Title: Methylcobaloxime: A Model for Vitamin B12
Aim: To create methyl(pyridine)cobaloxime(III) as a model for methylCobalamin, the methylated form of vitamin B12,
Introduction:
Using Co(I)Cl2 as a starting material, it is possible to produce a cyano(pyridine)cobaloxime like compound, Chloro(pyridine)cobaloxime whose chemistry can be investigated via means of methylation, as such using the Chloro(pyridine)cobaloxime as a model for replicating chemistry already known of cyano(pyridine)cobaloxime. These various Models for Vitamin B12 (right) all pivot on the known data that the octahedral cobalt central atom and it’s cyanide ligand are responsible for such methylation.
Cyano(pyridine)cobaloxime (left) was one of the first stable complexes …show more content…
Chloro(pyridine)cobaloxime – Molar Mass = 403.71gmol-1
Theoretical Yield = 21.8558665mmol
21.8558665mmol*403.71gmol-1 = 8.823431865g
[2.671g / 8.823431865g]*[100/1] = 30.271668%
Percentage Yield ≈ 30%
B) Chloro(pyridine)cobaloxime + MeI Methyl(pyridine)cobaloxime
Ratio of reactants to products: …show more content…
As is expected of a methyl-ethene system, the peak appears with a chemical shift of between 2.2 and 2.7ppm. This is due to the deshielding nature of the pi-system adjacent to the methyl groups, the deshielding character shifts the methyl peak downfield of the 1ppm region, the expected chemical shift of a methyl group. As the methyl group contains 3 equivalent hydrogens, each hydrogen within a methyl system will feel the effective magnetic field equally and they will all appear within the one singlet. This is due to the rapid rotation about the carbon-substituent bond, this rapid rotation it outside the resolution of the NMR and so fast that all Hydrogens are equivalent. Considering the Cobalt complex as a whole, it can be seen to be a highly symmetrical molecule. Within the plane of the two Dimethylglyoxime ligands, the two ligands can be divided into four quarter segments all symmetrical to one another, each segment would consist of an oh group, a nitrogen, an R-CH3 and half of the Pi-Bond. The Chlorine is the same from all sides, and the rapid rotation of the pyridine group about the cobalt metal centre is so fast that all Methyl Protons feel it equally and as such are equivalent, for this reason the four methyl groups appear as one
Aim: To create methyl(pyridine)cobaloxime(III) as a model for methylCobalamin, the methylated form of vitamin B12,
Introduction:
Using Co(I)Cl2 as a starting material, it is possible to produce a cyano(pyridine)cobaloxime like compound, Chloro(pyridine)cobaloxime whose chemistry can be investigated via means of methylation, as such using the Chloro(pyridine)cobaloxime as a model for replicating chemistry already known of cyano(pyridine)cobaloxime. These various Models for Vitamin B12 (right) all pivot on the known data that the octahedral cobalt central atom and it’s cyanide ligand are responsible for such methylation.
Cyano(pyridine)cobaloxime (left) was one of the first stable complexes …show more content…
Chloro(pyridine)cobaloxime – Molar Mass = 403.71gmol-1
Theoretical Yield = 21.8558665mmol
21.8558665mmol*403.71gmol-1 = 8.823431865g
[2.671g / 8.823431865g]*[100/1] = 30.271668%
Percentage Yield ≈ 30%
B) Chloro(pyridine)cobaloxime + MeI Methyl(pyridine)cobaloxime
Ratio of reactants to products: …show more content…
As is expected of a methyl-ethene system, the peak appears with a chemical shift of between 2.2 and 2.7ppm. This is due to the deshielding nature of the pi-system adjacent to the methyl groups, the deshielding character shifts the methyl peak downfield of the 1ppm region, the expected chemical shift of a methyl group. As the methyl group contains 3 equivalent hydrogens, each hydrogen within a methyl system will feel the effective magnetic field equally and they will all appear within the one singlet. This is due to the rapid rotation about the carbon-substituent bond, this rapid rotation it outside the resolution of the NMR and so fast that all Hydrogens are equivalent. Considering the Cobalt complex as a whole, it can be seen to be a highly symmetrical molecule. Within the plane of the two Dimethylglyoxime ligands, the two ligands can be divided into four quarter segments all symmetrical to one another, each segment would consist of an oh group, a nitrogen, an R-CH3 and half of the Pi-Bond. The Chlorine is the same from all sides, and the rapid rotation of the pyridine group about the cobalt metal centre is so fast that all Methyl Protons feel it equally and as such are equivalent, for this reason the four methyl groups appear as one