Difference between primary alcohols, secondary alcohols and tertiary alcohols

 Objectives:

1. To study the physical and chemical properties of alcohols

2. To identify the two unknown liquids from their experimental observations

3. To study the difference between primary alcohols, secondary alcohols and tertiary alcohols.

Introduction:

The hydrocarbon chains that attached with a hydroxyl group, OH^- to a carbon atom are known as alcohols. If the carbon atom is bonded to three hydrogen in addition to the OH^-, the alcohol is called methanol. Methanol, CH₃OH is the most simple alcohol molecule. The category of the alcohol is classified as three groups which are primary (1 ) alcohol, secondary (2 ) alcohols and tertiary (3 ) alcohol. If the alcohol bonded to one alkyl group, the alcohol is primary alcohol. The secondary alcohol is defined as the alcohol which one of the carbons is bonded to two alkyl groups and one hydrogen atom. If one of the carbons in alcohol is bonded to three alkyl groups is called tertiary alcohol.

All these alcohols share some similar characteristics but other characteristics are different owing to their different molecular structures. For physical properties, the size of alcohol determines its boiling point. Usually, the larger the size of the alcohol, the higher the boiling point. This is because the bigger the size of the molecules, the stronger the Van der Waals force between the alcohol molecules. So, more heat energy is needed to be absorbed in order to break down the intermolecular force between each alcohol molecules. Hence, the boiling point of the alcohols increases with the size of the alcohols. The solubility of the alcohol is depends on the size of molecules. Small alcohols are water soluble because the hydroxyl group can form hydrogen bond with water molecules. But, as the size of the alkyl group increases, the solubility of alcohol in water decreases as the hydrophobic property of alcohol increases. For example, if the carbon molecules in alcohol more than six per molecule, the particular alcohol definitely are not soluble in water. This is the result of the alkyl group disrupting the hydrogen bond among the water molecules. If the disruption becomes larger enough, the water molecules will repel the alcohol molecules effectively to reestablish hydrogen bonding.

The classification of an alcohol as primary, secondary or tertiary (see above) affects the chemical properties of the alcohols. Due to their different classes, the alcohols may give out different chemical properties when they react with the same compounds. Based on their chemical properties, we are able to differentiate among the classes of alcohols. Generally, Lucas test and chromic acid test is the two common tests that we always use to distinguish and categorize the classes of alcohols.

Lucas test

Water soluble alcohols are always tested by using the Lucas reagent to differentiate among the primary, secondary, and tertiary alcohols. This test depends on the appearance of an alkyl chloride as an insoluble second layer. Lucas reagent is the mixture of zinc chloride and hydrochloric acid. Zinc chloride is a Lewis acid which is added with hydrochloric acid to make its property becomes more acidic. The tertiary alcohol (water soluble) reacts with Lucas reagent almost immediately to form an alkyl chloride which is insoluble in the aqueous solution. The formation of a second layer liquid phase in the test tube almost as soon as the alcohol initially dissolves is indicative of a tertiary alcohol. The secondary alcohol reacts slowly with Lucas reagent, and it gives a second phase after heating for 10 minutes whereas the primary alcohol and methanol do not react with Lucas reagent under normal condition. The chemical equation is as shown below (if the reaction takes place)

ZnCl2

R-OH + HCl     ------------------>    R-Cl + H₂O

Chromic acid test

Chromic acid is a strong oxidizing agent which uses to oxidize the alcohols. This test is based on the reduction of chromium (VI) ions to chromium (III) ion. When chromic acid reacts with alcohols, the change in colour of the solution from red-brown to green is a positive test. Primary alcohols are oxidized to carboxylic acid by chromic acid. The Cr⁶⁺ in the chromic acid which is red-brown is reduced to green Cr³⁺; secondary alcohols are oxidized to ketones by the reagent with the same colour change. Tertiary alcohols are not oxidized at all by the chromic acid. Hence, this reaction can be used to distinguish tertiary alcohols form primary and secondary alcohols.For tertiary alcohols, the alcohols would not be oxidized by the reagent. Hence, this test is used to distinguish the tertiary alcohols from primary and secondary alcohols.

 

Reaction with sodium metal

The acidic properties of alcohol can be shown by adding the sodium metal into alcohol. The alcohols are weak acids when they react with sodium metal. The hydroxyl group can act as a porton donor to form an alkoxide ion. Alkoxide ions dissolved in alcohol are strong bases which can be prepared by the reaction of an alcohol with sodium metal. Hydrogen gas is released by the reaction.

2 R-OH + 2 Na  -------> 2 R-O^-Na⁺ + H₂

The hydrogen gas can be collected and tested by using a burning wooden splinter. A pop sound will be produced.

Apparatus: test tube, measuring cylinder, droppers

Materials: ethanol (C₂H₅OH), isopropyl alcohol (C₃H₇OH), t-butyl alcohol (C₄H₉OH), Lucas reagent (mixture of concentrated HCl and ZnCl), chromic acid (H₂CrO₄), sodium metal

Procedure:

A. Solubility of alcohols

1. In three separate dry test tubes, ethanol, isopropyl alcohol and t-butyl alcohol are added into each test tube.

2. Water is added to each tubes, the contents are mixed and observed.

3. The results are recorded in a table.

4. The above procedures are repeated with two unknown liquids and observations are being made.

B. Lucas test

1. Ethanol, isopropyl alcohol and t-butyl alcohol are added into separate dry test tubes and Lucas reagent is added at room temperature.

2. The tubes are closed with a cork, the tubes are shaken and the length of time it takes for the mixture to become cloudy or separate two layers.

3. The results are recorded.

4. The above procedures are repeated with the two unknown liquids and observations are being made.

C. Chromic acid test

1. Ethanol, isopropyl alcohol and t-butyl alcohol are added into separate dry test tubes.

2. A small piece of sodium metal is added and any reactions occur are noted.

3. The step 1 and step 2 are repeated with two unknown liquids and any observations are made.

Results:

Part A

Table 1 The solubility of alcohols and their observations

Alcohols	Observation	Reaction equations(if any)	Deduction/conclusion/discussion
Ethanol	Soluble in water to form colourless solution	C2H5OH (aq) + H2O (l) ---> C2H5O- (aq)+ H+(aq)	Soluble in water
Isopropyl alcohol	Soluble in water to form colourless solution	C3H7OH (aq) + H2O (l)  ---> C3H7O- (aq) + H+ (aq)	Soluble in water
t-butyl alcohol	Soluble in water to form colourless solution	C4H9OH (aq) + H2O (l) -----> C4H9O- (aq) + H+ (aq)	Soluble in water
Unknown A	Soluble in water to form colourless solution	-	Soluble in water
Unknown B	Soluble in water to form colourless solution	-	Soluble in water

Part B

Table 2 The reaction between alcohols and Lucas Reagent

Alcohols	Observation	Reaction equations(if any)	Deduction/conclusion/discussion
Ethanol	Yellowish solution remains unchanged after heating.		No reaction between ethanol and Lucas reagent.
Isopropyl alcohol	Yellowish solution remains unchanged within 10 minutes, but turns into cloudy solution after heating.	C3H7OH (aq) + HCl (aq)  -----> C3H7Cl (aq) + H2O (l)	The reaction between isopropyl alcohol and Lucas reagent occur after heating for 10 minutes.
t-butyl alcohol	Cloudy solution is formed immediately and two layers are formed. Upper layer is clear while lower layer is cloudy.	C4H9OH (aq) + HCl (aq) C4H9Cl (aq) + H2O (l)	The reaction between t-buytl alcohol and Lucas reagent is instant reaction.
Unknown A	Yellowish solution remains unchanged within 10 minutes, but turns into cloudy solution after heating.		Unknown A is isopropyl alcohol because it has the similar reaction with it.
Unknown B	Cloudy solution is formed immediately and two layers are formed. Upper layer is clear while lower layer is cloudy.		Unknown B is t-butyl alcohol because it has the same reaction with it.

Part C

Table 3 The reaction of alcohols with chromic acid

Alcohols	Observation	Reaction equations(if any)	Deduction/conclusion/discussion
Ethanol	The solution turns from colourless to green	3 CH3CH2OH + 2 CrO4- + 10 H+ à 3 CH3CHO + 2 Cr3+ +8 H2O	Ethanol is oxidized by the chromic acid.
Isopropyl alcohol	Two layers are formed. Upper layer is green while lower layer is black.	C3H7OH (aq) + H2CrO4(aq) -----> C2H6CO (aq) + Cr2(SO4)3(aq) + H2O(l)3 (CH3)2CHOH + 2 CrO4- + 10 H+ à3 (CH3)2CO + 2 Cr3+ +8 H2O	Isopropyl alcohol is oxidized by the chromic acid.
t-butyl alcohol	Two layers are formed. Then, the precipitate dissolves in solution to become reddish brown solution.	-	No reaction.
Unknown A	Two layers are formed. Upper layer is green while lower layer is black.		Unknown A is isopropyl alcohol because it has the similar reaction with isopropyl alcohol.
Unknown B	Two layers are formed. Then, the precipitate dissolves in solution to become reddish brown solution.		Unknown B is t-butyl alcohol because it has the similar reaction with t-butyl alcohol.

Part D

Table 4 The reaction between alcohols and sodium metal

Alcohols	Observation	Reaction equations(if any)	Deduction/conclusion/discussion
Ethanol	Bubbles of colourless gas released quickly. A pop sound is produced when tested with burning wooden splinter.	2C2H5OH (aq) + 2Na(s) ------> 2C2H5ONa (aq) + H2(g)	Many of hydrogen gas is being produced from the reaction between ethanol and sodium metal.
Isopropyl alcohol	Bubbles of colourless gas released slowly. A pop sound is produced when tested with burning wooden splinter.	2C3H7OH (aq) + 2Na(s) -----> 2C3H7ONa (aq) + H2(g)	Small amount of hydrogen gas is being produced from the reaction between isopropyl alcohol and sodium metal.
t-butyl alcohol	Bubbles of colourless gas released very slowly. A pop sound is produced when tested with burning wooden splinter.	2C4H9OH (aq) + 2Na (s) ------> 2C4H9Ona (aq) + H2 (g)	Less hydrogen gas is being produced from the reaction.
Unknown A	Bubbles of colourless gas released slowly. A pop sound is produced when tested with burning wooden splinter.		Unknown A is isopropyl alcohol because it has the similar reaction with isopropyl alcohol.
Unknown B	Bubbles of colourless gas released very slowly. A pop sound is produced when tested with burning wooden splinter.		Unknown B is t-butyl alcohol because it has the similar reaction with t-butyl alcohol.

Discussion:

This is because all of them, ethanol, isopropyl alcohol and t-butyl alcohol are short alkyl chain alcohols. Alcohols can soluble in water is because of the presence of the hydroxyl group (-OH) in the compounds. The hydroxyl group can form hydrogen bond with the water molecules and thus make it soluble in water.The solubility of alcohols in water are always depends on their structure and size. When the sizes of alcohols increase, the solubility of alcohols in water will decrease. This is because the bulky groups are highly hydrophobic and tend to block the water molecule from nearing alcohol and stabilize it. This is the result of the alkyl group disrupting the hydrogen bond among the water molecules. If the disruption becomes larger enough, the water molecules will repel the alcohol molecules effectively to reestablish hydrogen bonding. Usually, the number of carbon per molecule is more than six are not soluble in the water.

Based on the properties of solubility of alcohols in water, this information is not enough for us to differentiate clearly the classes of alcohols which are being used in the experiment. So, Lucas test is used to distinguish among the primary, secondary, and tertiary alcohols. For Lucas test, the formation of two layers (aqueous layer and cloudy layer) is known as a positive test. The second layer (cloudy layer) formed is alkyl chloride which is insoluble in the aqueous solution because all the alkyl halides molecules are insoluble in the water. The alkyl chloride produced from the reaction is not water soluble and causes cloudiness (emulsion) to form in the aqueous solution. When ethanol is added with Lucas reagent, the yellowish solution is still remains the same before and after heating. This is because primary alcohol does not react with Lucas reagent. Besides, isopropyl alcohol does not react with Lucas reagent before heating but it does turns to cloudy solution after heating for 10 minutes. The reason is the reaction between Lucas reagent and secondary alcohol is slow. For tertiary alcohol (t-butyl alcohol), the reaction of alcohol with Lucas reagent is very fast which can be known as an instant reaction. The reaction that takes place on the Lucas test is a SN₁ nucleophilic substitution. The alcohols with the properties of generating a stable carbocation intermediates will undergo the particular reaction. The OH group of the alcohol attracts the H in hydrochloric acid to form oxonium ion and leave the group (form water). The carbocation intermediate is formed and tends to react with Cl^- (nucleophile) to produce the alkyl halide product. The mechanism of SN₁ nucleophilic substitution is shown as diagram below:

The purpose of chromic acid test used in this experiment is to distinguish the primary and secondary alcohols from the alcohols group. In the reaction between alcohols and chromic acid, the chromic acid is being reduced which the chromium (VI) ions, Cr⁶⁺ reduced to become chromium (III) ion, Cr³⁺. The positive test for chromic acid is represented by the change in colour from orange to green-blue. In the test tube with ethanol, the colourless alcohol is turned to green solution because the chromium (IV) ions, Cr⁶⁺ are being reduced by ethanol. In the test tube containing isopropyl alcohol, two layers of colour are formed. The upper layer is green whereas the lower layer is black. This is shows that the isopropyl reacted with the chromic acid due to the secondary alcohol is readily to be oxidized. The black layer actually is the dark blue colour, it is hard to differentiate because the light is not enough in the laboratory. The t-butyl alcohol would not oxidized by the chromic acid since the tertiary alcohol is a highly oxidized alcohol. This is shown by the formation of reddish brown precipitate. The chemical equation for the reaction is shown as below:

Ethanol: 3 CH₃CH₂OH + 2 CrO₄^- + 10 H⁺ à 3 CH₃CHO + 2 Cr³⁺ +8 H₂O

Isopropyl alcohol: 3 (CH₃)₂CHOH + 2 CrO₄^- + 10 H⁺ à3 (CH₃)₂CO + 2 Cr³⁺ +8 H₂O

 

The general mechanism for the reaction is shown below:

Hydrogen ions will be attracted to one of the oxygen atom in the chromic acid that is double bonded to chromium. The lone pair electrons from hydroxyl group of alcohols will then attack the chromium ion and form a chromate intermediate. The hydrogen on the hydroxyl group of alcohols will then leave as hydrogen ion and combine with the oxygen atom that is previously attacked by a hydrogen ion. A water molecule will come and attack α-hydrogen on the alcohol and produce hydronium ion, forming C=O double bond. A water molecule will leave the chromate intermediate and the carbonyl product is formed. Since the oxidation of alcohol requires at least one hydrogen atom to be presence on α-carbon, thus tertiary alcohol cannot be oxidized because it does not have the α-hydrogen. This explains why t-butyl alcohol does not change the color of the chromic acid which is red brown to green. The precipitate formed at the beginning may be because of the formation of chromium trioxide precipitate.

The last test in this experiment is the reaction of sodium metal with alcohols. All the different classed of the alcohol are able to react with the sodium metal since all of them have OH group. According to the Lewis Bronsted Theory, a Bronted acid is a proton donor. In this case, alcohol acts as a proton donor which donates H⁺ into the solution while the sodium metal acts as a strong base. A strong base can deprotonates the alcohol to yield an alkoxide ion (R-O). The H in OH group will be substituted by sodium ion to form sodium alkoxide, R-O^-Na⁺. The reaction between an acid and an active metal in group I element definitely will produce hydrogen gas as the product as the metallic sodium reduces proton to form hydrogen gas. The evidence is shown by the release of bubbles of colourless gas from the reaction between sodium metal and alcohol. The release of hydrogen gas can be collected and tested with a burning wooden splinter. A pop sound will be produced as the hydrogen gas is an explosive gas. The chemical equation of alcohols and sodium metal is shown as the diagram 1 below:

2C₂H₅OH (aq) + 2Na(s) 2C₂H₅ONa (aq) + H₂ (g)

2C₃H₇OH (aq) + 2Na(s) 2C₃H₇Ona (aq) + H₂ (g)

2C₄H₉OH (aq) + 2Na (s) 2C₄H₉Ona (aq) + H₂ (g)

Diagram 1 The reaction between alcohols with different classes with sodium metal

The acidity of the alcohol can be determined by the rate of gas evolution in this experiment. The more the gas released from the reaction, the more acidic the properties of alcohol. The acidity of alcohols decreases as the number of carbon in bulky alkyl group that bonded to OH group increases. Since alkyl group is electron releasing group, the more bulky the alkyl group, the stronger the electron donor effect. Thus, the negative charge density on the O atom in and proton are less readily to be released. The second reason is the bulky groups are highly hydrophobic and tend to block the water molecule from nearing alcohol and stabilize it. The alcohols are arranged in order of increasing acidity of alcohol as below:

t-butyl alcohol < isopropyl alcohol < ethanol

Acidity increasing

 

Flickr Tags: physical and chemical properties of alcohols,primary alcohols,secondary alcohols and tertiary alcohols,alcohols and Lucas Reagent,alcohols with chromic acid,alcohols and sodium metal,solubility,mechanism

Produce cyclohexene through the acid catalyzed elimination of water from cyclohexanol

  

Objectives:

1. To produce cyclohexene through the acid catalyzed elimination of water from cyclohexanol

2. To understand mechanism involved in the reaction

3. To learn the technique of distillation

Introduction:

Dehydration is defined as a process of removing water from a substance. The loss of water from a molecule is called dehydration which is exactly opposite with the process of hydrolysis. Dehydration is an elimination reaction of an alcohol involves the loss of an OH from one carbon and an H from an adjacent carbon. Overall, this amounts to the elimination of a molecule of water, resulting in a pi-bond formation of an alkene or alkyne. In most of the dehydration of alcohol, heat and catalyze are needed in the reaction. Sulphuric acid (H₂SO₄) and phosphoric acid (H₃PO₄) are the most commonly used acid catalysts.

The dehydration process can be carried out by in two ways. The first way is heating a mixture of alcohol and dehydrating agent in a distilling flask and collecting the olefin (also known as alkene) from the mixture. The second way is passing the alcohol vapour through a heated tube packed with the dehydrating agent at 350°C. The collecting the olefin as it emerges from the tube. The chemical equation for dehydration of alcohol to form alkene and water is shown as below diagram:

For the dehydration of alcohol, the alkene is formed in the reaction. At the same time, the side products are produced from the reaction such as dicyclohexyl ether, polymer, mono and dicyclohexyl sulphate abd degradation products (carbon and carbon dioxide). In many cases, the phosphoric acid is used in the dehydration of alcohol instead of using sulphuric acid due to two reasons. The one of the reason is the lost of organic compound can be minimize through oxidation of phosphoric acid. In addition, the product is being produced without contamination with volatile decomposition products such as sulphurous acid.

Apparatus: Round-bottomed flask(50ml), take off distillation adapter, condenser, thermometer, electric flask heater

Materials: boiling chips, cyclohexanol, cyclohexene, 85% conc. Phosphoric acid, anhydrous magnesium sulpahte

Procedure:

1. Cyclohexanol and conc.(85%) phosphoric acid are added in a round bottomed flask and is mixed together by swirling.

2. Several boiling chips are added, the flask is clamped to a ring stand at electric flask heater height attached with a take off distillation adapter, a thermometer, a condenser and a small receiving flask as shown in the diagram below.

3. The mixture is heated and distillate boiling in the range 85°C-90°Cis obtained.

4. When the distillate is exhausted, the heat is increased gradually. Using the same receiver, the distillate boiling in the range of 90°C - 100°C is collected.

5. The two layers in the receiving flask are tested by adding the distilled water. With the aid of a 9-disposable pipette, the aqueous layer is drawn off and is being discarded.

6. The organic layer is dried up with anhydrous magnesium sulphate.

7. The drying agent is removed by filtering the mixture through a cotton wool plug wedged into the constricted part of a small funnel.

8. The filtrate is collected round bottom flask or small distilling flask. Boiling chip is added to the dried product and distil it through a take off distillation adapter packed with a few small wads of coarse steel wool.

9. The product boiling is collected in the range 3 below and 2 above the boiling point of cyclohexene (83°C) in a tarred bottle.

10. The yield in grams is calculated and the product is submitted to instructor.

Result and calculation:

Weight of cyclohexanol = 10.0060g

Weight of dry conical flask = 51.2460g

Weight of dry conical flask + weight of cyclohexene = 56.2281g

Experimental weight of cyclohexene = 4.9821g

Number of mole of C₆H₁₁OH =

= 0.09996 mole

1 mole of C₆H₁₁OH produces 1 mole of C₆H₁₂

0.09996 mole of C₆H₁₁OH produces 0.09996 mole of C₆H₁₂

Theoretical weight of C₆H₁₂ = 0.09996 mole X 84.096 g/mol

= 8.4066g

Yield percentage =

 

= 59.26%

Discussion:

In this experiment, the cyclohexanol solution is being used in the dehydration process. The cyclohexanol is a six carbon aromatic hydrocarbon which one of the hydrogen atoms, H is substituted by one hydroxyl group, OH^-. Due to the low melting point, the cyclohexanol appear in liquid form at room temperature. The dehydration process of alcohol will convert cyclohexanol which the hydroxyl group, OH^- will be removed to become cyclohexene. Cyclohexene is a six carbon aromatic hydrocarbon with a single double bond in the molecule.

The solution is added with concentrated phosphoric acid in a round bottomed flask and is mixed together by swirling. The phosphoric acid is added as catalyze as such increase the rate of reaction in dehydration without affects the particular chemical reaction. The boiling chips are added into the solution in order to prevent over boiling of the solution. The boiling chips are small, insoluble, and porous stones made of calcium carbonate or silicon carbide. There are a lot of pores inside the boiling chips which provide cavities both to trap air and to provide spaces to allow bubbles of solvent can be form. When boiling chips are heated, it will release tiny bubbles which can prevent boiling over. Boiling over of solvent will cause lost of solution which may lead to inaccurate result to be obtained.

The heating of mixture is carried out in a fractional distillation apparatus. As the mixture is heated, the alkene and water are produced as the products in the reaction. Besides, the side product and impurities of the reaction will be produced at the same time. The temperature of mixture when heating is fluctuated. During the temperature 80°C, the temperature of mixture drops suddenly by 2°C and the temperature remained at 78°C constantly for few seconds. This is because some impurities or side products are being produced in the mixture which may have the boiling point of 78°C. The temperature of mixture drops suddenly because the heat being is absorbed which used to break down the bonding of the side products. When the boiling point is reached, then temperature of mixture remains constant as the state of the side products are converted from liquid form to gaseous form. The temperature of mixture is increases until 107°C after all the side products are being converted.

The temperature of mixture is reached to maximum at 107°C. This temperature is known as the activation temperature which the cyclohexanol start to be dehydrated. The temperature of mixture drops to 83°C and remains constant. This phenomenon takes place because the cyclohexene with lowest boiling point will tends to be distilled first before the higher one. The temperature remains unchanged because the heat is being absorbed to break down the bond between cyclohexene molecules. In all the distillation process, some of the product will be lost since it is hold up in the apparatus which reduce the product yield. In order to maximize the yield, the mixture is continued to be heated at higher temperature range which more than the boiling point of cyclohexene. When the mixture is heated at 90°C-100°C, the water in the mixture will push over the products into the receiving flask along the condenser. The products produced are collected in the same receiving flask.

Then, the receiving flask containing cyclohexene, water and small amount of the impurities. Two layers liquid are present in the receiving flask, one drop of distilled water is added into flask in order to determine the location of aqueous layer. Since the water droplet mix with the lower layer, so the upper layer is determined as cloudy solution while the lower layer is aqueous layer. The upper cloudy layer is cyclohexene with some impurities and water inside. The lower aqueous layer is removed and discarded. But, that is not easy to remove all the water in the receiving flask. So, anhydrous magnesium sulphate is being added. The purpose of adding of anhydrous magnesium sulphate is used to remove residual water in the organic solvent. The magnesium sulphate in the granular form will be preferable. It is known as drying agent in the organic solvent which are not dissolves in the solvent but drying the solvent. The magnesium sulpahte clump together with the water droplets as it solidified them. In another words, it is reacts with water to form hydrates which is their preferred form when water is available. 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.

Water has been successfully removed from the organic compound mixture, so it is very important to not reintroduce water into the mixture. The final distillation of unpurified cyclohexene must be done very carefully in order to obtain purified products. The temperature of mixture is heated until approximate 85.5°C since this temperature is the boiling point of cyclohexene and hence the pure cyclohexene could be obtained rapidly. The weight of the cyclohenexe is 4.23g.

Error of source includes the condenser and Liebig tubes are not rinsed with a little of distilled water, a little of ethanol and acetone to speed up drying process. So that, the water in the desired products would not presents. This is because some of the water droplets is hold up and stick on the wall of condenser and the second distillation will produce the contaminated cyclohexene.

 

Flickr Tags: dehydration of an alcohol,cyclohexanol and cyclohexene,acid catalyzed elimination of alcohol

Dissolved oxygen (DO)

What is dissolved oxygen (DO)? 

Dissolved oxygen refers to the amount of oxygen level that present in the water or liquid sample. Dissolved oxygen level is very important to the aquatic life in a healthy ecosystem. Too much of oxygen may create algae bloom provided the high nutrient load present in the water, however too low of oxygen amount that present in water will make the fishes suffocating and eventually die. 

How do we measure DO in the water?

METHOD 1: Winkler titration method - conventional method

We can use Winkler titration to measure DO amount in the water sample that has been took and prepared manually from the site into water bottle. The Winkler titration method measures the DO amount in the water sample precisely, this is a method developed by Hungarian analytical chemist named Lajos Winkler in 1888. This method adding chemicals to the water sample to form an acidic solution by creating reaction with the oxygen that present in the water. Then we will add in the neutralizing agent to perform neutralization with titration and the amount of neutralizing agent required indicate the amount of the oxygen level in the water sample. Below is the diagram to show the Winkler titration in step by step. 

Although Winkler titration can deliver highly accurate result, but this method is laborious, time-consuming and highly susceptible to the interference. 

METHOD 2: Using in-situ DO meter

Alternatively, we can use DO meter that can provide instant result instead. A lot of researchers they prefer to use in-situ DO meter because the amount of oxygen in water sample may fluctuate across the time. The amount of oxygen may varies at different water level. The deeper the water level, the lesser the oxygen amount that present. The creation of in-situ DO meter ease the life of a chemist as winkler titration is a tedious procedure to perform DO measurement of water sample. Nowadays, the in-situ DO meter could deliver and perform almost similar result of DO as compared to Winkler titration and hence in-situ DO meter is the first choice of researchers. 

In in-situ DO meter, we do found two major types of technology that present in the market. The first type is called polarographic (aka Clark electrode) method. This is a method invented by Dr Leland Clark in 1956. In 1965, the first biological oxygen monitor is invented and this is a breakthrough technology in medical field. This technology enabled physicians to carry out open-heart surgery for the first time as the real blood oxygen measurements could be recorded during the operation! 

After years of study, another advanced technology called optical or luminescent technology in DO meter has took over the place of polarographic technology. Some researches has chose the DO meter with optical technology due to the easier maintenance and faster warm up time. You may see the differences in between polarographic electrode and optical technology as below. 

Comparison	DO meter with electrode	DO meter with optical
Technology	Polarization technology (old method)	Optical technology (advanced method)
Calibration frequency	Daily	Weekly
Warm up time	10 – 15mins	NO warm up time
Flow dependent	Flow dependent; need to stir manually while using the meter to get actual reading	Flow independent; no need to stir manually while using
Consumables	Membrane cap, filling solution	Membrane cap only
Maintenance	-change the membrance cap -refill the solution in cap -check the solution in cap from time to time -polish the electrode from time to time	- change the membrane cap once the coating of cap wear off; usually can last long up to 12 months or more, depends on usage
Measurement rate	Faster than optical method Estimated 8-25 seconds per sample, but subject to the sample condition	Slower than electrode method Estimated 40 seconds per sample, subject to the sample condition

What is pH?

pH value is a figure to indicate acid, neutral or alkaline. pH value being measured from 0-14, in which it does not have any unit for it. The pH is also being defined as the concentration of hydrogen ion, H+ ion in the aqueous solution. The higher the concentration of H+ in the aqueous solution, the more acidic the aqueous sample is. The pH value will tend to be lower than pH 7 and getting lower and lower and close to pH 0.

Normally, pH is being measured/taken by using the following method:
1. litmus paper
2. Color comparator
3. pH meter

The accuracy of pH reading

pH meter > Color comparator > Litmus paper

What is Water Quality Index?

Water Quality Index (WQI)

Water quality index (WQI) indicates the quality of water based on classification. The WQI will be classified based on the 6 parameters as below:

1. pH

2. Dissolved oxygen (DO)

3. Suspended solid

4. Ammonia nitrogen

5. Chemical oxygen demand (COD)

6. Biochemical oxygen demand (BOD)

Preparation of polyamide by condensation polymerization

Objective:

1. To study and understand the formation of polyamide (nylon) through condensation polymerization between surface technique

2. To observe the effect of external factors on molecular weight and polymer yield

 

Introduction:

Polymerization is a process of forming long and repeating organic polymer chains. Polymerization is categorized into different systems based on the mechanism and structure of polymer. Both structure and mechanism are usually required in order to clearly classify a polymer. Polymers were originally classified by Carothers (1992) into addition and condensation polymerization. However, the more recent terminology further classifies polymerization into step-growth, chain-growth and ring opening polymerization.

 

Condensation polymers are refers to the polymers that were formed from polyfunctional monomers by various condensation reactions with the elimination of some small molecules such as water. The type of end product resulting from the condensation polymerization is depends on the type of functional end group which can react. For example, the formation of polyester between a dicarboxylic acid and a diol in which two monomers are linked together and a water molecule has been produced.

Figure 1 Condensation polymerization between a dicarboxylic acid and a diol

This process can continue adding alternatively one molecule and then the other, building the polymeric chain of hydrocarbon units that joined by ester bond. This is molecule formed from the condensation polymerization is called polyester.

 

In order for a condensation polymerization to occur continuously, the small molecules need to be removed from the system either by heating or through vacuum. However, heating would cause degradation or breaking of polymer molecules from bigger size of polymers to smaller size of polymer molecules. As a result, this will cause the formation of low molecular weight polymer. In order to overcome this problem, polymerization should be carried out at low temperature either by using the method of polymerization in solvent or polymerization between two surfaces of solution (interfacial polymerization). Many polymers can be produced at lower temperature (instead of high temperature) by using Schotten-Baumann reaction of acid chlorides.

 

Interfacial polymerization is refers to the condensation polymerization between two reactants is carried out at the interface of two immiscible solvents in which each containing one of the reactants. Interfacial polymerization is important because it produces the polymer with high molecular weight in a short period of time which under normal conditions of temperature and pressure. This type of polymerization could be carried out at room temperature and the types of polymers that could be formed using this technique are polyurethane, polyamide and polyurea. Thus, the formation of nylon 6,10 which is a type of polyamide that can be produced by using this method of reaction between two surfaces of immiscible solvents. This technique is extremely useful to the reactants which are not stable under the polymerization conditions by other techniques. Among the techniques of condensation polymerization, interfacial polymerization has some added advantages over the other techniques.

 

Nylon 6,10 is produced from the two monomers named sebacoyl chloride (ClOC(CH­₂)₈COCl) and 1,6-diaminohexane (also known as hexamethylenediamine, H₂N(CH₂)NH₂). Sebacoyl chloride is a 10-carbon acid chloride with a –COCl group as functional group at each end. The other monomer is a 6-carbon chain with two amino groups, -NH₂ at each end. Nylon 6,10 is named based on the carbon number of two monomers, in which the first number come from amine monomer while the latter one from acid chloride monomer. These two compounds are polymerized via condensation with the loss of a molecule of hydrogen chloride, as shown in the equation below:

 

Nylon is a term to represent synthetic polyamides. Different nylons are described by a numbering system that shows the number of carbon atoms in the monomer chains. Nylons from diamines and dicarboxylic acids are designated by two numbers in which the first representing the diamine whereas the second representing the dicarboxylic acid. In order to generate a long chain polymer and to expedite the rate of kinetic reaction, the polymerization system should be stirred vigorously to make sure that a new surface of polymer is always produced in the system. This ensures that the polymers produced would give a high percentage of yields. Both hexamethylenediamine and sebacoyl chloride are not solvents in the polymerization process between the interface of solutions. In the stirring technique, the non equilibrium system that occurs eliminates the weakness that might occur in polycondensation process such as:

1. The necessity of high temperature

2. Longer time of reaction

3. Accurate stoichiometric balance of equivalent

Physical properties of nylon 6,10:

1. Bright yellow in colour

2. Less transparent

3. Melting point of 220 °C

4. Moderate crystallinity

5. High impact strength

6. Electrical insulating ability

 

Apparatus:

100ml beaker, 10ml graduated cylinder, glass rod, thin stick

 

Materials:

60% w/v hexamethylenediamine, sebacoyl chloride, sodium laurel sulphate (soap powder)

 

Procedures:

1. 3ml sebacoyl chloride solution and 100ml of hexane (solution B) were added into a 100ml beaker.

2. 0.01g of soap powder was added into the mixture (solution A) of 3ml of hexamethylenediamine, 50ml of water and 2 drops of phenolphthalein.

3. Solution A was added carefully into solution B so as not to disturb the interface between two solutions.

4. A layer of polymer was formed between solution A and solution B. The strand of polymer (nylon) was wrapped around the glass rod and nylon strand was winded onto the rod with a steady pace.

5. The polymer formed was rinsed with water.

6. Let it dry in the oven.

7. The weight of polymer was measured.

8. The steps above were repeated with 5ml of solution B and 3ml of solution A. The solutions were stirred.

9. The polymers formed were characterized and compared.

 

Results and calculations:

Table 1 Observation on polymers that made with different methods

Observations on polymers formed with different methods
Wrapping on glass rod	White, long and weak elasticity nylon was formed.
Stirring	A clump of nylon was formed.

Table 2 Weight of polymer obtained by wrapping on glass rod

Weight of empty watch glass	20.4850 g
Weight of (empty watch glass + polymer, Nylon 6,10)	20.9122 g
Weight of polymer (nylon 6,10)	0.4272 g

Table 3 Weight of polymer obtained by stirring

Weight of filter paper	0.4836 g
Weight of (filter paper + polymer, Nylon 6,10)	0.9099 g
Weight of polymer (nylon 6,10)	0.4263 g

 

Calculate percentage yield

Unstirred interfacial polymerization

Types of monomer	hexamethylenediamine	Sebacoyl chloride
Density (g cm-3)	0.84	1.121
Volume used (cm3)	3.00	3.00
Molecular weight (g mol-1)	116.00	239.00

 

Mass of hexamethylenediamine = 0.84 g/cm³ x 3cm³ x 60% = 1.512g

Theoretical number of mole of hexamethylenediamine = 1.512g / 116 g mol^-1

= 0.013 mol

Mass of sebacoyl chloride = 1.121 g/cm³ x 3cm³ = 3.363g

Theoretical number of mole of sebacoyl chloride = 3.363g / 239 g mol^-1

= 0.014 mol

Thus, limiting agent is hexamethylenediamine.

Molecular weight of repeating unit = 282 g mol^-1

Weight of nylon 6,10 obtained experimentally = 0.4272g

Actual number of mole of repeating unit = 0.4272g / 282 g mol^-1 = 0.0015 mol

Percentage yield = 0.0015 mol / 0.013 mol x 100%

= 11.54%

 

Stirred interfacial polymerization

Types of monomer	hexamethylenediamine	Sebacoyl chloride
Density (g cm-3)	0.84	1.121
Volume used (cm3)	3.00	5.00
Molecular weight (g mol-1)	116.00	239.00

Mass of hexamethylenediamine = 0.84 g/cm³ x 3cm³ x 60% = 1.512g

Theoretical number of mole of hexamethylenediamine = 1.512g / 116 g mol^-1

= 0.013 mol

Mass of sebacoyl chloride = 1.121 g/cm³ x 5cm³ = 5.605g

Theoretical number of mole of sebacoyl chloride = 5.605g / 239 g mol^-1

= 0.023 mol

Thus, limiting agent is hexamethylenediamine.

Molecular weight of repeating unit = 282 g mol^-1

Weight of nylon 6,10 obtained experimentally = 0.4263g

Actual number of mole of repeating unit = 0.4272g / 282 g mol^-1 = 0.0015 mol

Percentage yield = 0.0015 mol / 0.013 mol x 100%

= 11.54%

 

Table 3 Comparisons between two polymers that produced by different methods

Wrapping on the glass rod	Method to produce polymers	Stirring by using glass rod
0.4272g	Mass	0.4263g
11.54%	Percentage yield	11.54%
More elastic	Elasticity	Less elastic
Stronger	Stiffness (Malleability)	Weaker
Stronger	Ability to stick to surface	Weaker

 

Discussion:

In this experiment, the interfacial polymerization technique plays a very important role. This method is primarily used in the case of condensation polymers, in which two reactants have different solubility characteristics. Hexamethylenediamine and sebacoyl chloride were used in this experiment in order to demonstrate the condensation polymerization in the interface between two solutions.

 

Interfacial polymerization is different from the usual condensation polymerization. This is because the monomers (undergo this polymerization) tend to diffuse to the interface and will only react with the polymer chain end formed initially. The diacid chloride and diamine monomer molecules will react with the growing polymer chain end before they penetrate through the polymer film to create the growth of new polymer chains. Therefore, interfacial polymerization creates a high tendency to generate high molecular weight polymer in the polymerization process. In addition, stoichiometry automatically exists at the interface because both monomers diffuse from aqueous phase and organic phase respectively. The stoichiometry at the interface is perfectly controlled by mass transfer controlled rate of diffusion and consumption of each monomer species.

 

Sebacoyl chloride was placed in 100ml of hexane (solution B) while hexamethylenediamine was added with 50ml of water (solution A). The introduction of aqueous layer and organic layer was used to create an interface between two solutions since aqueous layer and organic layer will not mix together. Interfacial polymerization occurs instantly forming a thin film of solid nylon 6,10 at the interface. The thin film at the interface stops further reaction by preventing the monomers to meet each other. During the preparation of solution B, any apparatus used should be rinsed with hexane and thus they do not contain water. If water present, chloride ion, Cl^- of sebacoyl chloride tend to react with the hydroxide, OH^- in water to form sebacic acid as well as two hydrochloric acid molecules.

The reaction between sebacoyl chloride and water is very reactive since chloride ion is a group leaving group.

 

A small amount (0.01g) of sodium laurel sulphate (soap powder) was introduced into the solution B in order to lower the surface tension of solution A and hence hexamethylenediamine will easily solubilize in it. As a result, the condensation polymerization is encouraged to occur in the interface. Since hexamethylenediamine exists its base property in the presence of water, so two drops of phenolphthalein was added as an indicator to show the position of interface between the solutions and to determine the point where hexamethylenediamine has been used up. In alkaline solution, the colour of phenolphthalein is pink while it will be faded if the alkalinity of solution decreases.

 

The interface between solution A and solution B should not be disturbed during transferring the solution A into solution B. This is to make sure both reactants do not form a polymer clump in unstirred interfacial polymerization. The system contains two layers in which solution A with hexamethylenediamine is upper layer whereas solution B with sebacoyl chloride is the lower layer because sebacoyl chloride is denser than hexamethylenediamine. During the transferring process, condensation polymerization between hexamethylenediamine and sebacoyl chloride took place at the interface is shown as the equation below:

At the interface between the solutions, a bigger molecule was formed from the reaction between both monomers and two small molecules hydrochloric acid, HCl were generated. The bigger size molecule formed at the interface could be dimer, trimer, tetramer, oligomer and polymer.

 

In unstirred interfacial polymerization, the colour of phenolphthalein indicator was faded as more nylon was wrapped on the glass rod and finally it was decolourized. This is because the decrease in concentration of hexamethylenediamine in which the diamine was polymerized in the system. In addition, the condensation polymerization produced hydrochloric acid as by-product and it neutralized the acid present in the system. The increase of hydrochloric acid concentration caused the phenolphthalein indicator to be decolourized. In stirred interfacial polymerization, the pink colour of phenolphthalein decolourized immediately once both solutions were stirred. This method produced a polymer clump in the solution. As the time pass, the polymer clump was predicted to become smaller because the hydrochloric acid will hydrolyze the amide bond in the polymer molecule.

 

By comparing both polymers that produced from two methods, the nylon 6,10 in unstirred polymerization results a more elastic and stiffer polymer than the nylon 6,10 in stirred polymerization. Besides, its ability to stick on the wall of beaker and surface is stronger when compared to nylon obtained from the latter method. The molecular weight of nylon obtained in unstirred polymerization is predicted to have higher molecular weight since its mechanical strength is greater than the other one. The weight of nylon 6,10 obtained in the unstirred interfacial polymerization was 0.4272 g while the weight of nylon 6,10 obtained via stirred interfacial polymerization was 0.4263g. Both methods gave the same percentage yield of 11.54%.

 

Reference:

1. Odian, G. (2004). Principles of Polymerization (4^th Ed.). USA: John Wiley & Sons, Inc.

2. Wagh, S.J., & Jadhav K.T. (n.d.) A review on interfacial polycondensation:

types, applications andeffect of various parameters.

3. Ji, J., Childs, R.F., & Mehtra, M. (2001) Mathematical model for encapsulation by interfacial polymerization. 192 (2001), P 55-70.

Determination of chloride ion in natural water

Objectives:

1. To apply Mohr method to the measurement of chloride ion in natural water

2. To determine the chloride concentration in unknown water sample

3. To understand the titration theory of chloride with silver nitrate solution

 

Introduction:

Titration is a common analytical technique in which the concentration of a substance of solution can be determined by adding standard solution that reacts with the unknown. By using the stoichiometry of the reaction and the number of moles of standard solution needed, the concentration of analyte can be determined through titration.

Argentometric titration is a type of titration that based upon silver (I) ion, Ag⁺, for example, Mohr method. In Mohr method, the most important precipitating agent is silver nitrate in which this method determines the chloride concentration of a solution by titration with silver nitrate. In argentometric determination of chloride concentration, potassium chromate is used as an indicator by reacting with silver ions to produce brick-red silver chromate precipitate at the end point.

As the silver nitrate solution is slowly added, the chloride ion will react with silver ion to form silver chloride precipitate.

Ag⁺(aq) + Cl^-(aq) –> AgCl(s) K_sp = 1.70x10^-10

This reaction continues until all the chloride ion in the system has been used up to form precipitate and this approaches the end point of titration. After all the chloride ion has been precipitated, any more silver(I) ion will react with the chromate ion, CrO₄^2- to form a red-brown precipitate of silver chromate as shown in the following equation.

2 Ag⁺(aq) + CrO₄^2-(aq) –> Ag₂CrO₄ (s) K_sp = 9x10^-12

Mohr method can be used to determine the chloride ion concentration of water samples from many sources such as seawater, stream water, river water and estuary water. Two unknown water sample are tested by using the particular method. The Mohr method works well in a slightly acidic condition, hence the pH of the water sample should be adjusted to the range of 6~8 by adding dilute acid or dilute base as needed.

Apparatus:

Conical flask, burette, pipette, measuring cylinder, volumetric flask

Materials:

Deionized water, unknown A water sample, unknown B water sample, 0.075M silver nitrate solution, pH paper, sodium hydroxide solution, calcium carbonate, 0.25M potassium chromate solution

Procedures:

 

Discussion:

In this experiment, the titration of unknown sample was conducted in the pH range of 7 to 8. This is because at pH lower than 7, the chromate ion will be converted to dichromate ion.

CrO₄^2-(aq) –> Cr₂O₇^- (aq)

Eventually, this dichromate ion cannot form a brick red silver chromate precipitate with silver ion and hence end point of the titration cannot be detected. At pH higher than 8, the silver ion will tend to react with the excess hydroxide ion to form brownish silver hydroxide. Formation of silver hydroxide will cover the colour of silver chromate precipitate and hence end point also cannot be seen.

Ag⁺(aq) + OH^-(aq) --> Ag(OH) (s)

Mohr method can only roughly show the concentration of chloride ion in the water sample. This is because excess silver nitrate is needed to produce sufficient silver chromate precipitate to be observed in the solution with heavy white precipitate of silver chloride. Besides, the intense yellow colour of chromate solution causes the brick red silver chromate that formed is hardly to be observed.

 

Precaution steps:

1. All solution mixture that involved silver nitrate solution must be discarded into waste bottle because silver is heavy metal.

2. The burette must be rinsed with silver nitrate solution before titration starts.

3. Wear gloves when handling silver nitrate solution since it will cause skin staining and chemical burn.

4. Chromate solution needs to be used with care as chromate is a known carcinogen.

Flickr Tags: mohr method,chloride ion determination in water sample,titration,silver nitrate solution

Synthesis & characterization of a metal hydride complex

Objectives:

1. To synthesize a cobalt hydride complex

2. To deduce its chemical structure based on the spectral data

 

Introduction:

Metal hydride complexes are very crucial as the intermediates in many catalytic processes such as alkene oligomerization and hydrogenation. Covalently bonded metal hydride complexes are known for all the transition metals. The complexes often contain the metal in a low oxidation state with the ligands of phosphines, carbon monoxide or cyclopentadiene.

 

Isomerization of alkene is always a possibility in any homogenous catalytic reaction that involved alkene. Migratory insertion of alkene into the metal hydrogen (M-H) bond can occur in a Makovnikov addition or anti-Makovnikov manner. Alkene isomerization is a process that involves Makovnikov addition followed by a β-elimination which is shown in the diagram 1 below:

Figure 1: Isomerization of butene by hydride mechanism

In the Figure 1, it shows that the but-2-ene is synthesized from but-1-ene through the hydride mechanism in which metal hydride acts as the intermediate.

One of the well known processes of homogeneous catalytic reaction is hydrogenation of alkene. The metal hydride complex plays a very important intermediate in the hydrogenation of alkene, for example, Wilkinson’s catalyst in the hydrogenation of alkene. The following figure 2 shows how the metal hydride acts as an intermediate in the particular process.

 

Firstly, Wilkinson’s complex will dissociates one phosohine ligand to form a 14 electron complex. This followed by the oxidative addition of hydrogen to form a metal hydride intermediate. In the following step, alkene is added into the metal complex via ligand addition. Migratory insertion of alkene into the M-H bond leads to the formation of alkyl ligand that bonded to rhodium metal. Alkane is formed at the end of the process via reductive elimination between a hydride and an alkyl group. The catalyst is being reused in the hydrogenation process and the process is repeating again.

 

In this experiment, metal hydride complex is being synthesized and characterized by using proton nuclear magnetic resonance (¹H NMR) spectrophotometer in order to determine the number of proton present. IR spectrophotometer is also used in characterizing the metal complex.

 

Materials:

Sodium borohydride, ethanol, cobalt(II) nitrate hydrate, triphenylphoshate, methanol, dichloromethane

 

Instruments:

FT-IR spectrophotometer, NMR spectrophotometer

 

Apparatus:

Melting point apparatus, hotplate stirrer, magnetic stirrer bar, Buchner funnel, beaker, Erlenmeyer flask, Hirsch funnel

 

Procedure:

Part A: Preparation of Metal hydride

Part B: Characterization of Metal hydride

 

Results and calculation:

Table 1: Weight of hydridotetrakis(triphenylphosphito)cobalt(I)

Weight of empty sample vial	12.5925g
Weight of ( sample vial + metal hydride)	19.3553 g
Weight of metal hydride	6.7628 g

Calculating percentage yield of metal complex

One mole of Co²⁺ reacts with one mole of P(C₅H₅)₃.

Mole number of Co²⁺ = 1.5240g / 200.93g mol^-1

= 0.0076 mol

Mole number of P(C₆H₅)₃ = 6.8037g/ 261.97g mol^-1

= 0.026mol

Thus, Co²⁺ is the limiting reagent since P(C₆H₅)₃ is in excess.

Theoretical mass of hydridotetrakis(triphenylphosphito)cobalt(I)

= 0.0076mol x 1107.81g mol^-1

= 8.4194g

Percentage yield = 6.7628 g / 8.4194g x 100%

= 80.32%

 

Table 1: Integration value and number of proton present for each peak in NMR spectrum

Value of integration	= 60	= 1
Ratio	60	1
Number of protons present	60	1

Note: One triphenylphosphate contains 15 H

Since one triphenylphosphate contains 15H, so the presence of 60H indicates that there is four triphenylphosphate ligands binded to the metal hydride complex.

Table 2: Significant peaks of cobalt(II) nitrate hydrate in IR spectrum (Appendix I)

Significant signals	Wavenumber (cm-1)
Expected (from table)	Observed (from spectrum)
O-H stretch	3200-3550	3403
Asymmetric NO2 stretch	1450-1600	1629
Symmetric NO2 stretch	1260-1375	1384

Table 3: Significant peaks of triphenylphosphite in IR spectrum (Appendix II)

Significant signals	Wavenumber (cm-1)
Expected (from table)	Observed (from spectrum)
Aromatic C=C stretch	1400-1600	1481, 1590
=C-H stretch	3010-3100	3062, 3038
C-P stretch	700	absent

Table 4: Significant signals of hydridotetrakis(triphenylphosphito)cobalt(I) in IR spectrum (Appendix III)

Significant signals	Wavenumber (cm-1)
Expected (from table)	Observed (from table)
=C-H stretch	3010-3100	3067
Aromatic C=C stretch	1400-1600	1490, 1591
Co-H stretch	1745-1933	absent
C-P stretch	700	691

Note: The C-P stretch value is obtained from journal. (Kurita et al, 2003)

Electron counting of hydridotetrakis(triphenylphosphito)cobalt(I)

Hydride: donates 2 electrons, each triphenylphosphito donates 2 electrons

Hydride: -1, triphenylphosphosphito: neutral

Oxidation state of cobalt = Co(I), with d⁶

Total electron count = 2 + 4(2) + 6

= 16 electrons

 

Discussion:

The percentage yield of hydridotetrakis(triphenylphosphito)cobalt(I) is 80.32% in which the mass of product obtained experimentally is 6.7628 g.

In this experiment, sodium borohydride (white) was used to provide hydride to metal complex to form a cobalt hydride complex. Ethanol acts as a medium to allow the cobalt complex (greenish brown) can form in the solid state since the cobalt hydride complex is not soluble in ethanol. The product was washed with ethanol, water and methanol is to remove any other unreacted starting materials. This method can reduce the presence of impurities.

From the IR spectrum of triphenylphophate, the significant signals include aromatic C=C (1481cm^-1 and 1590cm^-1) and =C-H (3062cm^-1 and 3038cm^-1). However, the expected C-P stretch at 700cm^-1 did not present. According to the IR spectrum of product, the significant signals that present are =C-H- stretch (3067cm^-1), aromatic C=C (1490cm^-1 and 1591cm^-1) and C-P stretch (691cm^-1). However, there is another expected significant signal did not present in the IR spectrum which is the Co-H stretch (1745-1933cm^-1).

Based on the ¹H NMR spectrum of product, the integration of the first signal (at positive ppm value) shows 66mm while the second signal (at negative ppm value) shows 1.1mm. From the spectrum, number of protons present in the first and second signals is 60 H and 1 H respectively. Hence, the molecular structure of hydridotetrakis(triphenylphosphito)cobalt(I) is deduced as monocapped tetrahedral with the hydride as the face-capping ligand. The figure 3 below shows the structure of hydridotetrakis(triphenylphosphito)cobalt(I).

Figure 3: Molecular structure of hydridotetrakis(triphenylphosphito)cobalt(I)

The fomal oxidation state of cobalt in the metal complex is Co(I). This is because there is only one hydride carrying one negative charge bonded to cobalt while the other four triphenylphosphine ligands are neutral. Based on the electron counting in the calculation and result part, the total electron is 16. The metal complex is stable in 16 electrons because the four sterically bulky triphenylphosphine ligands blocked the vacant site and hence other ligand is not allowed to bind to cobalt. As a result, hydridotetrakis(triphenylphosphito)cobalt(I) has a total number of 16 electrons.

Precaution steps:

1. Make sure the sodium borohydride will never contact with acid or water since it will liberates hydrogen.

2. Dichloromethane is highly volatile and must be used in the fume hood only.

 

 

Flickr Tags: synthesis,characterization,metal hydride,IR,Makovnikov addition,β-elimination,hydridotetrakis(triphenylphosphito)cobalt(I)