ChemistryChemistry 212212 andand 300300 ““OrganicOrganic ChemistryChemistry II”II” WinterWinter SemesterSemester 20002000 Dr.Dr. RainerRainer GlaserGlaser

Examination #2 “Amines, Carboxylic Acids & Carboxylic Acid Derivatives”

Thursday, March 9, 2000, 9:00 - 9:50

Your Name: Answer Key

Question 1. Fish, Rotting Carcasses & Stinking Flowers 24 Question 2. Methamphetamine (News #19) 24 Question 3. Drug Delivery (News #19 & #20) 16 Question 4. Aspirin (VC #20 & web materials) 20 Question 5. Making Soap the Old-Fashioned Way 16 Total 100

Exam #2, Chemistry 212, WS00, Dr. Glaser — 1 — Question 1. Dead Fish, Rotting Carcasses & Stinking Flowers. (24 points)

Trimethylamine causes the bad smell of dead fish. Putrescine and cadaverine are responsible for the bad smell of rotting flesh and of “stinking flowers”. Let’s look at some syntheses of these amines.

Describe the Hofmann elimination of 2-aminobutane, CH3-CH(NH2)-CH2-CH3. This synthesis yields trimethylamine as a “side product”. Clearly indicate what other product is formed. Show important intermediates and show reagents for every step. (6 points)

Excessive methylation with MeI gives trimethylbutylammonium iodide Show the salt Treat with aqueous silver(I) oxide Gives trimethylamine and 1-butene as the major product

What substrate would you have to use for the synthesis of putrescine (1,4-butanediamine) by way of a

“Hofmann rearrangement” (e.g. using Br2, NaOH, H2O). Only supply the structure of the substrate and nothing else. (3 points)

H2N-CO-CH2-CH2-CH2-CH2-CO-NH2

What substrate would you have to use for the synthesis of cadaverine (1,5-pentanediamine) by way of a “Gabriel synthesis”. Just supply the structure of the substrate that would have to be reacted with the phthalimide. (3 points)

Br-CH2-CH2-CH2-CH2-CH2-Br

Exam #2, Chemistry 212, WS00, Dr. Glaser — 2 — Suggest a synthesis of putracine (1,4-butanediamine) from ethene. Show important intermediates and show reagents for every step. (6 points)

Bromination of ethene (210 material) Bimolecular nucleophilic substitution of both bromides using NaCN Reduction with LAH or other suitable reducing reagent (There are alternatives, e.g. via the azide reduction)

Suggest a synthesis of cadaverine (1,5-pentanediamine) starting from cyclopentene. Show important intermediates and show reagents for every step. (6 points)

Ozonolysis of cyclopentene (210 material) Reductive work-up (Zn/HCl, Me-S-Me) gives the 1,5-dialdehyde Reaction with ammonia gives the diimine Reduction of the imine using LAH or other suitable reducing reagent

Exam #2, Chemistry 212, WS00, Dr. Glaser — 3 — Question 2. Methamphetamine. (24 points)

The “news item” accompanying chapter 19 featured an article entitled “FEDERAL DRUG ENFORCERS REPORT RISE IN USE OF METHAMPHETAMINE” (Hearst Newspaper, June 26, 1998). Methamphetamine is a highly addictive and immensely destructive drug. Methamphetamine is especially hard to control since this drug can be synthesized easily by pretty much anybody starting from ephedrine or pseudoephedrine. Perspective drawings of ephedrine, pseudoephedrine and methamphetamine are shown.

· Mark every chiral carbon atom of each of the three compounds by a star, “*”. (3 points) · What is the stereochemical relationship between ephedrine and pseudoephedrine? Ephedrine and pseudoephedrine are ______(enantiomers, diastereoisomers, conformers). (2 p.)

Ephedrine Pseudoephedrine Methamphetamine

H HN CH3 HO HN CH3 H HN CH3 HO H H * Me * Me * Me * H * H H

· Configuration of the chiral C-atom of the methamphetamine molecule shown: __S__ (R or S). (2 p.)

· The amino-N atom ______(is, is not) a chiral center because the barrier to inversion of the N- atom is ______(low, high) and ______(can, cannot) be overcome at room temperature. (3 points)

· Provide an acceptable IUPAC name for methamphetamine: (2 points)

2-methylamino-1-phenyl propane

Exam #2, Chemistry 212, WS00, Dr. Glaser — 4 — A practical Synthesis of Methamphetamine is outlined below. This synthesis starts “from scratch” so to speak. Provide the structures of the two intermediates in the square boxes. The reaction that affords the secondary alcohol involves two consecutive reaction steps. Provide the reagents needed in these two steps in the boxes above and below the horizontal arrow. In the small box in the lower right corner, provide the molecule that is being eliminated upon reaction of the intermediate with methyl amine. In the box on the lower left, provide the organometallic reagent that will result in the methamphetamine after aqueous work-up. (10 points, 2 points for each intermediate and two boxes at arrows, 1 point each for two other boxes at arrows.)

Synthesis of Methamphetamine:

O OH (1) PhMgBr

H2C CH H2C CH (2) aq. workup CH3 Ph CH3

CrO3, H2SO4, H2O

HN CH3 O H C CH 2 H2C C Ph CH 3 Ph CH3

NaBH3CN N-Me H3C-NH2

H2C C - HOH Ph CH3

Which one of the following statements is true about the above synthesis? (2 points) · Yields racemic methamphetamine. · Yields an excess of R-methamphetamine if the R-isomer of the epoxide is employed. · Yields an excess of R-methamphetamine if the S-isomer of the epoxide is employed.

Exam #2, Chemistry 212, WS00, Dr. Glaser — 5 — Question 3. Drug Delivery. (16 points)

Salts are crystalline and they can easily be purified by recrystallization. Salts are stable over long periods of time and they are non-volatile. These are all good reasons to use salts for drug delivery. And you hve encountered quite a few examples. In lecture, we talked about the hydrochloride of ephedrine, a cold and allergy medicine, and in the news item accompanying chapter 19, you learned about the hydrochloride of cocaine, a local anesthetic. And in the news item accompanying chapter 20, you learned about the citrate of sildenafil and you know about its purpose. In this context, let’s review a few of the basics of ammonium salt formation.

Order the following bases in the order of increasing basicity (1= highest basicity, 3 = lowest basicity) (3) ammonia (2) methylamine (1) dimethylamine (3 points)

Order the following bases in the order of increasing basicity (1= highest basicity, 3 = lowest basicity) (3) acetonitrile (1) trimethylamine (2) pyridine (3 points)

Order the following acids in the order of increasing acidity (1= highest acidity, 4 = lowest acidity) (4) butanoic acid (1) oxalic acid (2) 2-chlorobutanoic acid (3) 3-chlorobutanoic acid (4 points)

The pKb value of ammonia is about 5 (it is 4.75). Which equilibrium reaction is described by the equilibrium constant Kb and what is the value of this equilibrium constant, Kb = 0.00001. Clearly state whether this equilibrium lies on the side of ammonia or of the ammonium salt. (3 points)

+ - NH3 + H2O <===> NH4 + OH equilibrium very much on the left

Salt formation between aniline and benzoic acid. Draw the structures of aniline, of benzoic acid and of their ammonium salt. Name the ammonium salt. Clearly state whether this equilibrium lies on the side of the amine or of the ammonium salt. (3 points)

+ - Ph-NH2 + HO2C-Ph <===> Ph-NH3 O2C-Ph equilibrium very much on the right Anilinium benzoate is one name, there are other possible names.

Exam #2, Chemistry 212, WS00, Dr. Glaser — 6 — Question 4. Aspirin. (20 points)

Aspirin is acetylsalicylic acid and it is a remarkable molecule indeed! Let’s first take a look at some structural aspects of aspirin. The two molecular models show different isomers of aspirin. (The lines between the atoms merely indicate connectivity; the lines do not imply single bonds.) Compare & contrast these two molecules. For each structure, state whether the acid is s-cis or s-trans. If there are any hydrogen-bonds, then identify the H-bond donor and the H-bond acceptor and indicate such H- bonding by drawing a dashed line between the H-bond donor and the H-bond acceptor. Below the models, provide a Lewis structure of each of the models and draw these Lewis structures “flat” (as opposed to perspective). (8 points)

H-bond H-bond donor donor

H-bond acceptor

H-bond acceptor

Lewis structure of the above model: Lewis structure of the above model:

H O O O O H O

CH3 O CH3 O

O

Exam #2, Chemistry 212, WS00, Dr. Glaser — 7 — Now, let’s take a look at the synthesis of acetylsalicylic acid. The synthesis of aspirin basically comes down to the question as to how to synthesize salicylic acid. Once the salicylic acid is made, its acylation is easily accomplished by ___Fischer___ esterification reaction between the phenol and acetic acid. So, let’s think about the synthesis of salicylic acid. (12 points)

One way to synthesize a carboxylic acid involves the synthesis of a nitrile and the subsequent hydrolysis of the nitrile to the carboxylic acid. Draw the structure of the nitrile that yields salicylic acid upon hydrolysis. Then show a sequence of reactions for the synthesis of that nitrile from ortho-aminophenol. Provide reagents for every reaction step. Draw the structure of important intermediates. The introduction of the cyano group by way of ______(electrophilic, nucleophilic, lipophilic) aromatic substitution of an aromatic _____diazonium____ ion is called ____Sandmeyer____ reaction. Structure of the nitrile precursor of salicylic acid:

Phenol with ortho cyano group.

Synthesis of the nitrile precursor from ortho-aminophenol:

Diazotization of amine with sodium nitrite and dilute HCl Reaction of diazonium salt with NaCN in the presence of Cu(I).

Exam #2, Chemistry 212, WS00, Dr. Glaser — 8 — Question 5. Making Soap the Old-Fashioned Way. (16 points)

Soap used to be made by saponification of fats. Provide brief definitions of “fat” and “soap”. You might want to provide the structures of a “fat” and a “soap” as part of your definition. (3 points each) Define “fat”:

Tri-ester of glycerol with three fatty acids.

Define “soap”:

Sodium salt (carboxylate) of a fatty acid (long chain carboxylic acid).

The term “saponification” is now used more broadly. “Saponification” describes any _base_-catalyzed ester hydrolysis. Provide the complete reaction mechanism of the saponification of methyl propanoate. Show all intermediates, show all lone pairs in every structure, show all formal charges, use curved arrows to indicate electron flow. Clearly show for each step whether this step is reversible or irreversible. (10 points)

Methyl propanoate: H3C-CH2-CO-OMe

Methyl propanoate plus NaOH gives tetrahedral intermediate, a sodium alkoxide in an equilibrium reaction.

Tetrahedral intermediate forms propanoic acid methoxide and ejects sodium methoxide in an equilibrium reaction.

Sodium methoxide deprotonates propanoic acid in an irreversible reaction.

Exam #2, Chemistry 212, WS00, Dr. Glaser — 9 — Exam #2, Chemistry 212, WS00, Dr. Glaser — 10 —