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:
% 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
Column Chromatography:
Four fractions were collected through column chromatography, which is consistent with the multiple spots shown on TLC.
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.
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 3. First NMR spectrum of purified (Z)-2-(4-chlorophenyl)-3-hydroxypent-2-enenitrile (CDCl3 solvent).
Analysis of the 1H NMR
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
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.
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.
% yield = (actual yield/ theoretical yield)100% = (0.204 g/ 0.222 g)100% = 92%
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.
-
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
Conclusion:
The desired product was not produced as indicated by TLC and NMR analysis.
Synthesis of compound 3 using reflux was not successful (previous entry). The procedure was adjusted appropriately.
| 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:
Results:
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 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.
-
Conclusion:
The desired product was not produced as indicated by TLC and NMR analysis.