19.13) Which of the following bases are strong enough to deprotonate CH 3 COOH?

1. 19.13) Which of the following bases are strong enough to deprotonate CH3COOH? a) F- pka of CH3COOH is 4.8. b) (CH3)3CO- b) has a pka of 18 c) has a pka of 50 d) has a pka of 38 c) CH3- d) NH2- pka of conjugate acid must be greater than that of the carboxylic acid being deprotonated. e) Cl-

2. 19.14) Rank the labeled protons in order of increasing acidity. Ha<Hb<Hc The more stable the conjugate base the more acidic the proton.

3. 19.15) Match each pka value with each carboxylic acid.(3.2, 4.9 and 0.2) • CH3CH2COOH • CF3COOH • ICH2COOH 4.9 0.2 3.2 Electron withdrawing groups make acids more acidic.

4. 19.16) Why is formic acid more acidic than acetic acid? The methyl group is electron donating and stabilizes the acid while destabilizing the conjugate base thus making it less acidic.

5. 19.17) Rank the compounds within each group in order of decreasing acidity. a) CH3COOH, HSCH2COOH, HOCH2COOH 3 2 1 b) ICH2COOH, I2CHCOOH, ICH2CH2COOh 2 1 3

6. 19.18) Rank each group of compounds in order of decreasing acidity. a) 2 1 3 b) 2 1 3

7. 19.19) Is the following compound more or less acidic than phenol? The more electron donating groups present, the less acidic a compound is. This compound has an additional hydroxy and alkyl group, both electron donating. So it is less acidic.

8. 19.22) Comparing CF3SO3H and CH3SO3H, which has the weaker conjugate base? Which conjugate base is the better leaving group? Which of these acids has the higher pka? CF3SO3H is the weaker conjugate base. CF3SO3H is the better leaving group because it is the weaker conjugate base. CH3SO3H, with the electron donating methyl group, has the higher pka and is thus a weaker acid.

9. Introduction to Carbonyl Chemistry; Organometallic Reagents; Oxidation and Reduction Introduction Two broad classes of compounds contain the carbonyl group: [1] Compounds that have only carbon and hydrogen atoms bonded to the carbonyl [2] Compounds that contain an electronegative atom bonded to the carbonyl

10. The presence or absence of a leaving group on the carbonyl determines the type of reactions the carbonyl compound will undergo. • Carbonyl carbons are sp2 hybridized, trigonal planar, and have bond angles that are ~1200. In these ways, the carbonyl group resembles the trigonal planar sp2 hybridized carbons of a C=C.

11. In one important way, the C=O and C=C are very different. • The electronegative oxygen atom in the carbonyl group means that the bond is polarized, making the carbonyl carbon electron deficient. • Using a resonance description, the carbonyl group is represented by two resonance structures.

12. General Reactions of Carbonyl Compounds Carbonyls react with nucleophiles.

13. Aldehydes and ketones react with nucleophiles to form addition products by a two-step process: nucleophilic attack followed by protonation.

14. The net result is that the  bond is broken, two new  bonds are formed, and the elements of H and Nu are added across the  bond. • Aldehydes are more reactive than ketones towards nucleophilic attack for both steric and electronic reasons.

15. Carbonyl compounds with leaving groups react with nucleophiles to form substitution products by a two-step process: nucleophilic attack, followed by loss of the leaving group. The net result is that Nu replaces Z, a nucleophilic substitution reaction. This reaction is often called nucleophilic acyl substitution.

17. Nucleophilic addition and nucleophilic acyl substitution involve the same first step—nucleophilic attack on the electrophilic carbonyl carbon to form a tetrahedral intermediate. • The difference between the two reactions is what then happens to the intermediate. • Aldehydes and ketones cannot undergo substitution because they do not have a good leaving group bonded to the newly formed sp3 hybridized carbon.

18. Preview of Oxidation and Reduction • Carbonyl compounds are either reactants or products in oxidation-reduction reactions.

19. The three most useful oxidation and reduction reactions of carbonyl starting materials can be summarized as follows:

20. Reduction of Aldehydes and Ketones • The most useful reagents for reducing aldehydes and ketones are the metal hydride reagents. • Treating an aldehyde or ketone with NaBH4 or LiAlH4, followed by H2O or some other proton source affords an alcohol.

21. The net result of adding H:¯ (from NaBH4 or LiAlH4) and H+ (from H2O) is the addition of the elements of H2 to the carbonyl  bond.

22. Catalytic hydrogenation also reduces aldehydes and ketones to 1° and 2° alcohols respectively, using H2 and a catalyst. • When a compound contains both a carbonyl group and a carbon—carbon double bond, selective reduction of one functional group can be achieved by proper choice of the reagent. • A C=C is reduced faster than a C=O with H2 (Pd-C). • A C=O is readily reduced with NaBH4 and LiAlH4, but a C=C is inert.

23. Thus, 2-cyclohexenone, which contains both a C=C and a C=O, can be reduced to three different compounds depending upon the reagent used.

24. The Stereochemistry of Carbonyl Reduction • Hydride converts a planar sp2 hybridized carbonyl carbon to a tetrahedral sp3 hybridized carbon.

25. Enantioselective Carbonyl Reductions • Selective formation of one enantiomer over another can occur if a chiral reducing agent is used. • A reduction that forms one enantiomer predominantly or exclusively is an enantioselective or asymmetric reduction. • An example of chiral reducing agents are the enantiomeric CBS reagents.

26. CBS refers to Corey, Bakshi and Shibata, the chemists who developed these versatile reagents. • One B—H bond serves as the source of hydride in this reduction. • The (S)-CBS reagent delivers H:- from the front side of the C=O. This generally affords the R alcohol as the major product. • The (R)-CBS reagent delivers H:- from the back side of the C=O. This generally affords the S alcohol as the major product.

27. These reagents are highly enantioselective. For example, treatment of propiophenone with the (S)-CBS reagent forms the R alcohol in 97% ee.

28. Reduction of Carboxylic Acids and Their Derivatives • LiAlH4 is a strong reducing agent that reacts with all carboxylic acid derivatives. • Diisobutylaluminum hydride ([(CH3)2CHCH2]2AlH, abbreviated DIBAL-H, has two bulky isobutyl groups which makes this reagent less reactive than LiAlH4. • Lithium tri-tert-butoxyaluminum hydride, LiAlH[OC(CH3)3]3, has three electronegative O atoms bonded to aluminum, which makes this reagent less nucleophilic than LiAlH4.

29. Acid chlorides and esters can be reduced to either aldehydes or 1° alcohols depending on the reagent.

30. In the reduction of an acid chloride, Cl¯ comes off as the leaving group. • In the reduction of the ester, CH3O¯ comes off as the leaving group, which is then protonated by H2O to form CH3OH.

31. The mechanism illustrates why two different products are possible.

32. Carboxylic acids are reduced to 1° alcohols with LiAlH4. • LiAlH4 is too strong a reducing agent to stop the reaction at the aldehyde stage, but milder reagents are not strong enough to initiate the reaction in the first place.

33. Unlike the LiAlH4 reduction of all other carboxylic acid derivatives, which affords 1° alcohols, the LiAlH4 reduction of amides forms amines. • Since ¯NH2 is a very poor leaving group, it is never lost during the reduction, and therefore an amine is formed.

36. Oxidation of Aldehydes • A variety of oxidizing agents can be used, including CrO3, Na2Cr2O7, K2Cr2O7, and KMnO4. • Aldehydes can also be oxidized selectively in the presence of other functional groups using silver(I) oxide in aqueous ammonium hydroxide (Tollen’s reagent). Since ketones have no H on the carbonyl carbon, they do not undergo this oxidation reaction.

37. For Wednesday, 20.1-20.16.

38. 20.1) What type of orbitals make up the indicated bonds? And in what orbitals do the lone pairs on the oxygen lie? a. sp3-sp2 b. sp2-sp2, p-p c. sp3-sp2 The lone pairs lie in sp2 hybridized orbitals.

39. 20.2) Which compounds undergo nucleophilic addition and which substitution? a) b) addition substitution c) d) substitution addition

40. 20.3) Which compound in each pair is more reactive toward nucleuphilic attack? a) b) c) d)

41. 20.4) What alcohol is formed when each compound is treated with NaBH4 in MeOH? a) b) c)

42. 20.5) What aldehyde or ketone is needed to synthesize each alcohol by metal hydride reduction? a) b) c)

43. 20.6) Why can’t 1-methylcyclohexanol be prepared from a carbonyl by reduction? Tertiary alcohols can not be made by reduction of a carbonyl because there are no hydrogens on the carbon with the -OH.

44. 20.7) Draw the products of the following reactions? a) b) c) d)

45. e) f)

46. 20.8) Draw the products when the following compounds are treated with NaBH4 in MeOH. a) b) c)

47. 20.9) What reagent is needed to carry out the reaction below? Two reagents are needed to carry out this reaction. First, the (S)-CBS reagent to produce the R-enantiomer. Followed by H2O to protonate the alcohol.

48. 20.10) Draw a stepwise mechanism for the following reaction.

49. 20.11) Draw an acid chloride and an ester that can be used to produce each product. a) b) c)

50. 20.12) Draw the products of LiAlH4 reduction of each compound. a) b) c)