CHEM 334 1st EditionFinal Exam Study Guide Lectures: 1-24Lecture 1 (January 13)Which of the following is a conjugated compound?a) // \ // \ b) // \ / \\ Answer: a) // \ // \Explanation:A conjugated compound have to have two or more adjacent π bonds separated by only one single bond. A conjugated compound must also have resonance. The molecule in a) is the only one that meets these criteria. Lecture 2 (January 15)From the following energy diagram, which product is the kinetic product and which is the thermodynamic product? E A prod. Int.Answer:B prod.A product is the Kinetic productB product is the Thermodynamic productExplanation:Under colder temperatures, the reaction is under kinetic control and will form the kinetic product. This means that the faster reaction will occur and form the product with the lowest transitions state energy. In the case of the diagram above, this is the A product. Under higher temperatures, the reaction will form the thermodynamic product. In this case, there is more energy available to cross over a higher transition state energy barrier and the more stable and lowest energy product is formed. In this case, this is the B product.Lecture 3 (January 20)Which of the following molecules are aromatic compounds? A B C D E F G H- + Answer:Aromatic: B, D, F, G, HNot Aromatic: A, C, EExplanation: For a compound to be considered aromatic it needs to meet Huckel’s criteria, meaning it must be cyclic, planar, have p orbitals on each atom within the ring, and much follow the π electron rule of πe’s=4n+2 where n much be an integer. All of the above molecules meet the first three criteria. However, deprotonated molecules, like A, gain a pair of π electrons. This makes the π electron count for A 4πe’s, which would make n in our equation 2/4, which is not an integer. Molecules in which a hydride is removed, such as in B and E, have a p orbital on positively charged atom, but do not gain π electrons. Therefore, molecule E also has only 4π electrons, as does molecule C, meaning that neither satisfy the π electron rule. N+SNNLecture 4 (January 22)Name the following disubstituted structures (use the prefixes ortho, meta, and para where applicable): A B C D EClBr Answer:A) Ortho-dimethyl benzeneB) Meta-bromotolueneC) Ortho-chloroanilineD) Para-bromo chloro benzeneE) 4-chloro-2-nitro tolueneExplanation:If there are two substituents right next to each other, the proper prefix is ortho. If the two substituents are separated by a single carbon, the proper prefix is meta, and if they are separated by two carbons, theproper prefix is para. Use the common name of the monosubstituted form of the molecule whenever possible, such as with molecule B and C. When there is no common name for the molecule, list the substituents in alphabetical order before “benzene”, as is demonstrated in molecule D. If you have more than two substituents and cannot use ortho, meta or para to describe location, as is the case with molecule E, specify location by number and continue to use common name for the molecule whenever possible.Lecture 5 (January 27)List the products of the following reactions:A) NBS hvB) NO2BrNH2ClClHBrC) KMnO4 D) KMnO4Answer: A) BrB) C) No reaction D) Explanation:BrOOHBromination is regioselective and will typically form a Markovnikov product, meaning that the bromine isattached to the more substituted carbon. An alkylbenzene reaction with KMnO4 results in oxidation to form benzoic acid, since KMnO4 is such a strong oxidizing agent that it removes the end carbon on the side chain, oxidizing the benzylic position to a carbonic acid and removing the alkyl portion. This reactioncan only take place if there is at least one hydrogen in the benzoic position, which is not the case in reaction A.Lecture 6 (January 29)Draw the product of the following reactions: A) Cl2 FeCl3B) O R Cl AlCl3Answer:A) ClB)Explanation:ORClClHA)Cl – Cl : Fe – Cl  Cl – Cl – Fe – Cl  Cl FeCl4  + Cl – FeCl3  B) O O O O O + O + + R Cl AlCl3  R Cl AlCl3  R R  R  Lecture 7 (February 05)Fill in the products of the following reactions:A. O Br2FeBr3 Br B. O Cl, AlCl3 O C... Cl + - Cl + -.. Cl Cl-HCCl-Al3 RCC+-+ N PHNH2 CH3Cl AlCl3Answer: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 8 (February 10)Complete the following reactions:PHA.O NaBH4 B. H NaBH4 C. LiAlH4 D. OH E.OH Answer: A.OH B.OH O OOOOOOOH2CrO4 PCCOC. OH OH D. O OH E. OExlanation:NaBH4 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 9 (February 12)Fill in the products of the following reactions:A) 1. R-MgBr