- Prepare solutions to run new PVOH based reactions
- Perform all calculations necessary for future PVOH reactions
- First PVOH reaction will be the synthesis of polyvinyl acetate
- Second PVOH reaction will be the attachment of 5-(4,6-Dichlorotriazinyl)Aminofluorescein which will be referred to as 5-DTAF
General Formula for reactions
Polyvinyl alcohol + [HAc]→Polyvinyl acetate
General Protocol for Polyvinyl acetate synthesis
- For a 1:1 ratio of polyvinyl alcohol OH groups to polyvinyl acetate dissolve ~1g PVOH inH2O
- Add ~1.36g HAc. Add ~2 drops of HCl to reaction mixture to catalyze reaction.
- Heat the reaction mixture to 40°C.
- Allow the reaction to stir for 1 hour
- Add sodium bicarbonate solution to the reaction mixture(titrate the reaction)
- Rotavap the remaining reaction mixture
- Overall, want to determine if have 1g of material how many moles of -OH groups. Once that is determined, will want to determine how many moles of -OH was reacted after the reaction.
Number of moles of OH groups in 1g PVOH:
MW of repeating polyvinyl alcohol unit [C2H3OH]: 44.06g/mol
Polyvinyl alcohol repeating unit picture
1gPVOH(1mol PVOH unit/44.06g) x (1mol OH/1mol PVOH unit)= 0.022696 mol OH/g
Dye attachment calculations:
- Want to react .01% of the OH groups with the dye
MW 5-DTAF dye: 495.28g/mol
(0.0227mol OH/g) x (.0001)= 2.27x10-6 moles dye
2.27x10-6 moles dye x(495.28g dye/1mol dye)=0.001124g dye
Polyvinyl acetate synthesis calculations:
- For all reaction mixtures, the total volume of the reaction mixtures will be 7mL
- HAc will be the abbreviation for acetic acid
Density of HAc= 1.05g/mL
(A.) 1:1 ratio
1g PVOH x (1mol PVOH/44g PVOH) x (1mol HAc/1mol PVOH) x (60g HAc/1mol HAc)= 1.36g HAc
1.36g HAc x (1mL/1.05g HAc)= 1.295mL HAc
(B.) 20% ratio
1gPVOH x (1mol PVOH/44g PVOH) x (0.2mol HAc/1mol PVOH) x (60g HAc/1mol HAc)= 0.273g HAc
0.273g HAc x (1mL/1.05g HAc)=0.26mL HAc
(C.) 40% ratio
1gPVOH x (1mol PVOH/44g PVOH) x (0.4mol HAc/1mol PVOH) x (60g HAc/1mol HAc)= 0.545g HAc
0.5195g HAc x (1mL/1.05g)= 0.5195mL HAc
(D.) 60% ratio
1gPVOH x (1mol PVOH/44g PVOH) x (0.6mol HAc/1mol PVOH) x (60g HAc/1mol HAc)= 0.818g HAc
0.818g HAc x (1mL HAc/ 1.05g)= 0.779mL HAc
(A.) Sodium bicarbonate solution preparation
0.2L x (1mol/1L) x (84.01g/1mol)= 16.802g sodium bicarbonate.
- Note: The actual amount of sodium bicarbonate used was 16.8046g. Also, it is not necessary to prepare this solution in a volumetric flask. Will standardize the solution to know the actual concentration.
(B.)Potassium hydrogen phthalate(PHP) needed
MW PHP= 204.24g/mol
1M PHP= x/0.1L
x=0.1mol x 204.24g/mol
x= 20.424g PHP needed
(A.) Sodium bicarbonate standard
- ~16.802g sodium bicarbonate was measured and placed in a 200mL plastic container.
- 200mL of deionized H2O was measured using a graduated cylinder and was poured into the plastic container.
- The plastic container was sealed and the bottle was swirled vigorously until all sodium bicarbonate was dissolved.
(B.)PHP solution preparation
- 20.424g PHP was measured and placed in a 120°C oven for 1 hour to dry
- Dried PHP was placed in a 200mL volumetric flask with a magnetic stir bar and 100mL of deionized H2O was added to the flask
- A rubber stopper was placed on the top of the volumetric flask and the mixture was allowed to stir for 72 hours.