b. What leaving group is involved? The leaving group in Compound 2 is -OH. -OH is a poor LG and It needs to be made into a better LG.

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1 For Prelab Questions: Look at the Lab Experiment on the Chem 12B Lab Information webpage for the lab procedure instead of searching online for an experimental procedure. Lab Activity Compound 3 is called a glycosylamine. What is this compound called, e.g., diol, acetal, eneamine, according to CHM 12B lecture? Compound 3 is an imine. 2. We looked how a nucleophile reacts with a carbonyl carbon in an aldehyde or ketone to form a tetrahedral intermediate. In the first step of the Maillard reaction, the amine group in a protein or amino acid reacts with an aldehyde group in a sugar to form an imine (Compound 3). a. Which compound is the tetrahedral intermediate? Compound 2 b. What leaving group is involved? The leaving group in Compound 2 is -OH. -OH is a poor LG and It needs to be made into a better LG. 3. a. See Compounds 1-8. The Compound (given Compounds 4, 5, and 6) that is not an enol is Compound 5. b. The enol forms (tautomerizes) into a ketone because the enol is less stable than the ketone. 4. You tested Question 1d by making toast. You added (sprinkled, spread) a small amount of base to a small area of the bread. a. What base in your kitchen can (did) you use? Baking soda (NaHCO 3 ) is a common base found in a household kitchen. Soap is another common base but you probably don t want to put this on your toast. Some of you used bleach for your base. Bleach is also an oxidizing agent so the bleach could oxidize something in the bread. Egg white is a base (ph approx. 8 according to uality/ ). Butter is not a base. Butter is a fatty acid or triglyceride. b. Did a base produce a higher yield of Maillard reaction products? Compare the color of the toast with the base and without the base (your control experiment). If the toast color is darker brown with the base, the base produced a higher yield of Maillard reaction products. c. Support your answer to 4b with numbers. Report the numbers you recorded in your color scale for the toast with base and without base.

2 d. Support your answer to 4b by uploading a photo of your toast to the shared Google doc "12B Lab Activity 12". Lab Activity 11 Lab 4b report 1. a. State whether you synthesized -bromoacetanilide. The amide group is an ortho, para director, so state whether you synthesized o-bromoacetanilide (2-bromoacetanilide) or p-bromoacetanilide (4-bromoacetanilide). Some students stated ortho or para (or 2- or 4-) and did not state whether you synthesized o-bromoacetanilide (2-bromoacetanilide) or p-bromoacetanilide (4-bromoacetanilide). b. Report the % yield. % yield = (actual yield/theoretical yield)x100 c. How do you know you made the ortho, meta, or para product? Compare your characterization data to the three possible substitution products. Best answers: We were able to characterize the product by melting point as our experimental melting point was between and which is close to the true melting point of para-bromoacetanilide 165 degrees celsius, and the true melting points of ortho-bromoacetanilide is 69 degrees C and meta-bromoacetanilide is 81 degrees C. Our experimental melting point ranges of degrees C for reflux and degrees C for the microwave reaction are closest to the melting point range of the para product: 168 degrees C. The ortho product has a MP of 102 degrees C, whereas the meta product has a MP range of degrees C, nowhere near our experimental MP values. Many students only reported the m.p. of one isomer and not the other two isomers. d. What is the purpose of adding 10% NaHSO 3 to the solution? (Hint: what type of reagent is NaHSO 3?) NaHSO 3 is a reducing agent and reacts with any unreacted Br 2 (oxidizing agent). Lab 4c report 2. a. State whether you synthesized -nitrosalicylic acid. The acid group is a deactivating group and meta director; the alcohol group is an activating group and ortho, para director, so the alcohol group should direct the position of EAS. State whether you synthesized 2-Hydroxy-3-nitrobenzoic acid (3-nitrosalicylic acid) or 2-Hydroxy-5-nitrobenzoic acid (5-nitrosalicylic acid). Some students stated 3- (or meta) and did not state whether you synthesized 3-nitrosalicylic acid. b. Support your answer to 2a with your characterization data.

3 Compare your experimental m.p. to the true m.p. Since there are three isomers, state these m.p. Many students only reported the m.p. of one isomer and not the other two isomers. c. Explain how your product supports the idea that the better activating group directs the EAS reaction. The acid group is a deactivating group and meta director; the alcohol group is an activating group and ortho, para director, so the alcohol group should direct the position of EAS. Best answer: The product we formed had the group attached in the ortho position to the highest activating group which was the Alcohol (OH). Lab 5 Prelab 1. a. Benzaldehyde reacts with allyl bromide acid and Zn to form _ 1-phenyl-3-buten-1-ol _. This reaction is a _addition reaction. (acid-base, substitution, elimination, addition) b. The function of benzaldehyde is Electrophile (carbonyl carbon) c. Calculate the theoretical yield in g of product g Zn = moles 0.21 ml benzaldehyde = moles = limiting reactant 0.21 ml allyl bromide = moles So moles of 1-phenyl-3-buten-1-ol are produced: moles x 148 g/mole = g = theoretical yield. Lab Activity Salicylic acid reacts with HNO 3 to form _ nitrosalicylic acid _. This reaction is a _ electrophilic aromatic substitution reaction. 2. a. The acid group in salicylic acid is a deactivating group. b. The group that directs the substitution is _ -OH group. 3. What is the function of calcium nitrate in this reaction? Calcium nitrate is the source of NO 2 +, which is the electrophile in this reaction. Calcium nitrate reacts with acetic acid: Ca(ONO 2 ) CH 3 COOH ---> 2 CH 3 COONO 2 + Ca(OH) 2 Think of CH 3 COONO 2 as (CH 3 COO - )(NO 2 + ). The electrophile is NO Calculate the theoretical yield in g of product. Reactants: 1.5 g Ca(ONO 2 ) 2 = moles

4 moles Ca(ONO 2 ) 2 produces moles CH 3 COONO 2 (see 1:2 mole ratio) = moles NO g C 7 H 6 O 3 (salicylic acid) = moles. Salicylic acid is the limiting reactant so moles of nitrosalicylic acid is produced moles of nitrosalicylic acid = 1.33 g nitrosalicylic acid theoretical yield = 1.33 g nitrosalicylic acid 5. How will you characterize the product? Melting point. The product is a solid. From Sigma-Aldrich: melting point of 3-nitrosalicylic acid = C melting point of 4-nitrosalicylic acid = C melting point of 5-nitrosalicylic acid = C In salicylic acid, Carbon 1 is bonded to the acid group Lab Activity 9 Lab 4a Report 1. Conclusion: a. The best electron donating group is because. Include numbers. The best electron donating group is the group that reacted with Br 2 by EAS the fastest. For Group A, the best electron donating groups are -OCH 3 and -OOCCH 3 because the anisole and phenyl acetate EAS bromination reaction occurred in instantly. For Group B, the best electron donating group are -OCH 3 and -OOCCH 3 because the anisole and phenyl acetate EAS bromination reaction occurred in instantly. b. The best electron withdrawing group is because. Include numbers. The best electron withdrawing group is the group that reacted with Br 2 by EAS the slowest. For Group A, the best electron withdrawing group is either -CH 3 -COCH 3 because the toluene or acetophenone EAS bromination reactions did not occur after 40 minutes. For Group B, the best electron withdrawing group is -CH 3 because the toluene EAS bromination reactions did not occur after 35 minutes. Lab 4b Prelab 2. a. This reaction is an EAS reaction. Is the acetamido group in acetanilide an activating group or deactivating group? activating group b. Would you expect the ortho, meta, or para substitution product? Ortho and para products since the acetamido group is an activating group. 3. Calculate the theoretical yield in g of product. The reaction of interest:

5 Br 2 is generated in situ from this reaction: 6 H Br BrO 3 ---> 3 Br H 2 O You will use 48% HBr = (48 ml HBr/100 ml solution) x (1.49 g/ml)/81 g/mole = 8.8 M HBr 0.3 ml of 48% HBr = liters x 8.8 M = moles HBr = moles Br mg of KBrO 3 = g/167 g/mole = moles BrO The BrO 3 is the limiting reactant (see coefficients in balanced equation) so moles of Br 2 is produced. This amount of B r2 reacts with acetanilide. Back to the first reaction: 200 mg of acetanilide = 0.2 g/135 g/mole = moles acetanilide moles of Br 2 So moles of the Br substituted acetanilide is produced. Theoretical yield of Br substituted acetanilide = moles x 214 g/mole = 0.32 g 4. How will you characterize the product? Melting point. The product is a solid. Look up the m.p. of Ortho-bromoacetanilide = o C Meta-bromoacetanilide = o C Para-bromoacetanilide = o C Lab Activity 8 1. Conclusion: a. Identify the fruit dye that produced the highest voltage DSSC. b. Rank the fruit dyes by voltage. Note: your fruit dye DSSC typically produces 0.1 to 0.5 V in sunlight or artificial light. c. A DSSC is a type of electrochemical cell. When sunlight of the right wavelength is _ absorbed by the dye to create an _ excited state of the dye, an electron is injected into the conduction _ band of TiO 2 where it moves by diffusion to the anode. 2. 1,3-cyclohexadiene is not aromatic. Which of the 4 criteria for aromaticity is not followed by this compound? The 4 criteria for aromaticity are: Conjugated Ring Atoms in the conjugated ring system are planar # of pi electrons in conjugated system = 2, 6, 10, 14, etc. (Huckel rule)

6 The last rule is not followed. 1,3-cyclohexadiene has 4 electrons in its conjugated system. 3. Toluene reacts with bromine to form _ ortho or para _-bromotoluene (or 2-bromotoluene or 4-bromotoluene). This reaction is a substitution reaction. (acid-base, substitution, elimination, addition). This reaction is an electrophilic aromatic substitution reaction that involves electrophilic addition to a pi bond followed by elimination (electrophile loss from a carbocation to form a pi bond). 4. Question 3e(i). What type of carbocation is in the most stable Intermediate A resonance structure? Tertiary carbocation Lab Activity 7 1. How many pi bonds are conjugated in lycopene? 11 Structure from 2. Use the Feiser-Kuhn rules to predict for lycopene. = M + n ( n) 16.5 R endo 10 R exo Where = wavelength of maximum absorption M = number of alkyl substituents / ring residues in the conjugated system = 8 n = number of conjugated double bonds = 11 R endo = number of rings with endocyclic (inside ring) double bonds in the conjugated system = 0 R exo = number of rings with exocyclic (outside ring) double bonds in the conjugated system = 0 = (8) + 11 ( (11)) 16.5 (0) 10 (0) = nm 3. Yellow is one of the colors in the tomato rainbow. a. The complementary color of yellow is: violet (see color wheel). b. The approximate wavelength of the color in 3a is: nm c. Based on your answer to 3b, which pi bond in lycopene reacts with Br 2? Use the numbering rules in nomenclature. See part d answers. The C=C bond in lycopene that reacts with Br 2 should match the calculated. d. Use the Feiser-Kuhn rules to calculate for your structure in 3c. Br 2 reacts at lycopene carbon 6 (1st C=C bond in conjugated system) and forms 1,2 product: = (7) + 10 ( (10)) 16.5 (0) 10 (0) = 459 nm Br 2 reacts at lycopene carbon 8 (2nd C=C bond in conjugated system) and forms 1,2 product:

7 = (7) + 9 ( (9)) 16.5 (0) 10 (0) = nm Br 2 reacts at lycopene carbon 10 (3rd C=C bond in conjugated system) and forms 1,2 product: = (6) + 8 ( (8)) 16.5 (0) 10 (0) = nm Br 2 reacts at lycopene carbon 12 (4th C=C bond in conjugated system) and forms 1,2 product: = (6) + 7 ( (7)) 16.5 (0) 10 (0) = nm Lab Activity 6 Part a. Structure Characterization: IR, NMR, and MS 1. In a H NMR spectrum, a peak is split into a triplet. This means there are 2 H's on the adjacent C or a -CH 2 or methylene group. 2. How many peaks would you observe in a C NMR spectrum of ethanol? 2 3. In Question 2, the molecular ion peak in the MS is 88. Give a number. 4. In Question 2, the MS, IR, H NMR, and C NMR show the compound is C 4 H 8 O 2. Give the name of the compound. Ethyl acetate or methyl propanoate Part b. Lab 2 report 5. Conclusion: State whether you synthesized phenacetin. Explain the evidence that supports your answer. Melting point: compare the melting point of your product to the true melting point of phenacetin (134 o C). Note the melting point of acetaminophen is 169 o C. IR: match the IR peaks to the bond types in phenacetin to the structure of phenacetin. You should observe a C-O bond from the ether group in phenacetin but not an O-H bond, which indicates acetaminophen. Your IR spectrum should also show a N-H peak, C=O peak, C=C peak, as well as C-H peaks for phenacetin. HPLC: did not use. 6. You would not get phenacetin by treating acetaminophen with sulfuric acid followed by ethanol because the carbon in the benzene ring that is bonded to the OH group is not an alpha carbon and can t react with the nucleophilic O in ethanol.

8 Lab 3 DSSC Prelab 7. What substance is responsible for the color in raspberries? Anthocyanins 8. Is this substance conjugated? Yes. There is only one C-C single bond between C=C double bonds. From 9. What fruit will you bring to lab to make your DSSC? Lab Activity 5 Part A. 1. a. In phenacetin, which C is the alpha C? Red circled carbon is the alpha carbon. 1. b. In phenacetin, what is the leaving group? Blue box is the leaving group. 2. a. Which reactant(s) would you use to convert phenacetin back to acetaminophen? Use an acid, e.g., HBr, H 2 SO 4, H 3 O +.

9 2. b. Elimination competes with substitution. What is the chemical formula of the smaller organic compound that forms? Part B 3. a. Which sugar source (grape juice, apple juice, etc.) produced the most ethanol per g of sugar source)? Include numbers. If there are two or more trials using the same sugar source, calculate an average volume of ethanol per gram of sugar source rather than looking at the sugar source with the highest volume/g. 3. b. Which sugar source (grape juice, apple juice, etc.) produced the least ethanol per g of sugar source)? Include numbers. If there are two or more trials using the same sugar source, calculate an average volume of ethanol per gram of sugar source rather than looking at the sugar source with the lowest volume/g. 4. The first step of glycolysis is a substitution reaction. At which atom in glucose does substitution occur? C-6 where C-1 is the carbon in the aldehyde group (carbonyl carbon). 5. The second to last step of glycolysis is a elimination reaction. 6. Distillation of my % ethanol (after fermentation) gave % ethanol.

10 Fermentation of sugar produces 10-15% ethanol. Distillation of an ethanol-water mixture should increase the % ethanol. Lab Activity 4 1. Which reactant is the limiting reactant, cyclohexanol or bleach? State the moles of cyclohexanol and bleach to support your answer. 3.8 g cyclohexanol (C 6 H 11 OH)x (1 mole/100 g) = moles 60 ml 0.7 M bleach (NaClO) = liters x 0.7 M = moles Limiting reactant is cyclohexanol. 2. The distillate should be cyclohexanone because the boiling point of cyclohexanol is higher than the boiling point of cyclohexanone. 3. What property distinguishes the product (cyclohexanone) from the reactant (cyclohexanol)? Compare cyclohexanol cyclohexanone boiling point, o C Density, g/ml Bond types O-H bond, C-H, C-C No C=O C=O bond, C-H, C-C No O-H Boiling points and densities are pretty close to each other. IR spectrum can clearly distinguish the reactant from product. 4. What are the reaction conditions to convert this unsaturated fat to a saturated fat? H 2 /Pt catalyst and heat. 5. Is this reaction an oxidation or reduction? Alkene alkane is a reduction reaction. Each vinylic carbon gains 1 H. Lab Activity 3 1. Acetaminophen reacts with in the presence of base to form phenacetin. This reaction is a reaction. Answer: C 2 H 5 I, substitution 2. What solvent is used in this reaction? Answer: ethanol

11 3. What is the function of NaOH? Answer: NaOH is a base and removes the acidic H in the alcohol group in acetaminophen. 4. What does ethyl iodide react with? Answer: the conjugate base of acetaminophen. 5. Calculate the theoretical yield in g of product. Answer: The acetaminophen is the limiting reactant. 0.7 g acetaminophen x (1 mole/151 g acetaminophen ) x (1 moles phenacetin / 1 mole acetaminophen) x (179 g phenacetin /mole) = 0.82 g 6. Question 3e. This method could be used to make phenacetin from acetaminophen. A substitution reaction will not occur at the C bonded to OH if the C is in a benzene ring (C 6 H 6 ). (i) In the 1st step, is ROH is acetaminophen or ethanol? Answer: ethanol. Protonate the ethanol with acid. Then, the O in the alcohol group in acetaminophen reacts at the alpha carbon in the protonated ethanol in a substitution reaction. If acetaminophen is protonated with acid, the O in ethanol cannot react at the C bonded to the O because a substitution reaction will not occur at the C bonded to OH if the C is in a benzene ring (C 6 H 6 ). Lab Activity 2 1. What will you use to get the fermentation reaction started? Answer: yeast 2. Fermentation is an anaerobic process. How will you prevent air and oxygen from coming in contact with the reaction mixture? Answer: Close (seal) the fermentation flask and use a trap (connect a sealed tube to allow the CO 2 gas fermentation product to escape into a test tube with mineral oil). 3. Write a chemical equation that represents the fermentation of sugar to produce ethanol. Answer: glucose is the sugar that undergoes fermentation. C 6 H 12 O 6 ---> 2 C 2 H 5 OH +2 CO 2 4. a. What sugar source will you be fermenting to make ethanol? Answer: sucrose or grape juice or apple juice or corn starch or potato starch 4. b. Calculate the theoretical yield in g of ethanol from the sugar source you use in Question 4a. Answer: Example: 50 g of corn starch (polysaccharide) will undergo a hydrolysis reaction to form 50 g of glucose (monosaccharide).

12 50 g C 6 H 12 O 6 x (1 mole/180 g C 6 H 12 O 6 ) x (2 moles C 2 H 5 OH / 1 mole C 6 H 12 O 6 ) x (46 g C 2 H 5 OH /mole) = 25.5 g C 2 H 5 OH 50 g potato starch (polysaccharide) forms 50 g glucose to produce a theoretical yield of ethanol of 25 g 51 g sucrose (disaccharide) forms 51 g glucose/fructose to produce a theoretical yield of ethanol of 26 g 32 g sugar (fructose or glucose) from 200 ml of grape juice (see label) produces a theoretical yield of ethanol of 16 g 22 g sugar (fructose or glucose) from 200 ml of apple juice (see label) produces a theoretical yield of ethanol of 11 g Lab Activity 1 Question 1. Preparation. a. List three properties that you want to know about the reagents in a chemical reaction. Answer: physical properties include state of matter, polarity and solubility (in your reaction solvent), boiling point (if you are heating or refluxing your reaction), melting point, Chemical properties include pka (if you are working with an acid or base) and safety/hazard information from a Safety Data Sheet (SDS) for reactivity and safety precautions. Question 2. Synthesis. a. A reaction occurs when reactants collide with sufficient energy for bonds to break and form. Name one way to make reactants collide with sufficient energy. Answer: raise temperature (heat reaction mixture) or raise the pressure (for gases). A catalyst increases the reaction rate by lowering the activation energy but a catalyst will not really make reactants collide with sufficient energy. b. Is a reflux like heating water without a lid or with a lid? Give reasons. Without a lid. When heated to its boiling point, the water evaporates and the steam contacts the cold lid and condenses and drips back into the water. The lid works like a condenser and prevents the water from escaping. Question 3. Workup. b. Name a solvent you can use for Lab Activity 1, Question 3a. Answer: Use a solvent that is not soluble (miscible) in water, e.g., hexane, ether, and dichloromethane. Question 4. Characterization. a. What are the two pieces of information that you can determine from measuring a melting point? Answer: identification of a substance - compare your experimental m.p. to the true m.p. And purity based on m.p. range - narrow range means higher purity.