Organic Chemistry Lab 2 Determining the Structures of Unknown Molecules Using NMR, IR and Mass Spectroscopy

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Experiment 3 Determining the Structures of Unknown Molecules Using NMR, IR and Mass Spectroscopy Xiaolei Gao, Ph.D. Background The theory of infrared spectroscopy, mass spectroscopy and NMR spectroscopy will be covered in the first two lab periods. Please read chapters 12 and 13 in McMurry for a complete discussion of these techniques. Procedure Part I The lab consists of three parts. The first part is a thought experiment. Consider the following four compounds: Using the knowledge of spectroscopy, you gained in the prior two classes, predict what the IR, NMR and mass spectrum of each would look like and fill in the following table (I completed part of table for you as an example): Compound # Carbon # Proton Approximate Chemical Shift # Signals Signals of Protons Main IR Stretch Expected Molecular Ion in M.S. One Possible Fragment in M.S. 1 5 5 1) The proton in OH appears 3200-3600 108 cm-1 randomly on 1 HNMR 2) The protons in CH2 appears @ ~4-5 ppm (singlet) 3) Protons on the benzene ring appear @~7 ppm (multiplet) 2 3 4 100 43 Part II In person students will come to the lab and carry out the IR experiment of the four unknown compounds. Remote students will click the link below or copy it to your browser, then scroll down the page to find "Virtual Lab 2: Infrared Spectroscopy". Follow the instruction there to complete the virtual lab (remember you can use your cursor to see different view of the lab). You need to carry out the virtual lab four times to obtain IR data for all four compounds. The credit of this virtual lab goes to North Carolina State University. https://sites.google.com/ncsu.edu/ncstatevrorganicchemistrylabs/home After you complete the in-person or virtual lab, report here which compound is which and explain why (Compound 1 was used as an example here to show you how this is done). Based on the IR result, I conclude that compound A is pentan-2-ol because IR of compound A show a strong and broad peak at 3200-3600 cm-1, which is an indication of the presence of alcohol functional group. I also observed a medium peak at 2850-3000 cm-1,but no peaks at 3000-3100 cm-1 .This indicates the presence of sp3 C-H, and absence to aromatic sp2 C-H. The only aliphatic alcohol within the four compounds is pentan-2-ol, therefore I conclude compound A is pentan-2-ol. Based on the IR result, I conclude that compound B is … because… Based on the IR result, I conclude that compound C is … because… Based on the IR result, I conclude that compound D is … because… Part III 1. Which of the four compounds has a MS as shown below? Explain all the major fragments, 2. The following 1HNMR belong to which of the four compounds in this experiment (s: singlet; d: doublet; t: triplet; m: multiplet)? Explain. 3. Which of the four compounds has the following 13CNMR? Explain. 2 5 6 1) Proton on Carbon 3200-3600 cm-1 88 #1  1 ppm (doulet) OH 2) Proton on Carbon H #2  4 ppm (multiplet) + 3) Proton on Carbon CH3 #3  1 ppm (quartet) 4) Proton on Carbon MW = 45 #4  1 ppm (multiplet) 5) Proton on Carbon #5  1 ppm (triplet) 4) OH proton appears randomly on 1HNMR 3 6 4 1) Proton on Carbon  1700 cm-1 120 #1  2 ppm O (singlet) 2) Proton on Carbon #4  7 ppm + (doublet) CH3 3) Proton on Carbon #5  7 ppm (triplet) MW = 43 4) Proton on Carbon #6  7 ppm (triplet) 4 5 4 1) Proton on Carbon #1  2 ppm (singlet) 1715 cm-1 100 O O 2) Proton on Carbon 43 #3  1 ppm (doublet) 3) Proton on Carbon #4  1 ppm (multiplet) 4) Proton on Carbon #5  1 ppm (doublet) Part II In person students will come to the lab and carry out the IR experiment of the four unknown compounds. Remote students will click the link below or copy it to your browser, then scroll down the page to find "Virtual Lab 2: Infrared Spectroscopy". Follow the instruction there to complete the virtual lab (remember you can use your cursor to see different view of the lab). You need to carry out the virtual lab four times to obtain IR data for all four compounds. The credit of this virtual lab goes to North Carolina State University. https://sites.google.com/ncsu.edu/ncstatevrorganicchemistrylabs/home After you complete the in-person or virtual lab, report here which compound is which and explain why (Compound 1 was used as an example here to show you how this is done). Based on the IR result, I conclude that compound A is pentan-2-ol because IR of compound A show a strong and broad peak at 3200-3600 cm-1, which is an indication of the presence of alcohol functional group. I also observed a medium peak at 2850-3000 cm-1, but no peaks at 3000-3100 cm-1 .This indicates the presence of sp3 C-H, and absence to aromatic sp2 C-H. The only aliphatic alcohol within the four compounds is pentan-2-ol, therefore I conclude compound A is pentan-2-ol. Based on the IR result, I conclude that compound B is benzyl alcohol because the IR of compound B shows a broad peak at 3200-3600 cm-1, which indicates the presence of an alcohol functional group. There was also a peak between 3100-3000 cm-1 which indicates the presence of aromatic sp2 hybridized carbons and a peak at 3000-2850 cm-1 which indicates the presence of alkane sp3 hybridized carbons. Since both carbons were shown on the IR of compound B, I can conclude that compound B is benzyl alcohol. Based on the IR result, I conclude that compound C is 4-methyl-2-pentanone because the IR of compound C shows a strong peak at 1700-1800 cm-1 which means there is a carbonyl group. The peak was shown to be a ketone saturated (C=O) at 1725-1705 cm-1. Therefore, this leads to compound C being either 4-methyl-2-pentanone or acetophenone. However, for the compound to be acetophenone, there would have been a peak at 3100-3000 cm-1 for the aromatic sp2 hybridized carbons. Since there was a peak at 2850-3000 cm-1 , this indicates the presence of alkane sp3 hybridized carbons, I can conclude that compound C is 4-methyl-2-pentanone. Based on the IR result, I conclude that compound D is acetophenone because the IR of compound D shows a strong peak at 1700-1800 cm-1 which means there is a carbonyl group. This peak was shown to be an aryl (C=O) at 1700-1680 cm-1. There was also a peak at 3100-3000 cm-1 which indicates the presence of aromatic sp2 carbons. Therefore, I can conclude that compound D is acetophenone. Part III 1. Which of the four compounds has a MS as shown below? Explain all the major fragments. This MS shown above is for benzyl alcohol because the base peak at 43 indicates alpha cleavage on carbon #2 to produce a fragment with a molecular weight of 43. Also, the parent peak at 100 matches the full molecular weight of benzyl alcohol as well as there are multiplets to match the aromatic ring in the compound. 2. The following 1HNMR belong to which of the four compounds in this experiment (s: singlet; d: doublet; t: triplet; m: multiplet)? Explain. in. The following 1HNMR belongs to 2-pentanol because the highest ppm is 4 ppm which would represent carbon #2 that has one hydrogen. The doublet represents carbon #1 with three hydrogens that is a CH3 near the OH group at about 1 ppm. The triplet represents carbon #5 with three hydrogens that appears at about 1 ppm. The multiplet with 4 hydrogens represents carbon #4 that lies between 1 ppm and 2 ppm. Lastly, the OH proton can appear randomly on the HNMR. Therefore, this OH proton is located between 2.5 ppm and 3 ppm as a singlet. 3. Which of the four compounds has the following 13CNMR? Explain. The following CNMR is shown for acetophenone for two main distinct peaks. The first one lies between 190 ppm to 200 ppm which indicates a C=O ketone. By knowing there is a C=O in the compound, this limits our options to acetophenone and 4-methyl-2pentanone. The second peak(s) lie between 120 ppm and 140 ppm which indicates the C=C of an aromatic benzene ring. Therefore, between our two options, the only one that has a benzene ring and a C=O is acetophenone. Experiment 3 Determining the Structures of Unknown Molecules Using NMR, IR and Mass Spectroscopy Xiaolei Gao, Ph.D. Background The theory of infrared spectroscopy, mass spectroscopy and NMR spectroscopy will be covered in the first two lab periods. Please read chapters 12 and 13 in McMurry for a complete discussion of these techniques. Procedure Part I The lab consists of three parts. The first part is a thought experiment. Consider the following four compounds: O O OH OH 2 Pentan-2-ol 1 Benzyl Alcohol 3 Acetophenone 4 4-Methyl-pentan-2-one Using the knowledge of spectroscopy, you gained in the prior two classes, predict what the IR, NMR and mass spectrum of each would look like and fill in the following table (I completed part of table for you as an example): Compound # Carbon # Proton Approximate Chemical ShiQ # Signals Signals of Protons Main IR Stretch Expected Molecular Ion in M.S. One Possible Fragment in M.S. 1 5 5 1) The proton in OH appears randomly on 1HNMR 2) The protons in CH2 appears @ ~4-5 ppm (singlet) 3) Protons on the benzene ring appear @~7 ppm (mul]plet) 3200-3600 cm-1 108 H 2C OH 31 2 5 6 1) The proton in OH appears randomly on 1HNMR 2) The proton in CH3 appears @ 1 ppm (3H, triplet) 3) Protons on COH appears @~3-4 ppm (1H, sextet) 4) CH3 aaached to C-OH appears @ ~1ppm (duplet) 5) Protons on CH2 aaached to C-OH appears @~1 ppm (2H, quartet) 6) Protons on CH2 appears @~0-1 ppm (2H, sextet) 3400-3600 cm-1 88 45 3 6 4 1) The protons in CH3 appears @ ~2 ppm (3H, singlet) 2) The protons in the benzene ring appears @~ 7 ppm (5H, mul]plet) 1660-1780 cm-1 120 43 4 5 4 1) The protons in CH3 1670-1780 100 O O cm-1 appears @ ~ 1 ppm (6H, duplet) 2) The protons 43 aaached to the CH2 appears @~2 ppm (2H, duplet) 3) The protons aaached to CH3 right next to the C==O appears @~ 2 ppm (3H, singlet) 4) The proton aaached to the CH appears @ ~0-1 ppm (1H, mul]plet) Part II Remote students will click the link below or copy it to your browser, then scroll down the page to find "Virtual Lab 2: Infrared Spectroscopy". Follow the instruction there to complete the virtual lab (remember you can use your cursor to see different view of the lab). You need to carry out the virtual lab four times to obtain IR data for all four compounds. The credit of this virtual lab goes to North Carolina State University. https://sites.google.com/ncsu.edu/ncstatevrorganicchemistrylabs/home After you complete the in-person or virtual lab, report here which compound is which and explain why (Compound 1 was used as an example here to show you how this is done) - Based on the IR result, I conclude that compound A is pentan-2-ol because IR of compound A show a strong and broad peak at 3200-3600 cm-1, which is an indication of the presence of alcohol functional group. I also observed a medium peak at 2850-3000 cm-1, but no peaks at 3000-3100 cm-1 .This indicates the presence of sp3 C-H, and absence to aromatic sp2 C-H. The only aliphatic alcohol within the four compounds is pentan-2-ol, therefore I conclude compound A is pentan-2-ol. - Based on the IR result, I conclude that compound B is benzyl alcohol because we are able to determine the presence of a hydroxide functional group on the compound. This is indicated by a strong broad peak at 3200-3600 cm-1. After observing a medium peak at 3000–3100 cm-1, we can determine the presence of an aromatic sp2 C-H and a sp2 C-H concluding that compound B is benzyl alcohol. - Based on the IR result, I conclude that compound C is 4-methyl-pentan-2-one because we are able to see a carbonyl group through a distinct peak at 1700-1800 cm -1. Since there is no peak at 3000-3100 cm-1, but a medium peak at 2850-3000 cm-1, this shows the presence of sp3 C-H as well as the absence to an aromatic sp2 C-H. - Based on the IR result, I conclude that compound D is acetophenone because an aryl group is shown through a distinct peak at 1700-1800 cm-1. An indication of an sp2 C-H and the aromatic sp2 C-H is observed through a medium peak at 3000-3100 cm-1. Part III 1. Which of the four compounds has a MS as shown below? Explain all the major fragments. - - From the chart below, the major peaks of fragmentation emerge from the alpha cleavage of the corresponding C-C bonds to the carbonyl group. The 4-methyl-pentan-2-one, (CH3) 2CHCH2COCH3 molecular formula should be fragmented as: (CH₃)₂CHCH₂-COCH₃ → C₄H₉ + C₂H₃O which gives m/z = 57 and 43. (CH₃)₂CHCH₂CO-CH₃ → C₅H₉O + CH₃ which gives m/z = 85 and 15. We can also observe the highest peak in the spectrum, which is the base peak at m/z = 43 and at m/z = 15, 57, and 85 all other peaks. Thus, the MS mentioned below belongs to 4-Methyl-pentan-2-one as the four compounds. 2. The following 1HNMR belong to which of the four compounds in this experiment (s: singlet; d: doublet; t: triplet; m: multiplet)? Explain. - The 1HNMR belongs to the Pentan-2-ol because the CH proton attachment to the hydroxyl oxygen group has a peak at ~3-4 ppm with (1H, multiplet) There is also a peak of (3H, triplet) at ~1ppm due to the coupling of the methyl group's neighboring two protons and at ~1.5ppm there's another peak with (3H, duplet). 3. Which of the four compounds has the following 13CNMR? Explain. - The compound that has 13CNMR is Acetophenone from the four compounds because it has a double-bond, carbon-carbon aromatic group and a carbonyl group. It also has six distinct atoms of carbon, which the 13CNMR has six peaks in total. Name Exp #4 Date: 2/17/2021 Lab Partner: n/a Title: Nucleophilic Substitution Reactions of Organic Halides 1. Abstract and Purpose: (3 point). When studying organic chemistry, there are many different types of reactions that occur. This lab will focus on nucleophilic substitution reactions known as SN2 and SN1, specifically examining the factors that affect the relative rates of SN2 and SN1 reactions of alkyl halides. These factors include alkyl halide structure, nature of the leaving group, properties of the nucleophile, and steric hindrance. The SN2 reaction is a one-step, concerted substitution process meaning it will make new bonds and break old bonds at the same time. Therefore, SN2 reactions do not proceed by an intermediate. Since there is no intermediate, the rate-determining step involves two components: the alkyl halide and the nucleophile. The SN2 reaction also gives inversion of stereochemistry at the reaction center. There are four main factors that can affect an SN2 reaction. However, the most important are the steric effects which means the more congested and hindered the alkyl halide, the slower the reaction will occur. The leaving group is another factor that can affect this type of reaction, where the weakest bases are the best leaving group for SN2 reactions as they can best stabilize the anion. The choice of solvent can affect the rate of an SN2 reaction such that polar aprotic solvents are ideal as they solvate the metal counterion of the nucleophile, making it more reactive. Lastly, the nucleophile has an effect as SN2 reactions would rather have larger atoms since they are more polarizable. The SN1 reaction is a two-step unimolecular process with one intermediate and two transition states. The rate depends upon only the concentration of the alkyl halide, therefore making the rate of this reaction independent of the nucleophile concentration. Name Lab Partner: n/a Exp #4 Date: 2/17/2021 Once again, many factors can affect an SN1 reaction. The alkyl halide must form a stable carbocation, in which the more stable carbocation, the faster the reaction will happen. This is due to lowering the activation barrier. Along with a stable carbocation, resonance structures can occur such that they provide a tremendous amount of stabilization to the carbocation. The leaving group can also have the same effect on an SN1 reaction as an SN2 reaction such that the weakest bases are the best leaving group as they can best stabilize the anion. Unlike SN2 reactions and as stated above, the nucleophile will not affect the rate of an SN1 reaction because it is not part of the ratedetermining step. Lastly, the choice of solvent can affect the rate of an SN1 reaction such that polar protic solvents are ideal as it stabilizes the carbocation intermediate as it forms, thus increasing the rate of the reaction. Overall, this lab will allow for the examination on all of the following factors that affect the relative rates of an SN2 or SN1 reaction. For the experiment on SN2 reactions, a solution of sodium iodide in acetone will be used. Sodium iodide is an excellent nucleophile that is soluble in acetone which is a polar aprotic solvent. For the experiment on SN1 reactions, a solution of silver nitrate in ethanol will be used. Ethanol is a polar protic solvent that will favor a stable carbocation. Silver halide salts are very insoluble and will precipitate from the solution, indicating if a reaction has occurred. Name Exp #4 Date: 2/17/2021 Lab Partner: n/a 2. Balanced equation: (2 point) SN2 Reaction SN1 Reaction 3. Reagent Table: (Add more rows when needed) (2 points) Name Silver Nitrate M.W. (g/mol) 169.87 Density * (g/mL) M.P.* (°C) 4.35 212 B.P.* (°C) 440 Amount (grams or mL) n/a Moles n/a Hazards/Precautions Oxidizing Solids, Corrosive to Metals, Skin Corrosion, Serious Eye Damage, Acute Aquatic Hazard, Chronic Aquatic Hazard Name Exp #4 Date: 2/17/2021 Lab Partner: n/a Neopentyl Bromide 151.04 1.199 n/a 105-106 n/a n/a Flammable Liquids, Skin Irritation, Eye Irritation, Specific Target Organ Toxicity Ethanol 46.07 1.60 -144 78.29 n/a n/a Flammable Liquids, Eye Irritation Benzyl Chloride 126.58 4.36 -43 177-181 n/a n/a Flammable Liquids, Acute Oral Toxicity, Acute Inhalation Toxicity, Skin Irritation, Serious Eye Damage, Skin Sensitisation, Germ Cell Mutagenicity, Carcinogenicity, Specific Target Organ Toxicity, Acute Aquatic Hazard Acetone 58.08 0.79 -94 56 n/a n/a Flammable Liquids, Eye Irritation, Specific Target Organ Toxicity 2-Iodobutane 184.02 1.598 -104 119-120 n/a n/a Flammable Liquids, Acute Aquatic Hazard 2-Bromobutane 137.02 1.255 n/a 91 n/a n/a Flammable Liquids, Acute Aquatic Hazard, Chronic Aquatic Hazard 2-Bromo-2methylpropane 137.02 1.22 -20 71-73 n/a n/a Flammable Liquids Name Exp #4 Date: 2/17/2021 Lab Partner: n/a 1-Chlorobutane 1-Bromobutane 92.57 0.886 -123 77-78 n/a n/a Flammable Liquids, Aspiration Hazard, Acute Aquatic Hazard, Chronic Aquatic Hazard 137.02 1.276 -112 100-104 n/a n/a Flammable Liquids, Skin Irritation, Eye Irritation, Specific Target Organ Toxicity, Acute Aquatic Toxicity, Chronic Aquatic Toxicity 5. Procedure, Observations and Data Procedure (3 point) Observations and Lab Data (4 point) Name Lab Partner: n/a Exp #4 Date: 2/17/2021 A summary of the procedure done with bullet points) Report all observations and all data that you collect in the (2 points) lab here Part I – Reactivity with NaI in acetone (SN2) Part A: Effect of Structure of the Alkyl Halide on Notes from Virtual Lab on Sn2 Reaction. • the Relative Rates of SN2 • 3 key features of an SN2 reaction is the position of the nucleophile attack, the Add approximately 1-inch depth of hot tap transition state, and the stereochemistry of water to a 100 mL beaker and place on the the product. hotplate at a setting of 2 to set up a water bath and stabilize the water temperature at 45°C. • 3 factors that affect the rate of an SN2 reaction is the structure of the alkyl halide, Clamp a thermometer in the water so that the temperature can be monitored. Make sure the the effect of sterics, and the identity of thermometer is not touching the bottom of the leaving group. beaker or else the temperature being recorded • • • The alkyl halides used were 1- will be of the glass beaker and not the water! bromobutane, 2-bromobutane, and 2- Measure 2 mL of 15% sodium iodide in methyl-2-bromopropane. acetone into each of three clean, dry 10-cm • test tubes. Make sure to mark the test tubes For the structure of alkyl halides, it was predicted that 1-bromobutane will react the respectively with number 1, 2, and 3. • fastest while 2-methyl-2-bromopropane Add 2 drops of 1-bromobutane to test tube 1 will react the slowest. and mix thoroughly by swirling. Be sure to note the start time. • • white and cloudy. Add 2 drops of 2-bromobutane to test tube 2 and mix thoroughly by swirling. Be sure to note the start time. The test tube with 1-bromobutane was • The test tube with 2-bromobutane was clear, therefore showing no reaction. Name Lab Partner: n/a • • Exp #4 Date: 2/17/2021 Add 2 drops of 2-methyl-2-bromopropane to • The test tube with 2-methyl-2- test tube 3 and mix thoroughly by swirling. bromopropane was also clear, therefore Be sure to note the start time. showing no reaction. Watch each test tube carefully and note the • time at which a precipitate appears and its Since 1-bromobutane produced white precipitate, also the only test tube to color. • produce a precipitate. After 5 minutes, place any test tubes that have no precipitate into the 45°C water bath. • fastest under the structure of alkyl halides. Again, watch for the formation of precipitates. • In conclusion, 1-bromobutane was the • For the steric effects, it was predicted that Heat for up to 6 minutes (11 minutes total) 1-bromobutane will react the fastest while then remove the test tubes and allow to cool 1-bromo-2,2-dimethylpropane will react to room temperature. As some acetone may the slowest. have evaporated during heating, there is the • possibility of a precipitate of unreacted The test tube with 1-bromobutane was white and cloudy. starting material, NaI, forming on the walls of the test tube. To guard against this “false • dimethylpropane was clear, therefore positive” result, gently shake these tubes for a showing no reaction. few minutes which should redissolve any NaI but leave other precipitates unchanged. Note • The test tube with 1-bromo-2,2- • Since 1-bromobutane produced white if precipitate disappeared upon shaking. precipitate, also the only test tube to Record all results. produce a precipitate. Empty the test tubes into the halide waste container and dispose the test tube to the white cardboard box for broken glasses. • In conclusion, 1-bromobutane was the fastest under the steric effects. Name Lab Partner: n/a Exp #4 Date: 2/17/2021 Part B: Steric Effects and the Relative Rates of • SN2 Reactions • For the effect of leaving group, it was predicted that 1-bromobutane will react the Measure 2 mL of 15% sodium iodide in fastest while 1-chlorobutane will react the acetone into each of two new clean, dry 10- slowest. cm test tubes. Make sure to mark the test • The test tube with 1-bromobutane was tubes respectively with number 4 and 5. • white and cloudy. Add 2 drops of 1-bromobutane to test tube 4 and mix thoroughly by swirling. Be sure to • clear, therefore showing no reaction. note the start time. • Add 2 drops of 1-bromo-2,2-dimethylpropane • Since 1-bromobutane produced white (neopentyl bromide) to test tube 5 and mix precipitate, also the only test tube to thoroughly by swirling. Be sure to note the produce a precipitate. start time. • The test tube with 1-chlorobutane was • Repeat the process as described in Part A. In conclusion, 1-bromobutane was the fastest under the steric effects. Part C: Effect on the Leaving Group on the Relative Rates of SN2 Reactions • Predictions of SN1 Reaction. Measure 2 mL of 15% sodium iodide in acetone into each of two new clean, dry test • • For Part 1 of the reactivity with AgNO3 in tubes. Make sure to mark the test tubes Ethanol, the 2-methyl-2-bromopropane will respectively with number 6 and 7. react the fastest. This is due to SN1 Add 2 drops of 1-bromobutane to test tube 6 reactions preferring a tertiary alkane. and mix thoroughly by swirling. Be sure to note the start time. • This reaction should produce a pale yellow precipitate. Name Lab Partner: n/a • • Exp #4 Date: 2/17/2021 Add 2 drops of 1-chlorobutane to test tube 7 • For Part 2 of the reactivity with AgNO3 in and mix thoroughly by swirling. Be sure to Ethanol, the 2-iodobutane will react the note the start time. fastest. This is due to having the best Repeat the process as described in Part A. leaving group. Part II – Reactivity with AgNO3 in Ethanol (SN1) • This reaction should produce a very bright Part A • yellow precipitate. Keep the hot water bath from part I and maintain at 45°C. • • Measure 2 mL of 1% silver nitrate in ethanol undergo since SN1 reactions do not react into each of three clean, dry 10-cm test tubes. with primary alkanes. Make sure to mark the test tubes respectively • Add 2 drops of 1-bromobutane to test tube 8 and mix thoroughly by swirling. Be sure to note the start time. • However, if this reaction were to go, it would produce a white precipitate. with number 8, 9, and 10. • The reaction with 1-chlorobutane will not • The reaction with benzyl chloride will be faster than 1-chlorobutane because the benzene ring helps stabilize due to the Add 2 drops of 2-bromobutane to test tube 9 resonance structures making better and mix thoroughly by swirling. Be sure to note the start time. • Add 2 drops of 2-methyl-2-bromopropane to test tube 10 and mix thoroughly by swirling. Be sure to note the start time. • Watch each test tube carefully and note the time at which a precipitate appears and its color. intermediates. Name Lab Partner: n/a • Exp #4 Date: 2/17/2021 After 5 minutes, place any test tubes that have no precipitate into the 45°C water bath. Again, watch for the formation of precipitates. • Heat for up to 6 minutes (11 minutes total) then remove the test tubes and allow to cool to room temperature. • Empty the test tubes into the halide waste container and dispose the test tube to the white cardboard box for broken glasses. Part B • Measure 2 mL of 1% silver nitrate in ethanol into each of three clean, dry 10-cm test tubes. Make sure to mark the test tubes respectively with number 11, 12, 13, and 14. • Add 2 drops of 2-bromobutane to test tube 11 and mix thoroughly by swirling. Be sure to note the start time. • Add 2 drops of 2-iodobutane to test tube 12 and mix thoroughly by swirling. Be sure to note the start time. • Add 2 drops of 1-chlorobutane to test tube 13 and mix thoroughly by swirling. Be sure to note the start time. Name Lab Partner: n/a • Exp #4 Date: 2/17/2021 Add 2 drops of benzyl chloride to test tube 14 and mix thoroughly by swirling. Be sure to note the start time. • Repeat the process as described in Part A. Name Lab Partner: n/a Exp #4 Date: 2/17/2021 6. Conclusions and discussions (4 points) Sn2 Reaction Conclusion The alkyl bromide that reacted the fastest with sodium iodide in acetone was 1bromobutane. The alkyl bromide that reacted the slowest with sodium iodide in acetone was 2-methyl-2-bromopropane. By looking at the structure of the alkyl bromides, it can be seen that 1-bromobutane is a primary alkyl bromide. Therefore, the SN2 reaction prefers primary structures. The 2-methyl-2-bromopropane was the slowest because the more congested and sterically hindered the alkyl halide, the slower the reaction will undergo SN2. Under the steric effects, the alkyl bromide that reacted the fastest with sodium iodide was 1-bromobutane compared to the slowest that was 1-bromo-2,2dimethylpropane. Even though both structure are primary, there was still a difference in reactivity due to steric effects. For 1-bromobutane, there are two hydrogens attached to the neighboring carbon. Unlike 1-bromo-2,2-dimethylpropane that has two methyl groups attached to the neighboring carbon. The two methyl groups are very bulking for iodide to go through, therefore it slows the rate of the reaction. The iodide has a much easier time to go through the two hydrogens in 1-bromobutane. For the effect of the leaving group, it was seen that the alkyl halide that reacted the fastest with sodium iodide in acetone was 1-bromobutane. The larger the leaving Name Lab Partner: n/a Exp #4 Date: 2/17/2021 group, the less electronegative the leaving group is, therefore it is the better leaving group under SN2 reactions. It is also known that the weakest bases are the best leaving group for SN2 reactions as they can best stabilize the anion. Therefore, the bromine is a better leaving group than the chlorine. However, if 1-iodobutane was used then it would react the fastest over the 1-bromobutane due to it being larger than bromine and chlorine. The reaction would take place as it is a primary structure, there are no bulky groups, and it has the best leaving group. SN1 Reaction Conclusion Since this lab was not completed in person or virtually, there can not be any definite conclusions. However based on the predictions made, it can be estimated what the conclusions would be. The alkyl bromide that would react the fastest with silver nitrate in ethanol would be 2-bromo-2-methylpropane. It is estimated that this would react the fastest due to it being a tertiary alkyl halide. For SN1 reaction, it prefers the tertiary structures compared to secondary or primary. Looking under the affects of the leaving group, it can be estimated that the alkyl halide that would react the fastest is 2iodobutane. This is because iodide is the largest compared to bromine, therefore it is less electronegative. Lastly, for SN1 reactions, benzyl chloride would react faster than 1chlorobutane. Even though they are both primary halides, the benzyl chloride allows for better intermediates as it produces resonance structures. The reaction of 1-chlorobutane would not undergo an SN1 reaction. Nucleophilic Substitution Reaction of Organic Halides Abstract Organic chemistry involves a variety of chemical reactions, which proceed at different rates. Nucleophilic substitution reactions are some of the organic reactions that proceed at different rates and compete with each other. There are two types of nucleophilic substitution reactions namerly SN1 and SN2 reactions. This lab experiment therefore focuses on the two types of the nucleophilic substitution reactions of the alkyl halides, deeply analysing factors that influence their relative rates of reactions. Factors that have been shown to affect SN1 and SN2 reactions include; chemical properties of the nucleophile, the structure of the alkyl halide, steric hindrance, and the leaving group. Background information. SN2 reactions organic reactions are one-step process reactions that proceed without intermediate. SN2 involves simultaneous bond breaking and bond formation. Given that SN2 reactions do not proceed by an intermediate, nucleophile and alkyl halide are the two components that influence the rate determining step of the SN2 reactions. The process of the SN2 reactions results in the formation of the stereochemistry at the center of the reaction. Steric hindrance is the most important factor that affects the rate of the SN2 reaction. The higher the concentration of the alkyl halide, the slower the rate of the reaction. Other factors such as the properties of the nucleophile ,which were mentioned at the beginning, may also influence the rate of the SN2 reaction. The polarization and the size of the nucleophilic atom can either increase or reduce the rate SN2 reactions. Larger atoms increases the rate of the SN2 reactions because they can be easily polarized. Nevertheless, the PH strength of the leaving group greatly determines the SN2 reaction. Strong bases are poor leaving groups and do not increase the rate of the SN2 reaction. Weak acids on the other hand, are good leaving groups and can increase the rate of SN2 reaction greatly. Thus an anion solution can stabilize the rate of the SN2. Finally, solvents can also lower or increase the rate of the SN2 reaction. The role of solvent in dissolving the metal counter ions are such that they can either make the reaction more reactive or lower its reactivity. For instance, polar aprotic solvents makes the nucleophile more reactive by dissolving the metal counter ions, making the SN2 more reactive.. As opposed to SN2 reaction, SN1 is a double step unimolecular reaction that proceeds in two distinct states: intermediate and two transition states. The rate of the SN1 reaction majorly depends on the concentration of the alkyl halide but not nucleophile species. Higher concentration of the alkyl halide increases the rate of SN1 ration while lower concentration reduces the rate of reaction. Additionally, the stability of the carbocation formed by alkyl halide determines the rate of the SN1 reaction. Stable carbocation increases the rate of the reaction while unstable carbocation reduces the rate of the reaction. Nevertheless, the occurrence of the resonance structure during the reaction may improve the rate of the reaction. This is because resonance structures improve the stability of the carbocation and lowers the activation barrier thus increasing the rate of the SN1 reaction. As aforementioned, nucleophiles do not have any effect on the reaction rate of the SN1 reaction. This is quite different from the SN2 reaction where the size and polarization of the nucleophile increases the rate of reaction. Along with the noted factors, the HP leaving group may stabilize or weaken the rate of SN1 reactions. As mentioned in SN2 reaction weak bases are good leaving groups that increase the rate of SN2 reactions. The same is applicable to SN1 reactions. Weak bases increase the stability of the ions thus increasing the rate of the SN1 reactions. The solvent used in the reaction process can either increase or reduce the rate of SN1 reaction. Just like SN2 reaction, protic solvents improve the stabilization of the carbocation intermediate, increasing the rate of the SN1 reaction. As mentioned in the introduction this lab experiment focuses on the above discussed factors that affect the rate of nucleophilic substitution reactions. Reagents used for the SN2 reaction are acetone and sodium iodide solution. Sodium iodide which is nucleophile is dissolved in acetone which is a good polar and aprotic solvent. For the SN1 reaction experiment, a solution of the silver nitrate dissolved in ethanol, which is a polar protic solvent, will be used. The protic nature of the ethanol will favor the formation of stable carbocation, resulting in a complete reaction of the reagents. The appearance of silver halide at the end of the reaction is an indication of the complete reaction of the process. Below are the balanced chemical equations for both SN2 and SN1 reactions respectively. Sn2 reaction Sn1 reaction. Name M.W. Densit M.P. y* * (g/mol) B.P.* Amoun Moles t Hazards/Precauti ons (°C) (grams (g/mL) (°C) or mL) Silver Nitrate 169.87 4.35 212 440 n/a n/a Oxidizing Solids, Corrosive to Metals, Skin Corrosion, Serious Eye Damage, Acute Aquatic Hazard, Chronic Aquatic Hazard Neopentyl Bromide 151.04 1.199 n/a 105-106 n/a n/a Flammable Liquids, Skin Irritation, Eye Irritation, Specific Target Organ Toxicity Ethanol 46.07 1.60 -144 78.29 n/a n/a Flammable Liquids, Eye Irritation Benzyl 126.58 4.36 -43 177-181 n/a n/a Chloride Flammable Liquids, Acute Oral Toxicity, Acute Inhalation Toxicity, Skin Irritation, Serious Eye Damage, Skin Sensitisation, Germ Cell Mutagenicity, Carcinogenicity, Specific Target Organ Toxicity, Acute Aquatic Hazard Acetone 58.08 0.79 -94 56 n/a n/a Flammable Liquids, Eye Irritation, Specific Target Organ Toxicity 2-Iodobutane 184.02 1.598 -104 119-120 n/a n/a Flammable Liquids, Acute Aquatic Hazard 2- 137.02 1.255 n/a 91 n/a n/a Bromobutane Flammable Liquids, Acute Aquatic Hazard, Chronic Aquatic Hazard 2-Bromo-2- 137.02 1.22 -20 71-73 n/a n/a methylpropane 1Chlorobutane Flammable Liquids 92.57 0.886 -123 77-78 n/a n/a Flammable Liquids, Aspiration Hazard, Acute Aquatic Hazard, Chronic Aquatic Hazard 1- 137.02 1.276 Bromobutane -112 100-104 n/a n/a Flammable Liquids, Skin Irritation, Eye Irritation, Specific Target Organ Toxicity, Acute Aquatic Toxicity, Chronic Aquatic Toxicity Procedure (3 point) Observations and Lab Data (4 point) Summary of the procedure (2 points) Part I – Reactivity with NaI in acetone Observation Notes from Virtual Lab on Sn2 Reaction. (SN2) 1. Features of an SN2 reaction include; the Part A: Effect of Structure of the Alkyl position of the nucleophile attack, the Halide on the Relative Rates of SN2 stereochemistry of the product, and the transition state. 1. 1-inch depth of hot tap water was 2. The alkyl halides used were 1added to a 100 mL beaker and bromobutane, 2-bromobutane, and 2placed on the hotplate at a setting of methyl-2-bromopropane. 2 to set up a water bath and stabilize 3. The key 3 factors affecting the rate of an the water temperature at 45°C. SN2 reaction are the effect of sterics, the 2. Thermometer was clamped in the structure of the alkyl halide, and the water so that the temperature can be identity of the leaving group. monitored. 4. The assumption was that the reactions of 3. 2 mL of 15% sodium iodide was the alkyl halide would differ; it was measured in acetone into each of predicted that 1-bromobutane will react three clean, dry 10-cm test tubes. the fastest while 2-methyl-2Make sure to mark the test tubes bromopropane will react the slowest. respectively with number 1, 2, and 5. There was no reaction in the test tube 3. with 2-bromobutane. Thus it was clear. 4. 1-bromobutane was added to the test tube 1 and mixed thoroughly by swirling. Be sure to note the start time. 5. Add 2 drops of 2-bromobutane to 6. The test tube that had 1-bromobutane appeared white and cloudy. 7. Since 1-bromobutane produced white test tube 2 and mix thoroughly by precipitate, also the only test tube to swirling. Be sure to note the start produce a precipitate. time. 6. Add 2 drops of 2-methyl-2bromopropane to test tube 3 and mix thoroughly by swirling. Be sure to note the start time. 7. Watch each test tube carefully and note the time at which a precipitate appears and its color. 8. After 5 minutes, place any test tubes that have no precipitate into the 45°C water bath. Again, watch for the formation of precipitates. 9. Heat for up to 6 minutes (11 minutes total) then remove the test tubes and allow to cool to room temperature. As some acetone may have evaporated during heating, there is the possibility of a 8. The test tube with 2-methyl-2bromopropane was also clear, therefore showing no reaction. 9. In conclusion, 1-bromobutane was the fastest under the structure of alkyl halides. 10. For the steric effects, it was predicted that 1-bromobutane will react the fastest while 1-bromo-2,2-dimethylpropane will react the slowest. 11. The test tube with 1-bromo-2,2dimethylpropane was clear, therefore showing no reaction. 12. The test tube with 1-bromobutane was white and cloudy. 13. The test tube with 1-bromobutane was white and cloudy. precipitate of unreacted starting 14. Since 1-bromobutane produced white material, NaI, forming on the walls precipitate, also the only test tube to of the test tube. To guard against produce a precipitate. this “false positive” result, gently shake these tubes for a few minutes which should redissolve any NaI but 15. 1-bromobutane was the fastest under the steric effects. 16. Since 1-bromobutane produced white leave other precipitates unchanged. precipitate, also the only test tube to Note if precipitate disappeared upon produce a precipitate shaking. Record all results. 10. Empty the test tubes into the halide 17. For the effect of leaving the group, it was predicted that 1-bromobutane will react waste container and dispose of the the fastest while 1-chlorobutane will test tube to the white cardboard box react the slowest. for broken glasses. 18. The test tube with 1-chlorobutane was clear, therefore showing no reaction. Part B: Steric Effects and the Relative 19. In conclusion, 1-bromobutane reacted Rates of SN2 Reactions faster under steric effect. 1. Measure 2 mL of 15% sodium Predictions of SN1 Reaction. iodide in acetone into each of two new clean, dry 10-cm test tubes. 1. For Part 1 of the reactivity with AgNO3 Make sure to mark the test tubes in Ethanol, the 2-methyl-2- respectively with number 4 and 5. bromopropane will react the fastest. This 2. Add 2 drops of 1-bromobutane to test tube 4 and mix thoroughly by is due to SN1 reactions preferring a tertiary alkane. swirling. Be sure to note the start time. 3. Add 2 drops of 1-bromo-2,2- 2. This reaction should produce a pale yellow precipitate. 3. For Part 2 of the reactivity with AgNO3 dimethylpropane (neopentyl in Ethanol, the 2-iodobutane will react bromide) to test tube 5 and mix the fastest. This is due to having the best thoroughly by swirling. Be sure to leaving group. note the start time. 4. Repeat the process as described in Part A. 4. This reaction should produce a very bright yellow precipitate. 5. The reaction with 1-chlorobutane will not undergo since SN1 reactions do not Effect on the Leaving Group on the react with primary alkanes. Relative Rates of SN2 Reactions Part C 6. However, if this reaction were to go, it 1. Measure 2 mL of 15% sodium iodide in acetone into each of two would produce a white precipitate. 7. The reaction with benzyl chloride will be new clean, dry test tubes. Make sure faster than 1-chlorobutane because the to mark the test tubes respectively benzene ring helps stabilize due to the with number 6 and 7. resonance structures making better 2. Add 2 drops of 1-bromobutane to test tube 6 and mix thoroughly by swirling. Be sure to note the start time. 3. Add 2 drops of 1-chlorobutane to test tube 7 and mix thoroughly by intermediates. swirling. Be sure to note the start time. 4. Repeat the process as described in Part A. Reactivity with AgNO3 in Ethanol (SN1) Part A 1. Measure 2 mL of 1% silver nitrate in ethanol into each of three clean, dry 10-cm test tubes. Make sure to mark the test tubes respectively with number 8, 9, and 10. 2. Keep the hot water bath from part I and maintain at 45°C. 3. Add 2 drops of 1-bromobutane to test tube 8 and mix thoroughly by swirling. Be sure to note the start time. 4. Add 2 drops of 2-bromobutane to test tube 9 and mix thoroughly by swirling. Be sure to note the start time. 5. Add 2 drops of 2-methyl-2bromopropane to test tube 10 and mix thoroughly by swirling. Be sure to note the start time. 6. Watch each test tube carefully and note the time at which a precipitate appears and its color. 7. After 5 minutes, place any test tubes that have no precipitate into the 45°C water bath. Again, watch for the formation of precipitates. 8. Heat for up to 6 minutes (11 minutes total) then remove the test tubes and allow to cool to room temperature. 9. Empty the test tubes into the halide waste container and dispose of the test tube to the white cardboard box for broken glasses. Part B 1. Measure 2 mL of 1% silver nitrate in ethanol into each of three clean, dry 10-cm test tubes. Make sure to mark the test tubes respectively with number 11, 12, 13, and 14. 2. Add 2 drops of 2-bromobutane to test tube 11 and mix thoroughly by swirling. Be sure to note the start time. 3. Add 2 drops of 2-iodobutane to test tube 12 and mix thoroughly by swirling. Be sure to note the start time. 4. Add 2 drops of 1-chlorobutane to test tube 13 and mix thoroughly by swirling. Be sure to note the start time. 5. Add 2 drops of benzyl chloride to test tube 14 and mix thoroughly by swirling. Be sure to note the start time. ● Repeat the process as described in Part A. Conclusion SN2 reaction The reaction of the alkyl bromide with sodium iodide in acetone proceeded to form two organic compounds: 1-bromobutane and 2-methyl-2 bromopropane. From the reaction, the fastest reaction of the alkyl bromide and sodium iodide resulted in the formation of the 1bromobutane while the lowest reaction formed the 2-methyl-2 bromopropane. From the analysis of the reaction it is clear that SN2 reaction proceeds in the primary alkyl halide. The slowest reaction could have been hindered by the statistically congested alkyl halide. This is in line with initial discussion that highlighted that steric alkyl halide and congestion hinders the rate of reaction of the SN2 reaction. The steric effect on the alkyl halide caused the difference in the reaction with the acetone. From the structural difference of the halides, two hydrogen atoms were attached to the neighbouring carbon atoms 1-bromobutane which was a result of faster reaction between sodium iodine and alkyl bromide. In the bromo-2-2-dimethylpropane, the two methyl groups were attached to the neighbouring atoms, hindering the reaction thus slowing down the rate of reaction. The two methyl groups prevented the passing of iodine because they are bulky. 1 bromobutane on the other hand had no bulky methyl groups making it easy for the iodine to go through. Hence the faster rate of reaction for the 1-bromobutane for SN2 reaction. The reaction alkyl halide with sodium iodide in acetone in terms of the leaving group was faster for the 1-bromobutane than 2-methyl-2-bromopropane. This was due to the fact the 1bromobutane was the largest leaving group. Larger leaving group increase the rate of reaction because larger leaving groups are less electronegative making them better leaving groups in Sn2 reaction. Additionally leaving groups are the weakest base and can therefore stabilize anion, resulting in a faster Sn2 reaction. From the result, bromine can be described as a better leaving group compared to chlorine. SN1 reaction This conclusion of the Sn1 reaction is based on assumption of what would have happened if the experiment was to be physically performed and analyzed. First, this experiment was never concluded virtually or physically due to technical issues. Back to the assumption, if the experiment were to be conducted and analyzed, it is hypothesized that 2-bromo-2-methylpropane would have reacted faster with the silver nitrate than the other alkyl bromide. This could have been due to the fact that 2-bromo-2-methylpropane is a tertiary alkyl halide. As mentioned before, Sn1 reaction proceeds better in secondary or tertiary halide compared to primary halide. Considering the effect of the leaving group, the reaction would have been faster in 2-iodobutane because it is the largest alkyl halide compared to bromine. This would have been the case as the larger the alkyl halide, the faster the rate of the reaction. The size of the alkyl halide is inversely proportional to electronegativity of the halide. Hence, less electronegative halide increases the rate of the Sn1 reaction. The reaction of benzyl chloride would be faster because it intermediates the further produce resonance structures which increase the rate of reaction. Chlorobutane on the other hand would not undergo Sn1 reaction due to bulkiness and much concentration of the electronegative. Х Experiment 3 Lab Report (... (doulet) 2) Proton on Carbon OH H KH CH3 #2 : 4 ppm (multiplet) 3) Proton on Carbon #321 ppm (quartet) 4) Proton on Carbon #41 ppm (multiplet) 5) Proton on Carbon MW = 45 #521 ppm (triplet) 4) OH proton appears randomly on 'HNMR 3 6 4 1) Proton on Carbon 1700 cm 120 #1 22 ppm (singlet) 2) Proton on Carbon #47 ppm (doublet) 3) Proton on Carbon CH3 #5:7 ppm (triplet) 4) Proton on Carbon MW = 43 #6 27 ppm (triplet) 1) Proton on Carbon 4 5 4 1715 cm 100 #12 ppm f (singlet) 2) Proton on Carbon 43 #31 ppm (doublet) 3) Proton on Carbon #4 = 1 ppm (multiplet) 4) Proton on Carbon #5 a 1 ppm (doublet) Part 11 In person students will come to the lab and carry out the IR experiment of the four unknown compounds. Remote students will click the link below or copy it to your browser, then scroll down the page to find "Virtual Lab 2: Infrared Spectroscopy". Follow the instruction there to complete the virtual lab (remember you can use your cursor to see different view of the lab). You need to carry out the virtual lab four times to obtain IR data for all four compounds. The credit of this virtual lab goes to North Carolina State University. https://sites.google.com/ncsu.edu/ncstatevrorganicchemistrylabs/home After you complete the in-person or virtual lab, report here which compound is which and explain why (Compound 1 was used as an example here to show you how this is done). Experiment 3- Elyse Rebelo Determining the Structures of Unknown Molecules Using NMR, IR and Mass Spectroscopy Xiaolei Gao, Ph.D. Background The theory of infrared spectroscopy, mass spectroscopy and NMR spectroscopy will be covered in the first two lab periods. Please read chapters 12 and 13 in McMurry for a complete discussion of these techniques. Procedure Part 1 The lab consists of three parts. The first part is a thought experiment. Consider the following four compounds: OH OH 2 1 Pentan-2-ol Acetophenone 4-Methyl-pentan-2-one Benzyl Alcohol Using the knowledge of spectroscopy, you gained in the prior two classes, predict what the IR, NMR and mass spectrum of each would look like and fill in the following table ( completed part of table for you as an example): Compound # # Carbon # Proton Approximate Chemical Main IR Stretch Expected Molecular One Possible Fragment in M.S. Signals Signals Shift of Protons lon in M.S. 1 5 5 3200-3600 cm 108 1) The proton in OH appears randomly HC on 'HNMR OH 31 2) The protons in CH2 appears @ --4-5 ppm (singlet) 3) Protons on the benzene ring appear @ 7 ppm (multiplet) 2 5 6 1) Proton on Carbon 3200-3600 cm 88 #1: 1 ppm (doulet) 2) Proton on Carbon OH #24 ppm KA (multiplet) 3) Proton on Carbon CH3 #31 ppm MW = 45 (quartet) 4) Proton on Carbon #41 ppm (multiplet) 5) Proton on Carbon #51 ppm (triplet) 4) OH proton appears randomly on 'HNMR 3 6 4 1) Proton on Carbon 1700 cm 120 #12 ppm (singlet) 2) Proton on Carbon #47 ppm (doublet) 3) Proton on Carbon CH3 #5 7 ppm (triplet) MW = 43 4) Proton on Carbon #6 27 ppm Х Experiment 3 Lab Report (... Part III 1. Which of the four compounds has a MS as shown below? Explain all the major fragments. LOO 80 60 40 20 0.0 0.0 20 40 60 80 100 120 This MS shown above is for benzyl alcohol because the base peak at 43 indicates alpha cleavage on carbon #2 to produce a fragment with a molecular weight of 43. Also, the parent peak at 100 matches the full molecular weight of benzyl alcohol as well as there are multiplets to match the aromatic ring in the compound. 2. The following 'HNMR belong to which of the four compounds in this experiment (s: singlet; d: doublet;t: triplet; m: multiplet)? Explain. 3H, d 3H, t 4H, m 1H, m 1H, s o in. The following 'HNMR belongs to 2-pentanol because the highest ppm is 4 ppm which would represent carbon #2 that has one hydrogen. The doublet represents carbon #1 with three hydrogens that is done Pupptiplet represents carbon Х Experiment 3 Lab Report (... 2. The following HNMR belong to which of the four compounds in this experiment (s: singlet; d: doublet;t: triplet; m: multiplet)? Explain. 3H, d 3H, t 4H, m 1H, m 11, s 5 4 in. The following "HNMR belongs to 2-pentanol because the highest ppm is 4 ppm which would represent carbon #2 that has one hydrogen. The doublet represents carbon #1 with three hydrogens that is a CH3 near the OH group at about 1 ppm. The triplet represents carbon #5 with three hydrogens that appears at about 1 ppm. The multiplet with 4 hydrogens represents carbon #4 that lies between 1 ppm and 2 ppm. Lastly, the OH proton can appear randomly on the HNMR. Therefore, this OH proton is located between 2.5 ppm and 3 ppm as a singlet. 3. Which of the four compounds has the following 13CNMR? Explain. ppm Int. 标记读 197.85 137.23 133.04 128.56 128.29 26.47 174 229 463 866 1 2 3 4 5 6 1000 259 Х Experiment 3 Lab Report (... in. The following HNMR belongs to 2-pentanol because the highest ppm is 4 ppm which would represent carbon #2 that has one hydrogen. The doublet represents carbon #1 with three hydrogens that is a CH3 near the OH group at about 1 ppm. The triplet represents carbon #5 with three hydrogens that appears at about 1 ppm. The multiplet with 4 hydrogens represents carbon #4 that lies between 1 ppm and 2 ppm. Lastly, the OH proton can appear randomly on the HNMR. Therefore, this OH proton is located between 2.5 ppm and 3 ppm as a singlet. 3. Which of the four compounds has the following 13CNMR? Explain. ppm Int. 标记歲 174 229 197.85 137.23 133.04 463 128.56 866 128.29 1000 26.47 259 1 2 3 4 + 5 + 6 200 180 160 140 120 100 80 60 40 20 0 CDS-00-674 pom The following CNMR is shown for acetophenone for two main distinct peaks. The first one lies between 190 ppm to 200 ppm which indicates a C=0 ketone. By knowing there is a C=0 in the compound, this limits our options to acetophenone and 4-methyl-2- pentanone. The second peak(s) lie between 120 ppm and 140 ppm which indicates the C=C of an aromatic benzene ring. Therefore, between our two options, the only one that has a benzene ring and a C=0 is acetophenone. 5 un 6 e 1) Th e around proton at 3400-3200 C2 is a cm-1 sextet and appears at around 4 ppm. 2) Th proton of the OH group appears randomly. 3) The proton at C3 is a quintet an d appears at around 1 ppm. 4) Th proton at C4 is a sextet and appears at around 1 ppm. 5) Th proton at C1 is a doublet e e 2 5 6 88 OH v m/z=73 1) The around proton at 3400-3200 C2 is a cm 1 sextet and appears at around 4 ppm. 2) Th e proton of the OH group appears randomly. 3) The proton at C3 is a quintet a n d appears at around 1 ppm. 4) Th e proton at C4 is a sextet and appears at around 1 ppm. 5) Th е proton at C1 is a doublet and appears at around 1 ppm. 6) Th e 3 6 4 120 CH2 1) One of the 1680 cm 1 aromatic protons is a doublet n d appears at around 8 a m/z=91 ppm. 2) Another aromatic proton is a triplet and appears at around 7-8 ppm. 3) The third aromatic proton appears as a triplet at around 7 ppm. 4) The protons of the methyl group appear at a singlet at around 2-3 5 4 e 1700 cm 100 IUO m/z=85 1) Th proton at с 3 appears as a doublet at around 2-3 ppm. 2) The proton at с 1 appears as a singlet at around 2-2.5 ppm. 3) The proton at C4 is a multiplet an d appears at around 2 ppm. 4) The protons at C5 and the methy! group are equivalent and appear as a doublet at around 1 ppm. Part II 88 nd 200 ОН 73 2 5 6 3200-3600 cm 88 OH I KH + CH3 MW = 45 1) Proton on Carbon #11 ppm (doulet) 2) Proton on Carbon #24 ppm (multiplet) 3) Proton on Carbon #31 ppm (quartet) 4) Proton on Carbon #41 ppm (multiplet) 5) Proton on Carbon #521 ppm (triplet) 4) OH proton appears randomly on 'HNMR 1) Proton on Carbon #12 ppm (singlet) 2) Proton on Carbon #47 ppm (doublet) 3) Proton on Carbon #57 ppm (triplet) 4) Proton on Carbon #67 ppm 3 6 4 1700 cm 1 120 O + CH3 MW = 43 N 5 6 3400-3600 88 45 cm-1 1) The proton in OH appears randomly on HNMR 2) The proton in CH3 appears @ 1 ppm (3H, triplet) 3) Protons on C OH appears @3-4 ppm (1H, sextet) 4) CH3 attached to C-OH appears @ 1ppm (duplet) 5) Protons on CH2 attached to C-OH OH + CH `CH₃ 3 appears @ 1 ppm (2H, quartet) 6) Protons on CH2 appears @ 0-1 ppm (2H, sextet) 3 6 4 1660-1780 120 43 1) The protons in CH3 cm- H -C-H appears @ 2 ppm (3H Singlet) 2) The protons in the benzene ring appears @7 ppm (SH multiplet) Lab Partner: n/a Title: Nucleophilic Substitution Reactions of Organic Halides 1. Abstract and Purpose: (3 point). When studying organic chemistry, there are many different types of reactions that occur. This lab will focus on nucleophilic substitution reactions known as SN2 and SN1, specifically examining the factors that affect the relative rates of SN2 and SN1 reactions of alkyl halides. These factors include alkyl halide structure, nature of the leaving group, properties of the nucleophile, and steric hindrance. The SN2 reaction is a one-step, concerted substitution process meaning it will make new bonds and break old bonds at the same time. Therefore, SN2 reactions do not proceed by an intermediate. Since there is no intermediate, the rate- determining step involves two components: the alkyl halide and the nucleophile. The SN2 reaction also gives inversion of stereochemistry at the reaction center. There are four main factors that can affect an SN2 reaction. However, the most important are the steric effects which means the more congested and hindered the alkyl halide, the slower the reaction will occur. The leaving group is another factor that can affect this type of reaction, where the weakest bases are the best leaving group for SN2 reactions as they can best stabilize the anion. The choice of solvent can affect the rate of an SN2 reaction such that polar aprotic solvents are ideal as they solvate the metal counterion of the nucleophile, making it more reactive. Lastly, the nucleophile has an effect as SN2 reactions would rather have larger atoms since they are more polarizable. The SN1 reaction is a two-step unimolecular process with one intermediate and two transition states. The rate depends upon only the concentration of the alkyl halide, therefore making the rate of this reaction independent of the nucleophile concentration. Once again, many factors can affect an SN1 reaction. The alkyl halide must form a stable carbocation, in which the more stable carbocation, the faster the reaction will happen. This is due to lowering the activation barrier. Along with a stable carbocation, resonance structures can occur such that they provide a tremendous amount of stabilization to the carbocation. The leaving group can also have the same effect on an SN1 reaction as an SN2 reaction such that the weakest bases are the best leaving group as they can best stabilize the anion. Unlike SN2 reactions and as stated above, the nucleophile will not affect the rate of an SN1 reaction because it is not part of the rate-determining step. Lastly, the choice of solvent can affect the rate of an SN1 reaction such that polar protic solvents are ideal as it stabilizes the carbocation intermediate as it forms, thus increasing the rate of the reaction. Overall, this lab will allow for the examination on all of the following factors that affect the relative rates of an SN2 or SN1 reaction. For the experiment on SN2 reactions, a solution of sodium iodide in acetone will be used. Sodium iodide is an excellent nucleophile that is soluble in acetone which is a polar aprotic solvent. For the experiment on SN1 reactions, a solution of silver nitrate in ethanol will be used. Ethanol 15 polar proti solvent that will favor stable Х Lab 4 Nucleophillic Substi... rate of an SN2 reaction such that polar aprotic solvents are ideal as they solvate the metal counterion of the nucleophile, making it more reactive. Lastly, the nucleophile has an effect as SN2 reactions would rather have larger atoms since they are more polarizable. The SN1 reaction is a two-step unimolecular process with one intermediate and two transition states. The rate depends upon only the concentration of the alkyl halide, therefore making the rate of this reaction independent of the nucleophile concentration. Once again, many factors can affect an SN1 reaction. The alkyl halide must form a stable carbocation, in which the more stable carbocation, the faster the reaction will happen. This is due to lowering the activation barrier. Along with a stable carbocation, resonance structures can occur such that they provide a tremendous amount of stabilization to the carbocation. The leaving group can also have the same effect on an SN1 reaction as an SN2 reaction such that the weakest bases are the best leaving group as they can best stabilize the anion. Unlike SN2 reactions and as stated above, the nucleophile will not affect the rate of an SN1 reaction because it is not part of the rate-determining step. Lastly, the choice of solvent can affect the rate of an SN1 reaction such that polar protic solvents are ideal as it stabilizes the carbocation intermediate as it forms, thus increasing the rate of the reaction. Overall, this lab will allow for the examination on all of the following factors that affect the relative rates of an SN2 or SN1 reaction. For the experiment on SN2 reactions, a solution of sodium iodide in acetone will be used. Sodium iodide is an excellent nucleophile that is soluble in acetone which is a polar aprotic solvent. For the experiment on SN1 reactions, a solution of silver nitrate in ethanol will be used. Ethanol is a polar protic solvent that will favor a stable carbocation. Silver halide salts are very insoluble and will precipitate from the solution, indicating if a reaction has occurred. 2. Balanced equation: (2 point) А A slow Nu- 의 B Nu- B + i Nu- B Substrate Transition State Product (Inverted) SN2 Reaction Nu slow fast SN1 Reaction Х Lab 4 Nucleophillic Substi... Name Silver Nitrate Neopentyl Bromide Ethanol Benzyl Chloride M.W. Density* M.P.* B.P.* Amount Moles Hazards/Precautions (g/mol) (g/mL) (°C) (°C) (grams or mL) 169.87 4.35 212 440 n/a n/a Oxidizing Solids, Corrosive to Metals, Skin Corrosion, Serious Eye Damage, Acute Aquatic Hazard, Chronic Aquatic Hazard 151.04 1.199 n/a 105- n/a n/a Flammable Liquids, 106 Skin Irritation, Eye Irritation, Specific Target Organ Toxicity 46.07 1.60 -144 78.29 n/a n/a Flammable Liquids, Eye Irritation 126.58 4.36 -43 177- n/a n/a Flammable Liquids, 181 Acute Oral Toxicity, Acute Inhalation Toxicity, Skin Irritation, Serious Eye Damage, Skin Sensitisation, Germ Cell Mutagenicity, Carcinogenicity, Specific Target Organ Toxicity, Acute Aquatic Hazard 58.08 0.79 -94 56 n/a n/a Flammable Liquids, Eye Irritation, Specific Target Organ Toxicity 184.02 1.598 -104 119- n/a n/a Flammable Liquids, 120 Acute Aquatic Hazard 137.02 1.255 n/a 91 n/a n/a Flammable Liquids, Acute Aquatic Hazard, Chronic Aquatic Hazard 137.02 1.22 -20 71-73 n/a n/a Flammable Liquids Acetone a 2-Iodobutane 2-Bromobutane 2-Bromo-2- methylpropane 1-Chlorobutane 92.57 0.886 -123 77-78 n/a n/a 1-Bromobutane 137.02 1.276 n/a n/a -112 100- 104 Flammable Liquids, Aspiration Hazard, Acute Aquatic Hazard, Chronic Aquatic Hazard Flammable Liquids, Skin Irritation, Eye Irritation, Specific Target Organ Toxicity, Acute Aquatic Toxicity, Chronic Aquatic Toxicity 4. Calculations: Shown each calculation for moles of reagents, limiting reagent, theoretical yield and percent yield. (3 points) Calculating moles of reagents: n/a Determining limiting reagent: n/a The limiting reagent is n/a. Calculating theoretical yield: n/a The theoretical yield is n/a. The actual yield is n/a. Calculating Percent yield: n/a Reaction percent yieldiena solution of sodium iodide in acetone will be used. Sodium iodide is an excellent nucleophile that is soluble in acetone which is a polar aprotic solvent. For the experiment on SN1 reactions, a solution of silver nitrate in ethanol will be used. Ethanol is a polar protic solvent that will favor a stable carbocation. Silver halide salts are very insoluble and will precipitate from the solution, indicating if a reaction has occurred. 2. Balanced equation: (2 point) A slow 1 Nu B + 1 Nu- D B Substrate Transition State Product (Inverted) SN2 Reaction Nu Nu -CY slow Z como fast х yic -Nu SN1 Reaction 3. Reagent Table: (Add more rows when needed) (2 points) Name M.W. Density* M.P.* B.P.* Amount Moles Hazards/Precautions (g/mol) (g/mL) (°C) (°C) (grams or mL) Silver Nitrate 169.87 4.35 212 440 n/a n/a Oxidizing Solids, Corrosive to Metals, Skin Corrosion, Serious Eye Damage, Acute Aquatic Hazard, Chronic Aquatic Hazard Neopentyl 151.04 1.199 n/a 105- n/a n/a Flammable Liquids, Bromide 106 Skin Irritation, Eye Irritation, Specific Target Organ Toxicity Ethanol 46.07 1.60 -144 78.29 n/a n/a Flammable Liquids, Eye Irritation Benzyl Chloride 126.58 4.36 -43 177- n/a n/a Flammable Liquids, 181 Acute Oral Toxicity, Acute Inhalation Toxicity, Skin Irritation, Serious Eye Damage, Skin Sensitisation, Germ Cell Mutagenicity, Carcinogenicity, Specific Target Organ Toxicity, Acute Aquatic Hazard Acetone 58.08 0.79 -94 56 n/a n/a Flammable Liquids, Eye Irritation, Specific Target Organ Toxicity 2-Iodobutane 184.02 1.598 -104 119- n/a n/a Flammable Liquids, 120 Acute Aquatic Hazard 2-Bromobutane 137.02 1.255 n/a 91 n/a n/a Flammable Liquids, Acute Aquatic Hazard, Chronic Aquatic Hazard 2-Bromo-2- 137.02 1.22 -20 71-73 n/a n/a Flammable Liquids methylpropane 1-Chlorobutane 92.57 0.886 -123 77-78 n/a n/a Flammable Liquids, Aspiration Hazard, Acute Aquatic Hazard, Chronic Aquatic Hazard 1-Bromobutane 137.02 1.276 -112 100- n/a n/a Flammable Liquids, 104 Skin Irritation, Eye Irritation, Specific Target Organ n CON box (98) - sdhani. OH ning appear Send Feedback to Lea 3/5/21 1.29 PM Part 1 5/5 218 Part II 6/6 Part III @-7 ppm (multiplet) 1) The proton on the hydroxyl group will appear randomly in the HNMR. OA it 1.2/3 The key peaks need to be explained are 100, 85. 58 ar compound 4 a ketone 2.1.53 Explain all peaks (ppm, splitting patterns) to supp conclusion 2) The methyl 3.253 group near the OH appears at approximately 1ppm (3H, d) 3) the proton Count number of carbon peaks in NMR and make accountable 6 CH3 6 attached to 5 COH appear 4ppm (1H, m 0 4) The CH2 group closest to the COH S. shows up at 1 ppm (2H, al. The other CH2 group shows up at "1ppm/2H, m) woryromoureur esdh... ?content_id wy Amazon.com: che other 3 mpty group .nows up at 1ppm (3H, t) 1)The protons on the aromatic rings show up at 7ppm (SH,m) 2)The methyl 4 group bonded to C=O shows up at wood 2 ppm (3H, ) B B 1695 120 - 43 1) The ethyl group bonded to CHO shows up at 2 CH 1715 100 0 ppm (3H) 2) The CH2 shows up as at2 ppm فال (2H.d) 4) the single proton that is ros attached to the two methyl groups shows up at Part 43 1.213 The compound 4 2.1.5/3 Explain all conclusion 3.25/3 Count number of carbon peat accountable los 2
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Explanation & Answer

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Experiment 3
Determining the Structures of Unknown Molecules Using NMR, IR, and Mass Spectroscopy

Compo
und #

1

#

Approximate

Main IR

Expected

One Possible

Carbon Proton

Chemical Shift of

Stretch

Molecular

Fragment in M.S.

Signals Signals

Protons

5

#

5

1) The proton
in OH
appears
randomly
on 1HNMR
2) The
protons in
CH2
appears @
~4-5 ppm
(singlet)
3) Protons on
the
benzene
ring
appear

Ion in M.S.
3200-3600
cm-1

108

@~7 ppm
(multiplet)
2

5

6

1) The proton
around
at C2 is a 3400-3200
sextet and
cm-1
appears at
around 4
ppm.
2) The proton
of the OH
group
appears
randomly.
3) The proton
at C3 is a
quintet
and
appears at
around 1
ppm.
4) The proton
at C4 is a
sextet and
appears at
around 1
ppm.
5) The proton
at C1 is a
doublet
and
appears at
around 1
ppm.
6) The proton
at C5 is a
triplet and
appears at
around 1
ppm.

88

3

6

4

1) One of the
aromatic
protons is a
doublet
and
appears at
around 8
ppm.
2) Another
aromatic
proton is a
triplet and
appears at
around 7-8
ppm.
3) The third
aromatic
proton
appears as
a triplet at
around 7
ppm.
4) The
protons of
the methyl
group
appear at a
singlet at
around 2-3
ppm.

1680 cm-1

120

4

5

4

1) The proton
at
C3
appears as
a doublet
at around
2-3 ppm.
2) The proton
at
C1
appears as
a singlet at
around 22.5 ppm.

1700 cm-1

100

3) The proton
at C4 is a
multiplet
and
appears at
around 2
ppm.
4) The
protons at
C5 and the
methyl
group are
equivalent
and appear
as
a
doublet at
around 1
ppm.

Part II
Based on the IR result, I conclude that compound B is benzyl alcohol because there is an observed
absorption at 3600-3200 cm-1 and 3100-3000 cm-1 which suggests the presence of an alcohol
group and an aromatic C-H stretching, respectively. These functional groups are both present in
benzyl alcohol, thus, compound B is benzyl alcohol.

Based on the IR result, I conclude that compound C is 4-methyl-pentan-2-one because the only
observed peaks belong to an alkane which is around 3000-2850 cm-1, and a peak at around 1700
cm-1, which is an indication of a C=O bond. The compound that has these two functional groups
are 4-methyl-pentan-2-one.

Based on the IR result, I conclude that compound D is acetophenone because there were
observed peaks at 3100-3000 cm-1 and 1700 cm-1 which correspond to aromatic C-H stretching
and C=O bond stretching. Acetophenone has an aromatic ring and a C=O bond, thus, compound
D is acetophenone.

Part III
1. Which of the four compounds has an MS as shown below? Explain all the major
fragments,

The molecular ion peak of the compound is at m/z = 100. Among the compounds given, 4-methyl2-pentanone has a molecular mass of 100. Thus, the MS spectrum belongs to 4-methyl-2pentanone. The fragment at m/z = 85 correspond to the loss of the C=O bond while the base peak
which is at m/z = 43 corresponds to the following fragment:

2. The following 1HNMR belong to which of the four compounds in this experiment (s: singlet;
d: doublet; t: triplet; m: multiplet)? Explain.

The NMR spectrum belongs to 2-pentanol. There are five peaks in the spectrum and among the

given compounds, 2-pentanol and benzyl alcohol are possible. 2-pentanol is expected to have 6
peaks, however, the protons of C3 and C4 can be observed as a single peak. Thus, only 5 peaks
are observed. This is shown by the multiplet peak at around 1.5 ppm with an integration of 4.
The triplet peak with 3 protons at around 1 ppm belongs to the protons of C5 while the doublet
at around 1 ppm also with the integration of 3 belongs to the protons at the C1 position. Also,
there is no observed peak at around 7-8 ppm, so, the compound does not contain an aromatic
ring. Thus, the compound is indeed 2-pentanol.

3.

Which

of

the

four

compounds

has

the

following

13CNMR?

Explain.

The compound is acetophenone. The distinct feature of the NMR spectrum is the peak at
around 200 ppm which is an indication of the presence of a C=O bond. There are only two
compounds that have a C=O bond: acetophenone and 4-methyl-pentan-2-one. The structural
difference between these two compounds is the presence of an aromatic ring in
acetophenone. In the spectrum, there are peaks at around 120-140 ppm which are an
indication of aromatic carbons. Thus, the compound is acetophenone.

Please view explanation and answer below.Please take a look at this one.

Experiment 3
Determining the Structures of Unknown Molecules Using NMR, IR, and Mass Spectroscopy

Compo
und #

1

#

Approximate

Main IR

Expected

One Possible

Carbon Proton

Chemical Shift of

Stretch

Molecular

Fragment in M.S.

Signals Signals

Protons

5

#

5

1) The proton
in OH
appears
randomly
on 1HNMR
2) The
protons in
CH2
appears @
~4-5 ppm
(singlet)
3) Protons on
the
benzene
ring
appear

Ion in M.S.
3200-3600
cm-1

108

@~7 ppm
(multiplet)
2

5

6

1) The proton
at C2 is a
sextet and
appears at
around 4
ppm.
2) The proton
of the OH
group
appears
randomly.
3) The proton
at C3 is a
quintet
and
appears at
around 1
ppm.
4) The proton
at C4 is a
sextet and
appears at
around 1
ppm.
5) The proton
at C1 is a
doublet
and
appears at
around 1
ppm.
6) The proton
at C5 is a
triplet and
appears at
around 1
ppm.

around
3400-3200
cm-1

88

3

6

4

1) One of the
aromatic
protons is
a doublet
and
appears at
around 8
ppm.
2) Another
aromatic
proton is a
triplet and
appears at
around 7-8
ppm.
3) The third
aromatic
proton
appears as
a triplet at
around 7
ppm.
4) The
protons of
the methyl
group
appear at a
singlet at
around 2-3
ppm.

1680 cm-1

120

4

5

4

1) The proton
at C3
appears as
a doublet
at around
2-3 ppm.
2) The proton
at C1
appears as
a singlet at
around 22.5 ppm.

1700 cm-1

100

3) The proton
at C4 is a
multiplet
and
appears at
around 2
ppm.
4) The
protons at
C5 and the
methyl
group are
equivalent
and appear
as a
doublet at
around 1
ppm.

Part II
Based on the IR result, I conclude that compound B is benzyl alcohol because there is an observed
absorption at 3600-3200 cm-1 and 3100-3000 cm-1 which suggests the presence of an alcohol
g...


Anonymous
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