Southern Methodist University Chemistry Department

CHEM 3118 – Organic Chemistry Laboratory Revision 2009-1

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Nucleophilic Aromatic Substitution Reactions: Synthesis of 2,4-Dinitrodiphenylamine

Nicholas Leadwater, Ph.D. and Cynthia McGowan, Ph.D.

Aromatic substitution reactions can be electrophilic or nucleophilic. The parameters which indicate a nucleophilic aromatic substitution are:

1) The attaching group is a nucleophile (of course!) 2) The aromatic system into which the substitution occurs must be deactivated with

electron withdrawing groups. 3) The aromatic system into which the substitution occurs must have a leaving

group, typically (in order of reactivity) fluoride, chloride, bromide or iodide.

W

L

:Nu-

W

Nu

L

:- W

Nu

+ :L-

W = electron withdrawing groupL = leaving group (F, Cl, Br, I):Nu

- = nucleophile

The electron withdrawing groups are necessary, because the aromatic system is attracting a nucleophile, and is therefore behaving as an electron deficient species. This is an unaccustomed role for an electron rich aromatic ring. Because of the need for a leaving group, the electron withdrawing groups do not direct the location of the substitution, as in electrophilic aromatic substitution reactions. For further discussion of nucleophilic aromatic substitution reactions, please refer to your text, Organic Chemistry, by Francis Carey, Chapter 23. For this reaction, we will be making use of a scientific microwave oven.

“While microwave ovens have been used in the kitchen for many years, it was not until 1986 that the first reports of thier application to organic synthesis appeared in the scientific literature. They showed that reations were greatly accelerated and product yeilds improved over conventional methods. Since that time the field has developed at a rapid rate, with a total of over 2,000 research articles published using microwave heating. Microwave technology has applications in chemistry as diverse as natural product synthesis, biotechnology, and nanoparticle preparation. In addition, sceintific microwave apparatus is now commerically available. It is possible to perform chemistry very safely, reporducibly and easily, as commercial microwaves are far superior to domestic microwave ovens. The tecnology has been adopted both by university resesearchers and those working in industry and

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it is becoming clear that for heating chemical reactions, the use of microwaves is going to be the method of choice in the future.”

-Nicholas Leadwater, Ph.D. The reaction that we will do today is the substitution of nucleophile aniline into the highly activated aromatic system, 1-bromo-2,4-dinitrobenzene.

Br

NO2

NO2

+NH2

NO2

NO2

NH

EtOH

microwave125C

Do not be confused by the fact that the nucleophile is an aromatic compound. Aniline does not have electron withdrawing groups or a leaving group, and the aniline ring will not react with nucleophiles. Experimental Procedure

Reagent MW (g/mol) Mass (g) Volume (mL) MP/BP (°C)

1-bromo-2,4-dinitrobenzene 247.00 0.148 71-73

aniline 93.13 0.225 0.220 184

ethanol 5.0

2,4-dinitrodiphenylamine (product) 259.22 159-161

Part A – Synthesis 1-Bromo-2,4-dinitrobenzene (148mg, 0.60 mmol) aniline (0.220mL, 2.40 mmol) and ethanol (5.0mL) are placed into a 10-mL glass microwave reaction vessel containing a stir bar. NOTE: The final volume must be at least 5 mL. The reaction vessel is sealed with a cap and then placed into the microwave cavity. The pressure device is put in place on top of the reaction vessel and the unit programmed using the ramp-to-temperature method to heat to 125°C over a 2-minute period and then held for 5 minutes.

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After the reaction is complete and the vessel has cooled, the pressure device can be removed and the vessel may be taken for the microwave cavity. CAUTION: The tube my still be hot to the touch. Part B – Isolation of the Product When the reaction vessel has cooled, it is placed in an ice bath to initiate crystallization. Once crystallization is complete, the product will be collected by vacuum filtration using a Hirsh funnel and washed with cold solvent. Allow the crystalline product to air dry on the Hirsh funnel for 3 minutes, and then oven dry at 120°C for 10 minutes. Determine a mass for your product. Part C –Product Characterization The melting range of the product is determined. Post Laboratory Questions:

1) a) Compare the melting range you obtained for your product with the literature value. What does this tell you about your product? b) How wide is your observed melting range? What does this tell you about your product?

2) What visual evidence, other than melting range, do you have that the starting materials have undergone a transformation?

3) What advantages do you find for the microwave technique over conventional

methods? (Hint: Review Dr. Leadwater’s article.)

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Name:_______________________________ Section:_________ Date:_________

Nucleophilic Aromatic Substitution Reactions: Synthesis of 2,4-Dinitrodiphenylamine

Pre-Laboratory Questions:

1) Calculate the theoretical yield of your product. Show calculations.

2) Indicate whether the following groups are electron withdrawing or electron donating:

-NO2 ____________________

-NH2 ____________________

-Br ____________________

O

CH3

____________________

3) Predict the product of the following reactions:

NO2

Br

NaOCH3

CH3OH

NO2

NO2

Cl

CH3NH2CH3OH

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Identification of Unknown Liquid and Solid CHEM 3118

Each student will be assigned an unknown organic liquid and an unknown organic solid. For the unknown liquid, the following parameters must be determined: • Boiling point • Refractive index • Infrared spectrum Additionally, the following parameters will be provided: • Mass spectrum • NMR spectrum For the unknown solid, the following parameters must be determined: • Melting range • Infrared spectrum Additionally, the following parameters will be provided: • Mass spectrum • NMR spectrum The identity of each unknown compound will be established using these parameters. Literature values and references for each parameter will be recorded. For reference spectra, only a reference is required; photocopies of reference spectra are not required. The unknowns will be issued in adequate amounts to complete all tests, and additional unknown will not usually be dispensed. However, should an unknown be entirely consumed, due to multiple test repetitions, spillage or loss, an additional aliquot may be issued with an appropriate point deduction from the final laboratory grade. The following tips and evaluation guidelines are provided by Professor E. R. Biehl. They are compiled from experiences with this experiment, and include many invaluable ‘hints.’ • Tip 1. A strong odor may give hints about compound identify: non-aromatic amines

have a fish odor, esters have fruity fragrance • Tip 2. Refractive index around 1.5 and above are usually aromatic compounds or

contain 2 or more halogens (Br, Cl, I). • Tip 3. All unknowns are listed in Aldrich chemical catalog.

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Identification of Unknown Liquid and Solid CHEM 3118

Evaluating IR Spectra - Some tips by ERB Remember important IR bands for the following families: Alkane/Alkene Stretches - around 3000 cm-1 (type of CH - slightly greater than 3000 (C=C-H) slightly less (alkyl hydrogens) Aromatic Compounds - 1600 and 1500 cm-1; substitution patterns on benzene ring are found in the 850-690 region - check your textbook for details.

Alcohols - 3500-330 cm-1 C-O stretch around 1050 to 1200 cm-1 (strong) One error to guard against is the recognition that solid alcohols look different than liquid alcohols. Note the following examples:

Fig. 1 Fig. 2 Fig. 3 Liquid alcohol Low M.W. solid alcohol High M.W. solid alcohol The absorbance for the O-H stretch in a solid alcohol peak is much sharper, and appears at slightly higher wavenumbers than the O-H stretch in a liquid alcohol. This absorbance is sometimes confused with the N-H stretch of a primary or secondary amine.

Ketones and Aldehydes (1725-1715 cm-1) conjugation lowers frequency about 30 cm-1 (e.g carbonyl attached to a benzene ring).Band is very strong. ALSO aldehydes have carbonyl hydrogen stretch around 2750 cm-1.

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Esters - around 1740 cm-1. Alkenes - remember you just made cyclohexene and should know the pertinent frequencies. Nitriles - 2250-2225 cm-1 Amine - two NH stretch frequencies in the 3400-3200 region for NH2, one for NH (secondary amine) and none for tertiary amines (that is, no NH group). Nitro - (look up spectra of nitro compounds in Aldrich IR references located in the lab to see if you can figure out where nitro stretching frequencies occur) Amides - RCONH2 C=O stretch 1690 cm-1. NH stretch similar to amines listed above.

Alkynes - hydrogen attached to triple bond around 3300 cm-1 (sharp) triple bond around 2100-2200 cm-1. CAREFUL: if symmetrically distributed, the band is quite weak (remember there must be a change in dipole moment during stretch - since triple bond is linear then the symmetrical stretch will have little dipole moment change during vibration.

Ethers - C-O stretch in the 1200- 1100 cm-1 region (same as that of alcohols, but alcohols easily distinguished from ethers in that only the former has a OH stretch frequency).

Alkyl halides - no reliable absorption peaks

Carboxylic acids - in the 3500-2700 cm-1 region, a complex broad band is observed that resembles the state of Texas. This is the OH band. Secondly, the carbonyl stretch is around 1720 cm-1 ; benzoic acids are lower by about 20-30 wavenumbers.

Aromatics - Your text book does not have sufficient information on important IR absorption bands for substituted benzene rings. So here are some. The region from about 690 to 850 cm-1 gives info on type of substitution in aromatic compounds. • monosubstituted bands are 690 and 730-760 cm-1 • ortho (1,2) appear at 700 and 760 cm-1 • meta at 710 and 790 cm-1 • para at 820-850 cm-1 • unsymmetrically trisubstituted 700 cm-1 and 830-870 cm-1 For some of you it will be important to know that conjugation lowers C=O frequency about 30 cm-1 for unsaturated unit attach to it. For example, if two units are attached to

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C=O the observed C=O stretch is observed a ca. 60 cm-1 from absorption frequency of the non-conjugated parent. These are 1740 cm-1 for esters, 1725-1710 cm-1 for aldehydes and ketones, 1690 cm-1 for amides.

Solids – Many of the Standard Spectra in the Aldrich Library of FT-IR Spectra were obtained using the “nujol” method. This method involves grinding the solid sample in mineral oil, which for this application is called “nujol oil.” This technique has the drawback that it introduces two interference peaks in the spectrum, due to the absorbances of the oil. They are the C-H stretch at 2900-3000 cm-1, and a rounded absorbance at 1460 cm-1 (C-C-C bend?) When comparing a KBr pellet spectrum to a nujol spectrum, these differences must be taken into account. Note: NUJOL C-H stretch 1460 cm-1

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Identification of Unknown Liquid and Solid CHEM 3118

Evaluating Mass Spectra - Some tips by ERB 1. Parent peak (P) - actual molecular mass of your unknown. Odd Nitrogen rule may be of help: Compounds containing an odd number of nitrogen atoms have an odd molecular mass (the corollary is a molecule containing zero or an even number of nitrogen atoms having an even molecular mass. Some compounds have very weak molecular ions and fragment to more stable ions. Ones to watch out for are: tertiary alcohols - lose water very easily, 1,4-dichloro- or 1,4-dibrombutanes - lose HCl or HBr. (The parent peak is not observed - the resulting P minus HX peak (X = Cl or Br) appears to be parent peak, but it isn't. 2. Isotope peaks - (P+1, P+2, etc) arise when one of more of the elements exist in nature in two or more isotopic forms. H, C and O exist mainly as 1H, 12C, and 16O. Thus the P+1 and P+2 isotopics peaks are small. However Bromine exist in nature as 79Br and 81Br: the relative abundance of each isotope is 50% each. - 79 and 81. Therefore we have equal chances of 79Br and 81Br being in a bromine containing compound. Consider CH3Br. Half will have a molecular mass of 12(C) + 3 (3H) + 79 or 94 molecular mass and 12(C) + 3 (3H) + 81 or 96. This gives rise to a P peak (94) and a P+2 peak (96) and they appear with equal intensity. By similar reasoning, Cl, which exists as 35Cl and 37C in a ratio of 3:1, respectively, leads to any chlorine containing compound having P and P+2 peaks in a ratio or 3:1 respectively. What if we have two bromines in a compound? We can use the binomial theorem to calculate the number of isotope peaks and the relative intensities (much like the N+1 rule in NMR) - using only the coefficients the solution. In general (A + B)n where n = the number of similar elements and A and B represents each isotope.) For two bromines we have (79Br + 81Br)( 79Br + 81Br) = 79Br79Br + 2 79Br81Br + 81Br81Br. This gives rise to a P, P+2 and a P+4 with relative intensities of 1:2:1. Consider chlorine and bromine: (79Br + 81Br)(335Cl+ 37Cl). Show that this leads to P, P+2 and P+4 peaks in ratio of 3:4:1. Two chlorine atoms is 9:6:1. 3. Most of the spectroscopic data packages include only the parent peak, isotope peaks, base peak (the largest peak), and the relative abundances of each. This is because mass spectra can now be “looked up on the internet.” In some cases, the abundances or base peaks were omitted, so that the compound cannot be identified without interpreting the spectra. For that reason, little or no fragmentation data is available.

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12

Identification of Unknown Liquid and Solid

CHEM 3118 Evaluating 1H-NMR Spectra - Some tips by ERB Remember where different types of hydrogens appear in spectrum- (1-2 ppm - alkyl groups away from electronegative atoms; around 2 ppm - hydrogen attached to carbon adjacent to carbonyl group, 3-4 = hydrogen attached to carbon adjacent to electronegative atom such as oxygen (CH-O) bromine or chlorine. 5-6 = alkenic hydrogens; 7-8 aromatic hydrogens, 9.5 = aldehydic CH, greater than 10 = CO2H) alcohol OH variable between 2.5 to 5.0. Esters (RCO2R') have interesting NMR spectra. Consider methyl acetate, CH3CO2CH3. The methyl group attached to oxygen appears around 3.7 ppm, whereas the methyl group attached to the carbonyl appears around 2.0 ppm. Integration especially of the aromatic hydrogens will tell you the nature of substitution on the benzene ring. For example, if there are 4 aromatic hydrogens, then you have a disubstituted benzene ring. Possibilities are 1,2 (ortho), 1,3 (meta) or 1,4 (para). If the groups are the same, then the hydrogens in the 1,4-case appear as a singlet ( all equivalent) if they are different (e.g. 4-chloronitrobenzene) then they appears as a doublet of doublets. meta and ortho are a little more complicated, but you can get ideas of what they look like from Aldrich FT-NMR library located in the laboratory. Mono substituted benzenes will appear as a broad singlet if the group attached is alkyl, but will become more complication if an electronegative group is attached (e.g. nitro, halogen, OH, OCH3, etc). It goes without saying that you should remember characteristic splitting patterns of simple alkyl groups. ALSO, if your unknown has a carbon chain containing roughly 4 or more carbons then virtual coupling is observed. The region between 1-2 is very complicated but you can usually determine the number of hydrogens in the band spread and if the chain is attached to an alcohol or amine, one can usually see a band associated with the CH attached to electronegative atom. Again, you might want to look in the Aldrich library for spectrum of say pentyl amine or 1-hexanol. ALSO I noticed that some NMR spectra contain impurity signals. You should check with an instructor or teaching assistant if there are questions. NOTE ALSO THAT THE LARGE SIGNAL AT ZERO PPM IS TMS!!

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Identification of Unknown Liquid and Solid CHEM 3118

Evaluating 13C-NMR Spectra - More Tips Organic chemistry texts are fond of saying that there should be a separate peak for each carbon in an organic molecule. However, for very symmetrical molecules, equivalent carbons occur fairly often. If carbons are completely equivalent, then they will present a single peak. Remember than integration cannot be used to determine the number of carbons in a 13C-NMR spectrum! A triplet of (usually) small peaks at 77ppm is the signal for the solvent in which the sample is dissolved, deuterochloroform (CDCl3.) This is absorbance is unavoidable, as any organic solvent is going to contain carbon. Ignore this signal. If a DEPT spectrum is provided, remember that these absorbances are comparable to the conventional 13C-NMR spectrum, except that:

• Carbons with two attached hydrogens (-CH2-) will appear upside down. • Carbons with one or three attached hydrogens will be unaffected. • Carbons with no attached hydrogens will disappear.

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Identification of Unknown Liquid and Solid CHEM 3118

Elements of the Final Report

Unknown reports must include the following: 1) your name 2) unknown sample numbers 3) must write a separate report for each unknown 4) unknown liguids - must report boiling point, refractive index, (density is optional) along with literature values ( e.g. bp = 60 °C; lit.[Aldrich Chemical Catalog, p. XXX, 1999) = 61 °C]). 5) Unknown solids - must report melting range along with literature value. 6) IR, NMR and MS for both liquids and solids - must report reference for IR and NMR (page number and spectrum number if appropriate.) 7) the name and structure determined for the unknown compound. 8) IR spectra must have pertinent absorptions indicated. 9) NMR spectra - equivalent sets of hydrogens must be indicated on the 1H spectra. All carbons must be identified on your 13C spectra. 10) Mass spectral data - I have indicated all major isotope peaks along with parent peak on your spectra. You should determine molecular weight, isotope information, and any other elucidation obtainable from the data. 11) You should write a concise story of how the spectrum and physical properties aided you in you unknown determination. Remember you can correctly identify your unknown, but lose many points for poor use of spectral data. For example, to not identity a peak around 3500-3300 cm-1 in, say, an unknown alcohol is unforgivable.

Electrophilic Aromatic Substitution Reactions Involving Multiple Directing Groups

Benzene reacts with bromine in the presence of Fe or FeBr3 to give bromobenzene and hydrobromic acid. In the reaction, a benzenoid hydrogen atom is substituted by a bromine atom. This is an example of an electrophilic aromatic substitution reaction. The mechanism for this bromination reaction is thought to consist of three steps, which are shown below. The first step involves the generation of the electrophilic bromonium ion Br+ (actually FeBr3

……Br……Br) which, in the rate-determining second step, adds to the benzene ring to give a resonance-stabilized benzene carbocation. In the third step, the aromatic ring is regenerated by the removal of the hydrogen atom attached to the bromine-substituted benzenoid carbon atom by FeBr4

-.

OFFICIAL MECHANISM OF ELECTROPHILIC AROMATIC SUBSTITUTION

STEP 1 FeBr3 + Br-Br → Brδ+…… Brδ- ̶̶̶ ̶̶ FeBr3

H

BrH

Br

H

Br

H

Br

+ ++ +

HBr

H

+

H

Br+

BrFeBr 3- Br

Brδ+…… FeBr

4δ- →

Another reagent which is often used to generate the electrophilic bromonium ion is a mixture potassium bromate (KBrO3) and hydrobromic acid (HBr.) Using these reagents avoids the necessity of using elemental bromine. Because it is a highly volatile corrosive liquid, elemental bromine can be difficult and dangerous to use.

STEP 2

STEP 3

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Bromination of a mono-substituted benzene derivative potentially leads to three products, that are termed ortho, (1,2) meta, (1,3) and para (1,4). However, two product distributions are generally observed. In the first case, certain substituents give rise mainly to ortho and para isomeric products. They are said to be ortho-para directors. With some exceptions (e.g. halogens are o,p-directing, but ring deactivating), ortho-para directing groups render the ring more reactive toward electrophilic substitution than benzene itself. They are said, therefore, to be ring activating groups. We can further divide these types of substituents into three categories:

Strongly activating groups, such as –OH, -NH2

Moderately activating groups, such as -OCH3 and -NHCOCH3.

Weakly activating groups, such as alkyl

Note that in the first two groups the atom attached directly to the benzene ring has a lone pair of electrons. These are, thus, available to participate directly in resonance with the benzenoid carbocation to give rise to a particularly important resonance structure as shown below.

H

BrH

Br

O

H+

+

OH

Particularly stable-

Contains more π bonds than A In the second case, certain substituents give rise to meta products. These substituents are invariably ring deactivating groups. Typical ring deactivating groups are: -NR3

+, -NH3+,

-CHO, -COR, -CN, -NO2. Note that the atom attached directly to the benzene ring have either a full positive charge or a partial positive charge due to being bonded to a more electronegative atom. These groups are deactivating groups since they are not capable of stabilizing the benzene carbocation intermediate. Further, carbocations formed by ortho- or para- addition can not be stabilized by these groups. In fact, they destabilize them due to the close proximity of the positive centers. ortho substitution para substitution

H

Br

+

+

NR3

Br H

+ NR3

+

Particularly unstable closeness of two positive charges

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However, addition to the meta position gives rise to a benzene carbocation in which the positive charge in any resonance structure is not located on the substituted site.

two positive centers never on adjacent atoms The effects of more than one substituent on electrophilic substitution are discussed on pp. 526 - 528 in Carey’s Organic Chemistry, 6th Edition. Two or more substituents exert a combined effect on the reactivity of the aromatic ring. Several rules of thumb have been developed for predicting where an electrophile will attack. You should review these. 3 In today’s experiments we will experimentally determine where bromination occurs in two pleasant-smelling tri-substituted benzenes, and then see if the application of these rules of thumb agree with experimental results. You will carry out the bromination of vanillin and isovanillin. Bromine will be generated in situ by the reaction of potassium bromate (KBrO3) and hydrobromic acid (HBr.) These pleasant smelling aromatic compounds contain three substituents: the meta-directing aldehyde (CHO) group, and the ortho, para- directing hydroxy (OH) and methoxy (OCH3) substituents. Since the aromatic rinds contain two strongly activated groups, no catalyst is necessary. As shown in the equations below, the possible mono-brominated products in the two reactions are 6-bromo-, 5-bromo-, and 2-bromovanillin, and 6-bromo-, 5-bromo-, and 2-bromoisovanillin, respectively.

Possible Products from Bromination of Vanillin

Vanillin 6-bromo 5-bromo 2-bromo NMR Jpara Jmeta Jortho

Coupling Constant 0-1 Hz 1-5 Hz 5-9 Hz

KBrO3/HBr

:NR3

Br

H

:NR3

Br

H

:NR3

Br

H

+

+

+

+

+

+

OH

CHO

OCH3

OH

CHO

OCH3

HBr

HOH

CHO

OCH3

HH

BrOH

CHO

OCH3

BrH

H+ +

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Possible Products from Bromination of Isovanillin

Isovanillin 6-bromo 5-bromo 2-bromo NMR Jpara Jmeta Jortho

Coupling Constant 0-1 Hz 1-5 Hz 5-9 Hz

Experimental Procedure

Bromination of vanillin or isovanillin.

1) Place 0.23 g (1.5 mmol) of vanillin (or isovanillin) and 3.0 mL of acetic acid into a 5 mL conical vial containing a stirring vane. Begin the solution stirring at room temperature.

2) Add 0.09 g of potassium bromate, followed by 0.30 mL of hydrobromic acid.

3) Stir the mixture 40 minutes.

4) Pour the mixture into 25 mL of water, add a stirring bar and stir for an additional

5 minutes.

5) Isolate the white solid obtained by vacuum filtration (Hirsh funnel.)

6) Wash with 0.5 mL of water, and oven dry at 80°C.

7) Weigh the product, determine percent yield and melting point, and obtain a proton NMR spectrum.

KBrO3/HBr

OCH

CHO

OH

OCH

CHO

OH

HBr

HOCH

CHO

OH

HH

BrOCH

CHO

OH

BrH

H+ +

3 3 3 3

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Product Identification – NMR. Obtain a proton NMR of each product and assign coupling constants to the benzenoid hydrogen atoms. To calculate the chemical shifts of each signal (remember it is being split into a double by a nearby hydrogen) subtract the smaller number from the larger number of the doublet and multiply the difference by the power of the instrument in megahertz, in this case, 500. Note that the position of the benzenoid hydrogen atoms in the 6-bromo-, 5-bromo-, and 2-bromo products are para, meta and ortho, respectively. Since there is a rather large difference between the coupling constants for ortho (~5-9 Hz), meta, (~1-5 Hz) and para (0-1 Hz) hydrogens, one can readily deduce the structure of the major product in this way. Product Identification – Melting Point. The structure of a newly synthesized product may also be obtained by comparison of the melting point of the product with literature values. This method necessitates that the possible structures have been synthesized, unambiguously characterized, and published. An extensive literature search has discovered the following reporting melting points for the possible bromination products:

CHO

OCH3OH

CHO

OCH3OH

CHO

OCH3OH

CHO

OCH3OH

Br

Br

Br

6-bromomp: 178°C

5-bromomp: 164°C

2-bromomp: 154-5°C

vanillinmp: 81-3°C

CHO

OHOCH3

CHO

OHOCH3

CHO

OHOCH3

CHO

OHOCH3

6-bromomp: 112°C

5-bromomp: unknown

2-bromomp: 202-7°C

Br

Br

Br

isovanillinmp: 113-5°C Based on these literature values, determine the structure of your product.

Post Laboratory Questions:

1. In the bromination of vanillin, does the ortho-para directing effect on the methoxy (OCH3) win out (giving 2- or 6-bromovanillin)? Or, does the combination of the ortho-para directing effect of the hydroxy (OH) group and the meta directing effect of the aldehyde (CHO) group control the substitution (giving 5-bromovanillin)?

2. If both 2- and 6-bromo products are possible, which one is the major product?

3. Write all resonance structures for the intermediate benzene carbocation leading to the observed product. Indicate which is the most important contributing structure.

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4. Do the predictions obtained by NMR analysis agree with the literature results, as determined by melting point?

5. In the bromination of isovanillin, does the ortho-para directing effect on the hydroxy (OH) win out (giving 2- or 6-bromoisovanillin)? Or, does the combination of the ortho-para directing effect of the methoxy (OCH3) group and the meta directing effect of the aldehyde group control the substitution (giving 5-bromoisovanillin)?

6. If both 2- and 6-bromo products are possible, which one is the major product?

7. Write all resonance structures for the intermediate benzene carbocation leading to the observed product. Indicate which is the most important contributing structure.

8. Do the predictions obtained by NMR analysis agree with the literature results, as determined by melting point?

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Preparation of Aspirin and Phenacetin Aspirin and phenacetin are two commonly used analgesics. They usually are synthesized be treating salicylic acid and p-phenetidine, respectively, in refluxing acetic anhydride. In today’s experiment, the analgesics will be prepared using microwave radiation. Microwaves have been used extensively in a variety of applications, including rapid synthesis of organic compounds. In today’s experiment you will be able to synthesize each compound using an irradiation time of a minute or less. The reactions of the day are shown below:

Both reactions proceed via a mechanism involving nucleophilic attack of N or O onto one of the carbonyl groups of the acetic anhydride reactant. In your report you will need to write the detailed mechanism for both reation with the arrows pointing in the right directions. Check Chapter 21, ‘Carboxylic Acid Derivatives’ in Wade for details.

H3PO4 cat.microwave

30% power 75 seconds

microwave

30% power30 seconds

p-phenetidine

acetic anhydride

phenacetinO

N H 2

O

O O

O

N H

O

+

salicylic acid acetic anhydride acetylsalicylic acid “aspirin”

O H C O 2 H

O

O O O C O 2 H

O+

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Experimental Procedure

Synthesis of Aspirin. Using a centrifuge tube, cool 10 mL dionized water in an ice bath. In a 100 mL beaker, combine 0.01 mol (1.38 g) of salicylic acid, 0.03 mol (2.6 mL) of acetic anhydride and one drop of phosphoric acid. Cover the mixture with a watch glass (concave side up!) and place the assembly in a microwave oven. Heat the assembly for 75 seconds using a power level of 3 (30%). Remove the assembly and allow the beaker to cool to room temperature. Place the beaker in the ice bath to aid in the precipitation of aspirin from solution. Recover the aspirin by vacuum filtration, rinsing with two 5 mL portions of chilled water, and then allow it to dry in a heated oven for ten minutes. Weigh the dried aspirin, determine its melting range, and calculate its percent yield. Determine the purity of the product by IR spectroscopy (KBr pellet.) Synthesis of Phenacetin. Using a centrifuge tube, cool 10 mL dionized water in an ice bath. In a 100 mL beaker, combine 0.01 mol (1.3 mL) of p-phenetidine, 7.5 mL of water and 0.03 mol (2.6 mL) of acetic anhydride. Cover the mixture with a watch glass (concave side up!) and place the assembly in a microwave oven. Heat the assembly for 30 seconds using a power level of 3 (30%). Remove the assembly and allow the beaker to cool to room temperature. Place the beaker in the ice bath to aid in the precipitation of phenacetin from solution. Recover the phenacetin by vacuum filtration, rinsing with two 5 mL portions of chilled water, and then allow it to dry in a heated oven for ten minutes. Weigh the dried product, determine its melting range, and calculate its percent yield. Determine the purity of the phenacetin sample by thin-layer chromatography using 6:4 pentane-ethyl acetate as the mobile phase. Run the chromatogram of your product along with known samples of p-phenetidine (starting material) and phenacetin (product.) Report Rf values for your product and the known samples. Post Laboratory Questions:

1. a) What are the by-products of the aspirin synthesis? b) What are the by-products of the phenacetin synthesis?

2. Sketch the reaction mechanism for each preparation (include the role of H3PO4 in the aspirin preparation).

3. Why do you think the aspirin synthesis requires a catalytic amount of acid while the phenacetin synthesis does not?

4. Compare the melting ranges obtained for each product to their literature values. What do these results tell you about the identity and purity of your products?

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Name:_______________________________ Section:_________ Date:_________

Preparation of Aspirin and Phenacetin

Pre-Laboratory Questions:

1. What is the structure of aspirin? Of phenacetin?

2. What is the non-microwave preparation of aspirin and phenacetin?

3. What is an analgesic?

4. How long will you irradiate the aspirin reaction? The phenacetin reaction?

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Synthesis of Terephthaloyl Nylon Read sections 26.1 and 26-7A in Organic Chemistry by Wade. Nylon is made industrially by condensing a diacid with a diamine. The most common reaction is the condensation of adipic acid and hexamethylene diamine to produce Nylon 66. This reaction requires somewhat exotic equipment and elevated temperatures. A room temperature reaction can be achieved by substituting adipoyl chloride, which is much more reactive than the corresponding acid. In today’s experiment you will make a slightly different polymer, which we will call Terephthaloyl Nylon, by treating terephthaloyl chloride with hexamethylene diamine in a rather mundane 100 mL beaker. When you mix the acid chloride (dissolved in diethyl ether) with the diamine (dissolved in water), a two layer liquid results (water at the bottom and ether at the top.) The reaction will occur (a film immediately forms) at the interface of the two liquid phases.

Reaction of the Day

Reagents and Properties substance conc’n quantity (mL) mol mass (g/mol) density terephthaloyl chloride 0.28M in ether 10 203.02 hexamethylene diamine 70% in H2O 0.6 116.20 0.89

hexamethylene terephthaloyl chloride +

OO

Cl Cl

O

Cl N

O

HN

O O

(CH2)6

H

(CH2)6

HNN

H

H + 2 HCl

NN

H

H

HH

“living” polymer chain etc. etc.

OO

Cl ClN

N

H

H

HH

25

Experimental Procedure: In a 100 mL beaker, dissolve 0.6 mL of 70% hexamethylene diamine (Caution! Caustic!) in 10 mL of water. Add 0.3 g sodium hydroxide. In a separate beaker, obtain 10 mL of 0.28 M terephthaloyl chloride dissolved in diethyl ether. Carefully pour the less dense of these two solutions (water or diethyl ether? If in doubt, check density in the Aldrich Chemical catalog) down the inside of the beaker containing the other solution with as little stirring as possible. A film forms at once at the interface. Push the film away from the wall of the beaker and pull out the mass that forms with a tweezers to raise it as a strand of continuously forming polymer. This can be wound on a rod or pencil to yield several feet of fiber. Place the resultant polymer on sheets of paper on the benchtop. DO NOT PERMIT THE LIVING POLYMER TO TOUCH THE BENCHTOP. (The impervious-looking benchtops are ‘epoxy resin’ hard plastic. Your polymer will react with the benchtop, and fuse with it.) Measure and record the length of the longest strand. Gather all of the polymer generated and blot the strands dry with paper towels. Measure and record the total mass of polymer formed. Post Laboratory Questions:

1) Describe briefly how you drew your strand, and whether this was an effective technique. Discuss what factors need to be considered to draw a good strand.

2) For this laboratory scale experiment, terephthaloyl chloride was used, rather than the less reactive terephthalic acid. Why is the acid chloride more reactive?

3) What was the purpose of the sodium hydroxide added to the original reaction mixture?

4) How do the fibers produced compare with commercially produced nylon, which is used (for example) in sports clothing?

Calculating theoretical and actual yields for polymers is impossible to do conventionally, because there is no absolute molecular weight for the product. The length of the polymer molecules is difficult to determine, and usually the molecules are of varying length. The mathematical solution to this difficulty is to base yield calculations on the “repeating unit” of the polymer, and use that as the molecular weight of the polymer. That is, determine a structure for which the polymer can be represented as: In this case, the molecular weight of the repeating unit, C14H18O2N2, gives a molecular weight of 246g/mol. Note that other repeating units may be selected, but all will lead to the same formula and molecular weight.

N

O

H

O

NH

N

O O

(CH2)6

H

NH

(CH2)6[ ] N

O

(CH2)6

H

N

O

H[ ]n

26

Calculating Retention Factors (Rf) for Thin Layer Chromatography

Upon the development and viewing of a TLC plate, the starting point and solvent front (the level the solvent reached when the plate was removed from the developing tank) are marked and all spots observed on the plate are circled in lead pencil. The location of each spot on the plate is then represented numerically by calculating a Retention Factor (Rf). This is accomplished by making the following measurements and calculations:

Retention factors are numbers between zero and one, representing the position of the spot on the TLC plate. Hence:

Rf ≈ 0 Rf ≈ 0.5 Rf ≈ 1.0 If the number obtained is not between zero and one, a mathematical error occurred. Note that this number is dimensionless, and the distances may be measured in centimeters, inches or barleycorns.

Chemistry 3117/3118 Southern Methodist University

a

b

Measure the distance from the starting point to the center of the spot on the TLC plate (distance a). Measure the distance from the starting point to the solvent front (distance b). Calculate the Retention Factor as: Rf =

a

b

27

How to get maximum efficiency from your Meltemp apparatus The Meltemp apparatus is the small piece of equipment you have been using to determine melting and boiling points. We have two models in the lab - one has a black body and the other has a gray body. They both work in essentially the same manner. Now that you will be doing more careful work with melting points, it would be a good idea to pick up some pointers concerning the Meltemp apparatus, particularly the correct power settings to use. Check out the graph below:

The y-axis corresponds to the expected melting temperature of whatever it is you are measuring. The x-axis corresponds to the power setting on the black-bodied Meltemp apparatus (the black-bodied apparatus has a power scale from 0 to 10 whereas the gray-bodied apparatus has a scale from 0 to 120. Obviously then, a setting of 5.0 on the black-bodied apparatus corresponds to a setting of 60 on the gray-bodied one). Normally, you want the apparatus to be heating at a rate of 2ºC/min at the expected melting point of your compound. So for example, if you have a compound that should melt at 150ºC, you find 150 on the y-axis and move over on the graph to the line that says “2º/min”. That corresponds to a power setting of approximately 3.7. If you go ahead and set the apparatus to 3.7, you can feel safe that the heating rate will be 2ºC/min at 150ºC.