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25th October 2016 @ 03:15

Goal: Synthesis of compound 3 by treating purified (Z)-2-(4-chlorophenyl)-3-hydroxypent- 2-enenitrile with a polymer supported methyl sulfonate reagent using microwave radiation.

 

See entry 5 for reaction scheme and procedure.

 

The crude yield was 89%. The product was purified using column chromatography (Biotage). After removing the solvent under pressure, the crude product was mixed with silica in DCM. The solvent was removed under pressure, resulting in a homogenous mixture of product and silica, which was then placed onto the automated Biotage column. The solvent system was 2:1 hexane to ethyl acetate.

 

Results:

 
Crude Yield:

% yield = (actual yield/ theoretical yield)100% = (0.198 g/ 0.222 g)100% = 89%

TLC:

TLC was done before the purification. A lot of spots showed up on TLC. These could be either impurities (even though the starting material was purified before being used for the reaction step) or compound fragments. By comparing the starting material and the product lanes, we observed two spots that only appear on the starting material lane and one spot that only appears in product lane. The Rf values for the three separate spots are shown below.

 

Silica plate

2:1 Hexane:EtOAc
Solvent front =  5.8 cm

Rf s.m. 1 = 2.4 cm/ 5.8 cm = 0.41
Rf s.m. 2 = 3.6 cm/ 5.8 cm = 0.62
Rf product 2 = 3.8 cm/ 5.8 cm = 0.66

Figure 1. TLC for reaction step #2 (L to R: (Z)-2-(4-chlorophenyl)-3-methoxypent-2-enenitrile, co-spot, (Z)-2-(4-chlorophenyl)-3-hydroxypent-2- enenitrile; 2:1 Hexane: EtOAc).

Column Chromatography:
Four fractions were collected through column chromatography, which is consistent with the multiple spots shown on TLC.

The fractions were not further analyzed, because we concluded that the starting material used in the reaction was not (Z)-2-(4-chlorophenyl)-3-hydroxypent-2-enenitrile, according to NMR (see previous entry).  


Conclusion: Reaction step #1 should be repeated and we should make sure that the target compound has been synthesized. Then, reaction step #2 should be carried out.  




Attached Files
25th October 2016 @ 03:09

Goal: Base-mediated condensation of 4-chlorophenylacetonitrile and ethyl proponoate in the presence of potassium tert-butoxide to synthesize compound 2. Purification of product with automated column chromatography (Biotage).

See entry 1 for reaction scheme and procedure.

 

The product was purified using column chromatography (Biotage). After removing the solvent under pressure, the crude product was mixed with silica in DCM. The solvent was removed under pressure, resulting in a homogenous mixture of product and silica, which was then placed onto the automated Biotage column. The solvent system was 2:1 hexane to ethyl acetate. The crude yield was 69%. The yield of the purified product was 48%.

Observations: The reaction mixture was initially brown, but turned dark red as the reaction proceeded. During the workup, the product turned yellow due to the addition of HCl.

After rotovapping, the product appeared to be a yellowish oil. The NMR of the product was taken, but solvent peaks (ethyl acetate) interfered with some of the product peaks (see explanation in the NMR section below). The product was rotovapped again, ending up as a yellowish powder.

 

Crude Yield:

% yield = (actual yield/ theoretical yield)100% = (0.897 g/ 1.300 g)100% = 69%

Yield (after Biotage):

% yield = (actual yield/ theoretical yield)100% = (0.624 g/ 1.300 g)100% = 48%

TLC:

TLC was done before the purification.

Silica plate

2:1 Hexane:EtOAc
Solvent front =  5.6 cm

Rf s.m. = 3.1 cm/ 5.6 cm = 0.55

Rf product 1 = 1.8 cm/ 5.6 cm = 0.14
Rf product 2 = 2.4 cm/ 5.6 cm = 0.43

Column Chromatogrpahy:

There were two fractions. NMR was taken for fractions to determine which fraction contained the product.

Figure 2. NMR of impurity (CDCl3 solvent).

Figure 2. NMR of impurity (CDCl3 solvent).

Figure 3. First NMR spectrum of purified (Z)-2-(4-chlorophenyl)-3-hydroxypent-2-enenitrile (CDCl3 solvent).


Figure 3. First NMR spectrum of purified (Z)-2-(4-chlorophenyl)-3-hydroxypent-2-enenitrile (CDCl3 solvent).
 
Figure 4. Second NMR spectrum of purified (Z)-2-(4-chlorophenyl)-3-hydroxypent-2-enenitrile (CDCl3 solvent).
Figure 4. Second NMR spectrum of purified (Z)-2-(4-chlorophenyl)-3-hydroxypent-2-enenitrile (CDCl3 solvent).

Analysis of the  1H NMR

One of the two fractions collected with Biotage resulted in a dirty NMR spectrum (Figure 2). We assumed that this fraction was an impurity. The other fraction contained the product (Figure 3). The integration of the the triplet (approximately 1.3 ppm) and the quarter (approximately 4.10 ppm) is greater than that expected. This might be due to ethyl acetate solvent peaks overlapping with the mentioned product peaks. Ethyl acetate peaks should appear at 1.26 ppm, 2.05 ppm, and 4.12 ppm. The product was rotovapped, was mixed with chloroform and was rotovapped once more (x2) to make sure that all ethyl acetate was removed. NMR was taken again. This time, the target peaks did not show up. This means that they were solvent peaks and that the reaction did not work out. The second spectrum seems really dirty and it is hard to analyze (Figure 4). Apart from the characteristic benzene peaks, a prominent peak at approximately 2.6 ppm shows up that might correspond to the starting material (Figure 5). However, the CH2 peak is more deshielded in the starting material and the benzene peaks are closer to one another and have a lower chemical shift. The rest of the peaks might be due to contaminated deuterated chloroform. It is possible that the target product was in the fraction that was considered to be an impurity. The NMR spectrum of each fraction was taken at different time points, and contamination could result in the dirty spectrum of that fraction.

Conclusion: The reaction was not successful. The TLC showed two distinct spots in the product lane which did not appear in the starting material lane. However, NMR suggests that the reaction did not proceed.



Attached Files
25th October 2016 @ 03:02

Goal: Condensation of (Z)-2-(4-chlorophenyl)-3- methoxypent-2-enenitrile with guanidine to synthesize pyrimethamine (Daraprim).

 

Figure 1. Synthesis of step 3 of the Daraprim scheme

Figure 1. Synthesis of step 3 of the Daraprim scheme


Table of Reagents

Reagents Amount in g or mL Amount in mmol Equivalent Additional comments

 

3-(4-chlorophenyl)-4-methoxyhexanenitrile
1.46 g 7 mmol 1 eq  
Guanidine chloride 2.0 g 21 mmol 3 eq Strong irritant
Sodium methoxide 1.1 g 21 mmol 3 eq Harmful, corrosive

Procedure:

Guanidine hydrochloride (3 eq) and sodium methoxide (3 eq) were dissolved in ethanol (15 mL) and stirred for 5 min. The cloudy, white solution was filtered to remove the NaCl precipitate and was added to a solution of 3-(4-chlorophenyl)-4-methoxyhexanenitrile (1 eq) in ethanol (5 mL). The reaction mixture was heated under reflux for 18 h.

 

Results:
The product was not analyzed, because NMR analysis revealed that the starting material used in the reaction step was not the target one, (Z)-2-(4-chlorophenyl)-3-hydroxypent-2-enenitrile.



Attached Files
25th October 2016 @ 02:56

Goal: Synthesis of compound 3 by treating (Z)-2-(4-chlorophenyl)-3-hydroxypent-2-enenitrile with a polymer supported methyl sulfonate reagent using microwave radiation.

 

See entry 5 for reaction scheme and procedure.

 

The reaction resulted in 92% yield, but the target compound was not synthesized according to NMR (Figure 1). TLC was taken immediately after synthesis. The TLC showed one spot on the product lane, indicating that the reaction went to completion (Figure 2). Interestingly, the starting material lane had two spots, even though itappeared as a single spot on a previous TLC (reaction step #1). We assume that the tautomers of the starting material (keto and enol) equilibrated during storage. During the methylation, the tautomers in their most stable forms (enol) reacted with the polymer-supported methyl sulfonate reagent leading to a single product.This explains the switch from a 2-spot starting material to a 1-spot product.

 
Observations: The reaction mixture was yellow in color. The product appeared to be a light brown liquid after rotovapping.
 
Crude Yield:

% yield = (actual yield/ theoretical yield)100% = (0.204 g/ 0.222 g)100% = 92%

 
Figure 1. NMR of (Z)-2-(4-chlorophenyl)-3-methoxypent-2-enenitrile (CDCl3 solvent; failed).
Figure 1. NMR of (Z)-2-(4-chlorophenyl)-3-methoxypent-2-enenitrile (CDCl3 solvent; failed).
 

Analysis of the  1H NMR

Comparing the H NMR of the starting material to the H NMR of the product of step #2 we noticed:

  • The H NMR of the starting material is expected to have 4 distinct peaks:

  • A doublet at ~7.5 ppm representing 2 hydrogens on the phenyl ring.

  • A doublet at ~7.3 ppm representing the other 2 hydrogens on the phenyl ring.

  • A sharp singlet at ~3.5 ppm representing the methoxy group.

  • A quartet at ~2.0 ppm representing the -CH2 of the ethyl group.

  • A triplet at ~1.0 ppm representing the CH3 of the ethyl group.

  • However, the H NMR of the product (Figure 1) has:

    • A doublet at ~8.0 ppm representing 2 hydrogens on the phenyl ring.

    • A doublet at ~7.5 ppm representing the other 2 hydrogens on the phenyl ring.

    • Two doublets at ~7.4 ppm and ~ 7.3 ppm which could be due to the benzene peaks of the the s.m. of reaction step #1 (trance amount).

    • A singlet at ~7.3 ppm due to d-chloroform.

    • A singlet at ~4.6 ppm, which is probably an impurity. This peak could correspond to the  target methoxy peak, but the integration is off by 2 hydrogens.

    • A singlet around ~4 ppm which could be due to the -CH2 of the s.m. of reaction step #1 (trace amount).

    • A multiplet at ~2.6 ppm that could potentially represent the -CH2 of the ethyl group. The more complex splitting pattern could be due to the presence of the alkene.

    • A triplet at ~1.0 ppm representing the CH3 of the ethyl group.

    • A couple of peaks around ~1.3 ppm due to water.

    • Peaks deduced to be impurities.

It seems that the product did not form. The spectrum looks more like that of the s.m.

TLC:

Silica plate

2:1 Hexane:EtOAc
Solvent front =  6.2 cm

Rf s.m. 1 = 2.8 cm/ 6.2 cm = 0.45

Rf s.m. 2 = 3.1 cm/ 6.2 cm = 0.50
Rf product = 2.4 cm/ 6.2 cm = 0.39


Figure 2. TLC for reaction step #2 (L to R: (Z)-2-(4-chlorophenyl)-3-hydroxypent-2- enenitrile, co-spot, (Z)-2-(4-chlorophenyl)-3-methoxypent-2-enenitrile; 2:1 Hexane: EtOAc).
 

Conclusion:

The desired product was not produced as indicated by TLC and NMR analysis.

 

 

Attached Files
25th October 2016 @ 02:42

Synthesis of compound 3 using reflux was not successful (previous entry). The procedure was adjusted appropriately.

Goal: Synthesis of compound 3 by treating (Z)-2-(4-chlorophenyl)-3-hydroxypent-2-enenitrile with a polymer supported methyl sulfonate reagent using microwave radiation.

Figure 1. Synthesis of step 2 for the Daraprim reaction scheme.
 
Table of Reagents

Reagents Amount in g or mL Amount in mmol Equivalent Additional comments

 

(Z)-2-(4-chlorophenyl)-3-hydroxypent-2-enenitrile
0.35g
(MW = 207.657 g/mol)
1.000 1 eq  
Polymer-supported ­methyl sulfonate 1.5g
(2.0- 2.5 mmol/g loading rate)
    Safer than other methylation agents
K2CO3 0.23g 1.000 1 eq  
MeCN 10 mL     Highly flammable

Procedure:

A mixture of (Z)-2-(4-chlorophenyl)-3-hydroxypent-2-enenitrile (0.35 g, 1.0 mmol) and potassium carbonate (0.23 g, 1.0 mmol) was dissolved in acetonitrile (10 mL). Polymer-supported methyl sulfonate (1.5g, 2.0- 2.5 mmol/g loading rate) was added to the solution. The reaction was carried out under microwave irradiation at 90 °C and 2 bar for 30 min.
 

Results:

The reaction resulted in 96% yield, but was not successful according to NMR (Figure 5). TLC was taken 2 days after synthesis and it showed three distinct spots on the product lane (Figure 3). We believe that the compound is air sensitive and was fragmented due to inappropriate storage conditions. The compound was stored at room temperature in a container that was not adequately sealed. Alternatively, the spots can be due to impurities. The NMR indicated the presence of s.m. peaks. The s.m. can tautomerize, which could also explain the multiple peaks seen on TLC.
 

Other observations: The reaction mixture was yellow in color. The product appeared to be a light brown liquid after rotovapping.

Crude Yield:

% yield = (actual yield/ theoretical yield)100% = (0.213 g/ 0.222 g)100% = 96%

 

TLC:

Silica plate

2:1 Hexane:EtOAc
Solvent front =  6.6 cm

Rf s.m. = 2.5 cm/ 6.6 cm = 0.38
Rf product 1 = 2.0 cm/ 6.6 cm = 0.33

Rf product 2 = 2.7 cm/ 6.6 cm = 0.41

Rf product 3 = 4.9 cm/ 6.6 cm = 0.74

Figure 3. TLC for reaction step #2 (L to R: (Z)-2-(4-chlorophenyl)-3-hydroxypent-2-enenitrile, co-spot, (Z)-2-(4-chlorophenyl)-3-methoxypent-2-enenitrile; 2:1 Hexane: EtOAc).

Figure 5. NMR of (Z)-2-(4-chlorophenyl)-3-methoxypent-2-enenitrile (CDCl3 solvent; failed).

Figure 5. NMR of (Z)-2-(4-chlorophenyl)-3-methoxypent-2-enenitrile (CDCl3 solvent; failed).
 

Analysis of the  1H NMR

Comparing the H NMR of the starting material to the H NMR of the product of step #2 we noticed:

 
  • The H NMR of the starting material is expected to have 4 distinct peaks (Figure 4):

    • A doublet at ~7.5 ppm representing 2 hydrogens on the phenyl ring.

    • A doublet at ~7.3 ppm representing the other 2 hydrogens on the phenyl ring.

    • A sharp singlet at ~3.5 ppm representing the methoxy group.

    • A quartet at ~2.0 ppm representing the -CH2 of the ethyl group.

    • A triplet at ~1.0 ppm representing the CH3 of the ethyl group.

  • However, the H NMR of the product (Figure 5) has:

    • A doublet at ~7.3 ppm representing 2 hydrogens on the phenyl ring.

    • A doublet at ~7.4 ppm representing the other 2 hydrogens on the phenyl ring.

    • A singlet at ~7.3 ppm due to d-chloroform.

    • A sharp singlet at ~4.6 ppm, which is probably an impurity. This peak could correspond to the  target methoxy peak, but the integration is off by 2 hydrogens.

    • A multiplet at ~2.5 ppm that could potentially represent the -CH2 of the ethyl group. The more complex splitting pattern could be due to the presence of the alkene.

    • A triplet at ~1.0 ppm representing the CH3 of the ethyl group.

    • A broad peak at ~2.0 ppm due to water.

    • Solvent peaks due to ethyl acetate (a singlet at ~ 2.05ppm, a quartet at ~4.12ppm, and a triplet at ~1.26ppm).

    • Peaks deduced to be impurities.

It seems that the product did not form. The spectrum looks more like that of the s.m.

Conclusion:

The desired product was not produced as indicated by TLC and NMR analysis.






Attached Files