CHEM 334 1st EditionExam# 1 Study Guide Lectures: 1 - 6Lecture 1 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 1 Which of the following compounds absorbs the shorter wavelength of light? A BAnswer:AExplanation:The longer a conjugated molecule is, the less energy is required to excite one electron to the lowest unoccupied molecular orbital (LUMO). Since energy increases as wavelengths get smaller, the less conjugated molecule (therefore the shortest one) which requires more energy, will absorb the shorter wavelength.Lecture 2 Fill in the product of the following reaction:Br2 ??Answer:Explanation:The product here favors the more substituted alkene, not the more stable radical Br2 Lecture 2 What is the major product for the following reactions?a) // \ // ??b) // \ // ??Answer:a) / \\ / \b) / \ //Explanation:At higher temperatures, the 1, 4 addition product dominates and at lower temperatures, the 1, 2 addition product dominates. Remember that the numbers 2 and 4 correspond to the carbon on the chain that the bromine attaches to, while the 1 corresponds to the carbon that the hydrogen attached to.hvBrhvBrH – Br 40⁰C H – Br -78⁰CBrBrLecture 2 From the following energy diagram, which product is the kinetic product and which is the thermodynamic product? E A prod. Int.Answer: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 2 From the following energy diagram, which product is the kinetic product and which is the thermodynamic product? E A prod. Int.B prod.B prod.Answer:B is both the kinetic and the thermodynamic productExplanation:The kinetic and thermodynamic product do not always have to be different. In this case, product B has the lowest transition state energy barrier and is therefore the kinetic product. However, product be is also the lowest energy and more stable product. Therefore, product B is also the thermodynamic product.Lecture 2 What are the products of the following reactions? A)B)C)D) oooooAnswer: A product)o B product) C product) O D product) oExplanation:These are all Diels Alder reactions. In these reactions, three π bonds are broken in order to make two single bonds and one π bond. Therefore, a six membered ring with at least one π bond will always be formed. If the dienophile contains a triple bond, then there will be two double bonds in the ring, one on either side.ooLecture 3 Draw the stereochemically correct products of the following Diels Alder reactions:A)B)Answer:A) OB)Explanation:The stereochemistry of the product matches that of the dienophile. So, for a cis substituted dienophile, like the one in reaction A, you will have a cis substituted product. For a trans substituted dienophile, like the one in B, you will have a trans substituted product. Remember also, for reaction B you will also get the enantiomer of this product (simply the mirror image).Lecture 3 Which of the following molecules are aromatic compounds? A B C D E F G H- + OOOS+NAnswer: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.Lecture 4 Name the following monosubstituted structures:ABC D E FAnswer:A) PhenolB) TolueneC) AnisoleD) StyreneE) AnilineF) Benzoic AcidLecture 4 Name the following disubstituted structures (use the prefixes ortho, meta, and para where applicable): A B C D EClNNOHNH2OHOOBrClNH2NO2BrAnswer:A) Ortho-dimethyl benzeneB) Meta-bromotolueneC) Ortho-chloroanilineD) Para-bromochloro 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 List the products of the following reactions:A) NBShvB)HBrAnswer:A) BrClBrB) Explanation:Bromination is regioselective and will typically form a Markovnikov product, meaning that the bromine