CHEM 334 1nd EditionExam # 2 Study Guide Lectures: 7 - 13Lecture 7 (February 05)Fill in the products of the following reactions:A. O Br2FeBr3 Br B. O Cl, AlCl3 O C. NH2 CH3Cl AlCl3 N PHAnswer:A. O N BrB. O O C. NH2 NH2 +Explanation:When you begin with two substituents on a benzene ring, the ortho/para directing substituent wins overmeta directing one. If both are ortho/para directing, then the strongest activator dictates the position of the new substituent.Lecture 7 (February 05)Name the following functional groups and indicate whether they are ortho/para directing or meta directing:A. OPHR OHB. OR OC. O RR N RD. O O OE. O ClF. O RG. O R RAnswer:A. Carboxylic acid, Meta directingB. Ester, Ortho/Para directingC. Amide, Ortho/para directingD. Anhydride, Ortho/para directingE. Acid chloride, Ortho/para directingF. Aldehyde, meta directingG. Ketone, meta directingExplanation:Electron donating groups or activating groups are all ortho/para directing. Electron withdrawing groups or deactivating groups are all meta directing with the exception of halogens. Even though halogens are deactivating, they are still ortho/para directing.Lecture 7 (February 05)Show the reagents you would use in the proper order to get the following product: ? Br SO3HAnswer: Br2 SO3FeBr3 H2SO4Explanation:Order is important in this case. If you added SO3 to the ring before Br2, then Br2 would be in the meta position compared to SO3, which would give you the wrong product.Lecture 8 (February 10)Complete the following reactions:A.O NaBH4 B. H NaBH4 C. LiAlH4 D. OH , thenOOOOOO OH2CrO4E.OH Answer: A.OH B.OH O OC. OH OH D. O OH E. OExlanation: PCCNaBH4 and LiAlH4 are reduction agents, which will reduce the amount of C=O bonds in a molecule in favor of more C=H bonds. PCC and H2CrO4 are oxidizers and oxidize a reaction in order to make more C=O bonds on a molecule.Lecture 8 (February 10)Fill in the steps for the following reaction: Br mg B.(2 parts) OH Et2Answer:A. MgBrB. 1. O2. H3OExplanation:Here you are forming a Grignard, which is highly polarized and can react with aldehydes, ketones, and esters. The ketone here binds to the site where the Grignard was and the double bond gets pushed up to the oxygen. We can then use an H3O acid workup to protonate the oxygen.Lecture 9 (February 12)Name the following molecules:A . B. C. D. A.OO O O Answer:A. 3-methyl butanal B. 2-heptanoneC. 2-butenal D. 3-hexynoalExplanation:The first part of aldehyde and ketone nomenclature is finding the longest carbon chain that contains the carbonyl. For an aldehyde, change the suffix from -e to –al. For a ketone, change suffix from –e to –one. When numbering, the aldehyde must be at the end of the chain. This means no number is needed to describe the position of the aldehyde(because it’s at the 1 position). For a ketone, the carbonyl gets the lowest number. For double bonded molecules, adden before the suffix. For triple bonds, add yn before the suffixLecture 9 (February 12)Fill in the steps for the following reaction to take place(to form a Grignard): Br A. Mg Et2OOHAnswer:A. Cl–Si B.BrB. Si–O Explanation:Since we cannot form an organometallic in the presence of an alcohol, we have to use a protecting group in order to make the alcohol into something we can make into a Grignard.Lecture 10 (February 17)Name the following structuresA. OB. O Answer:A. Cyclopentane carboxaldehydeB. CyclohenanoneExplanation:When naming an aldehyde ring, name the ring first and then add carboxaldehyde. When naming a ketone ring, add –one to the end of the name of the ring.Lecture 10 (February 17)Which group is more reactive, aldehydes or ketones?Answer:AldehydesExplanation:Aldehydes are more reactive than ketones due to sterics and electronics. Aldehydes always have a small H attached to the carbonyl, while ketones always have something larger attached to the carbonyl. Because the ketone has more sterics, it is harder for the nucleophile to attack. Alkyl groups are electron donating. The nucleophile attacks the electrophilic position. The ketone carbon is less electrophilic than the aldehyde carbon. This means that if we have an aldehyde and a ketone in a reaction, we know to react with the aldehyde.Lecture 11 (February 19)Write the mechanism for the following reaction:O HOR OH H+/HOR OR OR ORAnswer:Reaction 1.O: O+ O–H OH OR H–O+–R H–O–R O+–R H H H–O–R Reaction 2. OH H–O+–R +O O+–R O+–R OROR H OR R–O–H OR H–O–R ORExplanation:HReaction 1 shows the formation of a hemiacetal. For this reaction, the carbon-oxygen double bond isreplaced by an OH and an OR group bonded to the same carbon. While this reaction can be acid or base catalyzed, reaction 2 can only take place under acidic conditions. Reaction 2 shows the hemiacetal being reacted further to an acetal. Notice that these reaction are in equilibrium and can go forwards or backwards, depending on the amount of H2O in the reaction.