, a highly polar solvent (ethanol) is used to dissolve the alkyl halide. The chloride will ionize to the alkyl cation and chloride ion. The cation will react with the alcohol solvent to form the ether and HCl. In this case both products are soluble; however, if silver ion is present in the solution, insoluble AgCl will form and a precipitate will be visible. Primary halides do not react in this test, and secondary reacts only slowly with heating.

Apparatus: separatory funnel, Erlenmeyer flask

Materials: 2-methyl-2-propanol (t-butyl alcohol), conc. HCl, saturated aqueous NaCl, saturated aqueous NaHCO, anhydrous calcium chloride, silver nitrate (AgNO₃)

Procedure:

1. 5 ml of 2-methyl-2-propanol (t-butyl alcohol) is put in an Erlenmeyer flask, the flask is put over a stir motor with stir bar and commence stirring.

2. 13mL of concentrated HCl is added into the flask.

3. The mixture is stirred for 15 minutes.

4. The mixture is transferred to a separatory funnel and allowed to stand until two clear layers have separated.

5. The aqueous layer is removed and the organic layer is washed with 6mL of saturated aqueous sodium chloride solution, then with 6mL of saturated aqueous sodium bicarbonate solution and finally with another 6mL of saturated aqueous sodium chloride solution.

6. The organic layer is saved and dried with anhydrous calcium chloride.

7. The product is weighed and volume is measured to determine the yield.

Silver nitrate test: A few drops (1cm³) of your product is put into a small test tube. 2 drops of silver nitrate test solution is mixed. The appearance of a white precipitate indicates that a reaction has taken place between the alkyl halide and silver nitrate.

Result and calculation:

Observation: A white precipitate is formed after adding of five drops of silver nitrate.

Weight of conical flask = 46.6443g

Wight of conical flask + weight of t-butyl chloride = 48.7942g

Weight of t-butyl chloride = 2.1499g

Density of 2-chloro-2-mehtylpropanol = 0.7809 g/cm³

Weight = density x volume

Weight of t-butyl alcohol = 0.7809 g/cm³ x 5 cm³

= 3.9045 g

Number of mole of (CH₃)₃COH used = 3.9045 g / 145.072 g mol^-1

= 0.02691 mole

(CH₃)₃COH (aq) + HCl (aq) à (CH₃)₃CCl (aq) + H₂O (l)

1 mole of (CH₃)₃COH produces 1 mole of (CH₃)₃CCl

0.02691 moles of (CH₃)₃COH produces 0.02691 mole of (CH₃)₃CCl

Theoretical weight of (CH₃)₃CCl = 0.02691 mol x 163.522 g mol^-1

= 4.4004g

Percentage yield = ( experimental value/ theoretical value) x 100%

= ( 2.1499g / 4.4004g ) x 100%

= 48.86%

Discussion:

In this experiment, 2-methyl-2-propnanol (t-butyl alcohol), (CH₃)₃COH is converted to 2-chloro-2-methylpropane (t-butyl chloride), (CH₃)₃CCl. In order to synthesis t-butyl chloride from t-butyl alcohol, hydrogen chloride is used to react with it. During the reaction take places, the t-butyl alcohol undergoes first order nucleophilic substitution, S_N1 mechanism since t-butyl alcohol is a tertiary alkyl group. The tertiary alkyl group will not undergo second order nucleophilic substitution, S_N2 mechanism. This is because the tertiary alkyl groups are successively more hindered as compared to the primary alkyl, thus this resulting in successively slower S_N2 reactions.

The first order rate reaction where the rate of formation of t-butyl chloride is dependent only on the concentration of the alcohol, however it is independent of the amount of acid (HCL) used. The strong concentrated hydrochloric acid (HCl) added to the t-butyl alcohol is used to provide an acidic medium and hence this protonates the electron rich hydroxyl group (nucleophile) allowing it leave as a molecule of water as shown in the diagram below:

The above diagram shows that the hydroxyl group is substituted by chloride atom when HCl is introduced.

The S_N1 mechanism that t-butyl alcohol undergoes is shown in the diagram 4 below:

Diagram 4

In the diagram 4, the t-butyl alcohol acts as a nucleophile which attacks the proton from the hydronium ion in the solution. According to Bronsted-Lowry Theory, the t-butyl alcohol is considered as a base in this reaction. This is because it accepts a proton from the hydronium ion and hence t-butyloxonium ion is formed. In order to become a stable molecule, the bond between the carbon and oxygen of the t-butyloxonium ion breaks heterolytically. The breaking of bond between carbon and oxygen leads to the formation of a carbocation and a leaving group of water.

Diagram 5

As shown in the diagram 5, the carbocation is formed and it is acts as eletrophile which is the species lack of electron. Due to the lacking of electron, another nucleophile, chloride ion, Cl^-, tends to attack the carbocation and hence to achieve a stable molecule. The carbocation acts as a Lewis acid which accepts electron from the chloride ion, Cl^- to form t-butyl chloride. The formation of t-butyl chloride is synthesized via S_N1 mechanism is shown.

The addition of concentrated hydrochloric acid into the t-butyl chloride causes the formation of cloudy solution is formed when stirring. The reaction between t-butyl alcohol and hydrogen chloride is a simple reaction which can take place in the room temperature. Two layers are formed after transferring the mixture into a separatory funnel.

The upper layer is t-butyl chloride whereas the lower layer is the aqueous layer. A 6mL of saturated sodium chloride solution is introduced into the separatory funnel after the aqueous layer is being removed. The purpose of adding concentrated sodium chloride is to pull water away from the organic layer. In another word, the saturated aqueous solution of sodium chloride is an inexpensive drying agent that will remove the bulk of water from a wet organic solution. Saturated sodium chloride can also help decrease the solubility of an organic compound in an aqueous solvent.

Aqueous sodium bicarbonate solution is added into the organic to neutralize the acidic medium that caused by concentrated hydrochloric acid added. Since sodium bicarbonate is an alkaline solution. The neutralization process between sodium carbonate and hydrochloric acid could be shown in the following chemical equation.

NaHCO₃ (aq) + HCl (aq) --> NaCl (aq) + H₂O (l) + CO₂ (g)

The sodium chloride salt, water, and gaseous carbon dioxide are formed in the neutralization process. The two layers are formed second time due to the formation of water in the neutralization. The sodium chloride is highly soluble in aqueous layer which is being discarded together with aqueous layer. Another portion of 6mL of saturated sodium chloride solution is introduced into the separatory funnel, to isolate the organic layer from aqueous layer left and reduce the solubility of the organic layer in water. Finally, the drying agent, anhydrous calcium chloride is added to remove all the water droplets in order to obtain a dried organic layer. Excess anhydrous calcium chloride is highly recommended to be used to make sure that there is no water droplet inside. The chemical reaction of anhydrous calcium chloride with water is

CaCl (s) + H₂O (l) --> CaCl.nH₂O (s)

Anhydrous Hydrated

calcium chloride drying agent

The presence of tertiary alkyl halides can be tested by using silver nitrate test. Some of the product formed in the experiment is added with silver nitrate solution. The observation we obtained is a white precipitate is formed after addition of silver nitrate solution. This is because the t-butyl chloride containing tertiary alkyl group which reacts rapidly via S_N1 mechanism with the silver nitrate to form a precipitate of silver chloride.

As shown in the diagram above, a highly polar solvent (ethanol) is used to dissolve the butyl chloride. The chloride will ionize to the butyl cation and chloride ion. The butyl cation will react with the alcohol solvent to form the butyl ethyl ether via formation of C-O bond. The HCl is formed in this reaction too. In this case both products are soluble; however, if silver ion is present in the solution, insoluble AgCl will form and a precipitate will be visible. Primary halides do not react in this test, and secondary reacts only slowly with heating.