c. Oxidizing agent shown here oxidizes 2º alcohols to ketones and 1º alcohols to carboxylic acids. 3º alcohols DO NOT REACT.
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1 Exam 1 (Ch 17 and Review of CEM 331) Answer Key: 1. ne-step Questions: You need to know reagents for reagent arrows and to be able to draw products. I know a lot of them seem to look alike its your job to sort them out. SCl 2 is used to make chlorides from alcohols, PCl 3 is used to dehydrate. C 3 S 2 Cl is mesyl chloride (MsCl), p-c 3 C 6 4 S 2 Cl is tosyl chloride (TsCl). TMSCl is trimethyl silyl chloride, used with triethylamine, NEt 3, and is a protecting group for alcohols, like DP, dihydropyran, used with an acid catalyst, +, also forms a protecting group (TP group). PCC (Cr 3 with pyridinium hydrochloride) is used to oxidize, as are Cr 3, Na 2 Cr 2 7 and KMn 4 while LiAl 4 and NaB 4 are used to reduce. a. C 3 S 2 Cl is MsCl and converts an alcohol into a good leaving group without affecting the stereochemistry of the chiral center. No inversion of the stereochemistry. C 3 S 2 Cl pyridine Ms b. Take home message count your carbons people. You are starting with 8 carbons and ending with 10. Do the math. nly adding 2 carbons. Answer: C 3 C 2 MgBr. c. xidizing agent shown here oxidizes 2º alcohols to ketones and 1º alcohols to carboxylic acids. 3º alcohols D NT REACT. KMn 4, 3 + d. Tertiary alcohol is forming tertiary chloride. Reagent? NT thionyl chloride Use Cl. C 3 Cl C 3 e. Grignard reagents add one time to a ketone (yes that s a ketone not a cyclic ester the ring does NT open) and once to an aldehyde: C 3 MgBr f. Williamson ether synthesis adding three carbons to the alcohol. LEARN IT! 1. Na or K and 2. C 3 C 2 C 2 Br. Please do not write that backwards as C 2 C 2 C 3 Br. Carbon atoms should have four bonds
2 g. I think everyone knows that sodium borohydride only reacts with aldehydes and ketones. Look at the left side closely. That C 3 is NT attached to the carbonyl. That s a ketone on the left, not an ester Reduction of both the ketone and the aldehyde occurs. 3 C NaB 4 3 C 2. [10 pts] Name the following compounds according to IUPAC nomenclature: (from Problem Set online) a. #1 shown below was the problem on the exam. #2 was what most people were trying to name. See the structural difference. The first is a cyclohexanol with three other groups attached. The other is a cyclohexenol with two other groups attached. Numbering of ring #1 starts at alcohol position and you number around the ring so as to arrive at the lowest possible sum (go CCW). The double bond is not IN the ring so you do not give it lower numbers. It s a prefix called methylene and does not have any priority. [#2 would start its numbering at the alcohol and go CW so the double bond has the second priority and as low of a numbering as possible but its not the problem on the exam!] versus C 3 #1 #2 C 3 Answer: 3-methoxy-2-methyl-5-methylenecyclohexanol (automatically assumes is on #!). b. Most common mistakes: Couldn t count to nine (parent) or number from correct side (so ether is on lowest number). Answer: 4-ethoxy-2,7-nonadiene-5-ol. 3. [6 pts] In each of the molecules shown below, circle the two different functional group and identify what type each is (i.e. halide, alcohol, ketone, etc). See Problem Set online. aromatic ring alcohol alkene ester amide epoxide
3 4. [6 pts] For each transformation shown, label as an oxidation, a reduction or neither: [See Problem Set online] a. Adding to BT ends of the double bond - oxidation b. 2 bonds to become 2 bonds to N - neither c. 3 bonds to become 1 bond to and 2 bonds to (not shown) - reduction 5. [28 pts] Synthesis: Most everyone got the idea of synthesis although not knowing reagents or paying attention to structures caused some people some difficulties. a. I Na or K PCC b. Most people could make the Grignard reagent but then got confused when they needed to react it with a carbonyl (on the arrow) to form the alcohol product. You got too used to seeing the carbonyl in front of the arrow. Take off the mental blinders PBr 3 Br Mg MgBr c. Some of you don t like the Cuprate reagent yet which is why many of you got the last reagent wrong get used to it. More people got the Grignard reagent wrong because they
4 didn t recognize that a new CARBN was added, even though they drew is (because everyone knew what PBr 3 does, you were easily able to draw that alcohol). PCC C 3 MgBr this is a C! PBr 3 (C 3 ) 2 CuLi Br d. You were on your own for this problem. You needed to add three carbons to the left side. Everyone seemed to recognize a Grignard was required. That secondary ketone needed to be oxidized first but there s an aldehyde in the way. Reduce the aldehyde on the right end (you have to any way) and protect it FIRST, then deal with the left end. Reagents: 1. NaB 4 or LA 2. Protecting group (either TMSCl with NEt 3 or DP, + ) 3. PCC (so you don t accidentally remove the protecting group, PCC is the best choice), 4. Ethyl magnesium bromide, C 3 C 2 MgBr, and finally removes the protecting group (or F - for the TMS group). 6. [12 pts] Mechanisms take a lot of practice Thus, I suggest you start practicing + 7. [12 pts] Sequence Question: Identify the structures of the Compounds A, B, C and D formed in the following sequence of reactions. Word problems are always tougher if you don t know the NAMES a. 1,4-Pentanediol is reacted with one equivalent of trimethylsilyl chloride and triethylamine to form Compound A. 1 equiv. TMSCl, NEt 3 TMS Compound A
5 Note for future problems a backwards TMS group is NT an SMT group b. Compound A is oxidized using PCC and then reacted with methyl magnesium bromide, C 3 MgBr, to form Compound B. TMS Compound A 1. PCC 2. C 3 MgBr TMS Compound B c. The addition of fluoride anion to Compound B results in the formation of Compound C, which then can easily react with thionyl chloride and pyridine to form Compound D, C 6 13 Cl. TMS Compound B F - Compound C d. Compound D will react with Na or K to form the following cyclic ether in a Williamson ether synthesis reaction: Compound C SCl 2 pyridine Cl Compound D And Compound D does an intramolecular Williamson Ether Synthesis: Na Compound D 8. [8 pts] Thought Provoking Questions (a.k.a. Short Answers): Choose 2 of the 3 following questions. If you answer more than two, indicate which two you wish to be graded otherwise the first two will be chosen for you. a. Explain why a primary alcohol, like 1-pentanol shown below, cannot be converted into a primary chloride, using the reagent Cl. Cl The conversion of an alcohol to a chloride using Cl occurs via an SN1 type mechanism that requires the formation of a carbocation intermediate. The primary alcohol here cannot form a stable carbocation therefore it won t be able to react to form a chloride. Cl
6 b. In the following transformation, should the alcohol be converted into a halide or a tosylate to achieve the desired product? Explain your answer. 1.? 2. NaC 3 C 3 The second step of the reaction involves a salt that means you have a strong nucleophile which will react SN2 fashion. That will cause one inversion to happen in the sequence in step two. verall, you want one inversion to occur, therefore you don t want to do an inversion in the first step. Forming a halide does an inversion. Forming a tosylate does not cause an inversion. Form a tosylate in the first step! c. Explain why increasing the number of alkyl groups surrounding an alcohol decreases the acidity of the alcohol proton. The more alkyl groups, the more sterically hindered, the less able an alkoxide anion (anion of an alcohol) can be SLVATED to stabilize the anion.
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