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Amines

Building Bridges to Knowledge

Perspective, Photo of an unknown fence

MichaelleCadet

Amines can be primary, secondary, and tertiary. Primary amines have two hydrogen atoms attached to a nitrogen atom. Secondary amines have one hydrogen atom attached to a nitrogen atom, and tertiary amines have zero hydrogen atoms attached to a nitrogen atom. Another way differentiating a primary amine from a secondary amine from a tertiary amine is to identify the number of alkyl groups attached to the amine. A primary amine has one alkyl group attached to a nitrogen atom. A secondary amine has two alkyl groups attached to a nitrogen atom; and a tertiary amine has

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three alkyl groups attached to a nitrogen atom. The following structures respectively illustrate a primary amine, a secondary amine, and tertiary amine.

Amphetamine, an optically active psychostimulant, is an example of a primary amine. Amphetamine decreases fatigue and appetite. The IUPAC name for amphetamine is 1-phenylpropan-2-amine. The structural formula of amphetamine is represented by Compound I. Compound I has a chiral center identified as carbon atom “2.” The IUPAC name for compound I is 2-amino-3-phenylpropane.

Compound I Epinephrine (adrenaline) is an example of a secondary amine that has biological utility. Epinephrine is a hormone and a neurotransmitter. The complete name for the biologically active form of epinephrine is R-(-)-L-epinephrine and the IUPAC name for epinephrine is (R)-4-[1-hydroxy-2-(methylamino)ethyl]benzene-1,2-diol. Epinephrine, a catecholamine, is synthesized in vivo from phenylamine and tyrosine in neurons of the central nervous system and in the chromaffin cells of the adrenal medulla. Compound II represents the structural formula for epinephrine.

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Compound II Chlorphenamine is an example of a tertiary amine. Chlorphenamine is a synthetic drug used to treat and prevent allergies such as rhinitis, the inflammation of the nasal membrane. Chlorphenamine is also used to treat urticari, an allergy condition resulting in the formation of red itchy bumps. In addition, chlorphenamine is an antidepressant and an antihistamine. The IUPAC name for chlorphenamine is 3-(4-chlorophenyl)-N,N-dimethyl-3-pyridin-2-yl-propan-1-amine. Chlorphenamine is an optically active molecule with a chiral carbon on atom labeled “3” in in the following structure. Compound III is a representation of the structural formula of chlorphenamine.

Compound III Amines are weak bases, and the basicity of an amine increases with increase alkylation about the nitrogen atom. Other factors may increase or decrease the basicity of an amine in addition to alkylation around the nitrogen atom. Amines can be aliphatic or aromatic, i.e., the R groups of the primary, secondary, and tertiary amines could be aliphatic or aromatic. If the R group is aromatic,

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then the amine is referred to as aniline or an aniline derivative. If the R group is a phenyl group, then the compound is aniline. Aniline, an aromatic amine, is represented by compound IV.

Compound IV If the hydrogen atoms are replaced by alkyl groups, the resulting compounds would be N-substituted or N,N-disubstituted anilines. For example, compound V would be N-methylaniline and compound VI would be N,N-dimethylaniline.

Compound V Compound VI Aniline is a weaker base than alkyl amines, because the electron pair on the nitrogen is involved in resonance with the aromatic nucleus; therefore, reducing the availability of the electron pair to function as a Lewis base.

Basicity decreases with electron withdrawing groups in the ortho and para positions. Basicity increases with electron releasing groups on the aromatic nucleus. This discussion can be applied

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to tetrahydoquinoline (compound VII) and tetrahydroisoquinoline (compound VIII). Compound VIII, tetrahydroisoquinoline, is a stronger base than compound VII, tetrahydroquinoline, because the lone pair of electrons on the nitrogen in tetrahydroisoquiniline is not adjacent to the aromatic ring; therefore, is not involved in resonance.

Compound VII Compound VIII The lone pair of electrons in compound VII, tetrahydroquinoline, contributes to the resonance stabilization of the molecule; therefore, the lone pair of electrons is not readily available to function as a Lewis base. Piperidine (compound IX), pKb = 2.8) is more basic than pyridine (compound X), pKb = 8.8

Compound IX Compound X The reason piperidine (compound IX) is more basic than pyridine (compound X) is because the hybridization around the nitrogen in piperidine is sp3 and the hybridization around the nitrogen in pyridine is sp2. The more s-character in a hybridized orbital, the stronger the electrons are held by the system; therefore, the Lewis base character would be reduced. The electron pair on the sp2 hybridized orbital is more strongly held because the system has 33% s-character compared to 25% s-character for the sp3

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hybridized system. Therefore, piperidine is a stronger base than pyridine. Imidazole (compound XI), pKb = 7 is more basic than pyridine, pKb = 8.8, because the protonated imidazole is stabilized by resonance

Compound XI The nitrogen atom in imidazole without the attached hydrogen atom is protonated as illustrated below, and the protonated imidazole is stabilized by resonance as illustrated in the following equation.

Nomenclature of Amines The IUPAC nomenclature of amines uses alkylamines or alkanamines as part of the naming schema. For example, the IUPAC name for

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is propylamine or propanamine and, the IUPAC name for

Is cyclohexylamine or cyclohexanamine The IUPAC name for

Is N-methyl-1-methylbutylamine or N-methyl-1-methylbutanamine or N-methyl-2-pentanamine or N-methylpentan-2-amine. As indicated earlier, aniline is the parent IUPAC name for

The IUPAC name for

is 2-methyl-5-chloroaniline Nitrogen atoms with four groups attached are named as ammonium salts. The IUPAC name for

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is N-ethyl-N-methylcyclohexylammonium chloride. Imidazole (compound XI) is a constituent of histidine (compound XII) and histamine (compound XIII), two biologically important molecules.

Compound XII Compound XIII Histidine (compound XII) is an important amino acid involved in proton transfer, and histamine (compound XIII) stimulates the dilation of blood vessels. Amines can be easily dissolved in aqueous solution. This phenomenon can be used to separate amines from other compounds in plants. Also this property is key for preparing nitrogen containing drugs to be absorbed in the blood stream. The basic compounds in plants are referred to as alkaloids. A number of these biological compounds have interesting properties. Some are highly addictive and must be used with care. Quinine (compound XIV), an alkaloid from the cinchona bark, is used to treat malaria.

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Compound XIV Nicotine (compound XV), an alkaloid found in tobacco, is an extremely toxic compound,

Compound XV The pKb1 for compound XV is approximately 6 and the pKb2 is approximately 11. Why do you think this molecule has two pKb values? In addition, the pKb for triethylamine is 3.2. How can you rationalize this difference? Morphine (compound XVI), an opium alkaloid and a tertiary amine, is an addictive pain suppressant that must be used under careful clinical supervision.

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Compound XVI In an effort to synthesize an analgesic compound that would be less addictive than morphine, C. R. Wright, in 1874, acetylated morphine in an attempt to obtain diacetylmorphine (compound XVI). Compound XVI, commonly referred to as heroin, proved to be more addictive than morphine. The synthesis of heroin can be accomplished by a simple acylation reaction in which morphine is acylated with acetic anhydride.

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Heroin Epinephrine (compound XVII), also referred to as adrenaline, is a hormone secreted by the adrenal gland. Adrenaline is often referred to as a “fight or flight hormone.”

Compound XVII

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As indicated earlier, epinephrine, a secondary amine, is a neurotransmitter. Epinephrine is not isolated from a plant. Serotonin (compound XVIII), a neurotransmitter, is synthesized in the pineal gland. Excess serotonin in the brain can be attributed to mental disorders.

Compound XVIII Quaternary Ammonium Salts and Solubility Many quaternary ammonium salts, RR’R’’R’’’N+ X-, are soluble in nonpolar solvents. The R groups, RR’R’’R’’’, which may or may not be equal, are lipophilic, i.e., they are repelled by water (hydrophobic). This makes quaternary ammonium salts favorable substances for transferring materials between phases. Therefore, quaternary ammonium salts are phase-transfer agents. N-hexyl iodide can undergo an SN2 reaction with sodium thiomethoxide to form n-hexyl methyl sulfide and sodium iodide.

The reaction cannot be accomplished in an aqueous medium, because CH3CH2CH2CH2CH2CH2I is insoluble in water and NaCH3 is insoluble in n-hexyl iodide. However, the reaction can be facilitated by adding a small amount of a quaternary ammonium salt such as (CH3CH2)4N+ I- to the mixture. The quaternary

N

H

..

CH2CH2NH2HO

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ammonium salt facilitates the reaction between an anion and the n-hexyl iodide by transferring the -SCH3 anion from the aqueous phase to the organic phase as illustrated in the following illustrated. (1)

(2)

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(3)

Syntheses of Amines Amines can be synthesized in a variety of ways. For example, amines can be made through the alkylation of ammonia.

The reaction works relative well for synthesizing α-amino acids An effective synthesis for primary amines is the Gabriel Synthesis. The Gabriel Synthesis is a method that will allow the conversion of primary alkyl halides to amines without the problems encountered in forming amines from primary or secondary alkyl halides and ammonia. The reaction involves the extraction of hydrogen atom from phthalimide with strong base. The resulting imide undergoes an SN2 reaction with a primary alkyl halide, and the resulting N-substituted imide reacts with hydrazine to produce a primary amine.

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The following series of elementary steps rationalize the formation of the primary amine by way of the Gabriel synthesis.

(1) Reaction of the imide with a primary alkyl halide.

DMF is dimethylformamide, a solvent for the reaction. (2) The resulting N-substituted imide reacts with hydrazine to produce the desired primary amine.

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Following are the series of elementary steps that rationale the formation of the primary amine via the Gabriel Synthesis. (1) The initial step is the abstraction of the acidic hydrogen atom

of weakly acidic phthalimide (pKa = 8.3)

(2) The second step of the mechanism is a substitution

nucleophilic bimolecular displacement of the halogen atom of the primary alkyl halide to for the N substituted phthalimide.

(3) The third step is a nucleophilic attack of hydrazine on the

carbon atom of the carbonyl group resulting in the indicated

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change of a 2sp2 hybridized carbon atom to a 2sp3 hybridized carbon atom.

(4)

(5)

(6)

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(7)

(8)

(9)

(10)

Primary amines can be synthesized by the reduction of Alkyl Azides,

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Following is an example of the reduction of azides by lithium aluminum hydride.

The balance chemical equation (showing the stoichiometry) for this reaction is

The stoichiometry of the balanced chemical equation can provide insight into the mechanism of the reaction; therefore, knowing the stoichiometry of the reduction of azides by lithium aluminum hydride, suggest a series of elementary steps to rationalize the formation of the primary amines from the reduction of azides followed by the hydrolysis. Nucleophilic attack on an epoxide produces an alkoxide azide intermediate that can be reduced to a hydroxy primary amine. The following reactions illustrate this reaction schema. (1)

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(2)

Primary amines may be prepared by the reduction of nitriles. For example, 2-cyclohexylacetonitrile can be reduced with lithium aluminum hydride followed by hydrolysis to form 2-cyclohexylethylamine, a primary amine by the following sequence of reactions.

2-cyclohexyllacetonitrile

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Aniline can be prepared by the reduction of nitrobenzenes to aniline compounds by the use of tin in hydrochloric acid. For

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example, nitrobenzene can be reduced to aniline as represented by the following equation.

The balanced chemical equation (showing stoichiometries) for the reduction of nitro aryl compounds can be expressed by the following set of equations. (1)

(2)

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Reduction of primary, secondary, and tertiary amides produce primary amines, secondary amines, and tertiary amines respectively. Primary amines can be made by the reduction of primary amides followed by hydrolysis. For example, 2-phenylacetamide can be reduced to 2-phenylethylamine with lithium aluminum hydride followed by hydrolysis.

Secondary amines can be made by the reduction of secondary amides followed by hydrolysis.

Tertiary amine can be made by the reduction of tertiary amides followed by hydrolysis.

Aldehydes and ketones will react with ammonia to form imines. Imines can be reduced to primary amines by treating aldehydes or ketones with ammonia followed by reductive amination.

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The following sequence of reactions demonstrate how amines can be produced from the reaction of ammonia with aldehydes and ketones to produce imines followed by the reduction of imines to primary amines. (1)

imine (2)

and (1)

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(2)

Secondary amines can be prepared by reacting primary amines with aldehydes or ketones followed by the reduction of the resulting N-substituted imine with hydrogen in the presence of a metal catalyst. The N-substituted imine is referred to as the Schiff’s base. Reduction of the Schiff’’s base leads to a secondary amine. The reaction between 2-phenylacetaldehyde and cyclopentylamine is an illustration of the formation of an imine, N-cyclopentyl-2-phenylethanimine, a Schiff’s base can be reduced by catalytic hydrogenation to N-cyclopentyl-2-phenylethanamine. (1)

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(2)

Tertiary amines can be formed by reacting an aldehyde or a ketone with a secondary amine followed by reductive amination of the resulting iminium salt. For example, the reaction between 2-phenylacetaldehyde and cyclopentyl ethyl amine in the presence of hydrochloric acid is an illustration of the formation of the iminium salt N-cyclopentyl-N-ethyl-2-phenylethaniminium chloride. The iminium salt can be reduced by catalytic hydrogenation to ethyl cyclopentyl 2-phenylethyl ammonium chloride. Ethyl cyclopentyl 2-phenylethyl ammonium chloride when treated with base will produce N-cyclopentyl-N-ethyl-2-phenylethanamonium, a tertiary amine.

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The Hoffman Elimination Reaction The Hoffman Elimination reaction is a method that leads to the preparation of alkenes. Also, it is an analytical tool for structure

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determination. The reaction involves the formation of a quaternary ammonium hydroxide followed by basic elimination leading to the formation of an alkene. The interesting aspect of the Hoffman elimination is that it leads to the formation of the less stable alkene. The less stable alkene is formed from the abstraction of the less sterically hindered β-hydrogen atom. The Hoffman reaction begins with treating the amine with an excess of methyl iodide.

The resulting quaternary ammonium salt is treated with silver oxide in water to form a quaternary ammonium hydroxide that undergoes an elimination reaction upon the application of heat, forming the less stable alkene.

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A partial mechanism for the Hoffman elimination reaction shows that, in a bimolecular process, a hydrogen atom in the least hindered arrangement is abstracted by the hydroxide ion of the quaternary ammonium hydroxide to produce the less stable alkene.

In this example, the compound with the double bond in an exocyclic position is less stable than the compound with the double bond in the ring. The double bond in the ring would have resulted from the abstraction of a hydrogen atom in a more hindered position. Nitration of Aniline Nitration of aniline and aniline substitutes leads to the formation of tars. Protecting the amino group with an N-acetyl group can eliminate this problem. (1)

The N-acetylaniline can undergo nitration

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(2)

Aniline and aniline substitutes will undergo halogenation reactions too rapidly; therefore, the best approach is to make the N-acetyl aniline compounds first, and then halogenate the aromatic ring. For example, the bromination of N-acetyl aniline results in the formation of p-bromo-N-acetylaniline and o-bromo-N-acetylaniline

N-Nitrosamines N-Nitrosamines are carcinogenic compounds. N-Nitrosamines can be prepared by reacting secondary amines with nitrous acid. The nitrous acid is generated insitu from sodium nitrite and hydrochloric acid. For example, N-Nitrosoethylmethylamine can be formed by the following reaction.

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Some common nitrosamines are N-nitrosodimethylamine, produced during the tanning of leather.

N-nitrosopyrrolidine produced when bacon, cured with sodium, is fried.

N-nitrosonornicotine found in tobacco smoke.

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Nitrosamines are formed when nitrosating agents come into contact with secondary amines. Enzyme-catalyzed reduction of nitrates produces nitrites that combine with amines to produce N-nitrosoamines. Primary amines cannot form isolatable N-nitroso compounds, because alcohols and alkenes result from treating primary amines with nitrous acid. For example, n-butylamine reacts with sodium nitrite in hydrochloric acid to produce1-butene and n-butyl alcohol.

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The alkyl diazonium ion will diazotized to produce a carbocation and diatomic nitrogen:

The carbocation produce can rearrange to a more stable carbocation.

The more stable as well as the less stable carbocations can undergo elimination or substitution reactions.

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Electrophilic Aromatic Nitrosation Aromatic compounds undergo electrophilic nitrosation substitution reactions with the in situ generation of nitrous acid. These reactions occur when the aromatic ring contains a strong activator. Following is an example of nitrosation of an aromatic ring with a strong activator attached.

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Diazonium Salts can be prepared by reacting aniline like compounds with nitrous acid at 0oC. The nitrous acid is generated in situ via sodium nitrite and hydrochloric acid at 0oC. Following is balanced equation representing the formation of benzenediazonium chloride.

The net ionic equation for this reaction is

Following are the series of elementary steps that rationalize the formation of benzenediazonium. (1)

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(2)

(3)

(4)

(5)

(6)

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(7)

The sum of equations 1-7 gives the following net ionic equation:

The diazonium salt can undergo a series of reactions. For example, benzenediazonium chloride will react with Potassium iodide to form iodobenzene; water in a hydrolysis reaction to form phenol; fluoroboric acid, HBF4, followed by heat to form fluorobenzene; a series of Sandmeyer reactions to form bromobenzene, bromobenzene, and cyanobenzene; and hypophosphorous acid to form benzene. These reactions are represented by schema 10.1

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Schema 10.1 The mechanisms of these reactions follow different paths. For example, the mechanism of the Sandmeyer reactions (with copper(I) cyanide or copper(I) chloride or copper(I) bromide) follows the one-electron transfer mechanism suggested decades ago. Nonhebel and Waters confirmed this mechanism in a 1957 article published in the Proceedings of The Royal Society. A suggested format for this mechanism is outlined in the paper titled “Aryl Halides.”. There is evidence to suggest that the reactions proceed through free radical mechanisms. For instance, studies have suggested that the tetrafluoroborate formed when bezenediazonium chloride reacts with fluoroboric acid in the dark to form fluorobenzene follows a free radical mechanism. Diazonium salts react with activated aromatic compounds to form azo compounds. For example, benzenediazonium chloride will react with anisole to form the azo compound p-methyoxyphenylphenyldiazene.

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Azo compounds can be very effective dyes, because they have intense colors. For example, the azo dye formed from the reaction of sodium β-naphthoxide with benzenediazonium chloride has an intense orange-red color.

β-naphthoxide red-orange azo

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Problems Amines

1. Suggests syntheses for each of the following from benzene

and any other necessary organic and/or inorganic materials.

(a)

(b)

(c)

(d)

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(e)

(f)

2. Which of the following is more basic? Rationalize your

answer.

(a)

(b)

(c)

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3. Which of the following is the stronger acid, tetrahydropyrrole or pyrrole? Give a rationale for your answer.

4. Write the series of elementary steps (the mechanism) to rationalize the following conversion.

5. Phosphate esters, referred to as phospholipids, are

biologically important chemicals. Choline, constituent of phosphate esters, has a molecular formula

C5H14NO dissolves in water to form a solution with a pH that is considerably higher than 7. Choline can be synthesized from trimethylamine, water, and ethylene oxide.

(a) Suggest a structural formula for choline. (b) Write an equation showing the Arrhenius definition of

choline as a base. (c) An important derivative of choline is acetylcholine, a

neurotransmitter in the peripheral and central nervous systems, is an important biological molecule. Suggest a structure for acetylcholine.

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6. The proton magnetic resonance and C-13 magnetic resonance spectra of a compound containing 78.5% carbon, 8.4% hydrogen, and 13.1% nitrogen are indicated below. Suggest a structural formula for the compound.

H1 NMR

13C NMR

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7. Suggest structural formulas for A, B, C, D, E, F, and G in the following pathway:

A + heat → B (C9H15N) B + CH3I → C10H18NI + Ag2O → C (C10H19ON) C + heat → D (C10H17N) D + CH3I → C11H20NI + Ag2O → E (C11H21ON) E + heat → F (C8H10) F + Br2 → G (C8H10Br2) G + alc KOH → H (C8H8)

8. Vitamin B5,C9H17O5N, is a water soluble vitamin called

pantothenic acid. Vitamin B5 is needed to synthesize coenzyme Q and to metabolize proteins. Vitamin B5 reacts with sodium hydroxide, and then ethyl alcohol to produce C11H21O5N. C11H21O5N reacts with NaOH to give C6H11O4Na and sodium 3-aminopropionate. Acid hydrolysis of C6H11O4Na and sodium 3-aminopropionate produced C6H12O4 and 3-aminopropionic acid. Following is a pathway for the formation of Vitamin B5 :

(1)

(2)

NCH3

ONaBH4

H2OA

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(3)

(4)

(5)

9. An unknown compound exhibited the following H1 NMR, 13C

NMR, and partial mass spectra.

H1 NMR

C6H10O3H3O

+

C6H12O4 Δ

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13C NMR

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Partial Mass Spectrum

The unknown dissolves in sodium hydroxide to form a substances that gives a negative test for chlorine. Suggest structural formulas for the unknown that exhibits the given spectra and the compound that is formed when the unknown is treated with NaOH.

10. A popular anesthetic with the molecular formula C13H20O2N2

is insoluble in water and aqueous NaOH; however, the compound is soluble in dilute hydrochloric acid. When C13H20O2N2 is treated with sodium nitrite and hydrochloric acid under chilled conditions, C13H18O2N3Cl produced. The proton magnetic spectrum of the anesthetic is listed below. Boiling C13H20O2N2 in sodium hydroxide followed by acidification produces two separable compounds, C7H7O2N and C6H15ON. Spectra of these two compounds are indicated below.

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H1 NMR of C13H20O2N2

13C NMR of C13H20O2N2

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H1 NMR of C6H15ON

13C NMR of C6H15ON

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H1 NMR C7H7O2N

3C NMR of C7H7O2N

Suggest structures for C13H20O2N2, C13H18O2N3Cl, C7H7O2N and C6H15ON.

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11. Suggest a synthesis for for the following compound from benezene as the only organic starting material and any necessary inorganic material.

12. Compound H is an anti-inflammatory agent. Suggest a

synthesis for compound H from benzene as the only organic starting material and any other necessary inorganic reagents.

Compound H 13. The following reaction is known as the Hoesch reaction.

Suggest mechanistic pathways, i.e., the series of elementary steps to account for the formation of this compound. 14. Suggest a synthesis for 1,3-dihydoxybenzene (resorcinol or

m-dihydroxybenzene or m-hydroxphenol) from benzene as the only organic starting material and any other necessary inorganic reagents.

15. The synthesis of compounds K, L, M and N can be

accomplished by the following sequence of reactions.

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The molecular formula of G and H is C7H7ONa

G I

H J

I

J

H1 NMR compounds K and L

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C13 NMR compounds K and L

H1 NMR compounds M and N

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C13 NMR compounds M and N

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Suggest structures for compounds A, B, C, D, E, F, G, H, I, J, K, L, M and N.