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Some Families of Organic Compounds Tetrahedral Compounds A tetrahedral carbon is a carbon atom which has tetrahedral geometry. This geometry happens only when the carbon atom has 4 single bonds. **Definition: A chloroalkane is a compound in which one or more of the hydrogen atom...

Some Families of Organic Compounds Tetrahedral Compounds A tetrahedral carbon is a carbon atom which has tetrahedral geometry. This geometry happens only when the carbon atom has 4 single bonds. **Definition: A chloroalkane is a compound in which one or more of the hydrogen atoms in an alkane molecule have been replaced by chlorine atoms.** Monochloromethane Dichloromethane Trichloromethane Tetrachloromethane (chloroform) Chloroalkanes are used as solvents for removing oil and grease marks from machinery and in dry cleaning of clothes - Chloroalkanes are weakly polar -They are not soluble in water - They are soluble in non-polar solvents (cyclohexane) - Liquids at room temperature. Alcohols **Definition: A functional alcohol is an atom or group of atoms which is responsible for the characteristic properties of an organic compound or a series of organic compounds.** Alcohols may be imagined as derived from alkanes by replacing a H atom by an -OH group. The -OH is called the hydroxyl or hydroxide group and is the functional group in alcohols. General formula of alcohols = CnH2n+1OH Named by changing the -e in alkanes to -ol. **Definition: A primary alcohol is one where the carbon atom attached to the -OH group is attached to only one other carbon atom** Ethanol 1 **Definition: A secondary alcohol is one where the carbon atom joined to the -OH group is attached to two other carbon atoms.** Propan-2-ol **Definition: A tertiary alcohol is one where the carbon atom joined to the -OH group is attached to three other carbon atoms** 2-methylpropan-2-ol - Ethanol (C2H5OH) is the most common and best known alcohol. It is found in alcoholic drinks. - It is made by fermenting glucose using yeast: C6H12O6 → 2C2H5OH + 2CO2 - To prevent people from drinking industrial ethanol, methanol is added, as it is particularly toxic. This ethanol is now denatured and is called methylated spirits. - Alcohols have higher boiling points than their corresponding alkanes. This is due to hydrogen bonding between the alcohol molecules. - Small alcohol molecules are soluble in water due to the hydrogen bonding between the molecules. - Larger alcohol molecules (e.g., butanol) are not soluble in water as the effect of the hydrogen bonding decreases as the molecule gets bigger. - These larger alcohols are soluble in non-polar solvents like cyclohexane Planar Compounds A planar carbon is a carbon atom which has planar geometry. This geometry happens only when the carbon atom is unsaturated (contains a double- or triple- bond). Aldehydes Aldehydes form a homologous series of organic compounds containing the -CHO functional group. - Named from the parent alkane by changing the -e to -al. - Have dipole-dipole interatcions between molecules because of the polar carbonyl group - Boiling points are higher than their respective alkanes due to the dipole-dipole 2 forces - Boiling points are lower than their respective alcohols as alcohols have hydrogen bonds - Small aldehydes are soluble in water due to the polar carbonyl group Methanal 2-methylpropanal Benzaldehyde Ketones Ketones form a homologous series of organic compounds containing the >C=O functional group - Named by changing the final -e of the parent alkane to -one - The highly polar C=O carbonyl group is always located on one of the central carbons, and never at the end of the carbon chain in a ketone. - Boiling points are higher than their respective alkanes due to the dipole-dipole forces - Boiling points are lower than their respective alcohols as alcohols have hydrogen bonds - Small Ketones are soluble in water due to the polar carbonyl group - Used as solvents (Acetone – Removes nail varnish) Propanone Butanone Carboxylic Acids Carboxylic Acids are a family of organic compounds that contain the carboxyl group – COOH and have the formula CnH2n+1COOH - Named by changing the final -e of the parent alkane to -oic acid - The highly polar C=O carbonyl group is always located at the end of the carbon chain in a carboxylic acid - Boiling points are higher than their respective alkanes, alcohols, aldehydes and ketones due to the relatively strong hydrogen bonds between the carboxylic acid 3 molecules - Small carboxylic acid molecules are soluble in water due to the Hydrogen Bonding - The longer the carbon chain becomes, the less soluble the carboxylic acid is in water, as the effect of the polar -OH group decreases - Propanoic acid prevents mould and is used as a food preservative - Benzoic acid is used in antiseptic medicines - Ethanoic acid is vinegar Methanoic Acid Butanoic acid Esters Esters are a family of compounds with general formula RCOOR’ and a functional group -COO-. - Derived from carboxylic acids by replacing the H of the -OH group with an alkyl group - Prepared by the reaction between an alcohol and a carboxylic acid Ethanoic Acid Methanol Methyl Ethanoate An esterification reaction is an example of a condensation reaction **Definition: A condensation reaction is a chemical reaction in which two molecules combine to form a larger molecule with the loss of a smaller molecule such as water.** - To name an ester we must divide the structure by identifying the part derived from the carboxylic acid and the part derived from the alcohol - Name the alcohol-derived section first e.g., methyl, ethyl - Identify the carboxylic acid - Remove “-oic acid” from the name and replace it with “-oate” - Combine the name from step 2 with the name from step 3. - The polar carbonyl group (C=O) means that ester molecules have Dipole-Dipole intermolecular forces 4 - Small ester molecules are soluble in water due to the Dipole-Dipole intermolecular force - Used to give fruity flavourings to food - Used in the manufacture of perfumes and cosmetics Bonding in Benzene. Aromatic Compounds - Each Carbon atom in the benzene ring has four electrons in its outer shell - Each carbon atom uses three of these electrons to form sigma bonds by head- on overlap of atomic orbitals - Two of these sigma bonds are formed between each carbon atom and two adjacent carbon atoms - The third sigma bond is formed with an atom of hydrogen - Each remaining valence electron (6) is shared between all 6 carbons - These delocalised electrons give extra stability to the benzene molecule - Bond lengths between all carbon atoms are equal Organic Natural Products Many organic compounds are found to occur in nature Ibuprofen Vanillin (lab made) (lab made) Eugenol Menthol (Natural) (Natural) 5 Mandatory Experiment: To extract clove oil from cloves by steam distillation (1) 1. Set up the Quickfit apparatus as above 2. Note the mass of the cloves, and place them in the pear-shaped flask. Cover with a little warm water (about 5 cm 3). 3. Place an adequate supply of water in the steam generator, connect it to the rest of the apparatus and set it to boil. If Quickfit apparatus is used for steam generation make sure that you use anti-bumping granules in the steam generator. 4. If the level of the boiling water in the steam generator falls too low, the system will not work smoothly. Remove the heat, carefully loosen the safety valve, and top up the steam generator with hot water. Reconnect everything and resume heating. 5. Collect the distillate. It should have a pale milky appearance. Using the dropping funnel as a receiver at this point will facilitate the next stage of the separation. 6. After 20 to 30 minutes disconnect the steam generator to avoid the possibility of suck-back problems and turn off the heat under it. You will have probably collected between 40 and 50 cm 3 of distillate. Note the smell of the distillate. **Definition: A steam distillation is a separation process used to isolate compounds at temperatures below their composition temperatures. It is carried out by bubbling steam through the material and distilling off the immiscible liquids.** 6 **An emulsion is a dispersion of small droplets of one liquid in another liquid in which it is not soluble.** To isolate clove oil (eugenol) from an emulsion of clove oil and water by liquid-liquid extraction using cyclohexane 1. Disconnect the dropping funnel from the rest of the apparatus. 2. Add about 8 cm3 of cyclohexane to the distillate in the dropping funnel. 3. Stopper the dropping funnel and shake the mixture. Release any pressure build-up carefully after each shake by inverting the dropping funnel while holding the stopper and slowly opening and shutting the tap. 4. Run the lower aqueous layer off. Collect the top layer - this contains the clove oil and the cyclohexane. 5. Dry the organic layer by shaking with anhydrous sodium sulfate in a conical flask. If possible, allow to stand overnight, before removing the solid by filtration or decanting. 6. Separate the more volatile cyclohexane from the clove oil by placing the mixture in a small beaker (whose mass is known) on a water-bath in a fume cupboard. The cyclohexane evaporates, leaving the clove oil behind. (Note that if the cyclohexane is to be distilled off it will boil at 81 0C.) Prepare the students for a very low yield of clove oil. 7. Note the smell of the clove oil. Do not allow the clove oil to come in contact with your skin (2) - A small quantity of an oily substance is left in the conical flask - The oily substance has a strong smell of cloves 7 Exam Questions 2012 – HL – Section B – Question 8 8. Study the reaction scheme and answer the questions that follow. (a) Give the systematic (IUPAC) name for (i) the alcohol A, Propan-2-ol(ii) the ester B. Propyl methanoate (b) Alcohol A and propan-1-ol are structural isomers. Explain the underlined term. Compounds with the same molecular formula but different structural formula What is the structural difference between a primary alcohol and a secondary alcohol? Primary: One carbon attached to OH carbon. Secondary: Two carbons attached to OH carbon. Identify another pair of structural isomers from the reaction scheme. Propanal and propanone (c) Identify a compound in the scheme whose carbon atoms are all in tetrahedral geometry. A – propan-2-ol. (d) Name the reagent and catalyst used to bring about the conversions labelled R. Regent: Hydrogen. Catalyst: Nickel (e) Propanal is oxidised by Fehling’s reagent. Describe how this reaction is carried out. - Mix equal amounts of Fehling’s A and Fehling’s B in a test tube. - Add small amount of propanal. - Heat in water bath. - Red precipitate formed. Why does propanone not react with Fehling’s reagent? Propanone is not easily oxidised (f) Which compound in the scheme would you expect to have a fruity odour? B – propyl methanoate 8 References 1. Examlearn.ie 2. Examlearn.ie Types of Reactions in Organic Chemistry Substitution Reactions Halogenation of Alkanes **Definition: A substitution reaction is a chemical reaction in which an atom or a group of atoms in a molecule is replaced by another atom or group at atoms.** Alkanes undergo substitutions in their reactions with halogens in the presence of ultraviolet light. CH4 + Cl2 + CH3Cl + HCl - One of the hydrogen atoms in the methane has been replaced by a chlorine atom to form the new compound chloromethane. - If excess chlorine is used, further substitutions take place to form dichloromethane(CH2Cl2), trichloromethane (CHCl3), and eventually tetrachloromethane (CCl4) Mechanism of Monochlorination of Methane **Definition: The mechanism of a reaction is the detailed step-by-step description of how the overall reaction occurs.** This mechanism is known as free-radical substation. Initiation A chlorine molecule is broken down into two chlorine atoms in the presence of ultraviolet light Propagation A chlorine atom attacks a methane molecule to form hydrogen chloride and a species called methyl free radical 9 Propagation A methyl free radical attacks a chlorine molecule to form chloromethane and a chlorine atom The chlorine atom produced in step 2 can then react with another molecule of methane. This sets up a chain reaction. **Definition: A chain reaction is a reaction that continues on and on because a product from one step of the reaction is a reactant for another step of the reaction.** Termination When most of the reactants have been used up during Steps 2 and 3, there are only a small number of Chlorine radicals and methyl radicals left over. These combine to form Cl2, chloromethane and ethane Evidence Explanation Reaction takes place when exposed Suggests a free-radical mechanism to UV light whereby Cl2 is broken down into Cl radicals. For every photon of light absorbed, Chain reaction is taking place thousands of chloromethane molecules are formed Ethane is found in the products Two methyl free radicals must be combining to form ethane Free radicals speed up the reaction Only a reaction which uses free radicals would be sped up by adding free radicals Inhibitors such as oxygen slow down Shows a chain reaction is taking place the reaction 10 Esterification When a carboxylic acid is reacted with an alcohol, with H2SO4 as a catalyst, an ester is formed. This reaction is referred to as saponification and is used in the manufacture of soap. Mandatory Experiment: To prepare a sample of soap (1) (2) 1. Add 2.5 g of lard, 2.5 g of potassium hydroxide and 20 cm3 ethanol, along with a few anti-bumping granules, to the flask. Swirl to allow proper mixing. 2. Set up the reflux apparatus, making sure to grease all joints 3. Reflux the mixture for 20 minutes, using a water bath 4. Remove the ethanol by distillation. 5. Dissolve the residue in a minimum of hot water (approximately 15 cm 3). 6. Add this solution to the brine. The soap should precipitate out. 7. Filter the soap. 11 8. Test the soap for its lathering qualities by shaking a small sample of it with water. It is important to wash your hands afterwards to remove any potassium hydroxide still present. Addition Reactions **Definition: An addition reaction is one in which two or more molecules react together to form a single molecule.** - Always will occur on a double or triple bond - Geometry will change from planar to tetrahedral bonds - Used in the hydrogenation of vegetable oils - Use to form plastics Hydrogenation of Vegetable oils Polarisation The C=C double bond in ethane has a high concentration of negative charge. As the Br2 approaches the ethene, the electrons are repelled away from the ethene, polarising the Br2 Heterolytic fission The Br2 molecule splits into Br+ and Br- ions Carbonium ion formation The Br+ attacks the electron rich C=C double bond. This forms a carbonium ion 12 Ionic Addition The carbonium ion is then attacked by the Br- ion. This results in the formation of 1,2-dibromethane Evidence for this mechanism If this reaction is carried out in bromine water with Chlorine ions added, the normal 1,2-dibromoethane product is formed, but two other products form also; 1- bromo-2-chloroethane and 2-bromoethanol. These form because of the Cl - or the OH- Polymerisation Reactions **Definition: Polymers are long chain molecules made by joining together many small molecules.** Polymers are repeating structures consisting of thousands of monomers linked together. - (Poly)chloroetheane, PVC, is a common plastic **Definition: The repeating unit of a polymer is that part of the polymer whose repetition produces the complete polymer chain except for the end groups.** Polyvinyl chloride can be synthesised from ethene 13 Elimination Reactions **An Elimination reaction is one in which a small molecule is removed from a larger molecule to leave a double bond in the larger molecule.** Ethene can be produced from ethanol by means of an elimination reaction. It involves the removal of a molecule of water from a molecule of ethanol by passing the ethanol through aluminium oxide. Redox Reactions When a primary alcohol reacts with an oxidising agent, the primary alcohol is converted to an aldehyde. Two hydrogen atoms are removed from the primary alcohol If a secondary alcohol reacts with acidified sodium dichromate, a ketone is formed Aldehydes and ketones can be reduced back to alcohols in the presence of hydrogen and a nickel catalyst 14 Mandatory Experiment: To study the reactions of ethanal with (i) acidified potassium permanganate solution, (ii) Fehling’s reagent and (iii) ammoniacal silver titrate Oxidation by acidified potassium manganate(VII) 1. Using graduated disposable pipettes, place in a test-tube 2 cm3 of ethanal, 1 cm3of potassium manganate(VII) solution and 4 cm 3 of dilute sulfuric acid. 2. Warm the test-tube in a water-bath and shake gently. Observe and record any colour change in a table of data copied into your practical report book The potassium permanganate solution does not change colour, ketones are not oxidised by potassium permanganate Oxidation by Fehling's solution 1. Using a graduated disposable pipette, place in a test-tube 1 cm3 of Fehling's solution no. 1. 2. Using a separate pipette, add 1 cm3 of Fehling's solution no. 2 – swirl the contents so that the blue precipitate initially formed will dissolve. 3. Add 1 cm3 of ethanal, heat gently and shake. 4. Observe any change and record your observations. No red precipitate is observed. Ketones are not oxidised by Fehling’s solution Oxidation by Tollen's reagent 1. Using a graduated disposable pipette, place in a clean test-tube 3 cm3 of silver nitrate solution and 1 cm3 of sodium hydroxide solution. 2. Add aqueous ammonia solution drop by drop, with shaking, until the precipitate formed in stage 1 is just dissolved. 3. Add two or three drops of ethanal, mix by shaking and warm in a water bath. 4. Observe any change and record your observations, 5. Rinse out the test-tube with dilute nitric acid and then water. No change is observed, ketones are not oxidised by Tollen’s reagent 15 Organic Compounds that Act as Acids Reaction of Alcohols with Sodium The H on the –OH of an alcohol can act as an acid when reacted with the extremely reactive metal, sodium. This forms the sodium salt of the alcohol, and hydrogen gas. Acidic Nature of the Carboxylic Acid Group - Inductive effect: Carbon atom of the carbonyl group is slightly positive, the carbon atom tends to attract electrons from the oxygen atom at the -OH group. This electron pulling effect is called the inductive effect - Stability of carboxylate ion: When the H of the –OH dissociates to form a H+ we are left with a carboxylate ion – COO-. The negative charge is not localised to just one of the oxygens and is instead delocalised between the two of them. This delocalised structure gives the carboxylate ion extra stability, allowing the H+ to easily dissociate. To study the effects of ethanoic acid with (i) sodium carbonate, (ii) magnesium and (iii) ethanol 1. Test with magnesium: Coil the 5 cm clean strip of magnesium loosely, drop it into one of the test tubes, and swirl. Record your observations. Fizzing is observed in the test tube, a mild pop is heard. 2. Test with sodium carbonate Add 1 g of anhydrous sodium carbonate powder into the second test tube, and swirl. Record your observations. The lighted taper is extinguished and limewater turns a milky white colour 16 3. Esterification (i) With care add 2 drops of concentrated sulfuric acid to the third test tube. (ii) Add 1 cm3 of ethanol and warm gently. Carefully smell the reaction product. The contents of the test-tube have a fruity smell Synthesis of PVC from Ethene Step 1: Ethene and chlorine react to form 1,2-dichloroethane. Step 2: Heat is used to thermally crack the 1,2-dichloroethane into chloroethene and HCl. Step 3: The chloroethene undergoes a polymerisation reaction to form polychloroethene (PVC) 17 Chromatography and Instrumentation in Organic Chemistry Mass Spectrometry ** Definition - The Principle of Mass Spectrometry is that charged particles moving in a magnetic field are deflected by different amounts due to their masses. This separates the particles according to their masses.** There are five stages in Mass Spectrometry (3) 1. Vaporisation: The sample material is vaporised into a gas. 2. Ionisation: An electron gun fires high-energy electrons at the gaseous sample. This knocks electrons off the sample particles, leaving the sample as a group of positively charged ions. 3. Acceleration: A negatively charged plate attracts the positive ions. This accelerates the particles so that they travel at high speed through the spectrometer. 4. Separation: A magnetic field of a particular strength is used to deflect the particles. Particles that are too light are deflected too much and hit the side of the spectrometer. Particles that are too heavy are not deflected enough and hit the side of the spectrometer. Only particles that have a certain mass are deflected by just the right amount and make it through the spectrometer to the detector. 5. Detection: A detector senses the number of positive ions hitting it and displays the result on a mass spectrum. 18 Mass Spectrum Output The peaks shown in the mass spectrum have units of mass on the horizontal axis. The height of each peak represents the relative abundance of particles of that mass. We can see that the peak at 57 is roughly four times higher than the peak at 86. This means that there are 4 times as many particles of mass 57 as there are of mass 86. Uses of Mass Spectrometry -Drug testing and discovery -Food contamination detection -Pesticide residue analysis -Isotope ratio determination -Protein identification -Carbon dating. Chromatography ** Definition - Chromatography is a separation technique in which a mobile phase carrying a mixture moves in contact with a selectively adsorbent stationary phase.** Types of Chromatography Paper Chromatography – Stationary Phase – Chromatography Paper Mobile Phase – Water or Solvent 19 (4) 1. Spot of sample mixture is placed approx. 1 cm above the water line. 2. The mobile phase soaks up through the stationary phase and dissolves the materials in the mixture. 3. Each material dissolves to a different extent. 4. Less soluble materials will appear as a spot closer to the bottom of the paper. 5. More soluble materials will appear as a spot higher up the paper. 6. This separates the components in the mixture. Uses of Paper Chromatography Analysis of food colourings, dyes and indicators. Other Types of Chromatography Column Chromatography Stationary Phase – Silica Gel in Glass Tube Mobile Phase – Water, Ethanol, or other solvent Passing a solvent through a column is called elution. The solvent is known as the eluent. Uses: Separation of dyes in food colouring 20 Gas Chromatography Definition - The Principle of Gas Chromatography is that a mixture of components is carried by a gaseous mobile phase is separated based on their different interactions with a solid stationary phase and the gaseous mobile phase. (5) Stationary Phase – Coiled column filled with coated silica gel Mobile Phase – Inert Carrier Gas Uses -Measuring Alcohol levels in Urine Samples -Drug Testing in Athletes GC-MS Gas Chromatography is often used with Mass Spectroscopy. After separation by GC, each component is put through a Mass Spectrometer to identify each component. High Performance Liquid Chromatography (HPLC) The pump supplies the high pressure needed for the mobile phase. The reason the mobile phase needs high pressure is that the column is packed very tightly, and the elution would take too long if the pump wasn’t used. (6) 21 Stationary Phase: Coated Silica Gel Mobile Phase: Suitable liquid solvent (ethanol) under high pressure Uses -Testing for Growth Hormones in food -Measuring quantity of Caffeine in drinks Spectroscopy Infra-red Spectroscopy (IR) Organic compounds absorb infrared radiation. Different frequencies of IR light are absorbed by different types of bonds. So a C=O would absorb a different frequency of IR then an O-H or a C=C bond. By looking at how an organic compound absorbs IR light, we get an IR spectrum, which is used as a fingerprint for that compound. Uses -Identifying Plastics -Identifying Illegal Drugs -Used to analyse breathalyser results Ultraviolet Spectroscopy (UV) This instrument measures how a compound absorbs UV light of different wavelengths/frequencies. Like IR spectroscopy, this also results in a fingerprint for each compound. However, unlike IR, UV spectroscopy can also measure the concentration of the compound, making this a quantitative analysis. UV spectroscopy is commonly used with HPLC in order to identify each compound in a mixture, and their concentrations. 22 Mandatory Experiment: To separate rhe components of ink using paper chromatography 1. Add solvent to the bottom of the tank to a depth of about 10 mm. Cover the tank, and allow to stand for a few hours, to allow the tank to become saturated with solvent vapour. 2. Make a line with a pencil near the top of a rectangular sheet of chromatography paper, and another line about 3 cm from the bottom. 3. Place a small spot of each indicator and of the mixture of indicators at different points on the line near the bottom of the paper, using a capillary tube. Dry using a hair drier, and repeat. 4. Place the chromatogram in the tank, ensuring that the solvent level in the tank is below the line on which the indicator samples are spotted. Run the chromatogram, until the solvent reaches the line near the top of the paper. 5. Remove and dry. 6. Calculate and record the Rf values of each indicator. (Ammonia vapour is used to locate phenolphthalein – the paper is held in vapour coming from a boiling solution of dilute ammonia.) 7. Identify the indicators present in the mixture. 23 Exam Questions 2014 – HL – Section A – Question 2 A student prepared a sample of soap in the school laboratory. The experiment was carried out in the four stages illustrated on the previous page. At Stage 1, using a water bath, the student refluxed for approximately 20 minutes 4.45 g of glyceryl tristearate (an animal fat) together with an excess of sodium hydroxide pellets, anti-bumping material and about 30 cm3 of ethanol. The reaction shown in the following balanced equation took place. The apparatus was then allowed to cool and rearranged for Stage 2, distillation, again using a water bath. After distillation, the contents of the distillation flask were decanted or washed into a beaker containing brine – Stage 3. Filtration was used in Stage 4 to isolate the soap which was then thoroughly washed. (a) What is the purpose of refluxing in Stage 1 of the preparation? To allow time for reaction without losing volatile material Name the type of reaction that occurred during this stage. Saponification (b) What substance was removed by distillation in Stage 2? Ethanol (c) Explain the function of the brine in Stage 3. To precipitate the soap (d) Why was it necessary to wash the soap thoroughly in Stage 4? To remove sodium hydroxide How should the student have washed the soap? Brine (e) Draw the structure or give the name of the co-product of the reaction. Propane-1,2,3-triol Where was the co-product located at the end of the process? In the brine (f) Given that the sodium hydroxide was in excess, calculate the maximum yield in grams of soap that could have been obtained in this preparation. 4.45/890 = 0.005 mol of fat 0.005 x 3 = 0.015 mol of soap 0.015 x 306 = 4.59g (g) Suggest, with reference to its structure, how a soap like sodium stearate can dissolve both the non-polar oils and the ionic salts in sweat from the skin - C17H35 is non-polar and dissolves oils. –COO–Na+ attracted to salts in sweat. 24 2013 – HL – Section A – Question 2.In a practical examination, chemistry students were required to perform a number of tasks in a laboratory. They had access to all the necessary reagents and glassware and also to the required safety equipment and clothing. (a) How could a student have carried out a simple chemical test to confirm that a colourless liquid sample was ethanoic acid and not ethanol? Ethanoic Acid: Addition of acidified potassium permanganate will not give a colour change (b) A sample of ethene gas was supplied in a stoppered test tube. Describe fully how the gas could have been shown to be unsaturated. - Add a bromine solution to a test tube of ethene. - Colour change: Brown to colourless (c) Describe with the aid of a labelled diagram how a student could have used chromatography to separate a mixture of indicators. - Paper chromatography. - Apply mixture using dropper spot on paper slightly above eluent in beaker - Solvent moves up separating components (d) One of the tasks in the practical examination was to measure the melting points of two benzoic acid samples (A and B) and to use the results to determine which was the purer sample. The melting points obtained by one of the students were as follows: sample A = 117 – 120 C; sample B = 120 – 121 C. Which was the purer sample? A Justify your answer. Has a wider range The students were required to recrystallize the impure benzoic acid. What solvent should they have used for the recrystallization? Water Explain why this solvent is suitable. Very soluble in hot but slightly soluble in cold (e) The diagram shows a steam distillation apparatus assembled incorrectly by one of the students. Identify the flaw in the assembly and state how it should have been rectified. Have end of safety tube under water 25 References 1. Instituteforeducation.com 2. Scoilnet.ie 3. Chemguide.com 4. BYJU’s.com 5. Researchgate.net 6. Chemguide.com 26

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