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.

Determination of the Valency of Magnesium

Objective:

1. To study the quantitative relationship between the amount of reactants and products of a reaction.

2. To investigate the reaction of magnesium with hydrochloric acid

3. To identify the unknown X value in the chemical equation between magnesium and hydrochloric acid

4. To determine the valency of magnesium

Introduction:

Stoichiometry is the study of the quantitative relationship between amounts of reactants and products of a reaction. Stoichiometry canbe used to calculate the amounts of products with given the reactants and percent yield. In this experiment, a known starting mass of magnesium and the measured collection of hydrogen gas will be used to determine the reaction stoichiometry.

Magnesium is used to reacts with hydrochloric acid in order to produce hydrogen gases. One of the purpose of the experiments is to determine the value of X in the following equation:

Mg (s) + X HCl (aq) à MaCl₂(aq) + X/2 H₂(g)

The mass of magnesium is measured by the analytical balance before used in the experiment. A known amount of magnesium is reacted with a large amount of excess hydrochloric acid and hence the magnesium acts as a limiting agent in the particular reaction. As the amount of HCl is used in excess, this can ensure that all the magnesium will be reacted completely in the reaction. The formation of MgCl_x and H₂ are depends on the amount of magnesium used. The comparison of the amount of hydrogen gas produced with the amount of magnesium consumed will enable the X value to be determined.

Materials: magnesium ribbon, 0.5M hydrochloric acid, cotton, distilled water

Apparatus: 50ml burette, 25ml pipette, retort stand, analytical balance, watch glass, beaker, gauze, funnel, glass rod, thermometer

Procedure:

1. Upside down the burette and determine the volume of unmarked space in a clean and dry 50ml burette by pipetting 25ml of water into the vertically clamped burette.

2. Note the burette reading, drain the burette and repeat. Leave the water in the buretee for 10 minutes and check whether leaks occur.

3. A piece of magnesium ribbon is cleaned with steel wool. A piece of magnesium is cut off by scissors to a smaller size. Curl up the ribbon. Tare a watch glass on the four decimal balance and accurately weigh between 0.03g and 0.036g of the magnesium on the watch glass.

4. A small filter funnel with short stem (1cm-1.5cm) is covered with gauze and then is inverted. The filter funnel with gauze is placed on the watch glass over the magnesium.

5. The beaker is filled with tap distilled water until the level is approximately above the end of the funnel stem by 0.5cm-1.0cm. The burette is completely filled with 0.5M HCl and invert it(with supplied cork pressed flat over the open end) and place it in the water in the beaker.

6. Remove the cork and place the end of the burette over the stem of the funnel, ensuring that no air enters and clamp it into position.

7. Remove the excess of water with a pipette until the level is just above the stem if the funnel.

8. Add about 100ml of 0.5M of HCl to the beaker, use a glass rod to stir the water to ensure complete mixing such that the HCl reachest the magnesium. This may be helped by tapping the watch glass gently with a glass rod.

9. Stir the solution to initiate the reaction and then do not stir further so that the reaction proceeds unaided. At the completion of the reaction, tap the watch glas gently to dislodge any gas bubbles.

Result and calculation:

Mass of magnesium ribbon= 0.0356g

Initial reading of burette = 5cm³

Final reading of burette = 37cm³

Volume of unmarked place = 5cm³

Total volume of H₂ collected = 35cm³

Mg (s) + X HCl (aq) à MaCl₂(aq) + X/2 H₂(g)

Mole of magnesium = 0.0356g/24 g mol^-1

= 1.483x10^-3 mol

Actual mole of HCl used = 100cm³ x 0.5M / 1000

= 0.05 mol

Number of mole of HCl / number of mole of Mg

= 0.05 mol/ 1.483x10^-3 mol

= 33.72 > 1

Magnesium is a limiting agent.

Theoretical mole of hydrogen gas = 0.035dm³ / 24dm³ mol^-1

= 1.483x10^-3 mol

Number of moles of magnesium : number of mole of hydrogen gas

1.483x10^-3 mol : 1.483x10^-3 mol

1 : 1

Thus, X/2 = 1

X = 2

Discussion:

Magnesium is an alkaline earth metal which is categorized to the Group 2 element in the periodic table. All the alkaline metal reacts with acid will produce hydrogen gas. Since the HCl is added in excess, hence the magnesium is the limiting reactant in this reaction. Limiting reactant is the reactant which restricts the amount of products generated in the reaction. In this situation, the magnesium is used up completely to react with the excess hydrochloric acid.

From the calculation, the value of X in the chemical equation between magnesium and hydrochloric acid was determined. The value of X is 2. So that, the complete chemical reaction for magnesium and acid is Mg (s) + 2 HCl (aq) à MaCl₂(aq) + H₂(g). Based on the Las of Conservation of Matter, matter can be neither created or destroyed. An equation must have the same number of atoms of the same kind on both sides of the equation. Hence, 1.483x10^-3 mol of magnesium was used up completely to react with 2.966x10^-3 mol of acid to produce 1.483x10^-3 mol of magnesium chloride and hydrogen gas.

The reaction between magnesium and hydrochloric acid is an exothermic reaction. This reaction releases heat energy from the reaction to its surrounding which caused the solution is the beaker become hotter. The temperature cannot be detected significantly because the rise in temperature was just slight different. The temperature is only increased by few degree Celsius due to the large specific heat capacity of the water.

The magnesium ribbon used in the experiment was in the curved-shape. The curved shape magnesium ribbon was used instead of the flat magnesium ribbon because the former one can reacts faster than the latter one. The curved magnesium own larger surface area which increase the reaction proceed faster compared to the flat magnesium ribbon although they are in the same volume. As the surface area is increased, the magnesium atoms can have more surfaces to be exposed to the hydrochloric acid. Thus, the reactivity of the reaction increases.

Effect of temperature and concentration on equilibrium system

Objectives:

1. To study the effect of concentration on equilibrium

2. To study the effect of temperature change on equilibrium

3. To predict the direction of the net reaction in an equilibrium system by Le Chatelier’s principle

Introduction:

Chemical equilibrium applies to reactions that can occur in both directions. In a reaction such as:

aA(g) + bB(g) cC(g) + dD(g)

The reaction can happen both ways which is a reversible reaction. Some of the products are created the products begin to react to form the reactants. In a chemical reaction, chemical equilibrium is the state in which the concentrations of the reactants and products have no net change with time. Chemical equilibrium is a dynamic process that consists of a forward reaction, in which the reactants are converted into the products; and a reverse reaction, in which the products are converted into the product. Usually, this state results when the forward reaction proceeds at the same rate as the reverse reaction. The reaction rates of the forward and reverse reactions are generally not zero, but with the net change in concentration of the reactant and product are the same. The ideal equilibrium constant may expressed as where [ ] is the concentration of the chemicals

For the gaseous reactant, it is express as the equilibrium condition in the terms of partial pressures of reactants and products where P is the partial pressure

Factors that affect the equilibrium constant are the changes in the experimental conditions such as concentration, pressure, volume, and temperature disturb the balance and shift the equilibrium position so that more or less of the desired product is formed. The changes of outcome of these changes can be predicted by Le Chaptelier’s principle. According to the Le Chatelier’s Principle, changes the conditions of a chemical system at equilibrium such as temperature or pressure. The equilibrium shifts to the right or left to relieve the disturbance places on the system. To See how Le Chatelier’s Principle works, consider the reaction used to prepare hydrogen gas:

C(s) + H₂O(g) CO(g) + H₂(g) H = + 131 kJ

Le Chatelier’s Principle states that when a chemical system at equilibrium is disturbed, it retains equilibrium by undergoing a net reaction that reduces the effect of the disturbance. One of the ways is by adding or removing reactants or products, which is means that changing the concentration of reactants or products. In the hydrogen gas reaction if more steam (H₂O) was added, the reaction would shift to the right producing more hydrogen and carbon monoxide. The reaction shifts to cause more hydrogen to react. This shifts counteracts the stress of adding more hydrogen to the system. Likewise, by removing hydrogen and carbon monoxide, the equilibrium, shifts to the right, or if more of these products are added, the reaction shifts to the left.

Another factor of a chemical system is by changing the pressure. The key to this lies in the stoichiometry of the reaction. The effect of pressure on equilibrium depends on the number of moles of gas particles on the right and left sides of the balanced equation. In reactions where the numbers of moles of gas are equal, a change in pressure has no effect on the equilibrium. If Δn_gas = 0, there is no effect on the equilibrium position. Changes in the pressure are directly related to changed in volume. This is because a decrease in volume gives the same result as increasing the pressure, while an increase in volume is the same as a decrease in pressure. For example, this situation exists when equilibrium occurs in a cylinder with a movable piston and the volume is changed by moving the piston. Pure solids and liquids do not need to be considered when using Le Chatelier’s Principle because the concentrations of pure solids and liquids are constant during the course of the reaction.

The final factor to be considered is a change in temperature. Other than the three types of disturbances, only temperature changes alter K_c. To apply Le Chatelier’s Principle with temperature changes, the sign of H for the reaction needs to be known. A temperature rise will increase K_c for a system with a positive ΔH whereas a temperature rise will decrease K_c for a system with a negative ΔH. The increase of K_c will shift the equilibrium position to product side. Likewise, the decrease in K_c will shift the position of equilibrium to reactant side. If the heat is supplied to the endothermic reaction ( H is positive), the equilibrium will favor to the product direction. If the heat is removed from the exothermic reaction ( H is negative), the equilibrium will favor to the product direction. For the temperature is not favor to the reaction, the equilibrium position will shift to the reactants side.

Thymolphthalein and phenolphthalein are the acid base indicators. Both indicators are often written as HIn in shorthand notation. C₂₈H₃₀O₄ and C₂₀H₁₄O₄ are the formula of the thymolphthalein and phenolphthalein respectively. The thymolphthalein has transition range is at approximately pH 9.3-10.5. Below this pH 9.3, it is colorless whereas the colour is blue when above pH 10.5. For phenolphthalein, the colour of the indicator is pink within the range of pH 8.2 to pH12. The following diagram A and B shows the structure of the thymolphthalein and phenolphthalein

 

Apparatus: test tubes, thermometer ( 0-200 )

Materials: solid copper (II) nitrate, 0.4M cobalt (II) chloride, 0.1 M potassium chromate, 0.1 M potassium dichromate, 0.2 M HCl, 0.2M NaOH, thymolphthalein indicator, 0.1M ammonia solution, solid ammonium chloride, phenolphthalein indicator

Experimental procedure:

i) Effect of temperature changes on equilibrium

a) Investigate the equilibrium between a solution and a solid

Cu(NO₃)₂ (aq) Cu(NO₃)₂ (s)

1. Some solid Cu(NO₃)₂ is poured into a test tube and half of test tube is filled with distilled water. The test tube is shaked until form a saturated solution. If all the Cu(NO₃)₂ is dissolves, additional Cu(NO₃)₂ is added into test tube and shake it until a saturated solution with some excess solid is obtained.

2. The colour of solution at room temperature is noted. A waterproof pen is used to mark the level of the top of the solid.

3. Test tube is placed in a beaker with hot water at 60 for over 30 minutes. Observation is recorded.

4. Test tube is placed in an ice bath for another 30 minutes and observation is recorded.

b) Investigate the aqueous equilibrium of a cobalt complex

[Co[H₂O]₆]²⁺ (aq) + 4Cl^-(aq) [CoCl₄]^2- (aq) + 6H₂O (l)

1. Test tube is placed in 60 water bath for 15 min, and then observation is recorded.

2. Same test tube is placed in an ice bath for 15 min and observation is recorded.

ii) Effect of concentration changes on equilibrium

a) Dichromate/chromate solution

2CrO₄^2- (aq) + 2H⁺ (aq) Cr₂O₇^2- (aq) + H₂O (l)

1. 5 drops of 0.1M potassium chromate solution is placed into two small teest tubes and the colour is noted. Test tube is labeled as 1 and 2.

2. 5 drops of 0.1 M potassium dichromate solution is placed into another small test tube and the colour is noted. Test tube is labeled as 3.

3. 0.2 M HCl is added drops by drops into tube 1 and is shaked when adding, until a colour change. Observation is noted. The colour change is compared with tube 2 and 3 and observation is recorded.

4. 0.2m of NaOH is added drop by drop into tube 1 and is shaked when adding, until a colour change. Observation is noted.

5. Some acid is added into tube 1 and observation is noted.

b) Thymolphthalein equilibrium

Hln (aq) H⁺ (aq) + ln^- (aq)

1. Small amount of thymolphthalein solution is placed on a small piece of cloth with paint brush and leave it exposed to air.

2. The colour change of the blue stain is noted. Time is recorded it took to change colour.

c) Ammonium ion / ammonium solution

NH₃ (aq) + H₂O (l) NH⁴⁺ (aq) OH^- (aq)

1. 5cm³ of 0.1M ammonia solution is placed in two small test tube, 2 drops of phenolphthalein is added into each test tube.

2. One of the test tube is added with solid ammonium chloride, a little at time.

3. The colour of both test tube are compared and observation is recorded.

Result and calculation:

Part (i)

Table 1.1 Observation of copper(II) nitrate and its colour change at 60 and 0

Temperature,	Observation
Room temperature	Blue colour solution is formed after solid copper(II) nitrate dissolved.
60	Blue colour solution remains unchanged but the level of solid decreases.
0	Blue colour solution remains unchanged but the level of solid increases.

Table 1.2 Observation of colour change of cobalt(II)chloride at 60 and 0

Temperature,	Observation
Room temperature	Pink colour solution is formed
60	Colour of solution turns from pink to dark pink
0	Dark pink solution turns to light pink solution.

Part (ii)

Table 2.1 Comparison between colour of potassium dichromate and colour changes of potassium chromate solution with addition of HCl and NaOH

Table 2.2 Observation of colour change of thymolphthalein on cloth and time taken to decolourise

Solution	Observation
Thymolphthalein	The blue colour solution is turns to colourless after exposed to air for two to three second.

Table 2.3 Colour change of ammonium solution after with addition of ammonium chloride

Solution	Observation
Ammonium solution + phenolphthalein	The light purple colour is formed.
Ammonium solution + phenolphthalein + ammonium chloride	The light purple solution is decolourized.

Discussion:

In the experiment part (i) (a), the solid copper(II) nitrate is used to dissolve in a test tube filled with half distilled water. The solid copper(II) nitrate ionize in water to form copper(II) ions and nitrate ion. Hence, a blue solution is formed due to the blue copper(II) ion present in the solution. When excess solid is added, the solution will become saturated and do not allow any solid to dissolve, so excess solid will remain in the solution. This is shows that the equilibrium between solid and aqueous copper(II) nitrate is achieved. When the test tube is being placed in the water bath of 60 , the level of solid copper(II) nitrate tends to decrease to ionize in the solution but the blue intensity of the solution remains the same. The dissolution of copper(II) nitrate is an endothermic reaction. So, the equilibrium constant of the reaction increases as the temperature of the solution increases which allow more products are formed. The equilibrium position is shift to the left in the equilibrium equation below:

Cu(NO₃)₂ (aq) Cu²⁺ (aq) + NO₃^- (aq)

Eventually, more copper(II) ions and nitrate ion are formed at the high temperature due to the equilibrium effect. Oppositely, the level of solid copper(II) nitrate is increases at the 0 . When the test tube is placed in an ice bath, the low temperature causes the equilibrium constant of the solution decreases and hence the formation of solid copper(II) nitrate is increases. The equilibrium position shifts to the left at the condition of 0 .

Next, the cobalt(II) chloride is added with hydrochloric acid in this experiment in order to investigate its equilibrium at different temperature. The cobalt (II) chloride is added with hydrochloric acid, which a pink solution is formed. When the test tube is placed in a water bath with 60 , the solution turns from light pink to dark pink. This is because the cobalt(II) chloride dissolve in solution is an endothermic reaction which it tend s to form blue cobalt (II) chloride in the solution. The dark pink solution is formed instead of light pink because the dark pink colour of solution is the mixture of pink and blue. The equilibrium position shift to right in the equilibrium equation below:

[Co[H₂O]₆]²⁺ (aq) + 4Cl^-(aq) [CoCl₄]^2- (aq) + 6H₂O (l)

Since the equilibrium equation above is an endothermic reaction, the position of equilibrium tends to shift to right at high temperature, 60 in this experiment. The increase in equilibrium constant leads to more products is formed. So, the colour changes to dark pink due to the amount of cobalt(II) chloride formed is increases. On the other hand, the equilibrium position shifts to the left at the temperature of 0 . When the equilibrium shifts to left, the concentration of hydrated cobalt ion form and hence the solution forms at a light pink colour.

In the experiment part (ii) (a), the potassium dichromate solution is orange colour while the potassium chromate solution is appears in yellow colour. When there is addition of two drops of 0.2M hydrochloric acid,HCl, the colour of potassium chromate solution turns from yellow to orange due to the equilibrium effect. The equilibrium equation between chromate ion and dichromate ion is

2CrO₄^2- (aq) + 2H⁺ (aq) Cr₂O₇^2- (aq) + H₂O (l)

When HCl is being added, the acid will ionize to become H⁺ ion and Cl^- ion. The chromate ion, CrO₄^2- ion accept the excess H⁺ and shift to the right to become dichromate ion, Cr₂O₇^2- which is orange colour ion. But, when two drops of sodium hydroxide ion, NaOH is added, the OH^- ion is ionize and tends to combine with H⁺ to become water and hence the concentration of H⁺ ion in the solution decreases. The dichromate ion, Cr₂O₇^2- will release H⁺ ion to form chromate ion, CrO₄^2- in the solution, thus the colour change to yellow colour. When second time of HCl is added, the chromate ion, CrO₄^2- will accept the H⁺ ion to dichromate ion, Cr₂O₇^2- form if the H⁺ ion concentration is excess.

In the part (ii) (b) experiment, the thymolphthalein is undergoes reduction which its blue colour solution is decolorized. When the blue thymolphthalein is exposed to the air, the carbon dioxide in the air will react with the thymolphthalein since it is a basic solution. Initially, dilute sodium hydroxide solution is added into the thymolphthalein when preparing the thymolphthalein indicator in order to maintain it pH in basic condition. If hydroxide ion, OH^- is excess in the solution, the blue colour can be perceived. However, when thymolphthalein is exposed to the air, the sodium hydroxide will tends to react with carbon dioxide to form sodium carbonate and water as shown in the chemical equation below:

At the same time, the carbon dioxide will react with water in the solution to form carbonic acid, H₂CO₃. Due to the partial ionization of carbonic acid, the hydronium ion, H⁺ is produced which caused the condition become more and more acidic (pH value dropping).

The ionized H⁺ ion will react with the thymolphthalein which driving the equilibrium shift to the right to form colourless Hln in the solution.

Hln (aq) H⁺ (aq) + ln^- (aq)

The equilibrium between ammonia molecule and ammonium ion is studied in this experiment. When phenolphthalein is added into the ammonia solution, the colour present in the solution is light purple colour. This is because the ammonia solution is a weak acid with pH value of 11.6. The phenolphthalein turns from colourless to light purple when it is added into the solution with pH 11.6. When solid ammonium chloride is added, the ammonium chloride dissolves quickly in the solution to form ammonium ions and chloride ions. The ammonium chloride dissociates completely in the solution. The formation of ammonium ions in the solution combine with hydroxide ions and hence shifts the equilibrium position to left to form more ammonia and water as shown below:

NH₃ (aq) + H₂O (l) NH⁴⁺ (aq) OH^- (aq)

At the same time, the pH value of the solution decreases due to the concentration of the hydroxide ions is reduced by the reaction. As a result, the phenolthalein in the solution is decolourized due to the drops in pH of the solution.

Thermodynamic Study on reaction between an Acid and a Base

Objectives

1. To study the enthalpy chemistry

2. To study the exothermic reactions

3. To determine the calorimeter constant

4. To determine the enthalpy of reaction of acid-base reactions

Introduction

Enthalpy is defined as the total amount of total energy of a thermodynamic system. The enthalpy of a system includes the internal energy and the work done of the system. The work done is equals to the product of the volume of the system multiplied by the pressure exerted on it by its surrounding, as shown in the following:

H = U + w

H = U + pV

where H = Enthalpy

          U = Internal Energy

          W = work done

           p = Pressure

           V = Volume

Enthalpy of a thermodynamic system is usually measured in S.I. unit of Joule, J or kiloJoule, kJ. However, we often measure the change in enthalpy, ΔH instead of measuring the value of enthalpy of the system, H because the total enthalpy of the system cannot be measured directly.

Change in enthalpy, ΔH is defined by the following equation:

ΔH = H_Final – H_Initial

ΔH = Enthalpy change

H_Final = Final enthalpy of the thermodynamic system. In a chemical reaction, H_Final is the enthalpy of the products.

H_Initial = Initial enthalpy of the thermodynamic system. H_Initial is the enthalpy of the reactants in a thermodynamic reaction.

In the endothermic reaction, the magnitude of a thermodynamic reaction is shown in positive. By contrast, the exothermic reaction shows in negative value. Exothermic reactions involve heat energy transferred from the system into its surroundings, causing the temperature of the surroundings to rise. On the contrary, an endothermic reaction involves the energy acquired from the surrounding into the system.

The change in enthalpy of a thermodynamic system either endothermic or exothermic reaction is numerically equals to the magnitude of heat of the reaction which is ΔH = q. The heat of reaction is easily to be measured adiabatically by using a Dewar flask. The increase or decrease in temperature of the products produced by the reaction in solution is being measured and compared with the initial temperature of the system. The difference in the initial value of enthalpy and final value of the enthalpy is the change of the enthalpy. The Dewar flask is being used because its design is to preserve the heat from loss or minimize heat loss to the surrounding from the system. In addition, an isolated system is also needed in this experiment to obtain a more accurate data. Each Dewar flask used in the experiment has different calorimeter constant since each Dewar flask has difference in materials used. In order to measure the total amount of heat in a chemical reaction, the calorimeter constant ( ) must be firstly determined. The calorimeter constant is defined as the quantity of heat required to increase the temperature of the calorimeter and its content by 1 °C.

C_cal = ∆H / ∆T

The constant is measured by supplying the calorimeter and contents with a definite known quantity of heat. This can be done electrically or by adding a known amount of concentrated sulphuric acid.

Results & Calculation

Part 1 Calorimeter Constant

From Graph 1,

Δ T₁ = 31.0°C – 22.8°C

= 8.2 °C

From Table 4,

Volume of sodium hydroxide solution used = 34.2 cm³ – 15.0 cm³

= 19.2 cm³

Moles of sodium hydroxide used = 1M x (19.2/1000) dm³

= 0.0192 moles

H₂SO₄ + 2 NaOH àNa₂SO₄ + 2H₂O

Moles of sulphuric acid used = 1/2 x 0.0192 moles

= 0.0096 moles

Molarity of sulphuric acid solution = 0.0096 moles / (25/1000) dm³

= 0.384M

By using the values given, 0.384M in Graph 2, we can estimate the amount of heat liberated which is 2.775 kJ. Since the heat liberated by the dilution of sulphuric acid is then absorbed by the Dewar flask, so the value of the ΔH is in positive value.

From Graph 2,

C_cal = ΔH₁ / ΔT₁

= 2.775kJ / 8.2 °C

= 0.338 kJ / °C

Therefore, the calorimeter constant has a value of 0.338 kJ / °C.

Part 2 Enthalpy of Reaction

Part I

HNO₃ + NaOH à NaNO₃ + H₂O

From Graph 3,

ΔT₂ = 29.5°C – 22.0°C

= 7.5 °C

ΔH₂ = C_cal x ΔT₂

= 0.338 kJ / °C x 7.5 °C

= 2.535 kJ

The amount of heat absorbed by the Dewar flask after adding nitric acid to the mixture (50 cm³ of water + 50 cm³ of 1M sodium hydroxide) is 2.535 kJ. This means the total enthalpy of reaction, ΔH has a value of -2.535 kJ because the heat is being released from the reaction. However, the enthalpy changes during the dilution of nitric acid also need to be considered to obtain more accurate result.

Part II

From Graph 4,

ΔT₃ = 23.0°C – 21.5°C

= 1.5 °C

ΔH₃ = C_cal x ΔT₃

= 0.338 kJ / °C x 1.5 °C

= 0.507 kJ

Amount of heat absorbed by the Dewar flask after adding nitric acid to the 100 cm³ of water is 0.507 kJ. This means the dilution of nitric acid has liberated heat to the surrounding.

ΔH_dilution = -ΔH₃

= -0.507 kJ

ΔH_Reaction = ΔH₂ - ΔH_Dilution

= [-2.535 – (-0.507)] kJ

= -2.028 kJ

Therefore, the enthalpy of reaction of sodium hydroxide with nitric acid is -2.028 kJ.

Discussion

In this experiment, the calorimeter constant for the Dewar flask used has a value of 0.338 kJ / °C. In another word, this means that for every 0.338 kJ of energy absorbed by the Dewar flask, the contents of the Dewar flask will increase by 1 °C. In order to obtain an accurate result, the Dewar flask must be an isolated system. This is to ensure that all the heat released to or absorbed from the surrounding remain inside the Dewar flask but not from outside of the flask. However, ideal Dewar flask is does not exist in the world that makes sure no heat will be loss from it. This is because heat can travel through vacuum by radiation which is uncontrollable. So, the Dewar flask can only able to minimize the radiation of heat at the same time prevent conduction and convection of heat to the surrounding.

Besides, the top of the Dewar flask is open in this experiment to ease the measurement of temperature of the contents in the Dewar flask. This could cause the error in data because the system is an open system instead of an isolated system. The open system allows some of the heat loss to or absorbed from the surroundings which caused the change in temperature, ∆T to be inaccurate. During the titration, methyl orange was introduced into the solution before titration is carried out. The purpose of using methyl orange is to act as an indicator of the equivalence point of the titration. This is because it has a sharper end point and the change of colour is easily to be observed. The methyl orange is red in colour when the pH value is below 3.1 and yellow when above 4.4. The pH values which lie between 3.1 and 4.4 will show the mixture of the two colours.

From the experiment, both the reaction between sodium hydroxide with sulphuric acid and sodium hydroxide with nitric acid are exothermic reactions. This is because the temperature differences where the temperature of the mixture increased after reactions take place. The heat is released from the system to its surrounding which caused the surrounding temperature to rise. In part 2, we do not take the value of ΔH as the enthalpy of reaction between sodium hydroxide and nitric acid because dilution of nitric acid also can cause change in enthalpy. Instead, we take into consideration of enthalpy change of dilution of nitric acid to obtain a more accurate result.

Flickr Tags: enthalpy chemistry,enthalpy of reaction of acid-base reactions,determination of the calorimeter constant

Esterification: Methyl benzoate

Objectives:

1. To produce methyl benzoate by esterification

2. To learn the reaction mechanism involved in esterification

3. To demonstrate how an ester can be made by the interaction of a carboxylic acid and an alcohol with the presence of a sulfuric acid catalyst.

Introduction:

The ester group is an important functional group that can be synthesized in a number of different ways. The low molecular-weight esters have very pleasant odours and indeed are major components of the flavour and odour aspects of a number of fruits. Although the natural flavour may contain nearly a hundred different compounds, single esters approximate the natural odours and are often used in the food industry for artificial flavours and fragrances.

The esterification is a reaction between an alcohol and a carboxylic acid or a carboxylic acid derivative, water and ester will be formed as products in this process under reflux. In the chemical structure of carboxylic acid, R-COOR’, where R and R' are either alkyl or aryl groups. As shown in the diagram below, the esterification is also known as condensation process which water is produced.

In this case, each of the carboxylic acid will contributes hydroxyl group while each alcohol will contribute hydrogen atom to form a water molecule. So, the reaction involves the removal of water once the ester is formed. The ester linkage will appear as the bond that connected the one carboxylic acid and one alcohol in the ester molecule as shown as the following:

Generally, an acid will be used to act as a catalyst in esterification process.

The esterification is a reversible process which the equilibrium between reactants and products will be reached. The opposite of the esterification reaction is called hydrolysis. The addition of water to the ester link will cause breaking apart of the ester into their parent carboxylic acid and alcohol. The hydrolysis also requires the presence of a catalyst (either acid or base). The esterification is a slow process. The main reason is due to the water produced in the esterification will be used back in the hydrolysis which converts the ester to form parent alcohol and carboxylic acid in the reaction. As the ester is start to form in the reaction but at the same time the hydrolysis start to begin. An equilibrium is finally attained, all of the related reactants and products will present in the mixture formed via esterification and hydrolysis.

The mechanism of the formation of ester under acidic condition might be follows this steps below:

1)

However, the esterification only can be applied by using simple alcohol and carboxylic acid in acidic condition. In another word, the long chain alcohol cannot be used to react with carboxylic acid. We can use another method to produce ester by using carboxylic acid reacts with haloalkane in a basic condition. But the same problem may be encountered at the end, this method is only limited to the primary alkyl halides. So, in order to utilize the maximum ester, an alternative for the preparation of esters is to treat the alcohol with a reactive carboxylic acid derivative. For example, carboxylic acid anhydrides or chloride can be used.

(RCO)₂-O + R’’OH -----> RCOOR’’ + RCO₂H

RCOCl + R’’OH -------> RCOOR’’ +HCl

These reaction is irreversible and they react rapidly especially when catalyzed by a strong acid.

Apparatus:

Round bottomed flask (250ml), Liebig condenser, separating funnel, Bunsen burner. Thermometer

Materials:

Chloroform (trichloromethane), anhydrous sodium sulphate, benzoic acid, methanol, conc. sulphuric acid, anti bumping granules

Procedure:

1. Benzoic acid is placed into a 250ml beaker and then methanol is added .

2. Concentrated sulphuric acid is added whilst swirling the contents and washed with methanol.

3. Two or three anti-bumping granules are added to the mixture and fit in to the reflux set up.

4. The mixture is being refluxed for 1 hour, the mixture is poured into a separating funnel with 150ml of water after cooling.

5. The reaction flask is rinsed with chloroform two times and is added into the funnel.

6. The aqueous layer is being removed and water is used to wash the organic layer.

7. Anhydrous sodium sulphate is used to dry the solution and then is filtered out.

8. The organic solvent is distilled at three different ranges, which are 30-90°C, 91-190°C and 190°C above.

9. The ester is collected in a weighed flask and the distillation temperature range is noted.

10. The percentage of theoretical yield is calculated.

Results:

Weight of benzoic acid = 12.2017g

Weight of conical flask = 49.1914g

Weight of conical flask + weight of ester = 62.5712g

Weight of ester = 13.3798g

C₆H₅COOH + CH₃OH <-----> C₆H₅COOCH₃ + H₂O

Number of moles of C₆H₅COOH = 12.2017g / 122.118 g mol^-1

= 0.09992 mole

0.09992 mole of benzoic acid will produce 0.09992 mole of ester

Weight of C₆H₅COOCH₃ = 0.09992 mole x 136.144 g/mol

= 13.6035 g

Percentage yield = 13.3798g/ 13.6035g x 100%

= 98.3556 %

Discussion:

Benzoic acid and methanol are used the reactants with the presence of sulphuric acid in this experiment. The acid catalysed reaction between benzoic acid and methanol may be represented as:

This esterification using the benzoic acid and methanol is known as Fisher esterification. The concentrated sulphuric acid is added as a catalyzed in this experiment. The purpose of using catalyze is to speed up the esterification because it is a slow process. Concentrated sulphuric acid also serve as another function which is used to protonates the carboxylic acid and hence this will initiate the reaction to start. The esterification between benzoic acid and methanol is favourable in acidic condition; as a result more ester can be formed in this experiment.

After added with the concentrated sulphuric acid, another portion of methanol is introduced into the mixture. The adding of methanol is to ensure all the carboxylic acid could be reacted completely during reflux. Furthermore, anti-bumping granules are added to promote smooth boiling and to prevent bumping of the solvent. As mention at the above, the esterification is slow and reversible process so it is distilled for one hour and hence more ester could be generated. The esterification mechanism is take place as the diagram 1 below:

Diagram 1

The dissociation of sulphuric acid will produce hydrogen ion which can be used to protonates the carbonyl group in benzoic acid. The carbonyl group is protonated reversibly and caused the positive charge of carbonyl group to be increased. Thus this increases the reactivity of carbonyl group towards nucleophile. The C-O double bond is broken in order to stabilize the OH⁺ group to form hydroxyl group in the benzoic acid molecule. The methanol acts as a nucleophile attacks the benzoic acid.

Diagram 2

In the diagram 2, the methanol successful attacked the carbonyl group to form a new C-O bond to the carboxyl group in the benzoic acid to form a tetrahedral intermediate. This is called nucleophilic addition. The oxygen atom in the carboxyl group in benzoic acid is more electronegative due to its lone pair electron. The lone pair electron in the particular electron attracts the hydrogen atom from the methanol to form oxonium ion. Now, the oxygen atom in the methanol becomes unstable and hence the C-H bond will tend to be broken down. The electron between the C-H bond will delocalise to the oxygen atom of the methanol. The formation of oxonium ion in the carboxyl group in benzoic acid tends to be released from the intermediate to form water. Eventually, another hydroxyl group will donate the lone pair electron to the attached oxygen atom to form a more stable intermediate.

Diagram 3

The hydrogen atom in the ester intermediate will be attacked the acid in diagram 3 and the acid catalyze is regenerated. Thus, finally the methyl benzoate is formed.

During esterification, the hydrolysis process start to begin once the ester is being produced. The water produced in esterification is used back in the hydrolysis to hydrolyze the ester to form the carboxylic acid and alcohol. The process is a continuous reversible process until the equilibrium is reached. The hydrolysis process would be the following equation:

The hydrolysis process must under acidic or basic condition in order to break down the stable ester molecule. To avoid the hydrolysis process, the water could be removed from the mixture. According to Le Chatelier’s principle, the equilibrium position will shift to the product side if water is being removed. Another method to increase yield of ester is by adding more alcohol into the mixture. Hence, more ester could be generated when the amount of alcohol increased which shift the equilibrium to product side. The mechanism of hydrolysis is shown in the diagram 4 below:

Diagram 4

Water molecule acts as nucleophile to attack the carbonyl carbon of ester reversiblely. The C=O will be break and the oxygen atom attached to the carbonyl carbon form negative oxygen atom. The methoxy group tends to leave the tetrahedral intermediate to form a stable methoxide, so the negative oxygen atom donates the electron back to the carbonyl carbon. Finally, the methoxide acts as a nucleophile which attack the hydroxyl group bonded to carbonyl carbon and hence to form methanol. The oxygen atom now is negatively charge and it tends to get a proton from the environment. Thus benzoic acid is formed.

After reflux for one hour, the reaction mixture is introduced into the separating funnel with distilled water to carry out extraction. This is because the unreacted methanol can be dissolved in distilled water and hence it could be removed together with the aqueous layer. Small amount of the benzoic acid present in the mixture will dissolve in the water although it is highly insoluble in water. Distilled water can extracts the leftover of methanol and small amount of benzoic acid. Chloroform is added to extract the ester by dissolving the ester in the chloroform. After the aqueous layer is being removed, the anhydrous sodium sulphate is added as drying agent which will absorb water droplet and hence residual water can be removed completely.

Now, the residual ester still contains some methanol, benzoic acid and other side products. So, we use distillation to obtain the pure ester from the mixture. During distillation, the temperature of the distillate is kept constant at 63°C. This is because some of the methanol is still left in the mixture and its boiling point is around 65°C. So, the temperature remains at 63°C until all the methanol are completely distilled. For the second time of distillation, the temperature of distillate is keep changing in the range of 140°C to 180°C. This might be due to the mixture contain some impurities, so the temperature may fluctuate in the wider range of temperature. Another reason for the fluctuation of temperature may be side product formed during reflux is exist in the mixture. So, this contribute to the temperature to be fluctuated. The ester is being synthesized is 98.3556% but this figure is not reliable. The weight of the solution is very high due to the impurities present in the solution. The actual figure that an ester could be form is 60-70%, which according to the estimation of scientist.

 

Flickr Tags: Esterification: Methyl benzoate,mechanism of the ester formation,acid catalyzed esterification,benzoic acid and methanol

Synthesis of n-Butyl Ethyl Ether from 1-Butanol

Objectives:

1. To synthesize n-butyl ethyl ether from 1-butanol
2. To understand mechanism involved in the reaction

Introduction:

In this experiment, the procedure to generate n-butyl ethyl ether from 1-butanol is divided into two parts. The first part involves the formation of n-butyl bromide from 1-butanol. Alkyl halides are very useful intermediates in organic syntheses. The most common synthetic preparation of alky halides is the replacement of the hydroxyl group, OH of an alcohol by a halogen, HX. The displacement of a hydroxyl group by halide ion is successful only in the presence of a strong acid. 1-butanol is used to be converted into 1-bromobutane with adding of sodium bromide and sulphuric acid. The nucleophile for the reaction is Br^- ions. The nucleophile in this lab is generated from an aqueous solution of sodium bromide. The sulfuric acid acts as a catalyst in this reaction. The sulphuric acid protonates 1-butanol to produce suitable leaving group, OH, in S_N2 reaction. The chemical reaction is shown as below:

If this displacement reaction is attempted in the absence of an acid it is unsuccessful because leaving group would be a hydroxide ion which is a poor leaving group and a strong base.

In the second part of the experiment, the n-butyl bromide produced in the first part is being converted into n-butyl ethyl ether by using methanol and sodium hydroxide. The chemical reaction is shown as below:

C₄H₉Br CH₃(CH₂)₃-O-C₂H₅

The mechanism of synthesis of ether is also known as mechanism of Williamson ether synthesis. This mechanism involves one alkoxide reacts with alkyl bromide to form ether with two alkyl groups by using a strong base. This reaction can be used to produce both symmetrical or unsymmetrical ethers and also cyclic ethers. The following diagram is the mechanism of the Williamson ether synthesis.

 

The alkoxide ion functions as nucleophile and attacks the electrophilic C of the alkyl halide, displacing the bromide and creating the new C-O bond.

For the synthesis of n-butyl ethyl ether, both reactions are undergoes S_N2. S_N2 is known as second order nucleophilic substitution which is for bimolecular process. The kinetic rate of S_N2 is defined as

Rate = k [R-X][Nu^-]

The more the alkyl groups attached to the reacting carbon, the slower the reaction. The order os reactivity in S_N2 is shown in the following:

Tertiary alkyl > secondary alkyl > primary alkyl > methyl

--------------------------------->

Reactivity increasing

S_N2 involves the inversion of configuration (rearrangement of atoms in molecule) to form a transition state which is different with S_N1. First order nucleophilic substitution is a unimolecular process which form carbocation during the reaction.

Apparatus: Round bottomed flasks (50cm³ and 250cm³), Bunsen burner, condenser, thermometer, separating funnel

Materials: Sodium bromide, 1-butanol, conc. sulphuric acid, anti-bumping granules, 5% aqueous sodium bisulphate, distilled water, 10% aqueous sodium carbonate, anhydrous calcium chloride, sodium hydroxide, 95% ethanol, anhydrous magnesium sulphate

Procedure:

i) n-Butyl bromide

1. 27g of sodium bromide, 30cm³ of water and 20cm³ of 1-butanol are placed into a

250cm³ round bottom flask.

2. The mixture is cooled in an ice bath and 25cm³ conc. sulphuric acid is added with continuous swirling.

3. Two or three anti-bumping granules are added and attached with a gas trap to prevent HBr escaping, the round bottom flask is heated vigorously under reflux for 1.5 hours as shown in diagram 1 below.

Diagram 1

4. Distill the two layered mixture until the temperature reaches the boiling point of water.

5. The distillate is transferred to a separating funnel and shakes with an equal volume of 5% aqueous sodium bisulphate.

6. Allow the two layers to separate and wash the organic layer twice with 25cm³ water followed by 10% aqueous sodium carbonate (25cm³).

7. The product is dried with 5g calcium chloride and is filtered into 50cm³ round bottom flask.

8. Anti bumping granules are added and distilled, the material which boiled between 90-105 ̊C is collected.

9. The appearance of product is noted and weight is measured.

ii) Ethyl n-butyl ether

1. 4g of sodium hydroxide and 12cm³ 95% of ethanol are added into round bottom flask and is heated under reflux for 20 minutes.

2. 10cm³ n-butyl bromide is added into the mixture through the top of the condenser and the reaction is heated under reflux for 90 minutes.

3. After cooling, the mixture is transferred to a separating funnel and 50cm³ of water is added which has been used to rinse the reaction flask.

4. The mixture is shaked and the lower layer is removed.

5. The washing is repeated for two times with 20cm³ of water.

6. The organic layer is dried with anhydrous magnesium sulphate and the liquid is filtered into a 50cm³ round bottom flask.

7. The product is distilled slowly; the material which boiled in the range 90-96 ̊C is collected in a pre-weighed flask.

8. The density of the pure ethyl n-butyl ether is determined by pipetting 1cm³ liquid into a pre-weighed measuring cylinder and noting the weight difference.

Results and calculation:

Part I

Weight of conical flask = 78.2260g

Weight of conical flask + weight of n-butyl bromide = 89.8416g

Weight of n-butyl bromide = 11.6156g

Density = Mass/Volume

Mass = (Volume x Density) / Molecular mass

Number of mole of 1-butanol = (20cm³ x 0.81g/cm³) / 74.08g mol^-1

= 0.2187 mole

1 mole of NaBr reacts with 1 mole of 1-butanol to produce 1 mole of butyl bromide.

Thus, 0.2187 mole of butyl bromide is formed.

Theoretical weight of butyl bromide = 0.2187 mole X 136.972 g/mol

= 29.9558g

Experimental weight of butyl bromide = 11.6156g

Yield percentage of n-butyl bromide = 11.6156g/29.9558g X 100% = 38.78%

 

Part II

Weight of conical flask = 51.1825g

Weight of conical flask + weight of n-butyl ethyl ether = 58.0435g

Weight of n-butyl ethyl ether = 6.8610g

Weight of 5ml measuring cylinder = 14.7446g

Weight of 1ml n-butyl ethyl ether + weight of 5ml measuring cylinder = 15.6545g

Weight of 1ml n-butyl ethyl ether = 0.9099g

CH₃CH₂CH₂CH₂Br + NaOH + CH₃CH₂OH ---->

CH₃CH₂CH₂CH₂-O-CH₂CH₃ + NaBr + H₂O

Number of mole of n-butyl bromide = (10 cm³ x 1.2686 g cm^-3)/ 136.972g mol^-1

= 0.0926 mole

Theoretical weight of n-butyl ethyl ether = 0.0926 mole x 102.112g/mol

= 9.4556g

Experiment weight of n-butyl ethyl ether = 6.8610g

Yield percentage of n-butyl ethyl ether = 6.8610g/9.4556g x 100% = 72.08%

Density = mass/volume

Density of n-butyl ethyl ether = mass of 1ml of n-butyl ether / volume of 1ml of n-butyl ether

                                           = 0.9099g/1cm³

                                           =0.9099g/cm³

Discussion:

Alkyl halides can be prepared from alcohols by reacting them with a hydrogen halide, HX (X = Cl, Br, I). The mechanisms of acid-catalyzed substitution of alcohols are termed S_N1 and S_N2, where “S” stands for substitution, the “N” stands for nucleophilic, and the “1” or “2” for unimolecular or bimolecular, respectively. The purpose of this experiment is to synthesize n-butyl ethyl ether via an S_N2 reaction and to purify it using simple distillation where substances with different volatility and boiling points are separated from each other.

In the experiment, the primary alkyl halide n-butyl bromide can be prepared easily by allowing 1-butanol to react with sodium bromide and sulphuric acid. The sodium bromide reacts with sulphuric acid under reflux to produce hydrogen halides. The chemical reaction as shown in below indicates that the hydrogen halide is produced from the reaction.

2 NaBr + H₂SO₄ ------> 2 HBr + Na₂SO₄

The hydrogen halide produced is used to convert 1-butanol to become butyl bromide by undergoes nucleophilic substitution. Excess sulphuric acid serves to shift the equilibrium and thus to speed up the reaction by producing a higher concentration of hydrobromic acid. The sulphuric acid also protonates the hydroxyl group of 1-butanol so that water is displaced rather than the hydroxide ion OH^-. The acid also protonates the water as it is produced in the reaction and deacticvates it(water) as a nucleophile, hence the water keeps the butyl bromide from being converted back into the alcohol by nucleophilic attack of water.

Synthesis of butyl bromide from the 1-butanol is undergoes S_N2 mechanism. S_N2 is known as second order nucleophilic substitution for bimolecule. The essential feature of the S_N2 mechanism is that take place in a single step without intermediates when the incoming nucleophile, hydrogen bromide reacts with the 1-butanol from a direction opposite the group that leaves. As the bromide ion, Br^- comes in on one side and bonds to the carbon, the OH^- departs from the other side, thereby inverting the stereochemical configuration. The mechanism is shown in the figure 1 as below.

Figure 1

Since the hydrogen halide is polar molecule, the bromide ion is partial negative and the hydrogen is partial positive. The highly partial negative bromide ion acts as a nucleophile to attack the 1-butanol at the opposite side of the departing OH group. This leads to a transition state in which the new Br-C bond is partially forming at the same time the hold C-OH bond is partially breaking, in which the partial negative charge by both the incoming nucleophile and the leaving hydroxyl ions. The transition state for this inversion has the remaining three bonds to the carbon in a planar arrangement as shown in figure 1. The water is formed after the bromide successfully becomes part of the molecule.

After the reflux process, the distillate is introduced with same approximate same amount of sodium bisulphate, NaHSO₄. The purpose of adding of sodium bisulphate is used to absorb water from, the organic solvent since it has high hydrophobic property. Sodium carbonate (mild acid) is added to neutralize the acidic solution. Anhydrous calcium chloride is added as drying agent to absorb water droplets in order to purify the organic layer. An excess drying agent should be used to ensure that all the water in solvent is removed. If the water remains in the materials collected, it could interfere with the analysis. After filter out the drying agent, several anti bumping granules (boiling chips) are added to prevent over boiling during distillation. Distillation process is carried out to purify the butyl bromide in the range of temperature 90°C to 105°C. The pure n-butyl bromide is obtained in the distillation process.

The n-butyl bromide formed after complete distillation in the first part is used as the materials in the latter part of this experiment. Now, the n-butyl bromide is used to synthesis n-butyl ethyl ether. The synonym of n-butyl ethyl ether is known as ethoxy butane which has the molecular formula with C₆H₁₄O. This molecule is categories in the ether group which has the general formula of R-O-R’. Sodium hydroxide and ethanol are introduced together and is heated under reflux. The purpose of introduction sodium hydroxide and ethanol is to produce ethoxide ions. The chemical reaction is shown as the following equation:

Figure 2

During reflux, the ethoxide ions react with n-butyl bromide under the second time reflux to form n-butyl ethyl ether as product. In this reaction, the bromide ions are escaped from the alkyl bromide and combine with the sodium ions to form sodium bromide to achieve stable molecule.

Figure 3

The butyl bromide undergoes second order nucleophilic substitution, S_N2. The ethoxide functions as nucleophile which attacks the electrophilic C of the butyl bromide by displacing the bromide and creating a new C-O bond between ethoxide and butyl bromide. The bromide ion is being displaced and leaves the butyl group. As a result, an ether with butyl ethyl groups is formed which is known as n-butyl ethyl ether. The mechanism is shown in the figure 4 as below.

Figure 4

After the reflux, the mixture is transferred into a separatory funnel and water is introduced to rinse the mixture. This is because water is used to dissolve some unreacted NaOH and hence the NaOH can be removed by removing the water. In order to remove the water droplets, anhydrous magnesium sulphate is added. Distillation of the organic layer is carried out to purify the product. Pure n butyl ethyl ether is obtained through the distillation at temperature 90°C-96°C.The overall chemical equation for the synthesis of n-butyl ethyl ether is

CH₃CH₂CH₂CH₂Br + NaOH + CH₃CH₂OH  ----->

CH₃CH₂CH₂CH₂-O-CH₂CH₃ + NaBr + H₂O

In the experiment, the but-1-ene is being produced via E₂ elimination mechanism. E₂ elimination reaction also can be occurred because the ethoxide ions are strong base which initiate elimination reaction to compete with the substitution reaction. The mechanism of elimination of butyl bromide is shown in figure 5.

Figure 5

In figure 5, the mechanism shows that the bromide atom in butyl bromide attracts the electron from the C-Br bond to form partial negative and this causes the C₁ lack of electron. The hydroxide ion acts as nucleophile and attack the electrophile C₁ via E₂ mechanism to form a transition state. As the OH^- start to attack a neighboring H and begins to remove the H at the same time as the C-C double bond starts to form and the Br group start to leave. The water and bromide ion are leave and hence but-1-ene are formed via elimination. However, dibutyl ether also can be formed due to the strong sulphuric acid used. The strong acid causes the side reaction to the butyl alcohol which is dehydration and ether formation which is shown in the figure 6 below:

Figure 6

 

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