CHEM 345 Problem Set 07 Key

Transcription

1 CHEM 345 Problem Set 07 Key 1.) Fill in the appropriate reaction arrow. The starting material is on the left, the product is on the right. Use. Simple Ring Size. 5 and 6 are favored. 3 is not. That s it. or a. b. c. pka Table R=H or sp 3 C HCl H R H R H R R H R H R R H R H H R R H R R H H R SH HC

2 2.) This is a critical thinking problem. Using the base strengths of the anions, predict which reaction arrow you should use. The pka table on the previous page is one you should know. will have it on the first exam. After that Assume all of these products have approximately the same basicity (conj. pka ~15). ( ) Point the large arrow to the weakest base. f the numbers are close (say within six or so, it is reversible. f it is ten, then it can be reversible but does slow slowly. f it is greater than ten, not gonna happen) a. conj. pka ~ 15 H Me conj. pka ~ 15 b. Cl H conj. pka ~ -5 Me Cl conj. pka ~ 15 c. H conj. pka ~ 20 Me conj. pka ~ 15 d. H conj. pka ~ 45 Me conj. pka ~ 15 e. conj. pka ~ 10 C H Me C conj. pka ~ 15

3 3.) Draw mechanisms for the following reactions. Label the products as acetals, hemiacetals, imines, or cyanohydrins. A. cat. ah H H 2 H Hemiacetal B. H cat. H 2 S 4 H 2 H Hemiacetal

4 C. H cat. H 2 S 4 EtH Acetal

5 D. cat. AcH H 2 imine

6 E. cat. ah HC H Cyanohydrin C F. KC AcH H Cyanohydrin C

7 4.) Predict the products: A. cat. H 2 S 4 H B. KC H H C C. H 2 cat. AcH D. cat. H 2 S 4 H optically active H optically active E. cat. AcH H 2 F. cat. ah C HC H

8 5.) The following compounds come from a carbonyl (ketone or aldehyde). Fill in the boxes A. H cat. H 2 S 4 H ketone or aldehyde conditions B. cat. H 2 S 4 Ph H 2 Ph ketone or aldehyde conditions C. cat. H 2 S 4 H ketone or aldehyde conditions D. H cat. ah, HC or KC, AcH optically active C optically active ketone or aldehyde conditions E. ketone or aldehyde H cat. H 2 S 4 conditions H

9 6.) Draw the mechanisms for the following reactions. Show all curved arrows and intermediates: a. H 2 trace AcH s the product? Why or why not? Two Possibilities: The product is achiral, so it is neither optically active nor. r

10 6.) continued: b. Me Me H 2 trace AcH

11 7.) Predict the products: a.) H 2 Ph Ph trace AcH Major b.) H 2 trace AcH H 2 3 signals in 1 H MR

12 8.) Critical thinking: The following acetals when exposed to acid and water, aldehydes are formed. Which acetal is cleaved faster and why? Draw the mechanism of the formation of benzaldehyde from each of these acetals. Acetal A Acetal B

13 8.) Continued Acetal B is cleaved slower than A. When the first is protonated and leaves to form the carbocation, the is still tethered and will back before water can add. n the case of acetal A, the becomes completely detached and becomes less likely to come back as it diffuses away.

14 9.) Write the mechanisms for the following problems: a. abd 4 EtH H D b. H 1.) excess MeMgBr/Et 2 2.) H 3 + H H

15 a.) 10.) Using CD 3 D and abd 4 as your sole sources of deuterium, synthesize the following compounds. D abd 4 EtH H b.) D D 1.) Et 2 2.) H 3 + MgBr H D D PCC CH 2 Cl 2 c.) D D H 1.) Et 2 2.) H 3 + D D MgBr CD 3 D H H D D HBr Mg Et 2 D D H Br 1.) Et 2 2.) H 3 + MgBr PCC CH 2 Cl 2 CD 3 D d.) D D CD 3 Br acd 3 ah CD 3 D

16 e.) H 1.) Et 2 Mg 2.) H 3 + Et 2 MgBr Br CD 3 H H CD 3 CD 3 PBr 3 D 3 C Br Mg Et 2 D 3 C MgBr 1.) Et 2 2.) H 3 + H PBr 3 CD 3 f.) CD 3 D CD 3 1.) Et 2 2.) H 3 + Mg Et 2 H BrMg CD 3 Br CD 3 g.) CD 3 D PBr 3 D 3 C Br Mg Et 2 CD 3 D D 3 C MgBr 1.) Et 2 2.) H 3 + PBr 3 H CD 3 h.) D CD 3 D a D h. is just an acid base reaction. That is all. MgBr

17 11.) MR Problem: Predict the MRs of the following compounds:

18

19

20 12.) C 9 H 8. The MR is from the Sigma Aldrich Company.

21 13.) Chem 343 review: S 1 vs. S 2 vs. E2 The nucleophiles below (except for one) are all good enough nucleophiles to work in a S 2 reaction (except one). They have various base strengths which is inversely proportional to the strength of the conjugate acid. The basicity of a nucleophile as well as the hindrance of the electrophile provides a good measure whether a reaction will undergo S 1 or S 2 or E2. f the alpha carbon (the carbon connected to the good leaving group) is tertiary, then only nucleophiles whose conjugate acids have a pka less than 5 will give S 1 products in an appreciative amount. f the pka is above 5, then E2 becomes the major pathway. (At pka of five expect a complex mixture of E2 and S 1 products). f the alpha carbon is secondary, when the conjugate acid is 10 or below, then S 2 is the major product. When it is 15 or above, then E2 is the major product. f the alpha carbon is primary, the major pathway is S 2 except when a bulky base like tbu - is used, then E2 happens. f the alpha carbon is a methyl group, then E2 is impossible and S 2 is the only option besides no reaction. Even tbu - will react with Me in a S 2 reaction. Write next to each nucleophile what the approximate pka of its conjugate acid is. Then, react it with each of the four electrophiles. Write the major pathway and major product. (32 reactions total). ucleophiles Electrophiles conj. pka ~ 25 Br conj. pka ~ -5 C conj. pka ~ 15 conj. pka ~ 15 optically active S conj. pka ~ 10 conj. pka ~ 10 C conj. pka ~ 5 conj. pka ~ 10 Me

22 a. Br S 1 Br b. Br S 2 Br optically active optically active c. Br S 2 Br d. e. Br Me S 2 E2 MeBr f. E2 optically active g. E2 h. Me S 2

23 i. E2 j. S 2 Ph optically active optically active k. S 2 l. Me S 2 Me m. C E2 n. C E2 optically active o. C S 2 p. C S 2 Me

24 r. E2 s. E2 optically active t. S 2 u. S 2 v. S Me E2 w. S 2 S S optically active optically active x. S S 2 S y. S Me S 2 S

25 z. C E2 aa. C S 2 C optically active optically active ab. C S 2 C ac. ad. C Me S 2 E2 and S 1 Me C ae. S 2 optically active af. S 2 optically active ag. S 2 Me

26 14.) From the MR provided, determine the structure of the following compound. Assign the peaks Ha, Hb, Hc etc. accordingly. The MR spectra are from the Sigma-Aldrich Corporation. ntegrations are in bold above each signal. C 8 H DBE

27 15.) From the MR provided, determine the structure of the following compound. Assign the peaks Ha, Hb, Hc etc. accordingly. The MR spectra are from the Sigma-Aldrich Corporation. ntegrations are in bold above each signal. C 4 H 9 1 DBE

28 16.) From the MR provided, determine the structure of the following compound. Assign the peaks Ha, Hb, Hc etc. accordingly. The MR spectra are from the Sigma-Aldrich Corporation. ntegrations are in bold above each signal. All signals are shown. C 3 H DBE

29 17.) From the MR provided, determine the structure of the following compound. Assign the peaks Ha, Hb, Hc etc. accordingly. The MR spectra are from the Sigma-Aldrich Corporation. ntegrations are in bold above each signal. C 9 H DBE

30 18.) From the MR provided, determine the structure of the following compound. The MR spectra are from the Sigma-Aldrich Corporation. ntegrations are in bold above each signal. C 9 H 10 5 DBE

31 19.) Sometimes it is possible to combine two reactions in the same pot to get a new reaction. Draw the mechanism for the following reactions a.) AcH H H 3 b.) AcH KC H C

32 c.) H 4 Cl H 2 KC C

33 20.) ne way to make amines is to treat an imine with sodium borohydride in EtH. Draw the mechanism of the following two step sequence. AcH MeH 2 abh 4 EtH H

34 21.) Sodium cyanoborohydride is a weaker hydride source than sodium borohydride. Why is sodium cyanoborohydride weaker than sodium borohydride? The nitrile group is a strongly electronegative and electron withdrawing group. t pulls electron density from the boron which then pulls electron density from the hydrogens. The hydrogens then have less electron density and are thus less likely to attack a carbonyl. H H B a C Sodium Cyanoborohydride H t is a very handy reagent, because you do not need to preform the imine. The imine can be made in situ, so a two step procedure becomes a one step procedure. Draw the mechanism for the reaction. Hint: This goes through an imine reaction. n this mechanism try to avoid making a negative nitrogen. egative nitrogens are very basic and you should try to avoid them if possible. AcH MeH 2 abh 3 C H What goes wrong if abh4 is used instead? What major side reaction takes place? The aldehyde can be reduced by sodium borohydride to give an alcohol.

35 Chem 345 Reaction Sheets: Acetal formation cat. H 2 S 4 H Mechanism: cat. H 2 S 4 H Summary (Key words): Acetals are made under acidic conditions, so no basic compounds can be used in the mechanism. The mechanism consists of hemiacetal formation followed by an S1 reaction. The carbonyl oxygen is lost as water. Acetals are useful to protect ketones and aldehydes from nucleophiles such as Grignards. Acetal formation is reversible.

36 Chem 345 Reaction Sheets: Acetal Cleavage cat. H 2 S 4 H 2 H H Mechanism: cat. H 2 S 4 H 2 Summary (Key words): Acetals are cleaved by acid and water. The carbonyl oxygen comes from water. Acetals are useful to protect ketones and aldehydes from nucleophiles such as Grignards. All steps in the mechanism are reversible.

37 Chem 345 Reaction Sheets: mine Formation cat. H 2 S 4 H 2 Mechanism: cat. H 2 S 4 H 2 Summary (Key words): f a strong acid is used, then it must be catalytic. t is not necessary for the carbonyl before the amine attacks. Reaction generates water. All steps are reversible.

38 Chem 345 Reaction Sheets: mine Hydrolysis cat. H 2 S 4 H 2 H 2 Mechanism: cat. H 2 S 4 H 2 Summary (Key words): mines are slightly basic. t is not necessary for there to be an acid catalyst. mines are water sensitive and are readily converted to carbonyls. All steps are reversible.

39 Chem 345 Reaction Sheets: Cyanohydrin Formation AcH, KC or cat. ah, HC C H Mechanisms: AcH, KC cat. ah, HC Summary (Key words): Reaction is reversible. Cyanohydrins are base sensitive. Cyanide is a good nucleophile. t attacks the carbonyl directly.

40 Chem 345 Reaction Sheets: Cyanohydrin cleavage H cat. ah C Mechanism: H cat. ah C Summary (Key words): All steps reversible. Cyanohydrins are base sensitive.

41 Chem 345 Reaction Sheets: Reduction of a ketone with sodium borohydride abh 4 EtH H Mechanism: abh 4 EtH Summary (Key words): Addition of the hydride to carbonyl is irreversible. The hydrogen that attacks the carbonyl carbon comes from sodium borohydride. The proton that adds to the oxygen comes from the alcohol solvent. Sodium borohydride is basic. t can also deprotonate H s (cyanohydrins, hemiacetals)

42 Chem 345 Reaction Sheets: Reaction of a ketone with a Grignard Reagent 1.) MeMgBr, Et 2 2.) H 3 + H Mechanism: 1.) MeMgBr, Et 2 2.) H 3 + Preparation of a Grignard Reagent: Mg MeBr MeMgBr Summary (Key words): Addition of the carbon group to the carbonyl is irreversible. Always a two step procedure. Grignards are really strong bases. t will react with acidic hydrogens very quickly. This includes H s on cyanohydrins or hemiacetals.