4-cyano-N methyl benzamide

Prepararation 4-cyano-N methyl benzamide.

A mixture of 4-bromo-N-ethyl-3-methylnenzamide (3.76 g, 16.5 mmol, can be synthesized according to the following reference Oxford, A. W.; et al EP 533266 (1993)), K4[Fe(CN)₆]-3H₂0 (2.09 g, 4.94 mmol), Na₂C0₃ (1.05 g, 9.89 mmol), Pd(OAc)₂ (75 mg, 0.33 mmol), l,4-diazabicyclo[2.2.2] octane (74 mg, 0.66 mmol), and DMA (20 mL) was maintained under N₂ atmosphere at 126-130°C for 7.5 hr. Afterward, the mixture was cooled to room temperature, diluted with EtOAc, stirred for 20 min, and filtered through diatomaceous earth. The filtrate was concentrated under reduce pressure, and the residue (5.52 g) was stirred in a mixture of Et₂0 (5 mL) and hexanes (10 mL). Afterward, the solid was collected by filtration, and washed with Et₂0 to form the title compound (2.53 g). The mother liquor was concentrated under reduce pressure, and the residue was stirred in a mixture of Et₂0 (2 mL) and hexanes (4 mL) to form an additional amount of the compound (0.265 g) as a solid. Both batches were combined (2.80 g, 97%). 

¾ NMR (300 MHz, CDC1₃) δ ppm 2.59 (s, 3H), 3.03 (d, J = 4.9 Hz, 3H), 6.17 (br s, 1H), 7.61 (dd, J = 8.2, 1.5 Hz, 1H), 7.66 (d, J = 8.0 Hz, 1H), 7.73 (s, 1H).

http://www.google.com/patents/WO2014117274A1?cl=en

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3,5 difluoro-4 formyl-N-methylbenzamide.

 Preparation of 3,5 difluoro-4 formyl-N-methylbenzamide.

To an ice-cold solution of 3,5-difluoro-4-formylbenzoic acid (120 g, 645 mmol) in dichloromethane (1.5 L) and N,N-dimethylformamide (2.0 g, 27 mmol) was added oxalyl chloride (90 g, 709 mmol) drop-wise at a rate that allowed the mixture to not exceed an internal temperature of 10°C. The resulting mixture was stirred at the same temperature for 0.5 hr, warmed to room temperature, and stirred for an additional 1.5 hr. The solution was then cooled to 0°C, and aqueous methylamine (40%, 168 mL, 1.94 mol) was added drop-wise at a rate that allowed the mixture to not exceed an internal temperature of 7°C. Afterward, the mixture was quenched with aqueous HC1 (2M, 335 mL, 670 mmol) and warmed to room temperature. The organic layer was separated, washed with brine (500 mL), dried over MgSC>4, filtered, and concentrated under vacuum. The resulting residual solid was taken in MTBE (500 mL), and the resulting mixture was heated to reflux for 0.5 hr, cooled to room temperature, and stirred for 18 hr. Afterward, the mixture was cooled to 0°C, filtered, rinsed with pentane, and dried under vacuum to form the title compound (103 g, 80%) as a solid.

 ¾ NMR (300 MHz, CDC1₃) δ ppm 3.03 (d, J = 4.86 Hz, 3 H), 6.37 (br s, 1 H), 7.36-7.42 (m, 2 H), 10.36 (s, 1 H); 

MS m/z 200.06 [M+H]⁺ (ESI).

3,5-difluoro-4-formylbenzoic acid.

3,5-difluoro-4-formylbenzoic acid.

To a solution of 3,5-difluorobenzoic acid (291 g, 1.84 mol) in 2-methyltetrahydrofuran (4.35 L) was added TMEDA (604 mL, 4.03 mol) at room temperature. The resulting solution was cooled to -78°C. Afterward, /?-BuLi (2.5 M in hexane) (1.77 L, 4.43 mol) was added drop-wise, during which the temperature of the mixture remained at less than -65 °C. The mixture was then stirred at -78 °C for 1.5 hr. Anhydrous MeOCHO (239 mL, 3.88 mol) was added dropwise at a rate that allowed the temperature to be maintained at less than -65 °C. The resulting solution was allowed to warm at room temperature, and then maintained a room temperature while being stirred for 18 hr. The mixture was then cooled to 0-5°C, and excess base was quenched with 6M aqueous HC1 (2.2 L, 13.2 mol). The phases were then separated, and the aqueous layer was extracted 3 times with 2- methyltetrahydrofuran (3 x 500 mL). The combined organic phases were washed with saturated brine, dried over MgS0₄, filtered, and concentrated under vacuum. The residue was dissolved in ethyl acetate (350 mL) at reflux, and cooled to room temperature.

Hexanes (480 mL) were then added, and the resulting mixture was further cooled to -15°C. The solid was collected by filtration, rinsed with hexanes, and dried under mechanical vacuum to form the title compound (122 g, 35%) as a solid. 

H NMR (300 MHz, DMSO- d₆) δ ppm 7.63-7.70 (m, 2 H), 10.23 (s, 1 H); 

MS m/z (ESI) 187.17 [M+H]⁺.

A graph theory approach to structure solution of network materials from two-dimensional solid-state NMR data

An NMR crystallography strategy is presented for solving the structures of materials such aszeolites and related network materials from a combination of the unit cell and space group information derived from a diffraction experiment and a single two-dimensional NMR correlationspectrum that probes nearest-neighbour interactions.

 By requiring only a single 2D NMR spectrum, this strategy overcomes two limitations of previous approaches. First, the structures of materials having poor signal-to-noise in solid-state NMR experiments can be investigated using this approach since a series of 2D spectra is not required. Secondly, 

the structures of aluminophosphate materials can potentially be determined from ²⁷Al/³¹P solid-state NMRexperiments since this approach does not require the isolated spin pairs which have been important for determining structures of silicate materials by ²⁹Si solid-state NMR. Using concepts from graph theory, the structure solution strategy is described in detail using a hypothetical two-dimensional network structure. A collection of two-dimensional network structures generated by the algorithm under various initial conditions is presented.

 The algorithm was tested on a series of 27 zeolite framework types found in the International Zeolite Association’s zeolite structure database. Finally, the structure of the zeolite ITQ-4 was solved from powder X-ray diffraction data and a single ²⁹Si double quantum NMR correlation spectrum. The limitations of the strategy are discussed and new directions for this approach are outlined.

CrystEngComm, 2013,15, 8748-8762

DOI: 10.1039/C3CE41058G

http://pubs.rsc.org/en/content/articlelanding/2013/ce/c3ce41058g#!divAbstract

Ethyl propenoate...........your cock can teach you

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Ha: Split by Hb (J geminal = 0.5-3 Hz from above) and by Hc (J trans = 12 -18 Hz from above). Because they’re different we apply the compound N+1 rule and get a doublet of doublets.

Hb: Split by Ha (J trans = 12-18 Hz) and Hc (J cis = 7-12 Hz). Of the three It looks most like a doublet of doublets since it has the largest J values (peaks are more spaced out)

Hc: Split by Ha (J geminal = 0.5-3 Hz) and Hb (J cis = 7-12 Hz). Also adoublet of doublets, and very similar to Ha except the peaks are closer since J cis < J trans.

13C NMR

MASS

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Nmr problem...Our cock can interpret one

OUR COCK CAN INTERPRET ONE.......................................






Compound below has a composition a 68.2% C, 13.6% H and 18.2% O by mass.

• A notable feature of the ¹H spectrum is a multiplet of nine at d1.7. What is the cause of this multiplet?

A splitting pattern of nine peaks must be produced by eight adjacent protons (8+1)

¹H Spectrum

¹H Spectrum (expansion)

• Sketch the possible isomers of compound U, eliminating any which do not account for this multiplet of nine.

• Assign each of the peaks, giving reasons for the relative intensities and splitting patterns.

Peak  is due to protons in position d as it has a chemical shift of d3.6 (corresponds to a -OH containing functional group) and is a triplet (split by two adjacent hydrogens - H_c).
Peak  is due the proton in position b as it has a chemical shift of d1.7 (corresponds to a tertiary -CH- group) and is a nontet (split by eight adjacent hydrogens - H_a and H_c).
Peak  is due to protons in position c as it has a chemical shift of d1.4 (corresponds to a -CH₂ group) and is a quartet (split by three adjacent hydrogens - H_b and H_d).
Peak  is due to protons in position e as it has a chemical shift of d1.3 (corresponds to hydroxyl proton) and is a singlet (no splitting).
Peak  is due the proton in position a as it has a chemical shift of d0.8 (corresponds to a -CH₃ group) and is a doublet (split by one adjacent hydrogens - H_b).

¹³C Spectrum

• The ¹³C spectrum shows the reference peak and four peaks due to compound U. Explain.

Peak  is due to the carbon atom in position d as it has a chemical shift of d60.7 (corresponds to a -C-O- functional group) and is connected to an even number of protons (2).
Peak  is due to the carbon atom in position c as it has a chemical shift of d41.5 (corresponds to a secondary carbon group) and is connected to an even number of protons (2).
Peak  is due to the carbon atom in position b as it has a chemical shift of d24.6 (corresponds to a tertiary carbon group) and is connected to an odd number of protons (1).
Peak  is due to the two carbon atoms in position a as it has a chemical shift of d22.5 (corresponds to a primary carbon group). It can be seen that the carbon atoms are also connected to an odd number of protons (3).