Organic Chemistry Lab Manual - Introduction and Safety PDF

Lab #1: Introduction and Safety Laboratory Safety: 1) A laboratory coat must always be worn in the lab. This will protect you and your clothes from any contaminants that may be used during the laboratory experiments. 2) Shoes must have closed toes and heals. NO SLIPPERS, NO FLIP-...

Lab #1: Introduction and Safety Laboratory Safety: 1) A laboratory coat must always be worn in the lab. This will protect you and your clothes from any contaminants that may be used during the laboratory experiments. 2) Shoes must have closed toes and heals. NO SLIPPERS, NO FLIP-FLOPS, NO SANDALS. You may be asked to leave if you are not dressed properly to conduct an experiment. 3) In general, gloves should be worn during the lab exercises. These will protect your skin from corrosive chemicals. Gloves are NOT worn outside the lab. You do not wear gloves to go to the bathroom, to handle doorknobs, do not handle your pen, do not scratch your face... 4) NEVER pipette solutions with your mouth, eat, drink, smoke or apply cosmetics in the lab. 5) NEVER pour any chemical or solution into a sink without authorization. 6) NEVER use flames with or near volatile solvents 7) NEVER return reagents to stock bottles. 8) NEVER smell specimens or chemicals directly. 9) Use caution when using equipment for heating or melting reagents (e.g. Boiling water baths). 10) ALWAYS report any spillage or accident, however minor, to your instructor 11) Wash hands thoroughly with soap and water after handling laboratory reagents and especially prior to leaving the laboratory when the session is completed. 12) WHEN IN DOUBT ABOUT ANYTHING –ASK YOUR INSTRUCTOR. N.B: Your Instructor may decide to modify an experiment, change the reagents, or change the concentration of reagents based on logistics and experimental trial runs. It is your responsibility to pay attention to detailed instructions for any such modifications during the lab sessions. LAB #6: Reactions of Amines and Carboxylic acids AIM: To test and observe the reactions of some amines and carboxylic Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 1 of 19  acids. Reagents:  Phenylamine (Avoid skin contact. Toxic & harmful because of skin absorption)  Ethanoic acid (17.4M)  Red litmus paper  Sodium nitrite (10%)  Hydrochloric acid (Conc), 5% and 10% HCl  Propanol  Sodium hydroxide (Pellets)  Sodium carbonate salt (about 0.5g of powdered salt)  Lime water (Saturated aqueous solution of calcium hydroxide)  Benzene sulphonyl chloride Procedure Reactions of amine Tests Observation Inference a) Solubility in water Two layers formed: Phenylamine is non-soluble in water Test the solubility of Top layer was water and the telling us that it might be a non-polar phenylamine in water by shaking bottom layer was phenylamine compound. The red Litmus paper test 3 drops of phenylamine with Yellow droplet-like appearance showed no color change describing that 2cm3 of water. Dip a piece of red No red litmus change. Phenylamine is a neutral compound litmus paper inside and observe when added to water, as naturally any color change. phenylamine is a weak base. b) Solubility in HCl No layers formed. Soluble in Phenylamine is soluble in HCL, this Test the solubility of HCL confirms that phenylamine is a non- phenylamine in HCl by shaking Pale-yellow, translucent polar compound. 3 drops of phenylamine with solution 2cm3 of HCl. Compare with solubility in water Tests Observation Inference c) Nitrous acid test Upon the addition of the amine Before the mixing it was noted that Cool a mixture of sodium nitrite to sodium nitrite, two layers there was a formation of two layers Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 2 of 19 and a solution of the amine in formed: the top layer was indicating that amine has a lower and ice bath. The mixture must phenylamine and the bottom density than the Nitrous acid. It also be added in the ice bath. layer was bottom sodium nitrite. indicated that phenylamine is insoluble Add moderately concentrated After mixing and the addition of in sodium nitrite. hydrochloric acid to the mixture HCl, effervescence occurred and The effervescence observed (on above in the ice bath. Note the white gas was emitted. addition of HCl) highlighted that a gas effervescence. Crystallization (dark brown ppt) was being produced which we All additions for reaction c must formed in the bottom layer of the identified as Nitrogen gas be done in the ice bath. solution. (Diazotization reaction) After mixing the primary amine with sodium nitrite with HCl within the ice bath there was a formation of a dark brown precipitate indicated formation of a diazonium product. Tests Observation Inference d). Hinsberg test Transparent brown solution of The cloudy white solution formed upon Place about 10-15 drops of the amine turned cloudy white with addition of benzenesulphonyl chloride amine (0.5 – 0.8 g ,if solid) in crystal formation upon addition indicated that the amine can be primary or a test tube. Add 15 drops of of benzenesulphonyl chloride. secondary. There was a noted increase in benzenesulphonyl chloride and Heat was given off from the test temperature felt indicating that there was 10 – 15 drops of 10% NaOH tube upon mixing. an exothermic reaction solution. Stopper the test tube After the water bath, two layers After the water bath, the orange gel-like and shake it vigorously for a were noticed: the top layer was ppt. indicated the formation of a few mins. Remove the stopper thick and cloudy white, and the sulfonamide derivative (nucleophilic and warm the test tube in a bottom layer was a thin orange substitution reaction in which the amine water bath (60 – 70 0C) for gel-like ppt. replaced the chlorine in benzenesulphonyl about 1-2 mins and check for Upon addition of sodium chloride that produced the ppt.). reaction. hydroxide, the ppt. dissolved Upon the addition of sodium hydroxide, If ppt. forms in the alkaline completely. the sulfonamide derivative was soluble in solution, add 3-5 ml of a 10% When HCl was added, an orange sodium hydroxide and the precipitate and shake the test tube. If the gel-like ppt. in a transparent dissolved. This is because primary amines alkaline solution is clear, solution was reformed. form sulphonamides with an acidic acidify it with a 10% HCl to hydrogen. see if a ppt will be formed. Upon addition of HCl, the orange gel-like ppt. reappeared. Acidification re- protonated the sulfonamide, making it insoluble in acidic conditions. This behavior is characteristic of primary amine sulfonamides, which are soluble in base but insoluble in acid. This confirmed that phenylamine is a primary amine. Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 3 of 19  Reactions of carboxylic acids Test Observation Inference a). Add a few drops of ethanoic Blue litmus paper turned red This indicates that the ethanoic acid is acid to moist blue litmus paper acidic in nature. Test Observation Inference b). Add ethanoic acid to sodium Ethanoic acid has a strong The reaction between the weak ethanoic hydroxide (Note the smell of vinegar-like smell. acid and strong sodium hydroxide base ethanoic acid before and after When sodium hydroxide was is a neutralization reaction therefore, the NaOH is added) added, there was no detectable vinegar-like odor noticed before was smell, the solution was odorless. gone after the addition of the base. This indicated that ethanoic acid was neutralized forming sodium acetate. CH3COOH (aq) + NaOH (aq)  CH3COONa (aq) + H2O (l) Test Observation Inference c) Add ethanoic acid to sodium When ethanoic acid was added The effervescence observed indicated carbonate in a test tube and test to sodium carbonate, that a gas was formed in this case it was the gas given off with lime effervescence was noticed. carbon dioxide. This occurred because water, by using the delivery tube The gas produced led to the the carbonate decomposes releasing provided. formation of a white ppt. and carbon dioxide which is typical and turned the lime water milky. reactions between a carbonate and the acid. 2CH3COOH (aq) + Na2CO3 (aq)  2CH3COONa (aq) + H2O (l) + CO2 (g) When carbon dioxide reacted with lime water it produced the white ppt. calcium carbonate which turned the solution Milky. CO2 (g) + Ca(OH)2 (aq)  CaCO3 (s) + H2O (l) Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 4 of 19 d) Add 1ml of ethanoic acid to The initial solution was colorless Ethanoic acid and propanol mix well in about 1 ml of propanol and then and soluble. the presence of concentrated sulfuric add few drops of H2SO4. Heat the Upon heating, a mild fruity smell acid, which acts as a catalyst for the mixture in a boiling water bath was emitted. esterification reaction. for 5mins and pour the content After pouring the mixture into The reaction between ethanoic acid and into a beaker of water and smell. water, the resulting solution was propanol produced the ester propyl cloudy, and a stronger fruity ethanoate. smell was detected. The cloudy resulting solution from pouring the ester in water gave a stronger, fruitier apple/ pear smell which indicates that the ester formed was ethyl propanoate. Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 5 of 19 LAB #7: TITRATION OF AN AMINO ACID (GLYCINE) AIM: To study the titration curve of an amino acid (glycine). Theory: Titration curves are obtained when the pH of given volume of a sample solution varies after successive addition of acid or alkali. The curves are usually plot of pH versus the volume of titrant added. Amino acids are amphoteric molecules which can be titrated either against an acid or an alkali. They are weak Polyprotic Acids which exist as zwitterions at neutral pH. When an aqueous solution of an amino acid is titrated with an acid, it acts as a base, with a base, it acts as an acid. In this experiment we are finding out the titration curve of the amino acid Glycine. Glycine is a diprotic amino acid which means that it has two dissociable Protons, one on the α amino group and the other on the carboxyl group. In this fully protonated form; it can donate two protons during its complete titration with a base. In the case of Glycine, the R group does not contribute a dissociable Proton. Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 6 of 19 The dissociation of proton proceeds in a certain order which depends on the acidity of the proton: the one which is most acidic and having a lower pKa will dissociate first. So, the H+ on the α-COOH group (pKa1) will dissociate before that on the α- NH3 group (pKa2). Chemicals: 0.1M Glycine 0.1M NaOH Equipment: pH meter 250ml conical flask 5ml pipette Procedure: Pipette 25 cm3 of glycine into the conical flask provided and measure the pH. Now fill the burette to the zero mark with 0.1M NaOH. Using the burette, add 5ml of 0.1M NaOH to 25cm3 of glycine and record the pH of the resulting solution. Continue adding 5 ml of NaOH at a time and measure the pH for every 5ml added. Record your results as shown in the table below. Results Volume of NaOH added Total volume of NaOH pH of solution Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 7 of 19  0ml 0ml 1.60 5ml 5ml 1.92 5ml 10ml 2.12 5ml 15ml 2.40 5ml 20ml 2.70 5ml 25ml 3.22 5ml 30ml 6.56 5ml 35ml 7.82 5ml 40ml 8.64 5ml 45ml 9.94 5ml 50ml 10.42 5ml 55ml 11.26 QUESTIONS: 1) Using the data collected, draw a titration curve with pH on the vertical axis and the volume of NaOH added on the horizontal axis. pKa2 Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 8 of 19 What is the mathematical relationship between pKa and pI ? ( pK a1 +pK a2 ) pI= 2 pKa is the pH at midpoint, where the titration has been halfway completed or has reached the half of the equivalence point. Each pKa represents a step where one proton (p+) is lost from the molecule. pI, the isoelectric point, is the pH of an aqueous solution of amino acid (glycine) where the molecules on average have a net charge of 0. 2) Obtain the value of pI (isoelectric point) from the graph. We first look for an approximate range for our 2 pKa values to do this we look for a relatively flat point of the graph which we can see around the pH of 2.20 and the second pKa value being around pH 9.80, we will then add the 2 pKa Values together and then divide by 2 getting an approximate pI range about 6.0. pKa1= 2.20 pKa2= 9.80 pI= 6.0 3) What is zwitterion? A zwitterion is an ion that contains two functional groups and therefore possesses both positive and negative charges. The result of this would be a net charge of 0. Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 9 of 19  4) Draw the structure of glycine and state why glycine is unique compared to the other amino acids. ‌ Glycine is more unique than other amino acids because it is the simplest amino acid. Glycine is small to minimize sterical hinderance and allows for tight packing and turns within protein structures like collagen while other amino acids cannot. It has only one hydrogen on its side chain. It is also achiral and highly flexible because its central carbon is bonded to two hydrogens. This doesn’t allow for stereoisomerism simplifying its biochemical behavior as both a neurotransmitter and a key metabolic intermediate. Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 10 of 19 LAB #8 : ISOLATION OF CAFFEINE FROM TEA BY EXTRACTION AIM: To demonstrate the isolation of a natural product from a biological source, using extraction techniques. THEORY OF EXTRACTION Extraction with a solvent (Solvent extraction) is a separation technique most frequently employed to isolate one or more components of a mixture. The technique is based on the preferential solubilities of the components of the mixture for two different immiscible solvents. It involves the partial removal of a solute from one liquid in which it is less soluble to another immiscible liquid in which it is more soluble. In most cases, an organic solvent is used to remove an organic solute from an aqueous solution or suspension. If a solute is shaken with a mixture of two immiscible solvents at a fixed temperature, the solute will distribute itself in both solvents according to its solubility in each. The solute will partition itself according to the ratio: Kp = [solute]A/[solute]B Where Kp is the distribution coefficient, [solute]A is the concentration of the solute (in moles/L) in solvent A (usually the organic solvent) and [solute]B is the concentration of solute in (in moles/L) in solvent B (usually water). For example, 6.67g of phenol is dissolved in 100ml of water, while 8.33g will dissolve in 100ml of benzene, the distribution coefficient will be: Kp = (8.33g/100mL)benzene /(6.67g/100mL)water =1.25 Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 11 of 19 Solvent extraction is more efficient if a certain volume of the extracting solvent is used in several portions rather than one. Caffeine is an alkaloid that is present in coffee, tea, cola, chocolate any many non- prescription drugs. Caffeine is soluble in water because it has several polar and basic functional groups. This property makes it insoluble in aqueous base. Thus, by adding a weak base (e.g. sodium or calcium carbonate) to an aqueous solution of tea extract, one can decrease its solubility in water and increase its solubility in a less polar organic solvent ( such as CH2Cl2), into which the caffeine can be easily extracted, using a separatory funnel. Any neutral compound will be extracted into the organic phase, but unfortunately, there are a few of these present in tea. Any acidic compounds (such as tannic acid, a major component of tea) will be deprotonated and will remain in the aqueous phase. Therefore, Sodium carbonate serves two main functions: to place caffeine in a more basic environment so that it has a higher affinity for dichloromethane and to cause the tannins which are acidic to form phenolic salts in the aqueous solution. Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 12 of 19 PRECAUTIONS: Observe all laboratory precautions. PROCEDURE: 1. In a 400mL beaker, place 10 tea bags and approximately 100mL of water. Bring to boil on the hot plate and continue to boil for about 15 minutes. 2. Carefully remove the beaker from the hotplate, remove and discard the teabags, and dissolve 15g of Na2CO3 in the tea solution by stirring 3. Cool the tea solution in and ice water bath, then transfer it into a separatory funnel, using a funnel. 4. Add 20mL of methylene chloride via a funnel and shake the mixture gently. DO not shake to vigorously, or you will get an emulsion (i.e. a mixture consisting of droplets of one phase suspended into the other). If this happen let it stand for about 15 minutes to allow separation of the layers. 5. Extract the aqueous layer with another 20mL portion of methylene chloride 6. Combine the organic extracts in a 125mL conical flask, and dry them with about 1g of anhydrous MgSO4 7. Allow the solution to stand for about 10 minutes, swirling it occasionally to complete the drying. 8. Keep the aqueous phase in a beaker – do not discard until you are sure you do not need it anymore 9. Gravity-filter the methylene chloride solution into a small pre-weighed beaker. Add one boiling chip, and carefully evaporate the solvent to dryness in a hot water bath or a large beaker of water on a hotplate in the hood. Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 13 of 19  10.Remove the boiling chip, reweigh the beaker, and calculate the yield of crude caffeine. QUESTIONS: 1) Tea contains approximately 2% caffeine by weight. Assume that you started with 25g of tea leaves, calculate the percentage yield of your crude extract. Mass of beaker = 28.75g Mass of beaker with caffeine = 28.89g Mass of caffeine = 28.75g – 28.89g = 0.14g Actual yield = 0.14g Theoretical yield = 25g of tea leaves x 2% caffeine by weight =25g x 0.02 = 0.5g % yield = (actual yield / theoretical yield) x 100 = (0.14g / 0.5g) x 100 = 0.28 x 100= 28% Therefore, the percentage yield of the crude extract is 28%. 2) Suppose you forgot to add sodium carbonate to the tea solution prior to extraction. What effect would this have on your yield? If I forgot to add sodium carbonate to the tea solution before extraction, it will most likely result in a lower yield of caffeine. This is because Sodium Carbonate, which is a weak base, helps maintain a basic environment which does not allow caffeine to bind to the tannins found in Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 14 of 19  the tea and allows it to separate from those tannins much for efficiently therefore, without Sodium Carbonate the yield of caffeine would be much lower (Weaver, 2017, p. 2). The absence of the alkaline environment impedes the conversion of caffeine salts to its freebase form making caffeine transfer from the aqueous phase (Tea) to the organic solvent more difficult during extraction. As a result, a large amount of caffeine may remain in the aqueous phase resulting in the lower yield. 3) What are the advantages and disadvantages of using ether in solvent extraction? Ethers are highly soluble inorganic compounds but have a low boiling point which makes them easy to evaporate out. This would allow for the easy isolation of the solute, being caffeine. Ethers are also highly flammable and can cause an explosion posing serious health risks. Additionally, the miscibility of ethers in water is limited so when working with an aqueous solution this can be a disadvantage because it necessitates additional steps to ensure proper separation. 4) Why was the extraction with methylene chloride done twice with 20mL each time, instead of once with 40mL of methylene chloride? When the two extractions of 20 mL have reached equilibrium, there would be a higher yield of caffeine since each extraction’s yield would be at the maximum as opposed to a single 40 mL which would reach equilibrium only once. The smaller volume of solvent allows for better contact between the solvent and the caffeine being extracted. The increased surface area improves extraction efficiency by increasing the interactions between solvents and the organic compounds in the mixture. 5) Write equations for the reactions between ethanol and Na, phenol and NaOH and between ethanoic acid and Na2CO3. Ethanol and Sodium: 2CH3CH2OH(l) + 2 Na(s) → 2CH3CH2ONa(l) + H2(g) Phenol and sodium hydroxide: Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 15 of 19  C6H5OH(l) + NaOH(aq) → C6H5NaO(s) + H2O(l) Ethanoic acid and sodium bicarbonate: 2CH3COOH(aq) + Na2CO3(s) → 2CH3COONa(aq) + CO2 (g) + H2O(l) Conclusion: the isolation of caffeine from tea bags using extraction techniques with extraction solvent methylene chloride was accomplished successfully. The percentage yield of crude caffeine was extracted from the tea solution and calculated to be approximately 28%. In addition, the purposes of various materials such as methylene chloride and sodium carbonate were investigated, with the aim of maximizing caffeine yield and deprotonation. LAB # 9 : SAPONIFICATION Saponification is the hydrolysis of an ester with NaOH or KOH to give alcohol and sodium or potassium salt of the acid. Soap is now an essential everyday item and finds its importance in everyday life. But, how is soap made? The process of making soap is called saponification. Examples of a Saponification Reaction: In a saponification reaction, a base (for example sodium hydroxide) reacts with any fat to form glycerol and soap molecules. One of the saponification reactions taking triglyceride as an ester and sodium hydroxide as the base is as follows: In this reaction, triglyceride reacts with sodium hydroxide (a strong base) and glycerol is produced (an acid) along with soap (sodium palmitate). Similarly, potassium soap can be formed if a strong potassium base (like KOH) is reacted with an ester. This reaction is as follows: Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 16 of 19 It should be noted that for cleaning purposes, only potassium and sodium soaps are used. Based on the base used, soaps can either be hard soaps or soft soaps. In general, potassium soaps are soft and sodium soaps are hard. 1-Step Saponification vs 2-Step Saponification There can be either one-step saponification or two-step saponification process to convert triglycerides to soaps. The examples mentioned above are a one-step saponification process in which triglycerides, when treated with a strong base, split from the ester bond to release glycerol and soaps (i.e. fatty acid salts). On the other hand, in the two-step saponification process, the steam hydrolysis of the triglyceride yields glycerol and carboxylic acid (rather Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 17 of 19 than its salt). In the second step, alkali neutralizes fatty acids to produce soap. Effects of Saponification The effects of saponification can either be desirable or undesirable. Some effects of saponification are mentioned in the below-given points.  One of the major desirable effects of saponification is seen in fire extinguishers. Saponification is used by wet chemical fire extinguishers to convert burning fats and oils into non-combustible soap which helps in extinguishing the fire. Further, the reaction is endothermic and lowers the temperature of the flames by absorbing heat from the surroundings.  In an undesirable scenario, saponification damages oil paintings. In oil paintings, the heavy metals used in pigments react with the oil containing free fatty acids and form soaps. This way, the paintings get damaged gradually.  Soaps formed are used in everyday life like sodium soaps are used for laundry, potassium soaps are used for cleaning and lithium soaps are used as lubricating greases. There are various other soaps which are used for different purposes. Uses of Saponification  Wet chemical fire extinguishers: To extinguish cooking oils and fats, we use a saponification reaction. This is because cooking oils and fats have a flashpoint which is above 37 degrees which renders regular fire extinguishers useless.  Creating hard and soft soaps: By using different types of alkali in the process the type of reaction product can be altered between hard and soft.  Using KOH: We can obtain soft soaps  Using NaOH: We can obtain hard soaps  Materials Vegetable oil ( olive oil, coconut oil, palm oil) Sodium hydroxide solid (lye) Distilled water Sodium chloride ( common salt) Ethanol in Spray bottle Measuring cylinder Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 18 of 19 Beakers 250, 500 ml Glass stirring rods Balance Weighing boot Thermometer pH paper Hot plate with stirrer Proceadure Using a measuring cylinder place 66 ml of distilled water in a 250 ml beaker. Weigh 27 g of sodium hydroxide and stir into the distilled water in the beaker. Place a magnetic stir bar in the beaker. Place the beaker on the stirrer , allow to stir until the liquid ( lye water) is clear. Let stand to cool to about 500 C. To the lye water add 3 g of sodium chloride water. Using the 100 ml measuring cylinder place 47.5 ml coconut oil, 88ml olive oil and 65 ml palm oil in the 500ml beaker To the beaker of oil slowly add the lye solution Lye solution should be about 500 C using the glass rod while stirring. Add magnetic stir bar to the beaker and place on stirrer. Stir vigorously , using an industrial stirrer stir until there is medium trailing. This is the point that additives( scent, charcoal, oils) can be added. Pour the mixture into a container as directed. The mixture is then pour into a lined box to solidify overnight. After solidification the solid soap is cut into blocks with a knife. Organic Chemistry 2 Laboratory (CHEM 225) – F 2022 Page 19 of 19