Chem 201 Final. Beauchamp

Transcription

1 hem 201 Final Winter, 2018 Beauchamp ame KEY Problems Points redit 1. Functional Group omenclature (1 large structure) 0 2. Lewis tructures, esonance, Formal harge 18. yclohexane onformations, 2 substituents, ewman Projections, elative Energies 0 4. ewman Projections, onformational Energies, K eq alculation 0 5. tereochemical Analysis esonance tructures, tructure, ybridization, Angles, hapes 0 7. ipole Moments and Inductive versus esonance Effects Quantitative Acid/Base Equation, Identify onjugate Acid and Base and alculate K equilibrium, upply urved Arrows Acid / Base hemistry, Explanation, urved Arrows, Formal harge, Qualitative Equilibrium (7) /E Mechanisms, with all of the details, Templates Provided Various eactions, predict the products (20 reactions) Fill in all mechanistic details, curved arrows, lone pairs, formal charge, /E Mechanism, arbocation eactions Free adical ubstitution Problem Predict Possible Products, ow Much, tereochemistry and Provide a Mechanism For Major Product 0 Premidterm material = 18 Postmidterm material = 19 Total 61 This is a long exam. It has been designed so that no one question will make or break you. The best strategy is to work steadily, starting with those problems you understand best. Make sure you show all of your work. raw in any lone pairs of electrons, formal charge and curved arrows to show electron movement where appropriate. o your best to show me what you know in the time available. "hange your thoughts and you change your world." orman Vincent Peale

2 1. Provide an acceptable name for the following molecule. nly specify and where shown as. (0 pts) 2. Indicate all formal charges present in the following structures. Assume all electrons are shown as lines or dots. If other reasonable resonance structures are possible, draw the best other resonance structure using the proper arrow conventions. Indicate which resonance structure is better or if they are equivalent. (18 pts) The second resonance structure is better because it has an extra bond and has full octets and there is less charge. There are 2 other equivalent resonance structures. The second resonance structure is better because it has full octets and is neutral. The second resonance structure is better because it has an extra bond and full octets. There is an equivalent resonance structure with on the other nitrogen.

3 . raw all possible chair conformations of trans-1- t-butyl--ethylcyclohexane. Make the left most ring carbon 1 and number towards the front. how all axial and equatorial groups in the first chair. Which conformation is more stable? Provide a reason for your answer. raw a ewman projections of the most stable conformation using the 1 6 and 4 bonds to sight along. Point out any gauche interactions shown in your ewman projection. If the axial energy of a t-butyl group is 5.5 kcal/mole and the axial energy of an ethyl group is 1.8 kcal/mole and a t- butyl/ethyl gauche interaction is.0 kcal/mole, what is the ratio of the two conformations at equilibrium? how your work. ketch an energy diagram that shows how the energy changes (lower to higher) with the conformational changes and estimate the ratio of the two conformations at equilibrium. (0 pts) a. K = 10 2.T = 2 cal/mol-k T = 00 K chair 1 (more stable) axial ethyl = 1.8 chair 2 (less stable) trans-1-t-butyl-2-ethylcyclohexane axial t-butyl = 5.5 (larger group has larger axial energy and is less stable) total = 1.8 kcal/mole total = 5.5 kcal/mole ) = =.7 kcal/mole larger group (higher axial energy) is axial in right structure so is positive and K eq < 1. b. ewman projection ( 1 6 and 4 ) most stable, point out any gauche interactions with the substituent(s) chair 1 (more stable) Two gauche of with each side of the ring but we can only see one in the ewman projection. c. Energy diagram and relative percents (K eq =?) = 1.8 kcal/mole 99.8% t-bu =.7 kcal/mole 0.2% = 5.5 kcal/mole 5 K = 10 2.T K = K = K = K = (1)/(440) d. alculate an approximate difference between the two conformations. Use that value to estimate a K eq. (Assume = 2 cal/mol-k and T = 00 K.) Use energy values provided in the box. how your work. gauche = 0.8 2x axial Me = = 6. kcal/mole axial = 1.7 gauche = = 2.5 kcal/mole The energy table is not shown here. ) = = -.8 kcal/mole K = ) 180 = = 570 / 1

4 4. Use a ewman projection of the 4 bond of 2-methyl--phenylhexane to show the most stable conformation first. otate through all of the eclipsed and staggered conformations. Using the energy values provided in the table below, calculate the relative energies of the different conformations. Plot the changes in energy in the graph diagram provided. alculate a ratio of least stable to most stable based on values. int: raw a 2 structure first and bold the bond viewed in your ewman projection, then decide your line of sight. (25 pts) 2 tructure Approximate Eclipsing Energy Values (kcal/mole) ome were estimated by me. Me i-pr t-bu Ph Me i-pr t-bu Ph ewman projections: lowest PE Ph i-pr gauche /Ph = 0.2 gauche /i-pr = 0.2 gauche i-pr/ = 1.6 gauche / = 0.1 gauche / = 0.0 gauche /Ph = 0.2 Total = 2. Ph Ph 1.6 i-pr eclipsing /Ph = 1.7 eclipsing /ipr = 1.6 eclipsing / = 1.5 Total = 4.8 = G K eq = 10-2.T 2 = Ph 1 = i-pr 2-methyl--phenylhexane Approximate Gauche Energy Values (kcal/mole) ome were estimated by me. Me i-pr t-bu Ph Me i-pr t-bu Ph highest PE Ph Ph Ph Ph i-pr i-pr i-pr Total = Total = Total = 2.2 Total =.5 i-pr kcal/mole 4.8 kcal/mole 2.2 kcal/mole 6.4 kcal/mole.5 kcal/mole 6.7 kcal/mole 0 most stable least stable = = 4.5 kcal/mole ) = = 4.5 kcal/mole K eq = K eq = = = 1 / 1800 = (least) / (most)

5 5. Use the following set of Fischer projections to answer each of the questions below by circling the appropriate letter(s) or letter combination(s). int: edraw the Fischer projections with the longest carbon chain in the vertical direction and having similar atoms in the top and bottom portion. lassify all chiral centers in the first structure as or absolute configuration. (25 pts) 2 5 twist at top ( pts) twist twist at top, twist twist A at B bottom at top, at E bottom and bottom and bottom rotate 180 o 2 5 mirror meso a. Which are optically active? A B E b. Which are meso? A B E (17 pts) c. Which is not an isomer with the others? A B E d. Which pairs are enantiomers? AB A A AE B B BE E E e. Which pairs are identical? AB A A AE B B BE E E f. Which pairs are diastereomers? AB A A AE B B BE E E g. Which pairs, when mixed in equal amounts AB A A AE B B BE E E will not rotate plane polarized light? h. raw any stereoisomers of hexan-2,,4-triol as Fischer projections, which are not shown above. If there are none, indicate this. (5 pts) These are all of the stereoisomers. A= E enantiomers enantiomers enantiomers enantiomers 2 5 i. In the most recent rganic Letters, 2018, 20, , a new natural product was isolated, impterpenoid A, from a fungal source in the south hina ea. ircle all of the chiral centers. ow many stereoisomers are possible? how work. (5 pts) impterpenoid A There are 5 chiral centers. Maximum stereoisomers = 2 5 =2

6 6. raw two additional better 2 resonance structures of the given structure. Assume all nonhydrogen atoms have full octets except for carbon. Add in any necessary lone pairs and use proper curved arrows. Which structure(s) is(are) best and why? raw a structure for the given resonance structure. how bonds in front of the page as wedges, bonds in back of the page as dashed lines and bonds in the page as simple lines. how orbitals for pi bonds and lone pairs along with their electrons. Identify the hybridization, bond angles and descriptive shape for all numbered atoms in the given structure. (0 pts) best resonance has oxygen atoms with full octets and an extra bond, structure 2 and A B (5 pts) (15 pts) The best resonance has positive on either oxygen because there is an extra bond and full octets. Those are T the first structure. The second resonance structure is also shown here. Use the given (first) Lewis structure to answer this part. Atom hape ybridization Bond Angles # sigma bonds # pi bonds # lone pairs 1 linear sp linear sp tetrahedral sp linear sp trigonal planar sp

7 7. Explain what the following dipole moments suggest about inductive effects and resonance effects. You will need to draw additional structures to help your explanation. (15 pts) A B = 2. = 2.68 =.71 inductive = donating no resonance effect B (ethanal) is more polar than A (methanal), so the methyl (an group) must be inductively donating compared to hydrogen, which allows oxygen to steal away even more electron density. itrogen is more electronegative than carbon and is inductively withdrawing, which should make the amide () less polar than methanal (A). owever, nitrogen can use its lone pair to resonance donate to of carbon ( rd resonance structure), which should make the molecule even more polar. esonance must be more powerful than the inductive effect with methanamide (formamide) because polarity went way up (methanamide,, has a much larger dipole moment than methanal, A). inductive = withdrawing resonance = donating 8. nly the reactant acid and base are drawn below. ecide which is which and draw a mechanism to show formation of the conjugate base and acid. The two acids have pk a 's of 10 and 8 (K a values are and 10-8 ). Match the K a values with the proper acid, write a K equilibrium expression and calculate a quantitative K equlibrium value for the reaction. how your work. Provide an explanation for your value of K equilibrium. (15 pts) right K a = 10-8 K a = The inductive withdrawing effect of the oxygen makes the left acid more acidic. It's conjugate base (hydroxyl amine) is less basic. Ammonia is more basic and the ammonium ion is less acidic. The equilibrium shifts away from the stronger acid and stronger base (and towards the weaker acid and weaker base). The equilibrium should favor the right side, and it does (100 / 1). K equilibrium = [ 4 ] [ 2 ] [ ] [ ] = K a = 10-8 = 100 / 1-10 K a = 10

8 9. Using arrow-pushing mechanisms, write the expected products from the following reactions and indicate whether the equilibrium lies to the right or to the left. Also, very briefly explain your reasoning. (5 pts) f. g. c. d. e. b. a. F F F F F F right right left right left inductive effect of nitrogen helps stabilize the anion resonance puts minus on 2 oxygen atoms whicih is better than one oxygen atom and one nitrogen atom, oxygen is more electronegative left anion is spread over 2 nitrogen atoms (resonance) while right anion only is localized on one nitrogen atom the anion is delocalized on both sides, but the right anion is more stable from the inductive withdrawing effect of fluorine atoms the left nitrogen cation is resonance delocalized (2x and the ), while the right nitrogen cation is not delocalized left the ethyl carbanion is sp hybidized (lower % s) so is less able to stabilize the anion than the sp2 hybridized ethenyl carbanion (higher % s) oxygen and sulfur are both group 6 elements (Zeff = 6), but sulfur is larger and the anion electron density is more delocalized on sulfur right

9 10. Use (2,)--bromo-2-deuteriohexane to provide a simple, arrow-pushing mechanism for each of the following reaction conditions (show curved arrows, lone pairs & formal charge). Fill in the necessary details to clearly indicate any stereochemical features and/or conformational requirements. If reactants are not drawn in the proper orientation to show how the reaction must proceed, then redraw them in a more informative way that shows this. o not consider carbocation rearrangement possibilities. You can abbreviate (simplify) parts of the molecule that are not part of a reaction. (4 pts) a. raw a 2 structure and then a structure of the reacting molecule. A structure will be provided for the cost of the points of this part. ( pts) 2 2 structure 2 6 structure (,2)--bromo-2-deuteriohexane b. how the reaction (what kind?), indicate the absolute configuration(s) of the center in the product. (7 pts) mechanism configuration in product c. how all possible E reaction products (what kind?). Indicate if E, Z or neither. (1 pts) 2 5 2E 7 E2 mechanism 2E - is anti configuration in product

10 - is anti 2 5 2Z 7 E2 mechanism 2Z configuration in product a b a - is anti 2 5 E b 2 5 E2 mechanism E configuration in product 2 b - is anti b a Z 2 a E2 mechanism Z configuration in product

11 d. how the reaction (what kind?). You can use one intermediate to show all possible mechanistic possibilities. Indicate absolute configuration(s) of the center in your product(s). (10 pts) a b 2 a 2 7 a 2 2 b 7 b mechanism configuration in product 7 7 e. edraw the intermediate used in 8d above to show all possible E reaction products. Indicate if E, Z or neither. If multiple products are formed between two atoms, you can show a single mechanism and just draw the additional possible products. (10 pts) 2 other alkenes 2 5 2Z Z 2 - or needs to be parallel (on the top or the bottom) 2 5 E1 mechanism configuration in product E E possible alkenes 2 E 2 E

12 11. Indicate the major product in the following reactions. Indicate stereochemistry if part of the reaction. o T show mechanisms. (WK = workup = neutralize conditions) (0 pts) a. E2 > 2 k. 2. Al 4. workup 2 > E2 2 b. 2 > E2 l 2. Al 4. workup 2 > E2 2 c. d. e. a 1 equivalent acid/base. 2 h free radical substitution 1. a acid/base. 2 m n o Al 2 > E2 2 > E2 E2 > 2 f. h, free radical addition p only 2 g h i a equivalents 1. a 2. E2 > 2 1. a E2 twice acid/base > E2 q r s a 1 equivalent a 1 equivalent 2 only 2 2 E2 > 2 2 > E2 j 1. a 2. 2 > E2 t 2 1 > E1

13 12. Provide all missing arrow-pushing mechanistic details (curved arrows, lone pairs and formal charge) to explain the following transformation. Assume all nonhydrogen atoms have full octets unless a positive charge is written by a carbon atom. (15 pts) a. 4 carbons acid/base resonance acid/base acid/base acid/base acid/base resonance carbons 2 1. Provide a complete arrow-pushing mechanism for the following transformations (lone pairs, formal charge and curved arrows). (15 pts) / 1 E1 rearrangement 2 o carbocation resonance reaction top & bottom resonance o carbocation rearrangement o carbocation and resonance (more stable) (even more stable)

14 14. a. how all possible products when 2-chlorobutane is brominated with 2 /h? Indicate the approximate relative amounts of each product formed if the relative rates of reaction of a bromine atom with an sp - bond are: primary = 1, secondary = 80, tertiary = 1600 and - on a carbon with chlorine = Identify any stereoisomers as enantiomers, diastereomers or meso structures. pecify the absolute configuration of any chiral centers. (10 pts) 2 1 l chlorobutane l 2 2 / h relative amounts = A = (number of -) x (relative reactivity) l l l A = () x (1) = l 2 2 diastereomers A = (1) x (80) = 80 A = (1) x (80) = 80 l 2 A = () x (1) = enantiomers 2 A = (1) x (1600) = 1600 A = (1) x (1600) = 1600 b. Provide a complete arrow pushing mechanism to explain formation of the major product from the above reaction (show proper curved arrows, lone pairs as two dots and single electrons as one dot). learly label each distinct part of the reaction mechanism. alculate an overall for each step of your mechanism using the given bond energies. To make a bond is positive energy and to make a bond is negative bond energy. (15 pts) initiation 2 / h Me BE = 46 1 o o - 95 o - 92 l-- 90 Me o o - 68 o - 67 l a. propagation l BE = 90 2b. propagation free radical carbons are flat (sp2 hybridized) l BE = 46 BE = -67. termination = combination of 2 radicals to shut down chain reaction l Both and configurations are obtained. l = BE = overall = -19 -? -67 Talent is God given. Be humble. Fame is man-given. Be grateful. onceit is self-given. Be careful... John. Wooden