Organic Chemistry Chapter 14 Aldehydes and Ketones

Chapter 14 Aldehydes and Ketones

Bridge to the Huangpu River Cruise Boat (Shanghai, China) David Richardson

“Let us try to teach generosity and altruism, because we are born selfish. Let us understand what our own selfish genes are up to, because we may then at least have the chance to upset their designs, something that no other species has ever aspired to do.�

Richard Dawkins, The Selfish Gene

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The structure of aldehydes and ketones can be represented by the following general formulas.

Aldehydes have an alkyl, R, or aryl, Ar, group and a hydrogen atom that sandwiches a carbonyl group. A ketone has two groups (alkyl or aryl) sandwiching a carbonyl group.

Following are some examples of aldehydes and ketones:

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benzaldehyde

diphenyl ketone benzophenone

methyl phenyl ketone acetophenone

n-butanal

n-butyl ethyl ketone 3-heptanone

In naming aldehydes, one selects the longest continuous chain containing

,and drop the “e” of the longest continuous alkane containing the aldehydic group, and replaced the “e” with “al.” The aldehyde receives numerical preference over sidechain substituents and/or centers of unsaturation.

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Following these rules, the IUPAC name for

is 4,4-dimethylnonanal.

The IUPAC name for

is 4-methyl-4-phenyloctanedial When CHO, a formyl group, is attached to a ring, the ring name is followed by the suffix “carbaldehyde.�

For example, the IUPAC name for

Is cyclohexane carbaldehyde Some familiar aldehydes are formaldehyde (I), acetaldehyde (II), and benzaldehyde (III):

5

Suggest names for the following aldehydes. (a)

(b)

Answers 1. 3-bromo-4-ethoxybenzene carbaldehyde 2. 3-ethylcyclohexane carbaldehyde

Some familiar ketones are acetone (IV); acetophenone (V); benzophenone (VI):

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Ketones can be named by selecting the longest continuous chain in the carbon chain, including the carbonyl carbon, and naming the attached groups as substituents.

For example, the IUPAC name for

is 5-hexen-2-one.

The IUPAC name for

is 2,2-dimethyl-6-phenyl-3-hexanone.

Using IUPAC rules for nomenclature, suggest names for the following ketones.

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(a)

(b)

(c)

(d)

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Answers (a) 6-methyl-5-hepten-2-one (b) (E)-5-hepten-2-one (c) 2-methyl-2,6-diphenyl-3-hexanone (d) (Z)-5-hepten-2-one

The Nature of the Carbonyl Group The C=O bond distance is shorter than the C-O bond distance. The H-C-O bond angle in ketones and aldehydes is approximately 120o. The hybridization around the carbonyl carbon atoms of aldehydes and ketones is sp2.

The geometry around the carbonyl carbon atoms in aldehyde and ketones is trigonal planar, and the C=O bond is more polar than the C=C bond.

Physical Properties Aldehydes and ketones have higher boiling points than comparable molecular mass alkenes because the intermolecular forces between aldehydes and ketones are stronger than the intermolecular forces between alkenes.

Aldehydes and ketones have lower boiling points than comparable molecular mass alcohols because alcohols can molecular associate with each other.

However, carbonyl groups can molecularly associate with the OH of alcohols or phenols or carboxylic acids or water; therefore, carbonyl compounds are more soluble in water than comparable molecular mass alkenes. The solubility of carbonyl compounds decreases in alcohols.

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Reactions of Aldehydes and Ketones Aldehydes and ketones can form cyanohydrins. Cyanohydrins are compounds containing a nitrile group, CN, and a hydroxyl group, OH on the same carbon atom.

Cyanohydrins form when alkali metal cyanides react with the carbonyl group of aldehydes and ketones.

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Cyanohydrins convert to Îą-hydroxycarboxylic acids by acidifying cyanohydrins.

The net ionic equation for the reaction is

The following sequence of reactions represents a plausible mechanism for the acid hydrolysis of cyanohydrins.

(1)

11

(2)

+ H2O (3)

(4)

12

(5)

Natural Cyanohydrins Natural cyanohydrins exist as cyanogenic glycosides. Glycosides are acetals, geminal diethers, formed from natural sugars or sugar derivatives and HOR. Laetrile is an example of a glycoside.

Laetrile

The IUPAC name for laetrile is (2S,3S,4S,5R,6R)-6-[(R)-cyano(phenyl)methoxy]-3,4,5trihydroxyoxane-2-carboxylic acid.

The oxane ring or the glucuronic acid structure of Laetrile can be visualized by thinking of an intramolecular nucleophilic attack of the oxygen atom on the fifth carbon atom of glucuronic acid with the aldehydic carbonyl group of glucuronic acid to form a hemiacetal structure. The result would be a six membered heterocyclic ring (the oxane 13

ring). This process is illustrated by the following equation.

COOH ..

H O .. OH H H OH HO

O

H HOC HO

H

HO

H H

OH

H

O OH OH

H

H

The final step in the formation of Laetrile would be the formation of the glycosidic linkage at the anomeric carbon atom, and this is illustrated by the following equation.

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O

H HOC HO

H

HO

H

C

O +

OH

N:

C

HO H

OH H

H

O

H

C

HOC HO

H

HO

H

O O

N:

C H

OH H

H

Acetal Formation Acetals form from the reaction of alcohols with aldehydes in the presence of a mineral acid. The reaction sequence proceeds first through the formation of a hemiacetal, and then to the formation of the acetal.

The following is an illustration of the formation of acetals.

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The sequence of elementary steps that would account for this reaction are: (1)

(2)

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(3)

(4)

(5)

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(6)

(7)

Diols in the 1,2 and 1,3 positions form cyclic acetals (1,3-dioxolanes, 5-member rings, or 1,3-dioxane, 6-member rings) with aldehydes and ketones.

O

HO

O

+

C H

HO

O CH3

S

OH

H

+

H2O

O

O

acetal, 1,3-dioxolane (2-phenyl-1,3-dioxane)

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acetal, 1,3-dioxane (2-phenyl-1,3-dioxane)

Ketones can also undergo reactions with 1,2 and 1,3 diols analogous to aldehydes.

The resulting 1,3-dioxolane or 1,3-dioxane are ketals.

O

HO

O

+

C

HO

CH3

O CH3

S

OH

H3C

+

H2O

O

O

ketal, 1,3-dioxolane (2-phenyl-2-methyl-1,3-dioxolane)

HO

O

O

+

C CH3

HO

O CH3

S

OH

H3C

+

H2O

O

O

ketal, 1,3-dioxane (2-phenyl-2-methyl-1,3-dioxane)

The water produced in these reactions is removed azeotropically, and drives the equilibrium to the right, i.e., toward the 2-phenyl-1,3-dioxane, the 2-phenyl-1,3dioxolane, the 2-phenyl-2-methyl-1,3-dioxolane, or the 2-phenyl-2-methyl-1,3-dioxane.

Acetals/ketals readily hydrolyze in acid to form the aldehyde or the ketone. This reaction, for example, can be observed in the acid-catalyzed hydrolysis of 2-phenyl1,3-dioxane.

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The following is a sequence of elementary steps that rationalize this observation.

(1)

(2)

(3)

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(4)

(5)

(6)

Acetals/ketals as Protective Groups Like ethers, the acetal/ketal group is unaffected by many reactions; however, acetals/ketals can be converted to aldehydes or ketones by acid hydrolysis.

Assume an undergraduate senior chemistry student under the supervision of a faculty advisor is interested in synthesizing 1-phenyl-3-hexyn-2-one from benzylcarboxylic acid (phenylacetic acid).

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phenylacetic acid

1-phenyl-3-hexyn-2-one

The reaction can be approached by the following retrosynthesis, i.e., working backwards.

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Using an analogous synthesis as the one above, suggest a synthesis for p-acetylbenzyl alcohol from p-acetylbenzoic acid

p-acetylbenzy alcohol

p-acetylbenzoic acid

Answer The synthesis of p-acetylbenzyl alcohol from p-acetylbenzoic acid is a three-step process.

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Step 1 is the use of 1,3-butandiol as a protective group for the ketene carbonyl group.

Step 2 is the reduction of the acid with sodium borohydride.

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Step 3 is acid hydrolysis of the 1,3-dioxane ring in order to release the protective group.

The protective group is required, because sodium borohydride would reduce the ketone group as well as the carboxylic acid group.

Aldehydes and Ketones can react with primary amines as indicated in the following reaction to produce N-substituted imines.

The product, an N-substituted imine, is formed through a carbinolamine intermediate.

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The following is the mechanism for this reaction in an acidic medium.

The reaction is optimized at [H3O+] = 1 x 10-5 mol/L, i.e., at pH = 5.

At a pH less than 5, the primary amine will be protonated and hydrogen ions will be unavailable to protonate the aldehyde or the ketone. At a pH greater than 5, the carbocation in the rate-determining step (step 6) will not form because the solution will be too basic for the formation of a positive charge.

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(1)

(2)

(3)

(4)

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(5)

(6)

(7)

(8)

29

(9)

Imine-type compounds are useful in the traditional identification of Aldehydes and Ketones. These compounds are oximes, semicarbazones, phenylhydrazones, and 2,4dinitrophenylhydrazones (Chapter 20). The mechanisms of these reactions are similar to the addition and elimination of primary amines to carbonyl compounds to form imines.

Oxime

Semicarbazone

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Phenylhydrazone

2,4-Dinitophenylhydrazone

Reactions of Aldehydes and Ketones with Secondary Amines The reaction of aldehydes and ketones with secondary amines proceeds via the formation of carbinolamine intermediates, followed by dehydration to stable enamines (alkenyl-substituted amines).

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The following sequence of three elementary steps represents the mechanism that rationalizes the formation of the enamine.

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(1)

(2)

(3)

(4)

The enamine

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The Wittig Reaction The Wittig reaction is a method for converting aldehydes and ketones to alkenes via the use of a phosphorus ylides like methylenetriphenylphosporane.

The reaction is generally in an aprotic solvent like dimethyl sulfoxide (DMSO) or tetrahydrofuran.

DMSO The following reaction is an example of the Wittig reaction:

The following is a mechanism for the Wittig reaction:

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(1)

(2)

(3)

oxaphosphetane

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(4)

The phosphorus ylide is prepared from an alkyl halide by an SN2 reaction.

The resulting methyltriphenylphosphonium iodide crystallizes in high yield from nonpolar solvents. The salt is isolated, and converted to the desired ylide by a strong base such as the sodium salt of dimethyl sulfoxide or organolithium reagents.

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Phosphorus ylides are excellent precursors for the syntheses of alkenes. For example, the following is a retrosynthesis of 3-ethyl-4-methyl-3-hexene executed in three steps.

Step 3 is the reaction of the ylide with the desired ketone to produce 3-ethyl-4-methyl3-hexene.

Step 2 is the preparation of the ylide from the phosphonium iodide salt, 3pentyltriphenylphosphonium iodide, and a strong base, lithium ethide.

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Step 1 is the synthesis of 3-pentyltriphenylphosphonium iodide, the precursor to the ylide, from triphenylphosphine and an appropriate alkyl halide, 3-iodopentane.

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The Iodoform Reaction Methyl ketones, as well as methyl groups attached to carbon atoms forming secondary alcohols, react with iodine in sodium hydroxide to form a yellow precipitate. The reaction of methyl ketones with iodine in sodium hydroxide is the iodoform reaction.

The yellow precipitate is triiodomethane or iodoform, the name that gives rise to the nomenclature of the reaction The following is an example of the iodoform reaction showing the stoichiometry of the reaction. The stoichiometry of the reaction gives insight into the mechanism of the reaction.

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The following series of elementary steps explain how methyl ketones forms iodoform, the yellow precipitate.

(1)

(2)

(3)

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(4)

(5)

(6)

(7)

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(8)

(9)

(10)

(11)

yellow precipitate

The sum of the elementary steps 1-11 gives

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Ketones with methyl groups attached will react with iodine and sodium hydroxide to give a yellow precipitate, iodoform. Secondary alcohols with methyl groups attached will also react with iodine and sodium hydroxide to produce iodoform.

OH

(R or Ar)

+

C

3 I2

+ 3 NaOH

CH3 H O +

C (R or Ar)

3 H2O

+

3 NaI

+

CHI3

H

The formation of the yellow precipitate is a positive iodoform test. Similar reactions will occur when methyl ketones and secondary alcohols react with chlorine and sodium hydroxide. Also, similar reactions will occur when methyl ketones and secondary alcohols with methyl groups attached react with NaOH and bromine. However, the resulting chloroform and bromoform molecules are not precipitates with dramatic colors.

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The Cannizzaro Reaction The Cannizzaro reaction is named after its discoverer Stanislao Cannizzaro who discovered it sometime during the middle of the nineteenth century. The reaction involves the base-induced disproportionation of an aldehyde that doesn’t have a hydrogen atom attached to the alpha carbon atom. The products of the disproportionation reaction are the salt of a carboxylic acid, the oxidation product, and an alcohol, the reduction product.

The following is an example of the Cannizzaro Reaction.

The following is the mechanism of the Cannizzaro reaction.

(1)

(2)

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(3)

Stereoselective Addition to Carbonyl Groups Nucleophilic addition to carbonyl groups can give rise to stereoisomeric products.

Steric factors play a major role in the resulting product where the nucleophile approaches the carbonyl at the less hindered direction.

For example, treating 7,7-diethylbicyclo[2.2.1]heptan-2-one can give rise to two possible products- exo-7,7-diethylbicyclo[2.2.1]heptan-2-ol or endo-7,7diethylbicyclo[2.2.1]heptan-2-ol. The exo-7,7-diethylbicyclo[2.2.1]heptan-2-ol would be the predominant product because the two ethyl groups would hinder the approach of the đ??ľđ??ť!" anion from the endo position; consequently, the exo product would predominate.

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Therefore, the reduction product is stereoselective. The preferred attach is endo which leads to the exo product. Exo attack would lead to the endo product. Both products are theoretically possible, but, in this case, the exo product predominates.

If the reaction is enzyme-catalyzed, then one would expect that one of the two possible stereoisomers would be exclusively formed. For example, when pyruvic acid forms lactic acid (via the enzyme lactate dehydrogenase and its coenzyme, the

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reduced form of nicotinamide adenine dinucleotide, NADH) the predominate product is the dextrorotatory compound with an S-configuration, i.e., (S)-(+)-lactic acid exclusively.

The Baeyer-Villiger Oxidation of Ketones The Baeyer-Villiger reaction is the synthesis of an ester from a ketone, and Adolf von Baeyer and Victor Villiger reported the reaction in 1899. Ketones react with peroxy acids to insert an oxygen atom between the carbonyl group and the larger of two attached alkyl or aryl groups to form an ester.

For example,

Methyl chloride is one solvent used in Baeyer-Villiger oxidation reactions.

The series of elementary steps that lead to the formation of the ester includes the

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migration of the alkyl group with retention of configuration; therefore, if stereochemistry is possible, the reaction leads to the retention of configuration, i.e., the reaction is stereospecific.

(1)

(2)

Since the alkyl group migrates with retention of configuration, trans-(e,e)-4methylcyclohexyl methyl ketone would form trans-(e,e)-4-methylcyclohexyl acetate as the exclusive stereospecific product.

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Reformatsky Reaction Aldehydes and ketones can react with α-haloesters in the presence of metallic zinc to form β-hydroxyesters.

The following represents a reaction between methyl ethyl ketone and ethyl αbromoacetate in the presence of zinc to produce ethyl 3-hydroxy-3-methylpentanoate.

The following are the series of elementary steps that rationalize the formation of the product for the Reformatsky reaction.

Let’s use the above reaction as a model.

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1.

Zinc relinquishes two electrons to ethyl 2-bromoacetate to produce a zincbromoenolate.

2-bromoacetate

a zincbromoenolate

2.

The zincbromoenolate reacts with 2-butanone to give the zincbromo salt of ethyl 3methyl-3-hydroxypentanoate.

2-butanone

zincbromo salt of 3-methyl-3-hydroxypentanoate

In this example, acid hydrolysis of the zincbromo enolate would form ethyl 3-methyl-3hydroxypentanoate.

ethyl 3-methyl-3-hydroxypentanoate

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Problems

1.

Suggest IUPAC names for the following molecules. (a)

(b)

(c)

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2.

Suggest a mechanism for the following conversion.

3.

Suggest the product expected from the reaction of benzaldehyde with each of the following. (a)

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(a) dilute sodium hydroxide (b) concentrated sodium hydroxide (c) propanal and dilute hydroxide (d) butanal and dilute sodium hydroxide (f)

4.

Suggest a synthesis for the following molecule from the indicated starting material and any necessary organic and inorganic materials.

5.

Write a mechanism for your synthesis of the ester in problem 4.

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6.

Using 1-propanol as the only source of organic compounds and any other necessary inorganic reagents, suggest syntheses for the following.

(a)

(b)

7.

Suggest a synthesis for the following molecule from the indicated organic compound and any other available inorganic materials.

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8.

Suggest a synthesis for the following from the indicated starting material and any necessary inorganic and organic compounds.

9.

Compound I exhibits the following proton magnetic resonance spectrum.

1

H NMR of compound I

A constitutional isomer of compound I, compound II, exhibits the following proton magnetic spectrum.

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1

H NMR of compound II

When compound I was treated with lithium aluminum hydride hydrolysis, then aqueous acid at about 200oC, produced. Compound III has two

followed by

compound III, C10H12 was

isomers.

Following is the proton magnetic resonance spectrum of compound III, C10H12

1

H NMR of compound III

When compound II was treated with lithium aluminum hydride followed by hydrolysis, then aqueous acid at about 200oC, compound IV, C10H12 was produced. Following is

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the proton magnetic resonance spectrum of compound IV, C10H12. Compound IV has one isomer. Compounds III and IV are also constitutional isomers.

1

H NMR of compound IV

Suggest structural formulas for compounds I, II, III, and IV.

10.

A certain biologically active terpene, C10H16O follows the isoprene rule will form an oxime, a semicarbozide, and a 2,4-dintrophenylhyradazide. Upon ozonolysis, the terpene produces the following compounds.

Suggest a structure consistent with the experimental data and the isoprene rule.

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11.

A terpene that follows the isoprene rule decolorizes a dilute solution of potassium permanganate, and, upon, treatment with hot concentrated potassium permanganate in sulfuric acid, produces the following two organic acids:

Suggest a structure for the terpene that is consistent with the experimental evidence.

12.

Suggest a synthesis for the following conversion using appropriate inorganic reagents.

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13.

Suggest a series of elementary steps that would rationalize the following observations. (a)

(b)

14.

Predict the product or products expected in the following sequence of reactions.

15.

Compound I is treated with the reagents of the Clemmensen reduction to produce compound II. Treating compound II with the reagents for hydroboration-oxidation produces compound III. Compound III upon treatment with pyridinium chlorochromate produces compound IV.

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Compound I Suggest structural formulas for compounds II, III, and IV. 16.

Consider the following synthesis pathway:

O

O

C

C

CH3CH2

C9H16O4 +

CH2CH2

OH

+ 1,3-propandiol H3O

LiAlH4

+

C9H16O4

C9H18O3 H2O

C9H18O 3 H3O

C6H12O2

+

PDC

+

C6H12O 2

C6H10O2

Suggest structures of C9H16O4; C9H18O3; C6H12O2; and C6H10O2.

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17.

Treating compound 17A, C6H12O, with iodine in sodium hydroxide, results in the formation a yellow precipitate and compound 17 B, C5H9O2Na. Acidification of compound 17 B formed C5H10O2, compound 17 C. Treating compound 17C with LiAlH4 followed by hydrolysis produces compound 17 D. The infrared spectrum of compound 17 D exhibited a prominent stretching transmittance signal at 3333 cm-1.

Following are the proton magnetic resonance spectrum and carbon-13 magnetic resonance spectrum of compound 17 D. 1

H NMR Compound 17 D

13

C NMR

Compound 17 D

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Sulfuric acid converted compound 17 D to compound 17 E. Ozonolysis of compound 17 E gave compound 17 F and 17 G. Prominent mass spectrum peaks for compound 17 F occur at m/e 29 and 43. Prominent mass spectrum peaks for compound 17G occur at m/e 43 and 58.

Use these data to determine the structures of compounds 17 A, 17 B, 17 C, 17 D, 17 E, 17 F, and 17 G. Please support your answers with chemical equations and any given spectra data.

Provide a detailed mechanism for the conversion of 17 D to 17 E with sulfuric acid.

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Solutions to Problems Chapter 14

1.

Suggest IUPAC names for the following molecules. (a)

6-phenyl-3-hexanone (b)

7-phenyl-4-oxohexanal (c)

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5,7-diphenylheptanal

2.

Suggest a mechanism for the following conversion.

(1)

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(2)

(3)

(4)

(5)

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3.

Suggest the product expected from the reaction of benzaldehyde with each of the following. (a)

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(b) dilute sodium hydroxide No reaction a. (c) concentrated sodium hydroxide

b. (d) propanal and dilute hydroxide

(e) butanal and dilute sodium hydroxide

(f)

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4.

Suggest a synthesis for the following molecule from the indicated starting material and any necessary organic and inorganic materials.

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5.

Write a mechanism for your synthesis of the ester in problem 4.

(1)

(2)

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6.

Using 1-propanol as the only source of organic compounds and any other necessary inorganic reagents, suggest syntheses for the following. (a)

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(b)

7.

Suggest a synthesis for the following molecule from the indicated organic compound and any other available inorganic materials.

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72

8.

Suggest a synthesis for the following from the indicated starting material and any necessary inorganic and organic compounds.

73

74

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9.

Compound I exhibits the following proton magnetic resonance spectrum.

1

H NMR of compound I

A constitutional isomer of compound I, compound II, exhibits the

following proton

magnetic spectrum.

1

H NMR of compound II

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When compound I was treated with lithium aluminum hydride followed by hydrolysis, then aqueous acid at about 200oC, compound III, C10H12 was produced. Compound III has two isomers.

Following is the proton magnetic resonance spectrum of compound III, C10H12

1

H NMR of compound III

When compound II was treated with lithium aluminum hydride followed by hydrolysis, then aqueous acid at about 200oC, compound IV, C10H12 was produced. Following is the proton magnetic resonance spectrum of compound IV, C10H12. Compound IV has one isomer. Compounds III and IV are also constitutional isomers.

The structure of compound I is

The structure of compound II is

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Compound III may have a cis configuration or a trans configuration. These structures are

Compound IV does not have a nonsuperposable/nonsuperimposable mirror image.

1

H NMR of compound IV

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10.

A certain biologically active terpene, C10H16O follows the isoprene rule will form an oxime, a semicarbozide, and a 2,4-dintrophenylhydradize. Upon ozonolysis, the terpenes produces the following compounds

Suggest a structure consistent with the experimental data and the isoprene rule.

11.

A terpene that follows the isoprene rule decolorizes a dilute solution of potassium permanganate, and, upon, treatment with hot concentrated potassium permanganate in sulfuric acid, produces the following two organic acids:

Suggest a structure for the terpene that is consistent with the experimental evidence.

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12.

Suggest a synthesis for the following conversion using appropriate inorganic reagents.

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13.

Suggest a series of elementary steps that would rationalize the following observations. (a)

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(b)

(1)

(2)

82

(3)

14.

Predict the product or products expected in the following sequence of reactions.

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15.

Compound I is treated with the reagents of the Clemmensen reduction to produce compound II. Treating compound II with the reagents for hydroboration-oxidation produces compound III. Compound III upon treatment with pyridinium chlorochromate produces compound IV.

Compound I Clemmensen Reduction:

compound II

Hydroboration-Oxidation:

compound III

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Oxidation with Pyridinium Chlorochromate

compound IV

16.

Consider the following synthesis pathway:

C9H16O4 +

LiAlH4

C9H18O3 H2O

C9H18O 3 H3O

+

C6H12O 2

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C6H12O2

+

PDC

C6H10O2

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17.

Treating compound 17A, C6H12O, with iodine in sodium hydroxide, results in the formation a yellow precipitate and compound 17 B, C5H9O2Na. Acidification of compound 17 B formed C5H10O2, compound 17 C. Treating compound 17C with LiAlH4 followed by hydrolysis produces compound 17 D. The infrared spectrum of compound 17 D exhibited a prominent stretching transmittance signal at 3333 cm-1.

Following are the proton magnetic resonance spectrum and carbon-13 magnetic resonance spectrum of compound 17 D. 1

H NMR Compound 17 D

13

C NMR

Compound 17 D

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Sulfuric acid converted compound 17 D to compound 17 E. Ozonolysis of compound 17 E gave compound 17 F and 17 G. Prominent mass spectrum peaks for compound 17 F occur at m/e 29 and 43. Prominent mass spectrum peaks for compound 17G occur at m/e 43 and 58.

Use these data to determine the structures of compounds 17 A, 17 B, 17 C, 17 D, 17 E, 17 F, and 17 G. Please support your answers with chemical equations and any given spectra data.

Provide a detailed mechanism for the conversion of 17 D to 17 E with sulfuric acid.

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Mechanism of 17 D to 17 E: (1)

17 D

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(2)

(3)

(4)

17 E

90