Name______Key______215 HH W13-Exam No. 3 Page 2

I. (16 points) For each of the following pairs of compounds, predict which compound is more acidic. Compare the two underlined Hs for each pair and circle the compound that is more acidic for each pair.

The more acidic compound for each pair will be (circle one):

O O H H H H H (1) a. b. (5) b. H a. H H 2 2 OH O O OH (2) a. b. (6) a. O O H b. N H O H CH3O 2 O 2

H H O H O N H (3) a. (7) a. H b. N b. O H O H H 2 CH3O 2

O O O O O H a. b. O (4) (8) a. H b. H CO OCH H C OCH N 3 3 3 3 H N H H H H H 2 H 2 H

II. (12 points) (a) (6 points) In tne box provided below, draw the structures of the two potential conjugate acids of 4-dimethylaminopyridine (1) and explain, in several words, as to which of the two conjugate acids is more stable and would be expected to form preferentially. Use drawings of pertinent resonance form(s) and several words to explain your answer.

CH3 explanation: N N 4-dimethylaminopyridine (1) CH A CH3 CH3 3 H N N H N N structures of the CH3 CH3 conjugate acids B A H + Charge in A delocalized over two CH3 CH3 H N N N N N's by resonance. CH3 CH3 2 4

(b) (6 points) Recently, a similar pyridine compound 2 has been synthesized. In the box provided below, answer if compound 2 should be more basic than 4-dimethylaminopyridine (1) or not and then provide a brief explanation in using less than 20 words and pertinent resonance structure(s). explanation: The answer could be no here N N if the ring strain is mentioned 2 in the res structure on the H N N H N N right.

Is 2 more basic than 1? yes or no The conjugate acid of 2 is more stabilized by resonance as (curcle one that applies) the 4-RRN group is coplanar with the pyridine ring. 2 4 Name______Key______215 HH W13-Exam No. 3 Page 3

III. (28 points) The reaction shown below involves the base-catalyzed reaction of aldehyde 3 and α,β- unsaturated ketone 4 and consists of a Michael (or conjugate) addition, to form keto-aldehyde intermediate 6, followed by an intramolecular aldol condensation to α,β-unsaturated ketone 5. Draw in the box a step-by- step mechanism for this base-catalyzed reaction using the curved-arrow convention. You may use :B- and BH to represent a base and its conjugate acid, respectively.

H C H 3 CH3CH2ONa H3C H + O O + H2O CH3CH2OH H3C H3C 3 O 4 5 Mechanism:

H B H C O H3C 3 H B H H3C H H3C H C O 3 O 3 O H C 3 H 4 O intermedite structure: 2 pts each H3C O mech. arrows: 2 pts for each set H H3C H O H H H3C H3C O keto aldehyde B H C O 3 H C intermediate 6 O H3C 3 O H H3C O HO H B H H B E2 mech -3 pts

H3C H3C O O H C H3C 3 H3C O Carbanion struct instead of HO H B HO H H3C enolate struct acceptable. 5 28

IV (14 points) ) Draw in the box below a step-by-step, curved-arrow reaction mechanism for the following transformation [J. Org. Chem. 2013, 78, 9]. You may use :B- and BH to represent a base and its conjugate acid, respectively. Δ N O + HN O O Mechanism: HO intermedite structure: 2 pts each HN mech. arrows: 2 pts for each set N N H H O O O O O B O These deprotonation & protonation steps could be reversed.

N N Acceptable even N O H though it is a O 7-membered TS. H B O O O HO 14

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V (16 points) Draw in the boxes below a step-by-step, curved-arrow reaction mechanism for each of the following reactions. You need not rationalize the stereochemical outcome for these transformations.

(1) (10 points)

O O H SPh O + + LiSPh O Li + enatiomer

Mechanism:

SPh

O Li O O SPh SPh O H H O O O H O

intermedite structure: 2 pts each mech. arrows: 2 pts for each set 10

(2) (6 points) [Org. Biomol. Chem. 2013, 11, 44]

O O Na NO2 NO + 2 + NaBr Ph Br Ph O O + enatiomer Mechanism: intermedite structure: 2 pts mech. arrows: 2 pts for each set Na NO 2 O O2N O Br NO O Br 2

Ph Ph Ph O O O 6

Name______Key______215 HH W13-Exam No. 3 Page 5

VI. (15 points) Complete the following reactions by providing in each of the boxes the structure of the expected intermediate, or product. Indicate stereochemistry for the product/intermediate and if more than one stereoisomer is formed, draw one structure and write “+ enantiomer” or “+ diastereomer.”

(1) [Org. Lett. 2013, 15, 1004]

1. NaOH, H2O tetrahydrofuran (solvent)

+ 2. H3O (pH 1 ~ 2) O OH OH O O 3 (2) [Synthesis 2012, 51]

O 1. LiAlH4 tetrahydrofuran (solvent) N N 2. H2O workup Ph Ph O O 3 (3) [Org. Lett. in press] O OH O 1. LiAlH4 tetrahydrofuran (solvent) OH

2. aq NH4Cl H H 3 (4) [Org. Lett. 2013, 15, 1460]

CH3MgBr Br tetrahydrofuran Mg O O (solvent) O O CH O OCH3 OCH 3 N CH3O N 3 CH 3 CH3

aq HCl no need to show stereochem here 3 chelate intermediate

O O

CH3O H OCH3 + H N 3 Cl CH3

Name______Key______215 HH W13-Exam No. 3 Page 6

VII. (19 points) Complete the following reactions by providing in each of the boxes the structure of the expected intermediate, or product. Indicate stereochemistry for the product/intermediate and if more than one stereoisomer is formed, draw one structure and write “+ enantiomer” or “+ diastereomer.”

(1) [J. Am. Chem. Soc. 2012, 134, 19782]

* O 1. Na N[Si(CH3)3]2 O tetrahydrofuran (solvent) OCH2CH3 2. Cl H OCH2CH3 Cl I *Strong and non-nucleophilic base. + enantiomer 2 (2)

O NaOCH2CH3 O Ph Ph Ph HOCH2CH3 Ph C H O 19 20 C19H20O 3

(3) [J. Org. Chem. 2013, 78, 786]

CH3O CH3O N O K N + H2O CH3O CH3O O OH

O O 4

(4) [ Chem. Commun. 2013, 3149] H O N O H 1. LiCu(CH2CH=CH2)2 O N O tetrahydrofuran (solvent) H O O 2. aq NH4Cl workup CH3CH2O CH3CH2O + diastereomers 4

O O + enantiomer H H SO (5) OCH2CH3 2 4 O 1. Br Br H2O OLi OLi or Δ 2. H O+ workup OH O OCH2CH3 3 OCH2CH3

3 3 + 2 LiBr + CO2 + HOCH2CH3