Introduction: Objectives of the experiment.
The main objective of the experiment was to prepare N-(4-butoxyphenyl)acetamide by reacting paracetamol dissolved in sodium hydroxide with 1-bromobutane in presence of ethanol solution, followed by heating the reactants in order to increase the rate of reaction and form the product. In addition to that, the experimental yield obtained, another objective was to calculate the experimental yield that should have been obtained from the experiment when ideal, with no errors and hence, calculate the percentage yield. The calculation of the percentage yield is important for chemists as it enables them to find out whether the experimental procedure could be applied at industrial level in order to form the same product. This experiment also aimed at familiarizing the students with knowledge on how to carry out thin layer chromatography, in order to test for completion of a reaction. Moreover, the students were also to learn how to obtain an IR spectrum of the paracetamol and the product formed and use the information on data analysis. Lastly, the students were also to learn about SN1 AND SN2 reactions through the reactions of the haloalkanes with: silver nitrate in ethanol and with sodium iodide in acetone.
SN2 reactions are reactions that occur in one step with the nucleophile and the substrate being involved in the rate determining step.
The SN2 reaction only occurs if an empty orbital is ac since it is accessible since it is a backside attack reaction. Therefore, the SN2 reactions are affected by steric hindrance. Steric hindrance is the case whereby the leaving group or the attacking group is surrounded by bulky substrates that end up lowering the rate of the reaction. Therefore, for SN2 reactions, the primary haloalkanes are the most reactive followed by the secondary haloalkanes and finally the least reactive are tertiary haloalkanes. SN1 reactions are however reactions that occur in one step involving the loss of a leaving group that leads to the formation of a carbocation. This reaction is also called unimolecular because it does not involve the formation of an intermediary like the SN2 reaction.
SN1 reactions depend on the stability of the carbocations since, the first step as stated earlier involves the loss of a leaving group to form carbocations and hence, the rate of the reaction will be proportional to the stability of the carbocation. Carbocation stability increases with the increase in substitution of the carbon and hence, the rate of this reaction proceeds from tertiary haloalkanes being the fastest, to secondary haloalkanes and the least reactive are the primary haloalkanes. Resonance rate for the reaction will proceed in the same way. It is however, important to know that both SN1 and SN2 reactions do not occur for sp2 hybridized compounds. They only occur for alkyhalides and their related compounds such as tosylates and mesylates (James, 2012).
The Williamson ether synthesis experiment is an SN2 reaction between a deprotonated alcohol and an alkyl halide to form an ether. The reaction involves the breaking of a carbon bond (James, 2014). The synthesis in the Williamson ether experiment occurs at rates that depend on the number of atoms in the transition state. These rates are affected by the probability of the alkoxide approaching the carbon atom with the halide ion, as well as the resulting ring compound (Robert J. Ouellette and J. David Rawn, 2015).
In the reaction, an alkoxide is used in addition to the alcohol in the reaction because the conjugate base of the alkoxide is a better nucleophile than that of the alcohol, since, the alkoxide`s nucleophile has a higher electron density. This reaction only proceeds with alkylhalides because it is an SN2 reaction. The solvent used in the reaction is the conjugate acid of the alkoxide, in order to prevent the formation of a mixture of ether products, which would not be ideal (James, 2014). The reaction cannot be used with tertiary alkylhalides. The reaction is important as it has helped scientists to prove the structures of many ethers. In the experiment N-(4-butoxyphenyl)acetamide is prepared by reacting paracetamol dissolved in sodium hydroxide solution with 1-bromobutane in ethanol solution. The reaction is an SN2 reaction, which occurs in one step since it is bimolecular.
Part A: Preparation of the N-(4-butoxyphenyl)acetamide.
a) Preparation of N-(4-butoxyphenyl)acetamide
3.75 grams of paracetamol were put in a 100 ml round bottomed flask. Thereafter, 20 ml of ethanol was added to the flask and the mixture was swirled until it had completely dissolved. Sodium hydroxide was then prepared by taking 1.75 grams of solid sodium hydroxide and dissolving it in 2ml of distilled water. The resulting viscous solution of sodium hydroxide was thereafter added to the round bottomed flask that contained the paracetamol mixture. Afterwards, the contents were heated to allow for solubility. The reaction flask was then transferred into a fume hood where 1-bromobutane was added drop wise while stirring the contents. The reaction was thereafter left for reflux for one hour.
b) Checking of the reaction progress by thin layer chromatography (TLC)
The sample of the reaction was prepared by adding three drops of the reaction mixture into a sample dial, followed by dilution using three drops of ethyl acetate. Thereafter, a light line 1cm from the bottom edge of the chromatography paper was drawn. The sample of the reaction mixture, paracetamol and the sample product were then spotted on the chromatography paper and were then checked under ultra violet light to check if the stain was visible. A thin layer chromatography plate was then developed with ethyl acetate and hexane mixture in the ratio of 90:10.
Part B: SN1 and SN2 reactions of halohydrocarbons.
a) Reaction of with silver nitrate in ethanol solution.
2ml of 2% solution of silver nitrate were put in a clean test tube. Thereafter, two drops of 1-chlorobutane were added to the solution and shaken. The time taken for a white precipitate of silver chloride to form was then recorded. The reaction was thereafter redone while using 2-chlorobutane, 2-chloro-2-methylpropane, chloro benzene, benzyl chloride and 3-chloropropene. The time taken for each of the compounds to form the silver chloride precipitate was also recorded.
b) Reaction with sodium iodide in acetone solution.
2ml of 15% sodium iodide in anhydrous acetone was put in a clean test tube. Afterwards, two drops of 1-bromobutane were added to the solution in the test tube. The contents were mixed and the time taken for the formation of sodium bromide precipitate was recorded. The reaction was thereafter redone while using 2-bromobutane, 2-bromo-2-methylpropane and bromo benzene. The time taken for each of the compounds to for the sodium bromide precipitate was also recorded.
(The results of the experiment will be included in this section including the graphs if any, the chromatogram and the IR results from the experiment. Moreover, the calculations that are needed in the experiment`s analysis like the experimental yield and the percentage yield are also to be included in the results section. Any findings from the laboratory is included in this section of the report.)
Discussion of the results
(The discussion of the findings of the results are discussed in this section. The discussion should be depend on the results found from the experiment. Probable errors as a result of human errors while taking measurements or while carrying out the IR, that were likely to be present in the experiment and that may have caused the experimental results to deviate from the theoretical findings are also to be mentioned in this section. This analysis includes a comprehensive analysis of the results.)
1. Show that the SN2reaction mechanism is involved in the formation of N-(4-butoxyphenyl)acetamide.
2. Write mechanisms to show how an alkene and dibutyl ether can form as byproducts of the reaction.
During the Williamson ether synthesis, side reactions between ethanol and 1-bromobutane occur. During these side reactions, the alkene, ethane and dibutyl ether are formed.
2Br – + 2CH3CH2OH 2HBr + 2CH3CH2O
2CH3CH2O 2CH2CH2 + 2OH-
2OH- + 2CHCH2CH2CH2+ CH3CH2CH2CH2OCH2CH2CH2CH3 + H2O
CH3CH2OH + 2 CH3CH2CH2CH2Br 2HBr + CH3CH2CH2CH2OCH2CH2CH2CH3 + H2O
3. Explain why the sodium hydroxide is added to the paracetamol before the electrophile is added.
The sodium hydroxide is added to the paracetamol to deprotonate the paracetamol in order for the nucleophile formed to be able to attack the electrophile in an SN2 reaction. The reaction would not be possible without deprotonating the alcohol (University of Technology Sydney, 2018).
4. Between paracetamol and 4-acetamidophenolate which one is a better nucleophile and why?
Paracetamol is a better nucleophile as compared to the 4-acetomidophenolate. This is because the Williamson ether synthesis experiment is a SN2 reaction which is affected by steric hindrance of the molecule. In paracetamol, the OH group is not surrounded by bulky groups whereas in the 4-acetomidophenolate, the OH group is surrounded by bulky groups that cause steric hindrance, thus lowering the rate of reactivity (University of Technology Sydney, 2018).
5. Explain why 1-bromobutane is used instead of 2-bromo-2-methylpropane.
1-bromobutane is used as the electrophile because being a primary alkyl halide, it proceeds with a faster rate for SN2 reactions like the Williamson ether synthesis experiment. This is because primary alkyl halides have less steric hindrance to the leaving group. For 2-bromo-2-methylpropane there is steric hindrance as the leaving group is surrounded by bulky methyl groups as the molecule is a tertiary alkyl halide. This reduces the rate of the reaction and hence, the 1-bromobutane is used as the preferred electrophile (University of Technology Sydney, 2018).
6. This question requires an IR spectrum for the pure paracetamol that was carried out in the laboratory.
7. The IR spectrum of the paracetamol is compared with the IR spectrum of the product formed from the Williamson ether synthesis experiment.
8. What precipitate is formed when the alkyl halides react with
a) Silver Nitrate in Ethanol
Silver chloride precipitate that is white in color is formed when the haloalkanes are reacted with the silver nitrate in ethanol solution.
b) Sodium iodide in acetone
Sodium bromide precipitate is formed when the haloalkanes react with the sodium iodide in acetone solution.
9. For the experiment, give the order in which the precipitate (silver chloride) was formed from the haloalkane that took the least time to the one that took the longest time to form the precipitate and explain if they follow the SN1 mechanism.
The results that were obtained were fit in the SN1 reaction mechanism, which involves the loss of a leaving group first in order to form a carbocation. The carbocations are easily formed depending on their stability. Tertiary haloalkanes form the most stable carbocations and hence show the precipitate in a very short period of time. They are followed by secondary haloalkanes and lastly, the least reactive are the primary haloalkanes that form the least stable carbocations and hence, take a very long time before the formation of a silver chloride precipitate (University of Technology Sydney, 2018)..
10. For the experiment, give the order in which the precipitate (silver iodide) was formed from the haloalkane that took the least time to form the precipitate to the one that took the longest time to form the precipitate, and explain whether they follow the SN2 reaction mechanism (University of Technology Sydney, 2018)..
The results obtained were fit for the SN2 reaction mechanism which involves a single bimolecular reaction. The reactivity decreases with an increase in steric hindrance which increases from primary haloalkanes to secondary haloalkanes to tertiary haloalkanes. Therefore, the reaction occurs very fast in the primary haloalkanes that form the precipitate within a short period of time, to secondary haloalkanes and finally to the tertiary haloalkanes which take the longest time to form the precipitate (University of Technology Sydney, 2018).
11. Explain the difference in reactivity between the primary haloalkanes: 1-chlorobutane and allyl chloride.
1-chlorobutane forms a white precipitate whereas allyl chloride`s` solution retains the colorless color since, white precipitate is not formed. This is because 1-chlorobutane undergoes the SN2 reaction that leads to the formation of a silver chloride precipitate. Allyl chloride on the other hand is sp2 hybridized hence, it does not undergo the SN2 reaction and hence, does not form any silver chloride precipitate.
(The section is an overview of the results that seeks to explain whether the objectives outlines in the experiment were achieved and whether or not the experiment was successful.)
James, (2012) Comparing the SN1 and SN2 Reaction Mechanism. Master Organic Chemistry, Website Online, Available at: https://www.masterorganicchemistry.com/2012/08/08/comparing-the-sn1-and-sn2-reactions/ [Retrieved 29th April, 2018].
James, (2014) Williamson Ether Synthesis. Master Organic Chemistry, Website Online, Available at: https://www.masterorganicchemistry.com/2014/10/24/the-williamson-ether-synthesis/ [Retrieved 29th April, 2018].
Ouellette, R. and Rawn, D. (2015) Ethers and Epoxides: The Williamson Ether Synthesis. Organic Chemistry Study Guide, Website Online, Available at: https://www.sciencedirect.com/topics/chemistry/williamson-ether-synthesis [Retrieved 29 April, 2018].
University of Technology Sydney, (2018) Lecture Notes 4, Alkenes, Available at: file:///C:/Users/Downloads/2193608_1348087501_OrganicChemistry1week4Alkene-2%20(2).pdf [Retrieved 29 April, 2018].
University of Technology Sydney, (2018) Lecture Notes 5, Halo Hydrocarbons, Available at: file:///C:/Users/Downloads/2193609_1266341416_OrganicChemistry1week5halogenc%20(2).pdf [Retrieved 29 April, 2018].
University of Technology Sydney, (2018) Lecture Notes 6, Alcohols and Ethers, Available at: file:///C:/Users/Downloads/2193610_1249927985_OrganicChemistry1week6alcohols%20(4).pdf [Retrieved 29 April, 2018].