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Addition of HX to conjugated alkenes occurs via two modes:
- Direct addition (1,2 addition): HX adds directly across the ends of a C=C bond.
- Conjugate addition (1,4 addition): HX adds across the ends of conjugated system.
The distribution of the products depends on the reaction conditions (temperature).
At low temperature, the reaction is under Kinetic control (rate, irreversible conditions) and the major product is that from fastest reaction, that of the bromide with the secondary cation.
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- Rings + Unsaturation --- Hydrogenation If compound A C51H81BrN5O3P3 is hydrogenated to give compound B C51H101BrN5O3P3. How many rings does compound A have? Assume that P has a valency of 5. Would the answer be 4 rings? Formula -> unsat + rings = 1+C +N/2 - H/2 - X/2For the compounds given (A, B, C) 1. determine their relationship ( enantimers, idential/same compind, diatermeros, constitutional isomers) and give a short explanation why(a) Draw the molecular orbital picture for propa-1,2-diene, H2C=C=CH2. Hint: The three-dimensional geometry is shownin the chapter. (b) Draw the MO energy diagram for propa-1,2-diene. What is the HOMO? What is the LUMO?
- Consider the molecule 1-bromo-2-methylbutane. C3 and C4 should be drawn as Et as in theexample. This group is called an ethyl group and can be considered a sphere about twice the sizeof a methyl group. Draw the following Newman projections sighting down the C1C2 bond... a. The lowest potential energy conformation. b. The highest potential energy staggered conformation.Build a model of methylcyclohexane, and use the model to complete the following Newmanprojections of methylcyclohexane in the chair conformation: a. When the methyl group is in an axial or equatorial (circle one) position, the molecule is inits lowest potential energy conformation. b. Label one Newman projection above anti and the other gauche to describe the relationshipbetween the methyl group and C3 of the ring. c. In general, which is a lower PE conformation, anti or gauche? d. Explain how your answer to b and c provide an explanation for why it is more favorable fora large group to be in an equatorial than an axial position.Draw the energy diagram for cyclooctatetraene dianion , C8H8 (2-) , molecular orbitals. The polygon rule is helpful. Label each MO as bonding , antibonding, or nonbonding, and add the nonbonding line. Identify the LOMO, HOMO, LUMO and HUMO orbitals. Is this dianion aromatic , non aromatic or anti aromatic? Thank you!
- Answer the following questions for the MOs of 1,3-butadiene: a. Which are p bonding MOs, and which are p* antibonding MOs? b. Which MOs are symmetric, and which are antisymmetric? c. Which MO is the HOMO and which is the LUMO in the ground state? d. Which MO is the HOMO and which is the LUMO in the excited state? e. What is the relationship between the HOMO and the LUMO and symmetric and antisymmetric orbitals?iupac nme include stereochemistry if ncessaryDraw arrow-pushing mechanisms for the following reactions. Include all lone pairs and all formal charges.
- Using your model, construct an energy diagram to show the variation in the free energy of the molecule as the FRONT ATOM is rotated CLOCKWISE from 0º to 360º in 60º increments. In your energy diagram, you should clearly show the relative energies of each conformer.A trisubstituted cyclohexane with three substituents-red, green, and blue-undergoes a ring-flip to its alternate chair conformation. Identify each substituent as axial or equatorial, and show the positions occupied by the three substituents in the ring-flipped form.Consider the Newman projection below. a. Draw a full Lewis structure of this molecule with R1=Me,R2=Et , and R3=iPr . b. Given the sizes of these R groups (R3R2R1) , does the Newman projection above show thelowest potential energy conformation of this bond? If not, draw a Newman projectionshowing the lowest P.E. conformation (sighting down this same bond). c. To draw a Newman projection in the lowest P.E. conformation, the following rule of thumbusually applies: Place the largest group on the front carbon anti to the largest group on theback carbon. Is your answer to the previous question consistent with this rule of thumb?