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Reactions to know from Chapters 17, 18, and 19 1. Oxidation of aldehydes O O K2Cr2SO4 H2SO4 H Butanal OH Butanoic acid Aldehydes will oxidize in the presence of oxygen only [O], the above oxidizing agent and catalyst will work, but not necessary. Ketones do not oxidize 2. Reduction of aldehydes and ketones Aldehydes are reduced to primary alcohols, ketones are reduced to secondary alcohols a) Using H2 and a metal catalyst (Pd) O OH Pd + H2 2-pentanol 2-pentanone This is a non-selective process, it will reduce double bonds as well as ketones and alcohols b) Using NaBH4 as a reducing agent NaBH4 O OH H2O 3-cyclopentenone 3-cyclopentenol NaBH4 is a moderate strength reducing agent that will selectively reduce aldehydes and ketones to alcohols, but leave alkene double bonds intact. Because the H- produced by NaBH4 attacks the partial positive carbonyl carbon of the aldehyde or ketone. 3. The addition of alcohols to aldehydes and ketones OH O + HO O H Hemi-acetal Alcohol Aldehyde OH O + HO O Alcohol Ketone Hemi-acetal Starting with either an aldehyde or a ketone, you can see that the hemiacetals formed are characterized by having a carbon bonded to an OH- group and an OR- group. Here, the oxygen of the alcohol attacks and bonds with the carbonyl carbon of the aldehyde or ketone If the alcohol group and the carbonyl carbon of an aldehyde or ketone are on the same molecule, a cyclic hemi-acetal can be formed. (shown below) O 5 3 4 1 2 H OH OH group and carbonyl group on the same molecule Redraw 3 5 H 1 4 1 4 C O 2 3 2 5 O H A cyclic ring forms O OH Alcohols can react with hemi-acetals to form acetals H+ O O OH O OH hemi-acetal Alcohol acetal Here, the OH- group of the hemi-acetal is protonated to H2O+ . The H2O+ leaves, and is then replaced by the OR- group of the alcohol. Notice that for the acetal, it is characterized by having a carbon bonded to two OR- groups. 4. An acid-base reaction of a carboxylic acid O O + H2O + NaOH O- Na+ OH Sodium propanoate Propanoic acid Here, the hydroxide ion takes the acidic proton from the carboxylic acid leaving it with a negative charge. Salts of carboxylic acids are more soluble in water than the neutral acid form. 5. Reduction of a carboxylic acid Strong reducing agent O LiAl H4 OH OH H2O LiAlH4 is a much stronger reducing agent than LiBH4. It is necessary to reduce carboxylic acids, which are somewhat resistant to reduction. Carboxylic acids are reduced to primary alcohols, there is no aldehyde intermediate. LiAlH4 is a selective reducing agent that leaves alkene double bonds intact. 6. Fisher esterification of a carboxylic acid O O H2SO4 + OH Carboxylic acid + H2O O HO Alcohol Ester In the final product (ester) the OH- group of the carboxylic acid has been replaced by the OR- group of the alcohol 7. Decarboxylation of a carboxylic acid O Extreme heat C + CO2 OH O O O Mild heat + CO2 OH If a beta carbonyl carbon is present, the reaction takes place much easier (mild heat as opposed to extreme heat) 8. Preparation of an amide O O O + Heat + H2N + H2O +H3N N H O- OH The first step is an acid base reaction between the carboxylic acid and the amine. In the second step, heat is applied expelling a water molecule (O from the carboxylic acid, 2H from the amine), and the amide bond is formed. 9. a) Hydrolysis of an ester with an acid catalyst H3O+ O O + + H2O HO OH O Propanoic acid (carboxylic acid) Ethyl propanoate (ester) Ethanol (alcohol) 9. Hydrolysis of an ester with sodium hydroxide. O H2O O + + NaOH HO O-Na+ O Sodium propanoate (carboxylic salt) Ethyl propanoate (ester) Ethanol (alcohol) 10.Hydrolysis of a carboxylic anhydride O O O O + + H2O O Acetic anhydride (carboxylic anhydride) OH Acetic acid (carboxylic acid) HO Acetic acid (carboxylic acid) 11. Hydrolysis of an amide O O + HCl + H2O OH H2N N H N-ethylpropanamide (amide) Propanoic acid (carboxylic acid) This is one way you break down proteins into amino acids! ethanamine (amine)