Organic 233:Lab 7-11
Organic 233:Lab 7-11 233
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Date Created: 05/02/16
Robin Cordero Chemistry 233: Final IR and primer and Labs 711 Multiple Choice: 20 points: Prelab Quizzes Short answer: 80 points: Homework (general theory, as well as mathematical problems). Questions about the procedure (general procedure and why you did certain steps at certain times) Study Question, Prelab slides, Textbook to review the chemical theory of IR and labs 711) Material (make sure you know them for the exam) Robin Cordero Safety Hazards Wear gloves and goggles at all times Toxic liquids Silica gel Methylene Chloride Bromine Flammable Liquids Ethyl acetate/butyl Hexanes 2butanone Hot Injection ports on the GC Corrosive /Severe burns (relief burn use 5% of sodium bicarbonate) Hydrochloric acid Sulfuric Acid Bromine Lucas Reagent Iodoethane Caustic Liquid Sodium Hydroxide Potassium Hydroxide Methanolic solution of solutions of sodium borohydride Heat Concentrated acid and water mix H2 gas Solutions of sodium borohydride in methanol Lab 7: Preparation of Alkyl Halides by Substitution reactions. NaI and AgNO tests 3 Robin Cordero Purpose: 2chloro2methylbutane is prepared from 2methyl2butanol and concentrated HCl. 1 bromobutane is prepared 1butanol and sodium bromine in sulfuric acid . Both product are isolated by simple distillation. The substitution of the products is verified by performing the NaI and AgNO 3 classification tests and comparing them with each other and a secondary alkyl halide to note the differences between substitution patterns of alkyl halides. SN1 Reagent: 3>2>1 Carbocation formation stabilization intermediate Rate: K(substrate) Product: Racemization (enantiomers) Stepwise mechanism: Loss of the leaving group and then nucleophilic attack Solvent: HCl acidic Polar proticlower transition state energy by solvating the charged species. Polar solvents lower the transition state energy by solvating the charged species.The increased number of attractive, dipoledipole intermolecular interactions stabilize the transition state,which lowers its energy. If the transition state energy is lowered then so isΔH‡ for the ratedetermining step,which increases the rate of that step(and therefore the overall reaction). SN2 Reagent: 1>2>3 Carbocation not stable , high in energy Rate: k(substrate)(nucleophile) carbocation (which is dependent on the concentration of the alcohol) Concerted Mechanism: Simultaneously Leaving group and Nucleophile attack Product: Inversion Solvent: H2SO4 and NaBr Polar aprotic Evidence involves backside attack by nucleophile Optically active (chiral)(4 different groups attach to carbon) alcohol undergo inversion of stereochemistry in the product (Sn2) R and S configuration (optically rotations) Rclockwise, Scounterclockwise Robin Cordero Hydrogen needs to always be in the back Atomic number Procedure Synthesis of 1bromobutane (SN2) 1 . Measure out 11.1 g NaBr and transfer to a round flask. Add 10 mL of water, 1butanol and boiling stone 2. Mix the contents of the flask by swirling gently , and then place the flask in an icewater bath. 3. Slowly add 10 mL of concentrated sulfuric acid to the chilled reaction mixture while continuing to swirl. 4. Assemble the reflux apparatus. Warm the flask gradually until heating under the reflux , and continue heating for 45 minutes Reflux: the state of a boiling liquid whose vapor are continually returned by condensation; typically used to hold reactions at the boiling point of the solvent for extended periods of time without evaporation. 5. After the 45, remove the water hoses from the Hempel column and drain the water from the outside of the column, Reassemble the apparatus for simple distillation 6. Distill the product mixture rapidly and keep everything you collect until 115. Do not collect the material above 115. Keep your receiving flask in an icewater bath at all times. 7. Transfer the distillate to a separatory funnel and wash with water. Note the 1bromobutane (organic) is the denser, bottom layer. 8. Separate the two layer and return the organic layer to the separatory funnel. Wash the organic layer With H2SO4 Two x 5 mL of 2M NaOH Water Saturated sodium chloride remove residual water from the organic layer, The salt works to pull excess water from the organic layer to the aqueous in order to dilute the salt concentration( and also salts are more soluble in water than organic solvents Note: concentrated sulfuric is denser than 1bromobutane , so the organic layer will be the top layer for this washing step. Sodium hydroxide , water, and brine are less than 1bromobutane , and so the organic layer will be in the bottom layer for these washing steps. 9. Transfer the cloudy 1bromobutane to a clean flask and dry using sodium sulfate (add until no longer clumping, this sucks all the water out) and becomes clear. 10. Transfer the dried 1bromobutane to a preweight and weight final product. 10. Perform the silver nitrate and sodium test from your test in the test tubes. Robin Cordero If your product is impure you may experience a precipitate in the silver nitrate. However it is not a positive test unless the precipitate (AgBr) is yellow. NaI and AgNO3 helps distinguish 1,2,3 alkyl halides. AgNO3 (silver nitrate) Positive test: AgCl will precipitate in EtOH (solvent) Ag+ is strong enough lewis acid and can remove Cl to form a carbocation Mechanism is Sn1 like 3>2>1 Precipitate: AgCl NaI (sodium Iodide) Positive test: NaCl will precipitate in acetone (solvent) Na+ is not a strong enough lewis acid and cannot remove Cl to form carbocation Instead mechanism is concerted and Sn2 like Therefore, the reactivity of alkyl chlorides is 1>2>3 Precipitate: NaCl or NaBr Synthesis of 2chloro2methylbutane (Sn1) Place 10 mL of 2methyl2butanol and 25 mL of concentrated (12 M) hydrochloric acid in the separatory funnel. Swirl the contents of the separatory funnel gently without the stopper on the funnel to mix the reactants. After swirling the funnel for about 1 min, stopper and then carefully invert it. Release the excess pressure by opening the stopcock with the funnel inverted. Do not shake the funnel until the pressure has been equalized. Now close the stopcock and shake the funnel for several minutes, with intermittent venting. Allow the mixture to separate into two distinct layers.Separate the layers and determine which is the organic layer. Wash the organic layer sequentially with 10mL portions of saturated aqueous sodium chloride and cold saturated aqueous sodium bicarbonate. Sodium bicarbonate neutralizes excess HCl from the reaction , despite CO2 built up, since carbonate is not a strong enough base to promote the formation side product (alkene), and therefore only one organic product is produced. if (NaOH) Advantage water and salt are the by products so no built up of gas. The disadvantage of sodium hydroxide is that hydroxide is a strong enough base to promote the competing elimination reaction with the product of the initial reaction On initial addition of the bicarbonate solution, vigorous gas evolution will normally occur; gently swirl the unstoppered separatory funnel until this stops. Stopper the funnel and carefully invert it; vent the funnel immediately to release gas pressure. Shake the funnel gently with frequent venting and then shake it vigorously with frequent venting. Separate the organic layer, and again wash it sequentially with 10mL portions of water and saturated aqueous sodium chloride. Carefully remove the aqueous layer, transfer the 2chloro2methylbutane to an Erlenmeyer flask, and dry the product over several spatulatips full of anhydrous sodium sulfa Swirl the flask occasionally for 10–15min until the product is dry; add further small portions of anhydrous sodium sulfate if the liquid appears cloudy.Using a Pasteur or a filtertip pipet, carefully transfer the crude product to the roundbottom flask, add a stir bar, and equip the flask for short path distillation. Distill the product, collecting it in a tared receiver Robin Cordero cooled in an icewater bath. Because of the relatively small quantity of product, it may be difficult to obtain an accurate boiling point, so you should collect the fraction having a boiling point greater than about 75 °C (760 torr). Lab 8: Preparation of Alkenes by E1 and E2 elimination reaction. Baeyer and Bromine tests for unsaturation Purpose: 2methyl1butene and 2methyl2butene are prepared from 2methyl2methyl butanol and H2SO4 by E1 mechanism and from , and from 2chloro2methylbutane and KOH by an E2 mechanism. Alkene formation is verified by performing the Baeyer and bromine test for unsaturation. The distribution of products is determined by GC analysis. E1Sn1, Strong base KOH heat Robin Cordero Most substituted alkene, no stereochemistry needed 3>2>1 Polar aprotic Rate: Alkyl halide E2Sn2, strong base no heat, leaving group must be anticoplanar Zaitsev: substituted alkene, small base Steric factor Stereoelectronic Hoffman: bulky base 1>2>3 Polar protic Rate: alkyl halide (base) Procedure E2 reaction: Synthesis of 2methyl1butene and 2methyl2butene 1. Measure out KOH and 1propanol and transfer the reagents to an appropriately sized flask. Add one boiling stone. 2. Warm the mixture gently until the potassium fully dissolve. Cover the flask with a watch glass to prevent evaporation of 1propanol. After dissolution of the basa cool the flask to room temperature , first by removing the flask from the heat mantle, then in an ice bath. 3. Slowly add 2chloro2methylbutane (product from lab7) to the cooled reaction mixture. 4. Assemble the reflux apparatus. Grease the joints before assembly to avoid freezing from the caustic KOH solution. The Hempel column should be filled with raschig rings to prepare for the fractional distillation later 5. Turn the water. Heat the reaction under the reflux for 1 hour 6. Allow the reaction to cool below its boiling point with the reflux condenser and water on. When cooled to room temperature , remove the water hoses from the hempel column and drain the water from the outside of the column. Reassemble the apparatus for fractional distillation 7. Distill the product mixture , collecting all distillate that boils below 45. Keep your receiving flask in an icewater bath at all time 8. Weight final product. Then perform the bayers and bromine test. The GC analysis E1 Reaction: Synthesis of 2methyl1butene and 2methyl2butene 1. Measure out 2 ml of 2methyl2butanol and transfer to a 25 ml flask 2. Slowly add 10 ml of H2S04 to the reaction 3. Assemble the fractional distillation apparatus. The hempel should be filled with raschigs rings 4. Heat the mixture and collect all the distillate that boils below 45.keep receiving flask in an ice water bath at all times Robin Cordero 5. Weigh final product. Then perform the baeyer and bromine. Then GC analysis (checks the purity of your product) Baeyer Test Procedure 1. Prepare 3 test tubes by adding 95% ethanol to each 2. To one add 12 drops of your product mixture of E1 reaction; to another 12 drops of your product of E2; to the third add nothing 3. Add a solution of KMnO4 to the 3 test tubes: A positive test is indicated when the color solution turns brown instead of purple Baeyer’s test is an oxidation reaction between an alkene and potassium permanganate. The reaction is a syn addition that goes through a cyclic manganate ester intermediate, which leads to a diol formation, as seen in figure 1. Since the permanganate is purple, the diol is colorless, and side product, manganese dioxide, are light brown, disappearance of the purple and appearance of brown constitute a positive test for unsaturation. The problem with the Baeyer test is that permanganate also oxidized other functional groups such as aldehyde, so this test is not selective for carboncarbon multiple bonds. Bromine Test 1. Prepare 3 test tubes by adding 95% ethanol to each 2. To one add 12 drops of your product mixture of E1 reaction; to another 12 drops of your product of E2; to the third add nothing 3. Add a solution of Br2 in dichloromethane to the 3 test tubes dropwise until the light orange just persists. Positive tests are indicated by a rapid disappearance of the orange color to yield a colorless solution. The bromine test is an addition reaction of molecular Br2 across a double or triple bond. The reaction is an antiaddition that goes through a cyclic bromonium ion intermediate, followed by attack of the ion by bromide to form a vicinal dihalide, as shown in figure 2. Because the bromine is orangered in color and the vicinal dihalide is colorless, disappearance of the orange color constitutes a positive test for unsaturation Dehydration of alcohols (E1) Rate determining step is the formation of the alkyloxonium Acid catalyst (H+) Dehydrohalogenation of alkyl halides (E2) Strong Base (OH) Factor that favor elimination Strong bases Robin Cordero Steric hindrance High Temperature Lab 9: Preparation of Alcohol: reduction of fluorenone and lucas test for alcohol Purpose: 9fluorenol (2) is prepared by reducing 9fluorenone (1) using sodium borohydride. The reduction is specific for the carbonoxygen double bond. The product us purified by recrystallization and characterized by melting point analysis and IR spectroscopy. Alcohol formation is verified by performing the lucas test. The test fro 9fluorenol (secondary alcohol) will be compared to known alcohols to note the difference between primary, secondary, and tertiary alcohols. Reduction: Is the gain of electrons (): become smaller Gain of electrons Lose bonds to oxygen Gain bonds to hydrogen Decrease oxidation # of C Oxidation: Loss of electrons () : becomes bigger Lose of electrons gains bonds to oxygen lose bonds to hydrogen Increase oxidation # of C NaBH4 acts as a source of nucleophilic hydride ion (H), which is delivered to the electrophilic carbon of the carbonyl group on the ketone. Sodium borohydride contains 4 equivalent of hydride. Robin Cordero Solvent (in this case, water) acts as a proton (H+) source to neutralize the alkoxide intermediate. Acid will neutralize the base (in this case, sodium hydroxide) that is produced over the course of the reaction. Homo and Lumo: The smaller the Homolumo energy gap, the longer the wavelength of light and vice versa. The more conjugated a molecule is the smaller the Homolumo gap, and the more likely it is that the molecule is colored. Amount electron density around the carbon atom changes, since carbon makes covalent bond and therefore does not formally gain or lose electrons. When reaction occurs in which a carbon bound to a more electronegative atom is replaced with a carbon bound to a less electronegative atom, the electron density around carbon increases, and the reaction is reduction. Carbon 1. For every atom bound to C that is more electronegative than C, add +1 2. For every atom bound to C that is less electronegative than C, subtract 1 3. For every time C is bound to another C, add 0 Reducing Agents Metal hydrides: Sodium Borohydride NaBh4 ( imines, aldehyde, and ketone) , reduce pi bonds Lithium Hydride: imines, aldehydes, ketone, carboxylic acid, esters, and amides Molecular hydrogen (H2) with a metal catalyst , (ketoneto alcohol, iminesamines, nitro group amines) Thin layer chromatography Monitor the reaction progress (Rf values) Rf values and polarity (carbonyl and alcohol) The Lucas Test The Lucas test for alcohols is a classification test that helps distinguish between primary, secondary, and tertiary alcohols. The mechanism proceeds by using a mixture of ZnCl and 2 l Robin Cordero that increases the reactivity of the alcohol toward the acid, because ZnCl 2is a strong Lewis acid that reacts with the lone pairs electrons of the oxygen with the alcohol. The ZnClOH complex dissociates and yield a carbocation formation, which quickly reacts with the halide to form an alkyl halide product. The rate determining step of the reaction is carbocation formation, and therefore will increase with carbocation stability. For this reason, tertiary alcohol reacts faster than secondary alcohols, and primary alcohols being unreactive at all with the Lucas test. Since the alkyl halide is not soluble in aqueous solution, whereas the starting alcohol is, a change forms a clear to cloudy solution constitutes a positive test. It is important to note the change in turbidity and the rate which the solution becomes cloudy. Tertiary alcohol reacts instantly, whereas secondary made take log 520s to react. One of the drawbacks of the Lucas test is that the starting alcohol needs to be soluble in the Lucas reagent, so this test has limitations to the type of alcohol that may be tested. In addition, smaller alcohol (6 or less) tend to be the most successful with the Lucas test. Sn1 mechanism RDS is carbocation formation 3>2>1 Procedure: Reduction reaction 1. Add your 9fluorenone to flask containing methanol (EtOH). Swirl and heat the flask gently until all the 9fluorenone dissolves. 2. Cool the reaction to room temperature. Weigh the sodium borohydride and add it immediately to the reaction flask in one portion. Swirl the flask vigorously to dissolve the reagent. Do not stop the reaction with NaBH4 in methanol Hydrogen gas is evolving and it can be dangerous if pressure build up in a stoppered flask! 3. Allow the reaction to stand at room temperature for 20 min, with occasional swirling. Your reaction should become colorless over their period. If , after the first 10 minutes, the reaction is not colorless add a small portion of sodium borohydride no more than half the amount you initially added. Make sure you swirl the reaction mixture again at this point. 4. Before you work up your reaction, take a TLC on a silica plate using 1:9 ethyl acetate:Hexanes as your eluant to confirm that your reaction is complete. Your TLC plate should contain 3 lanes: 1) pure fluorenone; 2) pure 9fluorenone + reaction mixture (cospot); 3) reaction mixture. Examine the TLC plate under UV light and circle the spots you observe, Calculate the Rf values 9fluorenone and 9fluorenol. 5. Workup the reaction by adding the calculated amount of 3M sulfuric acid to the reaction mixture. Heat the flask gently and swirl for 57 min to dissolve the solid that is formed upon acid addition. Use a watch glass to minimize solvent loss. If all the solid won’t dissolve after this time add more methanol, in 0.51 ml portions, with heating until all the solid dissolves. Robin Cordero 6. Remove the flask from heat and cool to room temperature and then an ice bath until the solid precipitates, 510 minutes. 7. Filter the solid and wash thoroughly with water, at least 200 ml. Test the water coming out the buchner funnel after washing using pH paper to ensure neutral. Dry the product attached to vacuum. 8. Recrystallize the final product using the minimum amount of hot methanol possible. Heat 1020 mL of methanol and add the hot solvent to your product until it just dissolves. Keep the solution of 9fluorenol and methanol on your heat mantle, covered with a watch glass, during the dissolution. Once all the 9fluorenol is dissolved, remove the solution from the heat and let it cool first to room temperature and then inan ice bath. Filter the final product and dry attached to vacuum. 9. Weight final product and calculate percent yield. Make sure you do this before any test or characterization. 10. Characterize the product by melting point analysis and IR spectroscopy. Part 11: Confirmation of Alcohol formation by Lucas classification test Lucas test procedure 1. Prepare 3 test tubes by adding 1 ml of the lucas reagent to each 2. To one test tube add 5 drops of the tertiary alcohol. A positive test is indicated by the solution becoming cloudy .Note how quickly (if at all) this occurs. If the solution does not become cloudy immediately, shake the test tube for up to 5 min. If the solution does not become cloudy in that time the result is “no reaction”(negative test) 3. Repeat the procedure for the secondary (your product) and primary alcohol. Note that your product is a solid , so you will need to prepare a solution of 9fluorenol in 0.3 ml of ethanol before performing the test (add small spatula full of 9fluorenol to the ethanol and make sure it fully dissolves) 4. Compare the result of the test for primary, secondary, and tertiary alcohols, What do the results tell you about the mechanism of the test? Lab 10: Williamson Ether Synthesis: preparation of phenacetin from acetaminophen. Robin Cordero Product: Phenacetin (2) is prepared by the williamson ether synthesis using acetaminophen (1, active ingredient of Tylenol) and Iodoethane in the presence of a base. The phenolic hydrogen is sufficiently acidic to be deprotonated by potassium carbonate to allow the reaction to occur. The product is purified by recrystallization and characterized by TLC , melting point analysis, and IR spectroscopy. William Ether synthesis Ether are found in natural products such as melatonin (antioxidant) Fluoxetine, antidepressant, tamoxifen (cancer drug), analgesics The reaction proceed via two steps. The first step is deprotonation of the alcohol by a suitable base to form an alkoxide ion. The second step is an Sn2 substitution reaction where the alkoxide acts as the nucleophile and the alkyl halide acts as the electrophile. Limitations; Acidbase reaction: Alcohol are no highly acidic species, with Pka values of 1618. For this reason it is important to choose a base that is strong enough to facilitate the deprotonation; otherwise an alcohol is not sufficiently nucleophilic to proceed with step 2 of the reaction. Phenols are better acids than aliphatic alcohols, with pka value of 10 and therefore weaker bases may be used for their deprotonation. The mechanism is an Sn2 substitution reaction, and therefore any criteria or limitation that applies to Sn2 reaction, also applier to the williamson ether synthesis. Most important is the choice of alkyl halide. Sn2 reactions occur best with sterically hindered alkyl halides, because the nucleophile has to attack the electrophile carbon at the same time as the carbonhalide bond is broken. Therefore , methyl and primary alkyl halide react more quickly than secondary alkyl halides, and tertiary alkyl halides do not react via Sn2 mechanism. In addition , there is always competition between substitution and elimination. Because alkoxides are good bases they can perform elimination reactions to convert the alkyl halide to an alkene. The more sterically hindered the alkyl halide the more likely is for elimination to be favored over substitution. Robin Cordero Therefore , the williamson ether synthesis is often inefficient when secondary alkyl halides are used and tertiary alkyl halides do not yield ether products. Other factors that increase the rate of Sn2 reactions should also be considered when performing the williamson ether synthesis in order to optimize the reaction. The reaction rate is dependent on 1) carbonoxygen bond formation , which is dependent on the strength of the nucleophile , and 2) carbonhalogen bond breakage, which is dependent on the leaving group.The nucleophile will always be an alkoxide but their is a choice of electrophile. Recall from substitution reactions that weaker base make better leaving groups , and so alkyl iodides react faster than alkyl bromides, which in turn react faster than alkyl chlorides, all else being equal. Also recall from substitution reaction that solvent choice can affect the rate of the reaction. Sn2 reactions proceed more rapidly when polar aprotic solvents are used because protic solvents, hydrogen bond and solvate the nucleophile , thus hindering its approach to the nucleophile. Therefore, solvents such as dimethyl sulfoxide, dimethyl formamide, acetonitrile, and acetone are common for the williamson ether synthesis Acetaminophen and Phenacetin as Analgetics Mechanism of reaction of acetaminophen: two pathways Inhibitor of the cyclooxygenase family of enzymes, produce prostanoids (are involved in the inflammatory response in our bodies. Metabolized into Nacetylpbenzoquinone imine which acts on cation channel receptors in the spinal cord to suppress signal transduction , ultimately leading to pain relief. Phenacetin: Older analgesics (pain relief and reduce fever) Phenacetin is metabolized into acetaminophen (active painkiller) by our bodies Found to cause cancer and kidney (acetaminophen is no carcinogenic) Williamson Ether synthesis procedure 1. Crush 1.3 grams of acetaminophen using a mortar and place it in a flask 2. Add 2.5g of K2CO3 and 15 mL of 2butanone and boiling stone. (deprotonation and Sn2) NaOH is acceptable is an appropriate substitution Polar aprotic solvents only (not alcohol group), propanol will not be suitable because it can hydrogen bond and solvate the nucleophile , which hinders its approach on the electrophile. 3. In the fume hood, add 1 ml of iodoethane to the reaction mixture, and then assemble the apparatus for reflux. 4. Reflux for an hour. Robin Cordero 5. Cool the reaction mixture to below its boiling point and vacuum filter the solids. Wash the solids 2x with 5 mL of ethyl acetate Since ether is organic and immiscible in water it is an appropriate substitution. It also maximize the amount of organic product in the filter. 6. Take a TLC of the reaction mixture before workup. Your TLC plate should contain 3 lanes. 1) pure acetaminophen,; 2) pure acetaminophen + reaction mixture (cospot); 3) reaction mixture. Elute the TLC plate with of 4:1 ethyl acetate: methylene chloride. Examine the TLC plate under UV light and circle the spots you observe. Save your TLC plate until the purification of phenacetin is complete. 7. Transfer the filtrate to a separatory funnel and extract the solution with 20 mL of 5% NaOH and then 20 mL of water NaOH is used to deprotonate any unreacted acetaminophen , thus making it ionic and more soluble in the aqueous layer. If water is used then the acetaminophen will remain in the organic phase and will not be removed for the product phenacetin. Acetaminophen has a phenol , which when deprotonated by NaOH the phenol oxygen becomes negatively charged, and therefore the ionized acetaminophen is more soluble in water than it is in organic solvent. Phenacetin , on the other hand, has no functional group that can be deprotonated by sodium hydroxide, and therefore it will remain in the organic layer during the extraction. 8. Transfer the cloudy organic layer to clean flask and dry using sodium sulfate (Na2SO4 until the solid no longer clumps). Swirl the flask as add the Na2SO4the phenacetin solution should become clear 9. Decant and dried phenacetin solution to a flask and remove the solvent using rotatory evaporator. Removes solvents from compounds that are solids at room temperature (ethyl acetate and 2butanone from product) Solution is under vacuumlowering the pressure above the liquid decrease its boiling pointevaporate faster Hot water bath heats the liquid , bringing it closer to the boiling point Advantages: centrifugal and frictional forces created by rotating the flask cause the solution to spread out over the flaskincrease surface area for evaporation. Rotation helps prevent “bumping” (to help this, fill your evaporation flask more than half way) 10. Recrystallize the resulting solid using hot ethanol. Remember that for recrystallization you want to use the minimum amount of hot ethanol possible to just dissolve the solid, and that you should keep both the ethanol and the beaker with your solid + ethanol hot at all times during the dissolution process. Robin Cordero 11. Once all the phenacetin is dissolved , remove the solution from heat and let it cool first to room temperature and then in an ice bath. Scratch to induce crystallization if none occurs in the ice bath. Filter the final product and dry attached to vacuum. 12. Weight your final product and calculate percent yield. Make sure you do this before performing any tests or characterization. 13. Characterize your product by TLC, melting point analysis, and IR spectroscopy. For the TLC , prepare a tiny amount of solution of your product in ethyl acetate and use the same solvent system as you did for the reaction mixture. Calculate the Rf values of all spots all TLC plates. Alcohol are not sufficiently nucleophilic for the williamson ether synthesis to occur, so if cyclohexanol is not deprotonated then no appreciable amount of ether will be formed. Williamson ether NaOH, NaH (deprotonate the alcohol) ACN, acetone, DMF, DMSO (solvent, polar aprotic) 1. Deprotonation 2. Sn2 reaction Lab 11: Synthesis and Chemiluminescence of luminol The peroxide decomposes to form the dianion of 3diaminophthalic acid in an electronically excited triplet state 16T, a species having two unpaired electrons with the same spin quantum number, i.e., 1/2 or 1/2. As seen in Figure 20.14, 16T undergoes intersystem crossing (ISC) to an excited singlet state 16S*, which has two unpaired electrons but with opposite spin quantum Robin Cordero numbers. Relaxation (decay) to the singlet ground electronic state 16S of the dianion then results in fluorescence in the form of bluish green light. Preparation of Luminol Setting Up:Combine 1 g of 3nitrophthalic acid and 2 mL of an 8% aqueous solution of hydrazine in the test tube and carefully heat the mixture until the solid dissolves. Add 3 mL of triethylene glycol and a boiling stone and clamp the vial in a vertical position. Insert a thermometer in The carbonyl compound of the carboxylic acid acts as the electrophile, and the nitrogen of the amine acts as the nucleophile. This reaction will not occur under normal conditions due to the weak electrophile/weak nucleophile pairing. to the solution, securing the thermometer with a clamp. Approximately 230 degrees Celsius is required to overcome the activation energy of this process. The high activation energy is required because of the low pka of the carboxylic acid hydrogens in the presence of basic nitrogens. This heat prevents an acid base reaction from occurring, and increases the rate of the proton transfers in the reaction. This heat also overcomes the weakness of the electrophile and drives off water, causing the reaction to progress forward. A very high boiling point solvent must be used to solve this issue, in this case ethylene glycol, which has a boiling point of about 260 degrees Celsius. Ethylene glycol is also an appropriate solvent because of its extremely polar nature, and allows the polar reactants to dissolve in it. Also, carboxylic acid functional groups dissolve best in very polar solvents. Reaction: Bring the solution to a vigorous boil to remove excess water. During this time, the temperature should be around 110–120 °C. After the water has evaporated, the temperature should rise to 215 °C in a 3 to 4min period. Maintain the temperature at 215–220 °C for 2 min and then remove the test tube from the heat source and allow the solution to cool to about 100 °C. While the test tube and its contents are cooling, bring about 15mL of water to boiling in the Erlenmeyer flask. Slowly add the hot water to the test tube, stir the contents with a glass stirring rod, cool the mixture to room temperature, and collect the solid nitrohydrazine by vacuum filtration.Return the damp solid to the uncleaned test tube, add 5 mL of 3 M aqueous NaOH, and mix the contents until the solid is dissolved. Add 3 g of fresh sodium dithionite to the solution, rinsing any solid adhering to the walls of the test tube into the solution with a few drops of water, and heat the resulting mixture to just below boiling for 5 min, taking care to avoid bumping. Robin Cordero Reduces nitrophthalhydrazide to an amine group. When dissolve in basic solution luminol become dianionic Workup and Isolation Add 2 mL of glacial acetic acid to the reaction mixture, cool the test tube in a beaker of cold water, and collect the crude luminol by vacuum filtration. A second crop of product may separate from the filtrate upon standing, but do not combine it with the first crop, which is to be used in the chemiluminescence experiment. It is used to neutralize the dianion to form luminol To decompose any diothinite Chemiluscences Setting Up: Prepare stock solution A by placing the crude luminol (40–60 mg dry weight, but which need not be dry) in a 50mL Erlenmeyer flask and dissolving it in 2 mL of 3 M aqueous NaOH and 18 mL of water. Prepare solution B in the second 50mL Erlenmeyer flask by combining 4 mL of 3% aqueous potassium ferri cyanide, 4 mL of 3% aqueous hydrogen peroxide, and 32 mL of water. Combine 5 mL of solution A with 35 mL of water in the 125mL Erlenmeyer flask. Reaction: Working in a lowlight environment, simultaneously pour the diluted por tion of solution A and all of solution B into the 250mL Erlenmeyer flask. Swirl the flask and record what you observe. Test what happens if you add additional small portions of the aqueous NaOH solution and crystals of potassium ferricyanide to the reaction mixture. In a luminol reaction, the excited electron jumps to a triplet state, and flips the spin of the electron for it to return back to the singlet state, then the ground state. This example of fluorescence gives off a blue green glow when done correctly. Luminol is used by detectives to detect the presence of blood. The iron in the blood serves as the oxidizing agent that allows the chemiluminescent reaction to occur. Because fresh supplies of blood are not available in lab, potassium ferricyanide is used as the oxidizing agent because it has very similar iron components to those in blood. If the reaction with luminol and blood were to glow, it is a sign that blood is present. All of the presumptive blood test examined here are useful to the modern forensic investigator, However, the KastleMeyer and Leucomalachite green test use carcinogenic reagents, while polilight exhibitd poor sensitivity, making their use undesirable. The luminol test is by far the most sensitive technique, and is easy to perform, particularly when application over a large Robin Cordero area is required. When the luminol technique is not appropriate , the Hemastix test is a simple and effective alternative.
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