CHEM 302 Organic Chemistry I Exam III 12-December-2006 Answers

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1 CEM 302 rganic Chemistry I Exam III 12-December-2006 nswers 1) (18 pts) Compound, C 9 16, was found to be optically active. n catalytic reduction over a palladium catalyst, 2 equivalents of hydrogen were absorbed to yield compound. zonolysis of gave two compounds. ne was identified as acetaldehyde, C 3 C; the other compound, C, was an optically active dialdehyde, C y formulating the reactions involved, determine the structures of,, and C. ere is the reaction scheme, which should help you get the structures: 2 2 C 9 16 Pd (C 9 20 ) 3 C C (dialdehyde) + 3 C We see from C and acetaldehyde that we need 2 acetaldehydes to make up the necessary formula. lso, since C is a dialdehyde, both ends must be C and it must be optically active. This leaves as the only possible structure for C: Given this structure, both and follow. (Note that can have either the Z or E conformation around the double bond; ozonolysis destroys any stereochemistry there):

2 2) (20 pts) Provide all necessary reagents (i-v) and missing reaction products (-E) in the following sequences: 1) NaN 2 /N 3 2) C 3 I 2 Pd/CaC 3 /Pb i ii C a) I gave you reagent i: NS/CCl 4, since we hadn t formally covered it yet. Reagent ii: 3, then 2 2 / + (ozonolysis with oxidative work-up). Structures and are below: Ts I iii iv v C b) These reactions have anti-markovnikov written all over them. ere are your reagents: iii: 3 ; 2 2 / - ; iv: TsCl/pyridine (or similar base); v: NaI/acetone Ph 2 acetone D 2 /C/Pt 2 E c) Simple S N 1 reaction (hydrolysis) then reduction: ere are D and E:

3 Ph Ph D E 3) (22 pts) Provide syntheses for the following products. You may start from methylenecyclohexane, 1, inorganic reagents of your choice (N carbons!), and sodium hydride (Na), diazomethane, tert-butyl alcohol, carbon tetrachloride, chloroform, chlorophyll, clorox, chlordane, chloramphetamines, dimethyl sulfide, trifluoroperacetic acid, and a single frog s toe. Please pay attention to how you write these answers: means add chlorox and a frog s toe all at once, whereas means add chlorox, and then, in a second step, add a frog s toe. X chlorox frog's toe X 1) chlorox 2) frog's toe

4 (a) (b) (c) (d)??? (e) (f) (g) (h) (i) (j) (k)

5 3) Well, this is just going to be sort of a laundry list of reagents. a) C 2 N 2 /hν or b) CF 3 C c) KMn 4 / - d) 2 / 2 e) 3, then 2 2 / - f) We don t yet know /RR, so we can use: 3, then 2 2 / -, then P 3 g) The short way to this one is to make products (e) and (f), then mix (e) with Na, finally adding (f) to it. h) 2 /Pt 2 i), then tert-butoxide (from tert-butanol + potassium) j), then tert-butoxide (from tert-butanol + potassium), then 2, and finally, two equivalents of tert-butoxide. k) 3, then 2 2 / + 4) (20 pts) When (R)-3-bromocyclopentene, 1, is treated with, trans-1,2- dibromocyclopentane, 2, and both cis- and trans-1,3-dibromocyclopentane, 3, are formed: Develop a mechanism that accounts for the following observations: i) The cis-1,3-dibromocyclopentane formed is optically inactive. ii) The trans-1,3-dibromocyclopentane formed is optically active. iii) The trans-1,2-dibromocyclopentane is formed as a racemic mixture. iv) There is no cis-1,2-dibromocyclopentane formed. bservation (d) simply tells us that we get a cyclic bromonium ion, so that there is no top-side attack. That is not a surprise. The cis-1,3-dibromide should be optically inactive, as there is a mirror plane. The trans-1,3-dibromide must be optically active, as nothing disturbs the first center. ere is a mechanism:

6 First, the 1,3-products: top attack mirror - bottom attack optically active Now, the 1,2-products: - mirror images - likely. ttack from above or below (1,3-products), or left or right (1,2-products) equally 5) (20 pts) Calculate the equilibrium constant for the reaction of methane with Cl 2 to give C 3 Cl and Cl. Use the following information for your calculations: ond dissociation energies: C 3 (104.8 kcal/mole), Cl Cl (59 kcal/mole); C 3 Cl (85 kcal/mole); Cl (103.2 kcal/mole) S rxn = x 10-3 kcal/mole K T = 0ºC, R = cal/mole K

7 Recall that Η = Σbonds_broken - Σbonds-made. ere bonds broken are: Cl-Cl and C 3 -; bonds made are: -Cl and C 3 -Cl. Thus our = ( ) ( ) = = kcal. lso recall that G = T S = -RT ln(k th ). Proceeding, we get: G = 24.4 kcal (273.15K)(0.29 x 10-3 kcal/k) = kcal = -(1.987 cal/mole K)(273.15K) ln (K th ). Solving for ln (K th ) = , which gives K th = 3.87 x