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Showing posts with label LAHORE. Show all posts
Showing posts with label LAHORE. Show all posts

Saturday 30 August 2014

Benzaldehyde to Benzilic Acid


Benzilic acid






Starting material












Benzoin: Thaimine hydrochloride (1.52 g, 0.45 mmol), water (2mL) and 95% ethanol (15 mL) were combined in a 50-mL Erlenmeyer flask and swirled until dissolved and homogeneous.  Aqueous sodium hydroxide (4.5 mL) was added and swirled until the solution appeared pale yellow.  Finally, pure benzaldehyde (4.5 mL, 4.41 mmol) was added to the flask and the mixture was stored for two days.  The crystals that formed at room temperature were placed in an ice bath and then filtered under vacuum.  The crystals were washed with 5-mL portions of ice-cold water and left to dry.  To isolate the pure product, the crude material was crystallized with 95% ethanol (24 mL).  The pure product of benzoin showed the following physical characteristics: 2.07 g (44.3 % yield) mp: 129-132°C (lit: 135-135 °C).  1H NMR (CDCl3, 300 MHz) δ: 7.79 (d, J=1.5, 2H) 7.25 (m, 2H), 7.24 (m, 2H), 7.19 (m, 2H), 7.14 (m, 2H), 5.82 (d, J= 1.2, 1H), 3.92 (s, 1H) ppm.  13C NMR (CDCl3, 75Hz) δ: 199.2, 139.2, 139.1, 134.0, 129.2, 129.1, 128.7, 128.5, 127.8, 76.2 ppm.  IR 3403, 3003, 1761, 1203 cm-1.
The final product of benzoin contained 13C NMR peaks at 199.2 ppm accounting for the carbonyl group and eight peaks in the range of 139.0-127.8 ppm representing the alkene bonds as well as the carbons of the aromatic rings.  Finally, a peak at 76.2 ppm represented the carbon with the alcohol group attached.  Regarding the1H NMR spectra, four multiplet peaks appeared in the range of 7.79 and 7.14 ppm representing the hydrogens surrounding the aromatic rings.  A peak at 5.82 ppm accounted for the hydrogen attached to the carbon containing the alcohol group.  A peak at 3.92 ppm represented the hydrogen of the alcohol group.  An impurity of ethanol appeared at 4.42 ppm.  Finally, the IR spectra displayed a peak at 3403 cm-1 representing the C-H stretches, a peak at 3003 cm-1, accounting for the alcohol group, and a strong peak at 1761 cm-1representing the carbonyl group.  Overall, the spectra confirmed the condensation of benzoin.


Benzil: Benzoin (2.51g, 1.18 mmol), concentrated nitric acid (12 mL, 28.8 mmol), and a stir bar were placed into a 25-mL round-bottom flask with a water condenser and heated in a hot water bath at 70 °C for one hour.  After heating and magnetically stirring, the mixture was added to 40 mL of cool water and stirred until crystallized into a yellow solid.  Vacuum filtration was used to collect the crude product.  The pure product was collected through recrystallization by using 95% ethanol (20 mL).  The product displayed the following properties: 1.91 g (76.8 % yield) mp: 89-92 °C (lit: 95 °C).  1H NMR (CDCl3, 300 MHz) δ: 7.86 (m, 4H), 7.56 (m, 2H), 7.53 (m, 4H) ppm.  13C NMR (CDCl3, 75Hz) δ: 192.0, 132.3, 130.4, 127.3, 126.5 ppm.  IR 3010 (w), 1668 (s) cm-1.
The 13C NMR produced a peak at 192.0 ppm representing the two carbonyl groups.  Four peaks appeared between 132.3 and 126.5 ppm accounting for the carbons within the aromatic ring and the alkene bonds.  The 1H NMR displayed three multiplet peaks at 7.86, 7.56, and 7.53 ppm representing the hydrogens around the aromatic ring that coupled with the surrounding hydrogens.  Finally, the IR spectrum produced a C-H stretch peak at 3010 cm-1 and a carbonyl peak at 1668    cm-1.  This data proved the success of the oxidation of benzoin to produce benzil.


1H NMR

1H-NMR: Benzil300 MHz, CDCl3delta [ppm]mult.atomsassignment7.50m (d)4 H2-H, 2'-H, 6-H, 6'-H7.64m (dd)2 H4-H, 4'-H7.95m (dd)4 H3-H, 3'-H, 5-H, 5'-H




13C NMR


13C-NMR: Benzil
75.5 MHz, CDCl3
delta [ppm]assignment
129.0CH arom.
129.8CH arom.
133.0CH arom.
154.8C quart. arom.
194.5C=O
76.5-77.5CDCl3


IR
IR: Benzil[KBr, T%, cm-1][cm-1]assignment3064arom. C-H valence1660C=O valence, ketone1594, 1579arom. C=C valence
pure product chromatogram

GC: pure productcolumnDB-1, L=28 m, d=0.32 mm, film=0.25 µminleton column injection, 0.2 µLcarrier gasH2, 40 cm/soven90°C (5 min), 10°C/min --> 240°C (30 min)detectorFID, 270°Cintegrationpercent concentration calculated from relative peak area





Benzilic Acid: Benzil (2.10 g, 1.0 mmol), 95% ethanol (6 mL), and a boiling stone were added to a 25-mL round-bottom flask with a reflux condenser and heated until the solid benzil was dissolved.  Aqueous potassium hydroxide (5 mL, 18.2 mmol) was added dropwise to the flask and the mixture was boiled for 15 minutes.  The mixture was cooled, transferred to a beaker, and placed in an ice-water bath until crystallized.  The crystals were isolated through vacuum filtration and washed with 4-mL portions of cold 95% ethanol.  The solid was transferred to a 100-mL flask of hot water (60 mL) and mixed until completely dissolved.  Concentrated hydrochloric acid (1.3 mL) was added drop-wise until a permanent solid was present and a pH of 2 was maintained.  The solution was cooled in an ice bath and the crystals were filtered through vacuum filtration and washed with 2, 30-mL portions of ice-cold water.  The remaining crystals were identified by the following properties: 0.41 g (17.4% yield) mp: 151-152 °C (lit: 150 °C).  1H NMR (CDCl3, 300 MHz) δ: 7.47 (m, 6H), 7.26 (s, 4H), 2.18 (s, 1H) ppm.  13C NMR (CDCl3, 75Hz) δ: 175.8, 141.4, 128.3, 128.2, 127.4, 82.0 ppm.  IR 3399 (s), 2889 (s, b), 1718 (s), 1177 (s) cm-1.
The melting point corresponded to the known melting point of 150 °C. The 13C NMR spectra displayed a weak peak at 175.8 ppm, which accounted for the carbonyl group within the carboxylic acid.  Four peaks at 141.4, 128.3, 128.2, and 127.4 ppm represented the carbons within the aromatic rings.  Finally, a peak at 82.0 ppm represented the carbon attached to the alcohol group.  The 1H NMR spectrum produced a peak at 7.47 and 7.26 ppm representing the two groups of equivalent hydrogens attached to the aromatic rings.  A peak at 2.18 ppm represented the hydrogen of the alcohol group.  A peak did not appear at 12 ppm that would have represented the hydrogen of the carboxylic group, which means the reaction was not carried to completion.  In the IR spectrum, a hydroxyl peak appeared at 3399 cm-1.  A broad peak appeared at 2889 cm-1 representing the carboxylic acid functional group of compound.  Finally, a peak at 1718 cm-1 represented the carbonyl group and a peak at 1177 cm-1 accounted for the carbon-oxygen bond in both alcohol groups.
raman
1 Bruggink, A.; Schoevaart, R.; Kieboom, T. Org. Proc. Res. Dev.20027,622-640.
2 Pavia, L; Lampman, G; Kriz, G; Engel, R. A Small Scale Approach to Organic Laboratory   Techniques, 2011, 266-269.
3 Lachman, A. J. Am. Chem. Soc., 192244, 330-340.



extras




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 Punjabi sweets on a leaf in Lahore Pakistan, 20 Rupees or about 30 cents. I wasn't sure if one is supposed to eat the leaf too but I ate it anyways. 10/07


 
 

 

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Wednesday 23 July 2014

4-METHYL BENZALDEHYDE EXAMPLE IN SPECTROSCOPY

Example 

C8H8O
MW 120

Calculate the degree of unsaturation: the answer is 5. If the degree of unsaturation is 4 or greater, look for an aromatic ring, which has a degree of unsaturation of 4 (3 double bonds plus 1 ring). In addition to an aromatic ring, the molecule can have a carbon-carbon double bond, a carbonyl, or another ring.

IR Spectrum

Look for a carbonyl, since the degree of unsaturation indicates that the compound could have a double bond and we know that the molecule has an oxygen. There is a band at 1703, suggesting an alpha, beta unsaturated aldehyde or ketone(1710-1665). To see if the compound might be an aldehyde, look for bands in the region 2830-2695. In the spectrum below, note the two bands in this region, suggesting that the compound is indeed an aldehyde.

The IR can also help determine whether or not the compound is an aromatic (although the NMR is a better diagnostic method for this). Look for the C–H stretch in aromatics from 3100-3000. Note that compounds that are not aromatic show C-H stretch from 3000-2850.




From the IR alone, we have determined that the molecule probably has aromatic, alkyl, and aldehydic functional groups. Now let's look at the NMR.

Proton NMR Spectrum

Since the IR spectrum indicates an aldehyde, look for this functionality in the NMR spectrum. The aldehydic proton appears in the NMR from 9-10, usually as a small singlet. The singlet of one proton in the spectrum below verifies the presence of an aldehyde:




image of 4-methylbenzaldehyde





Aromatic protons show up from 6.5-8.5 ppm. The four protons in the region 7.3-7.8 ppm indicate four aromatic protons; thus the aromatic ring has two substituents. Benzylic protons - hydrogens on a carbon adjacent to an aromatic ring - show up from 2-3 ppm; the three protons in the singlet peak at 2.4 ppm are likely benzylic protons. Since there are three protons, it is a methyl group.

Therefore we know that the molecule is a di-substituted aromatic ring with a -CHO group and a -CH3 group. Double check that we have the proper number of carbons: 6 (in ring) plus 1 (aldehyde) plus 1 (methyl) equals 8, the number given in the molecular formula.




These hydrogens correspond as follows with the peaks in the NMR:





The molecule is 4-methylbenzaldehyde: a para substituted aromatic. In the organic chemistry teaching laboratory courses, you are not held responsible for proper assignment of ortho-, meta-, or para-substituted rings. However, para substituted rings almost always show a symmetrical pattern in the aromatic region, as in the spectrum above. (See Below

13 C NMR

WATCH OUT FOR INTERPRETATION OF 13 C NMR

Summary and answer

Example  is 4-methylbenzaldehyde:





MORE
IR Spectroscopy
Aromatics show a lot more bands in an IR spectrum than do alkanes, alkenes, and alkynes. The C-C stretching vibrations show up as several bands in the region 1600-1585 and 1500-1400.
One of the most “telling” bands is the presence of a band just to the left of 3000. Alkyl groups show up to the right of 3000, aromatic C–H stretches to the left of 3000.
  • C–H stretch from 3100-3000
  • overtones, weak, from 2000-1665
  • C–C stretch (in-ring) from 1600-1585
  • C–C stretch (in-ring) from 1500-1400
  • C–H “oop” from 900-675

NMR Spectroscopy

Aromatic protons show up from 6.5-8.5 ppm. Benzylic protons are from 2–3 ppm.



Monosubstituted rings will have 5 protons in the region 6.5-8.5 ppm; disubstituted rings will have 4 protons; trisubstituted rings will have 3 protons (and so on). Examples of the NMR of aromatics of mono-, di-, and tri-substituted aromatics are shown below. When interpreting the spectrum of an aromatic compound, remember to count the number of protons in the aromatic region to determine how many times the ring is substituted.




If an aromatic ring has more than one substituent, careful analysis of the shifts and splitting pattern of the protons in the aromatic region reveals the positions on the ring of the different substituents. However, the the shifts depend on the substituents and the splitting is not really first order. Most sophomore-level courses do not cover NMR spectroscopy in the depth required to analyze the aromatic region. Therefore, you will not be able to designate the exact ring substition in many cases.

 In the specific case of disubstituted aromatic rings, para-substituted rings usually show two symmetric sets of peaks that look like doublets. The para-substitution NMR aromatic region pattern usually looks quite different than the patterns for both ortho- and meta-substituted aromatic rings.

Examples of orthometa, and para substitution are illustrated in the NMR spectra of different isomers of chloronitrobenzene, below. The CDCl3 peak is pointed out in each spectrum. (The samples were run using CDCl3as the solvent, and a small contaminant of this deuterated solvent is CHCl3, which shows up at 7.24 ppm. This is used to calibrate the spectrum.) Note the symmetry of the para substituted chloronitro benzene. (Click on each full-size image to view details of the region from 6.5-8.5 ppm.)


click on each NMR spectrum below to see an expanded view of the aromatic region2-chloronitrobenzene
3-chloronitrobenzene
4-chloronitrobenzene


SOME INTERESTING INFO

Trimethyl 2-Hydroxy-2-(2-methoxy-2-oxoethyl)-4-(4-methylphenyl)-6-oxo-1,3,5-cyclohexanetricarboxylate
Edmont V. Stoyanov
Faculty of Pharmacy, Medical University of Sofia, Dunav 2, BG-1000 Sofia, Bulgaria
Tel.: (+359 2) 988 3142; Fax: (+359 2) 987 9874; E-mail: estoyanov@yahoo.com

Aromatic aldehydes react with dimethyl acetonedicarboxylate in molar ratio 1:2 with spontaneous intermolecular Michael addition to give polysubstituted cyclohexanones [1]. We report now the synthesis of an analogous product from 4-methylbenzaldehyde.

To a solution of 4-methylbenzaldehyde (1.20 g, 10 mmol) and dimethyl acetonedicarboxylate (3.48 g, 20 mmol) in 25 ml ethanol, 0.3 ml piperidine was added. The reaction mixture was left to stay at room temperature for 3 days. The separated crystals were filtered off, washed with cold ethanol, recrystallized from dioxane and air-dried. Yield: 3.18 g (71 %). 

Colorless crystals, m. p. 149-150 ºC (dec.) from dioxane.

1H NMR (300 MHz, d6-DMSO): 
2.23 (s, 3H, PhCH3), 2.43 (d, 1H, J=17.0 Hz, HCH), 
2.96 (d, 1H, J=17.0 Hz, HCH), 3.44 (s, 3H, OCH3), 
3.52 (d, 1H, J=12.2 Hz, H-3), 3.56 (s, 3H, OCH3), 
3.60 (s, 3H, OCH3), 3.66 (s, 3H, OCH3), 
3.94 (t, J1=J2=12.2 Hz, H-4), 4.38 (d, 1H, J=12.2 Hz, H-5),
4.67 (s, 1H, H-1), 5.40 (s, 1H, OH), 
7.06 (d, 2H, J=8.0 Hz, 2' and 6'H arom.), 
7.19 (d, 2H, J=8.0 Hz, 3' and 5'H arom.).


13C NMR (75 MHz, d6-DMSO): 
41.4, 42.8, 51.1, 51.5, 51.6, 51.7, 54.3, 61.2, 62.8, 66.3, 74.3, 
128.1 (2xC), 128.9 (2xC), 136.2, 
136.6 (2xC), 167.7, 168.2, 169.5, 169.8.


FT IR (KBr, cm-1): 3511, 2953, 1729, 1516, 1495, 1364.


ESI MS [FIA in MeOH, CH3COONH4/CH3COOK]: 468.2 [M+NH4]+, 489.2 [M+K]+

Reference
1. Haensel, W.; Haller, R. Arch. Pharm. (Weinheim Ger.) 1970303, 334-338.



lahore, pakistan
Map of lahore city .



Image result for lahore pakistan.
.




















 Punjabi sweets on a leaf in Lahore Pakistan, 20 Rupees or about 30 cents. I wasn't sure if one is supposed to eat the leaf too but I ate it anyways. 10/07


 
 

 

 New Food Street Lahore


..............