Name Key 215 F12-Exam No. 2 Page 2

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1 ame Key 15 F1-Exam o. Page. (9 points) For each of the following sets of molecules, rank the molecules in order of to least acidic. ompare the underlined s for each set. (a) (b) (c). (11 points) Draw in the box below the structures of the three possible conjugate acids of methylguanidine and explain as to which of the three conjugate acids is most stable and would be expected to form preferentially. Use drawings of pertinent resonance forms and several words to explain your answer. Three possible conjugate acid structures: methylguanidine 1... The most stable conjugate acid is: 1 (circle one that applies) Explanation: n conjugate acid, there are three major resonance contributors shown above. These are either identical or close in the structure and energy and the positive charge is delocalized among three nitrogen atoms over a wider range, thus making conjugate acid quite stable.

2 ame Key 15 F1-Exam o. Page. (18 points) Draw in the box below a step-by-step mechanism for the a -catayzed formation of sodium enolate from diethyl malonate (1) and epoxide, using the curved-arrow convention. a a 1 enantiomer a 1 pts for each intermediate; pts for each set of mechanistic arrows The other enolate or carbanion form acceptable. V. (1 points) Treatment of aldehyde with a followed by acidic work-up results in the formation of ester 5. Draw in the box below a step-by-step, curved-arrow mechanism for this transformation. a a a more electrophilic than the ester = carbon. 5 pts for each intermediate pts for each set of mechanistic arrows 5 ote: - is far more acidic than R-. 1

3 ame Key 15 F1-Exam o. Page V. (1 points) The a-mediated intramolecular condensation reaction of yields sodium enolate 7 [J. rg. hem. 01, 77, 98]. Provide in the box below a step-by-step mechanism using the curved-arrow convention for this transformation. a a TF a 7 The other enolate or carbanion form acceptable. pts for each intermediate; pts for each set of mechanistic arrows V. (10 points). There are, in principle, three different intramolecular aldol condensation reaction products that could form from dione 8 by its treatment with % a/ /. (1) ( points) Draw in the boxes below the structures of these three potential aldol condensation products. a % a/ () ( points) n the box below indicate as to which of these three compounds is expected to be the major product from the reaction and provide a brief explanation for your choice.,, or (circle one) would be the major product from the reaction ( points). Explanation for your choice ( points): oth Enones and have the highly strained cyclobutene ring. ccordingly, as the aldol condensation reaction is thermodynamically driven, the most stable enone should be the major product.

4 ame Key 15 F1-Exam o. Page 5 V. (17 points) omplete the following reactions by providing in each of the boxes the structure of the starting compound, intermediate, or product. ndicate stereochemistry for the product/intermediate and if more than one stereoisomer is formed, draw one structure and write enantiomer or diastereomer. (1) [J. rg. hem. 01, in press] heat () [Eur. J. rg. hem. 011, 19] K () a (1 mol equiv) a LD (1 mol equiv) a Li cceptable even w/o a cceptable even w/o a/li mono-anion or mono-enolate di-anion or di-enolate The other enolate and/or carbanion form(s) acceptable. r r Lir ar enantiomer

5 ame Key 15 F1-Exam o. Page V. (0 points) omplete the following reactions by providing in each of the boxes the structure of the intermediate or product. ndicate stereochemistry for the product/intermediate and if more than one stereoisomer is formed, draw one structure and write enantiomer or diastereomer. (1) [rg. Lett. 01, 1, 0] Si 1. LMDS (1.1 mol equiv)* -0. Si *Li[Si( ) ] : a strong, bulky base () [J. rg. hem. 01, 77, 805] a (catalytic) () [rg. Lett. 01, 1, ] Mgr l, rmg 1 1 chelate intermediate eed not to indicate stereochem here. l () [rg. iomol. hem. 01, 10, 80] K-( ) (catalytic) enantiomer

A Grignard reagent formed would deprotonate H of the ethyl alcohol OH.

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