12A Lab Activity Notes Lab Activity 12 In this experiment, RX reacts with a base. The methoxide ion (CH 3 O - ) is a small, strong base. The tert-butoxide ion ((CH 3 ) 3 CO - ) is a large, strong base. 1. Question 4. See Table 3. For the 2-bromopentane + ethoxide (EtO - ) reaction, explain the product composition. In other words, if a reaction produced more substitution product, explain why more substitution product formed than elimination product. Include substrate, nucleophile size and strength, leaving group, and solvent. The data in Table 3 show the 2-bromopentane + ethoxide (EtO - ) reaction produces 70.8% alkene (elimination product) and 29.2% ether (substitution product). 2-bromopentane is a secondary (2 o ) RX. Ethoxide (EtO - ) is a large, strong base. More elimination product forms because the large, strong base has easier access to the H bonded to the beta carbon than the alpha carbon. 2. Questions 5c and d. Based on the several reactions in Lab Activity 12, what factor(s) make a substitution or elimination reaction occur? a. RX reacts with a base. What type of RX usually produces more substitution product? Depends on size and strength of base. b. A small, strong base reacts with RX. More _ substitution product forms if _ primary RX reacts. See Table 3. More substitution product forms for the following reactions: 1-bromopentane + MeO - ---> 1.8% alkene (elimination product) + 98.2 ether (substitution product) 1-bromopentane + EtO - ---> 2.8% alkene (elimination product) + 97.2 ether (substitution product) 1-bromopentane + t-buo - ---> 25.8% alkene (elimination product) + 74.2 ether (substitution product) 2-bromo-2-methyl butane + MeO - ---> 47.3% alkene (elimination product) + 52.7 ether (substitution product) c. A large, strong base reacts with RX. More elimination product forms if _ secondary and tertiary RX reacts. See Table 3. More elimination product forms for the all the reactions except: 1-bromopentane + MeO - ---> 1-bromopentane + EtO - ---> 1-bromopentane + t-buo - ---> 2-bromo-2-methyl butane + MeO - ---> d. If a weak base reacts with RX, I would expect more elimination product to form if _ tertiary RX reacts. The alpha C is a better electrophile than a H bonded to a beta carbon. When a primary or secondary RX reacts with a weak base, the substitution product is favored because the more electrophilic alpha C is more reactive than a H bonded to a beta carbon and can react with a weak base (weak nucleophile). So when a tertiary RX reacts with a weak base, the elimination product is favored e. Based on the Table 3 data, when RX reacts with a base, more product forms. Can t say based on the data. See Table 3. Based on the 8 reactions, 4 reaction produced more substitution product and 4 reactions produced more elimination product. f. In general, when RX reacts with a base, more elimination product forms.
Lab Activity 11 1. Lab 5 Report, Part A. Question 3a. Provide experimental evidence that you made the desired product. a. % yield of 1-bromobutane % yield = (actual yield/theoretical yield) x 100 theoretical yield of 1-bromobutane = 1.48 g b. Describe IR spectrum and interpretation (match each peak to a bond type and each bond type to the structure). IR gives us information about structure. Match each IR peak to a bond type. Then, match the bond type to the structure of your product. 1-bromobutane has the following bond types: C-H bonds: IR peaks for C-H stretch at approximately 3000 cm -1 and C-H bend at 1450 and 1375 cm -1. C-Br bond: IR peak for C-Br between 500-600 cm -1. C-C bond - not interpretatively useful. NOTE: Several students are not reading and interpreting an IR spectrum correctly. The energy in wavenumbers plotted on the x-axis of an IR spectrum tells you the bond type, not the % transmittance plotted on the y axis. c. Describe GC retention time and compare to the retention time of 1-bromobutane. GC gives us information about a specific substance. Each substance takes a different amount of time (retention time) to pass through the GC column. NOTE: the retention time in a gas chromatogram is not related to bond stretch or bend, like in IR. Compare the retention time(s) of the peak(s) in your sample to the 1-bromobutane and 1-butanol retention times to determine if you made the desired product. Using the GC with a polar column, N 2 moving phase, oven temperature 85-150 o C with 10 o C/min ramp and 6.5 min run time, 1-bromobutane has a retention time of 2.2 min and 1-butanol has a retention time of 3.1 min. 2. Lab 5 Report, Part A. Question 4. Conclusion: summarize your results in one sentence. 1-bromobutane was produced in % yield = by reacting 1-butanol with NaBr and H 2 SO 4 by reflux for 50 min/microwave for min; evidence for 1-bromobutane was shown by a C-Br peak in the IR spectrum and retention time of 2.2 minutes by GC that matched the 1-bromobutane sample. 3. Lab 5 Report, Part B. Question 1. Provide experimental evidence about the mechanism of the t-butyl chloride to t-butanol reaction. This reaction involves a 3 o RX so we predict the reaction mechanism to be S N 1. A 3 o carbocation should form. The mechanism is shown below:
In this reaction, the ph should drop to around 3 to 5. The color should change from blue to green to yellow to red. If this reaction occurred by a S N 2 mechanism, OH - should react at the alpha C to replace the Cl.- The ph will drop to about 7 due to less OH - in solution. HCl does not form so the ph will not drop below 7. 4. Lab 5 Report, Part B. Question 2. Provide experimental evidence about the mechanism of the 2-chlorobutane to 2-butanol reaction. If the reaction mechanism is S N 1, you should see the color change from blue to green to yellow to red. This indicates the ph dropped from about ph 10 to ph 3 to 5 due to the formation of HCl. If the reaction mechanism is S N 2, you should see the color change from blue to green. This indicates the ph dropped from about ph 10 to ph 7 or 8 due to less OH- in solution. 5. a. Lab 6 Prelab. Question 1g. What property of the product distinguishes it from the reactant? How will you characterize the product? 2-methyl-2-butanol has a boiling point of 102 o C. 2-methyl-1-butene has a boiling point of 31.2 o C. 2-methyl-2-butene has a boiling point of 38.6 o C. You will characterize the product by GC. b. Lab 6 Prelab. Question 1f. Calculate the theoretical yield of product. 2-methyl-2-butanol + sulfuric acid 2-methyl-1-butene or 2-methyl-2-butene + H 2 O + sulfuric acid 0.80 g 2-methyl-2-butanol/88 g/mole = 0.0091 moles limiting reactant 0.5 ml concentrated H 2 SO 4 (1.84 g/ml)/98 g/mole = 0.0094 moles 0.0091 moles of C5H10 produced (molar mass = 70 g/mole) = 0.64 g = theoretical yield Lab Activity 10 1. Part A. Distillation Question 5. a. You have a mixture of 2-propanol and 2-pentanone. Would you use a simple distillation or fractional distillation to separate this mixture? Give reasons. 2-propanol has a boiling point of 81-83 o C and 2-pentanone has a boiling point of 101-105 o C. The difference in boiling point is approximately 20 o C so a fractional distillation should be used to separate this mixture. A simple distillation can be used to separate liquids with a boiling point difference of 40 o C or greater.
2. Part A. Distillation Question 5. b. Graph 1 shows 2-propanol and 2-pentanone distillation data with temperature in o C on the y-axis and volume of distillate in ml on the x-axis. Which substance is being distilled in the first 4 ml of distillate? What property of this substance is shown in this graph? 2-propanol (the lower boiling point component) is distilled in the first 4 ml of distillate. Note the temperature matches the boiling point of this substance. The property of this substance is shown in this graph is boiling point. 3. Part B. Chromatography Question 2. A mixture of a polar substance and non-polar substance is separated by GC with a non-polar column. The substance that should show a longer retention time is. A non-polar substance is more attracted to the non-polar stationary phase column than a polar substance so the non-polar substance has a longer retention time. 4. Part B. Chromatography Question 3. You run a GC experiment for the Lab 5 substitution reaction.you run a standard sample of 1-butanol, which shows a retention time of 3.0 minutes, and 1-bromobutane, which shows a retention time of 2.1 minutes.you run your reaction mixture, which shows a big peak at 2.0 minutes and a small peak at 2.95 minutes. Did you make the desired product? Give reasons. For Lab 5, 1-butanol ---> 1-bromobutane. The big peak with a 2.0 minute retention time matches the retention time of 2.1 minutes for the 1-bromobutane standard. The small peak with a 2.95 minute retention time matches the retention time of 3.0 minutes for the 1-butanol standard. Bigger peak means more substance so you made the desired 1-bromobutane product. Lab Activity 9 1. Intro to Reactivity a. See Question 2. 1-bromobutane reacts with NaI. What observation in Lab 4 tells you this reaction occurred or did not occur? If this reaction occurred, the white NaBr precipitate forms. b. See Question 2. In 1-bromobutane, carbon 1 is the alpha carbon. Is the alpha carbon a nucleophile or electrophile? Alpha carbon is an electrophile. c. See Question 2f. The iodide ion reacts at an alpha carbon. Which carbon is easier for the iodide ion to get to? Give reasons. The iodide ion has easier access to the alpha carbon in 1-bromobutane. There are fewer atoms blocking the iodide to the alpha carbon in 1-bromobutane (1 o RX) compared to 2-chloro-2-methylpropane (3 o RX). d. See Question 7. Cyclohexene reacts with Br. What observation in Lab 4 tells you this reaction occurred or did not occur? If this reaction occurred, the red Br 2 color disappears. e. See Question 7. Cyclohexene reacts with Br 2. Which atom or bond in ethylene behaves like a nucleophile? The C=C pi bond is a nucleophile. f. See Question 6. Propylene reacts with HCl. Which carbocation intermediate forms? A 2 o carbocation forms, which is more stable than a 1 o carbocation. 2. Pre-Lab 5. Part A. Calculate the moles of each reactant (NaBr and 1-butanol) in this reaction. Identify the limiting reactant. Calculate the theoretical yield of product. Overall reaction: C 4 H 9 OH + NaBr + H 2 SO 4 C 4 H 9 Br + H 2 O + NaHSO 4 1:1 mole ratio of C 4 H 9 OH to NaBr 1.33 g NaBr x (1 mole/103 g) = 0.013 moles
0.8 g 1-butanol x (1 mole/74 g) = 0.0108 moles limiting reactant Moles of C 4 H 9 Br product = 0.0108 moles C 4 H 9 OH x (1 mole C 4 H 9 Br/1 mole C 4 H 9 OH) = 0.0108 moles C 4 H 9 Br Theoretical yield of C 4 H 9 Br = 0.0108 moles x 137 g/mole = 1.48 g 3. Pre-Lab 5. Part B. Calculate the moles of each reactant (NaOH and t-butyl chloride) in this reaction. Identify the limiting reactant. Calculate the theoretical yield of product. Overall reaction: C 4 H 9 Cl + NaOH C 4 H 9 OH + NaCl 1:1 mole ratio of C 4 H 9 Cl to NaOH 1.5 ml of 0.1 M NaOH = 0.0015 liters x 0.1 moles/liter = 0.00015 moles NaOH 1 ml of 5% by volume C 4 H 9 Cl = 1 ml solution x (5 ml C 4 H 9 Cl/100 ml solution) = 0.05 ml C 4 H 9 Cl x 0.84 g/ml x (1 mole/92.5 g) = 0.00045 moles limiting reactant Moles of C 4 H 9 OH product = 0.00045 moles C 4 H 9 Cl x (1 mole C 4 H 9 OH /1 mole C 4 H 9 Cl ) = 0.00045 moles C 4 H 9 OH Theoretical yield of C 4 H 9 OH = 0.00045 moles x 74 g/mole = 0.034 g 4. Reflux Question 2b. What is the reason for doing a reflux instead of just heating? If you heat a mixture to boiling, it will evaporate and you will have to add more solvent periodically. If you do a reflux, the solvent that evaporates condenses when it comes in contact with the cold condenser and drops back into the reaction flask. 5. Reflux Question 5. In Lab 5, you will reflux a mixture of NaBr (s), H 2 O (l), butanol (l), and concentrated H 2 SO 4 (aq). What is the approximate temperature at which this reaction mixture boils? Give reasons. Approximately 100 o C. Compare the boiling points of the liquids. The lowest boiling point liquid is water. Lab Activity 8 NMR. See UCLA Web Spectra (https://webspectra.chem.ucla.edu/#problems) Beginning Compound 1. The answer is ethyl acetate. 1. a. How many non-equivalent protons does ethyl acetate have? In other words, how many peaks would be observed in a H NMR spectrum of this compound? 3 non-equivalent protons b. What is the ratio of non-equivalent protons in this compound? 3:2:3 c. Explain why one methyl group shows a singlet but the other methyl group shows a triplet. The splitting of a main peak into sub-peaks depends on the number of non-equivalent H s on an adjacent carbon. Methyl group bonded to the C that is double bonded to O - the C adjacent to this methyl group does not have any H s bonded to it so the NMR peak for the 3 H s in this methyl group will not be split singlet. Methyl group bonded to the CH 2 group - the C adjacent to this methyl group is the CH2 group. This CH2 group has 2 H s bonded to it. Using the n + 1 rule, the NMR peak for the 3 H s in this methyl group will be split into 2 + 1 = triplet.
2. Picospin NMR instrument operation. a. What are the two variables to adjust to obtain a "good" NMR spectrum? 1. Auto tx offset 2. Phase correction b. What information does the NMR peak area give you? Peak area is proportional to the number of equivalent protons. Lab 3 report. 4. How many chirality centers are in limonene? 1 5. How many stereoisomers does limonene have? Apply 2 n rule so 2 1 = 2 stereoisomers. 6. Limonene does not have any meso compounds because _ only 1 chirality center and no plane of symmetry. 7. a. What does the energy from the microwave oven do to the limonene in lemons or oranges? Hint: what is the state of matter of limonene before you turned on the microwave oven? Microwave heats up limonene to change phase from liquid to gas. b. What is the function of the frozen ice core? When the limonene gas comes in contact with the frozen ice core, the gas condenses to a vapor. c. Do you want the frozen ice core to melt in this experiment? No d. Report the volume of limonene you extracted. Lab Activity 7 Lab 2 Report 1. a. Report the % recovery of benzoic acid. % recovery = (mass recovered/original mass) x 100 Original mass = 0.36/3 = 0.12 g b. Report your experimental melting point of benzoic acid. Report a melting point range - this range gives you information about the purity of your sample. c. Report the true melting point of benzoic acid. 122 o C 2. a. Report the % recovery of 4-tert-butylphenol. % recovery = (mass recovered/original mass) x 100 Original mass = 0.36/3 = 0.12 g b. Report your experimental melting point of 4-tert-butylphenol. Report a melting point range - this range gives you information about the purity of your sample. c. Report true melting point of 4-tert-butylphenol. 97-101 o C 3. a. Report the % recovery of dimethoxybenzene. % recovery = (mass recovered/original mass) x 100 Original mass = 0.36/3 = 0.12 g b. Report your experimental melting point of dimethoxybenzene. Report a melting point range - this range gives you information about the purity of your sample. c. Report the true melting point of 1,4-dimethoxybenzene. 1,4-dimethoxybenzene has a melting point of 54-56 o C
4. Were you able to separate the acid, base, and neutral compounds by an acid-base extraction? Support your answer with your data and results. For each solid you isolated, did the melting point match the true melting point? Include numbers. If so, then you were able to separate and isolate the specific component of the mixture. Example: We were able to separate the benzoic acid-4-t-butylphenol-dimethoxybenzene mixture by acid-base titration because we isolated three separate solids. \We recovered 50.0% of the first solid. This solid had a m.p. of 120-125 o C, which indicates the solid is benzoic acid. We recovered 40.0% of the second solid. This solid had a m.p. of 90-105 o C, which indicates the solid is-4-t-butylphenol (although it is not very pure). We recovered 4.0% of the third solid. This solid had a m.p. of 52-55 o C, which indicates the solid is dimethoxybenzene. 5. Give the chemical formula of the substance you used to separate the 4-tert-butylphenol from the dimethoxybenzene. NaOH was used to separate the 4-tert-butylphenol from the dimethoxybenzene. The 4-tert-butylphenol and dimethoxybenzene were dissolved in ether. Add NaOH (aq) to this ether solution and you saw two layers form. NaOH (aq) reacts with the -OH group in 4-tert-butylphenol to form its conjugate base. The conjugate base of 4-tert-butylphenol is more soluble in water than ether so this ion is extracted into the aqueous layer while the dimethoxybenzene remains in the ether - separation! Stereochemistry - Go to the shared 12A Lab Activity 7 Google Doc to show your answers to Question 5. Lab Activity 6 1. From Question 1e. Test Tube 1 (the top layer) should be the tert-butyl methyl ether layer because tert-butyl methyl ether is less dense than water (0.74 g/ml for tert-butyl methyl ether vs. 1.0 g/ml for water). 2. From Question 1e. Tube 1 should contain 4-tertbutylphenol and tert-butyl methyl ether. 3. From Question 1f. Test Tube 2 (the bottom layer) should be the aqueous layer because water is more dense than tert-butyl methyl ether (0.74 g/ml for tert-butyl methyl ether vs. 1.0 g/ml for water). Note: you need to state the relative densities of the layers. 4. From Question 1f. Tube 2 should contain the conjugate base of benzoic acid (benzoate ion). You want to convert the benzoate ion back to benzoic acid by adding acid (use 12 M HCl).The acid reacts with unreacted NaHCO 3 and benzoate ion. 5. From Question 1f. Dropwise add enough 12 M HCl until the solution is neutral. Then add one more drop of 12 M HCl. You should see a solid form. This solid is _ benzoic acid. Lab Activity 5 1. You mix HCl and NaOH and NaHCO3. HCl reacts with. NaOH. The strongest acid reacts with the strongest base. NaOH is a stronger base than NaHCO 3 (see pk a table). 2. You dissolve the benzoic acid, 4-tert-butylphenol, and dimethoxybenzene mixture in ether. Benzoic acid does not react with 4-tert-butylphenol because.
The O in the alcohol group in 4-tert-butylphenol is too weak of a base to react with benzoic acid (see pk a table). In this question, the alcohol behaves like a base. The pka of the conjugate acid of an alcohol is -2. 3. An OTC analgesic contains 250 mg aspirin, 250 mg acetaminophen, and 50 g sucrose per tablet. To make sure the amounts on the drug label are accurate, you are asked to determine the amount of each ingredient in one tablet. a. Which compound is an acid? What functional group makes it an acid? Report the pka. Aspirin. The acid functional group makes it an acid with a pk a of approximately 5. b. Which compound is a base? What functional group makes it a base? Report the pka of the conjugate acid of this base. Acetaminophen. The alcohol functional group makes it a base. The pka of the conjugate acid of ROH is approximately -2. 4. a. Which solvent would you use to dissolve a mixture of aspirin, acetaminophen, and sucrose? Ethanol b. Give reasons for you choice in Question 4a. Find a solvent in which all three compounds are soluble. Aspirin, acetaminophen, and sucrose are soluble in ethanol. Sucrose is soluble in water; aspirin and acetaminophen are slightly soluble in water. Sucrose is not soluble in ether or dichloromethane. 5. a. Which substance (reagent) would you add to a mixture of aspirin, acetaminophen, and sucrose to separate the aspirin from the acetaminophen and sucrose? Based on pk a table, use a base that is strong enough to react with aspirin but not the acetaminophen, e.g., NaHCO 3. b. Give reasons for you choice in Question 5a. NaHCO 3 is a strong enough base (below) to react with aspirin but is too weak to react with acetaminophen. Lab Activity 4 Lab 1 Aspirin Lab Report. Conclusion summarize your results in one sentence. Include numbers. a. mass of aspirin in tablet in g. Report number only; do not include text. b. mass of aspirin recovered in g. Report number only; do not include text. c. % recovery of aspirin. Report number only; do not include text. % recovery = (mass of aspirin recovered/mass of aspirin in tablet) x 100 (NOTE: many students used a wrong % recovery formula) Ideally, your % recovery should be high but not greater than 100%. d. melting point range of your aspirin sample. Report number only; do not include text. True melting point of aspirin = 136 o C Report a melting point range. This range indicates purity. The narrower the range, the more pure your solid. e. If your % recovery was less than 90%, identify the specific step in your experiment that accounts for the loss of aspirin. Step 1: measure mass of aspirin tablet and crush tablet. Step 2: dissolve tablet in water. Step 3a: add dichloromethane to aspirin-water solution, mix, allow two layers to form (liquid-liquid extraction). The aspirin should move from the water to the dichloromethane. In other words, dichloromethane extracts the aspirin from the aqueous solution. If all of the aspirin is not extracted into the dichloromethane, you will lose aspirin in this step. Step 3b: separate the two layers. E.g., use a pipet. Incomplete separation could lead to loss of aspirin.
Step 4a: add MgSO 4 drying agent to dichloromethane to remove any water present. Aspirin could stick (adsorb) to the MgSO 4 pellets and would result in losing some aspirin. Step 4b: separate the solid drying agent from the dichloromethane solution. Step 5: Separate the aspirin from the dichloromethane by evaporating the dichoromethane, which leaves solid aspirin in your container.. Step 6: Dry the aspirin in the oven. Step 7. Measure mass of aspirin recovered. Step 8: measure melting point of aspirin. f. You add aspirin (crushed) to water in Test Tube 1. Then, you add dichloromethane and see two layers form. You use a pipet to remove the lower layer and place this liquid in Test Tube 2. (i) The aspirin is in Test Tube 2 (ii) MgSO4 should be placed in Test Tube 2 Lab 2 Prelab. After benzoic acid, 4-tert-butylphenol, and dimethoxybenzene are dissolved in ether and NaHCO 3 is added, what is the lower layer solvent and component (benzoic acid, 4-tert-butylphenol, or dimethoxybenzene)? What is the upper layer solvent and component (benzoic acid, 4-tert-butylphenol, or dimethoxybenzene)? a. lower layer solvent: water b. lower layer component(s): benzoic acid Based on pka, the benzoic acid is a strong enough acid to react with NaHCO 3 in an acid-base reaction but the 4-tert-butylphenol and dimethoxybenzene are too weak of an acid to react with NaHCO 3. c. upper layer solvent: ether d. upper layer component(s): 4-tert-butylphenol and dimethoxybenzene Lab Activity 3 Lab 1, Part C Prelab. 1. Then add an organic solvent probably dichloromethane. Not ethanol or methanol because they mix with water! Shake well and separate using a separating funnel. (Isolate the aspirin from the other ingredients.) a. Is this a solid-liquid extraction or liquid-liquid extraction? liquid-liquid extraction - the aspirin-water solution is a liquid. You will use dichloromethane as the second extraction liquid. b. Two layers should form. The top layer is water. Compare density of water (1.0 g/ml) to dichloromethane (1.33 g/ml). c. Is aspirin more soluble in water or in dichloromethane? Give numbers to support your answer. Aspirin is more soluble in dichloromethane. At 25 o C, 1 g of aspirin dissolves in 100 ml of water. Aspirin is not very soluble in water. At 25 o C, aspirin is soluble in dichloromethane (approximately 1 g of aspirin dissolves in 25 ml dichloromethane). d. You separate the two layers. The aspirin should be in the dichloromethane layer. See Question 1c. 2. Drying Agents: If you are using a solid drying agent, how can you tell when a solvent is dry? The drying agent should be free flowing when the mixture is shaken. Lab Activity 2
Part A. You would recrystallize benzoic acid using water as the solvent rather than methanol. Explain why. Benzoic acid is soluble in water at high T but insoluble at low T. If benzoic acid is dissolved in hot water to make a saturated solution, it will start to crystallize (precipitate) out of solution as the solution cools. Benzoic acid is soluble in methanol at low, medium, and high T. If benzoic acid is dissolved in hot methanol to make a saturated solution, it will not crystallize (precipitate) out of solution as the solution cools. Part B. 1. Ethanol is more soluble in water than in hexane. Predict whether the distribution coefficient = [ethanol (water)]/[ethanol (hexane)] is greater than 1 or less than 1. Give reasons. [ethanol (water)] is greater than [ethanol (hexane)] because ethanol is more soluble in water than in hexane so the distribution coefficient is greater than 1. Electronegativity refers to the ability of an atom in a bond to attract electrons toward itself. Atoms in a compound have an electronegativity but compounds do not have an electronegativity. Part B. 2. Ethanol is miscible in hexane. Add water to this solution and mix. Two layers should form. Is water the top layer or bottom layer? Give reasons. Water is the bottom layer because it is most dense liquid. Density of water = 1 g/ml Density of ethanol = 0.8 g/ml Density of hexane = 0.7 g/ml Lab Activity 1 Determine the structure and identity of a compound from an IR spectrum. For each IR peak, determine the bond type, e.g., C-H bond. Relate the bond type to the structure. Look at structure of the four choices: Cyclohexane, Cyclohexanol, Cyclohexanone, methyl cyclohexane. Identify the bond types in each compound. Each compound has C-H bonds (approximately 3000 cm -1 for the C-H stretch and 1450 cm -1 for the C-H bend, and C-C bonds (not interpretatively useful in IR) so each spectrum should show peaks at 3000 and 1450 cm -1. 1. a. What is the identity of the substance in IR spectrum 1? Cyclohexane. Peaks at approximately 3000 cm -1 for the C-H stretch and 1450 cm -1 for the C-H bend. Peaks at 1450 and 1375 cm -1 is the C-H bend in a -CH 3 group. Since the peak at 1375 cm -1 is not observed, there is no -CH 3 group so this IR spectrum is not methyl cyclohexane. 1. b. What is the identity of the substance in IR spectrum 2? Cyclohexanol Peak at 3200-3500 cm -1 means O-H bond. 1. c. What is the identity of the substance in IR spectrum 3? Cyclohexanone Peak at 1700 cm -1 means C=O bond. Peak at 3200-3500 cm -1 means O-H bond but this peak is small and is probably due to water in the sample. 1. d. Which bond type is represented by the IR peak at approximately 1700 cm -1? C=O bond.
Make sure you identify the specific atoms in the bond. It is not enough to say double bond. 2. Cyclohexene and toluene have a C=C bond. The C=C bond peak is at a higher energy in cyclohexene than toluene. Is the C=C bond in cyclohexene stronger, weaker, or the same strength as the C=C bond in toluene? Give reasons. Higher energy (higher cm -1 ) means more energy is required to stretch or bond a bond. More energy to stretch a bond means a stronger bond. So the C=C bond in cyclohexene is stronger than the C=C bond in toluene. Note: peak intensity has to do with the size of the peak (big or small). Peak energy refers to wavenumbers (cm -1 ).