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Organic Chemistry Peer Tutoring Department CHEM 51B University Organic Chemistry Peer Tutoring Department CHEM 51B University of California, Irvine Professor Pronin Aubrey Taduran ([email protected]) http://sites.uci.edu/ochemtutors ​ ​ ​ Crystal Wong ([email protected]) Georgene Vasquez ([email protected]) Midterm 2 Review Packet Key 1. Draw the starting reagents for the first reaction and the product(s) for the second reaction. Determine the mechanism performed in each reaction. Explain how the second mechanism was determined. Alcohol dehydration can be determined from the strong acidic reagent (H2SO4). Because ​ ​ ​ ​ the produced alcohol is a 2° alcohol, we can conclude that the dehydration reaction underwent an E1 mechanism. 2. Predict the product(s) for the following reaction. Indicate the major and minor product if applicable. The alcohol molecule has two differently positioned beta hydrogens that allow the dehydration of alcohols in acid. The two differently positioned beta hydrogens result in two different products. 3. Rank the following alcohol molecules from fastest to slowest (1-fastest, 3-slowest) rate of reaction with HCl. Determine mechanism each alcohol would undergo when reacting to HCl and the common byproduct that will result from the reaction. The common byproduct created through the conversion of alcohols to alkyl halides with HX is H2O. ​ ​ ​ ​ 4. Provide the product and mechanism of the following reaction. 5. Provide the product and mechanism of the following reaction. 6. Complete the table below for converting alcohols to alkyl halides. HX HX PBr SOCl ​3 ​2 Alcohol (CH3OH, 1°, 2°, 3°) CH3OH, 1° 2°, 3° CH3OH, 1°, 2° CH3OH, 1°, 2° ​ ​ ​ ​ ​ ​ ​ ​ Mechanism (SN1, SN2) SN2 SN1 SN2 SN2 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ 7. Determine the starting reagents for the following reaction. Note that the stereochemistry of the molecule is retained. 8. Provide the product and mechanism of the following reaction. This is a Williamson Ether Synthesis reaction. This reaction undergoes SN2 which forms an ether at the end. 9. Draw the products of the following reactions and provide a mechanism. With ethers only strong acids such as HBr and HI are the ones that can attack. The oxygen gets protonated by the the acid and then gets cleaved at the more substituted side. Now since we have a carbocation the anion attacks it in a SN1 fashion. The other equivalent of HI attacks the alcohol in an SN2 nucleophilic attack and the alcohol gets cleaved. 10. Draw the following products of the following reaction, show stereochemistry of applicable. Since HBr is a strong acid, the protonation of the epoxide occurs first then the anion its going to attack the epoxide on the most substituted side. 11. Let’s take the reaction in the previous reaction. What would happen to the reaction if it -. was a nucleophile CN ​ instead of HBr? Show the mechanism. ​ This time, since it is a negatively charged nucleophile, it will attack the epoxide in the least substituted side. Because of this, the oxygen will now have a negative charge but the water in the reaction will help protonate the oxygen to make alcohol. 12. Determine if the following reaction is a possible hydrohalogenation product. Why or why not? This is not a possible hydrohalogenation product because with an unsymmetrical alkene it follows the Markovnikov Rule where the hydrogen in the HCl bonds to the less substituted side. The correct product would be 13. Determine the reaction and provide the starting material that would produce the product shown below. The following is a hydration reaction which means that there is alkene in the starting material. 14. Provide a mechanism for the following reaction. Use curved arrow to show the flow of electrons. Show intermediates, stereochemistry, and by-products. The nucleophile attacks the carbon with the greater δ+ in the halonium ion intermediate because it is better stabilized. 15. Draw products of the following reaction. Show stereochemistry if applicable. Anti-addition → enantiomer products. Even though this example is symmetrical, it’s important to know that water attacks the most substituted carbon belonging to the bridged halonium ion intermediate. 16. Propose starting materials that are expected to produce the following compound as a major product upon treatment with the regents shown over the arrow. Show stereochemistry, if applicable. In hydroboration oxidation, BH3 attacks the alkene and H and BH2 add in syn, with BH2 on the least substituted carbon of the alkene = anti-Markovnikov. The -OH replaces the BH2. 17. Provide a mechanism for the following reaction. Use curved arrow to show the flow of electrons. Show intermediates, stereochemistry, and by-products. First, the alkyne attacks the H in H3O+ and it is added in Markovnikov addition, going to the C that is least substituted/with the most H’s. Then, water attacks the carbocation and adds to the same side as the H. The alkene attacks another H in H3O+ → formation of an enol. Tautomerization occurs and produces a ketone product. 18. Which compound(s) produce an aldehyde after hydroboration-oxidation? Only compound 1 can produce an aldehyde because it is a terminal alkyne. Ketones are produced when the triple bond is internal (compared to alkenes producing alcohols during hydroboration oxidation). 19. Propose a synthesis for the following reaction: First, we recognize that a vicinal dibromide is necessary to become an alkyne via a double elimination. Second, we know we can eliminate the alkyl bromide, turning it into an alkene. Then the alkene can be halogenated by Br2 to get the vicinal dibromide we are looking for. Finally, we perform a double elimination of our alkyl bromides via tert-butoxide to get our alkyne. .
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