Question

If you had 200 mg of 2,5-bis(bromomethyl)-1-methoxy-4-(2-ethylhexyloxy)benzene (monomer 2) How much potassium tert-butoxide do you need (from a 2M solution of THF)? This is from a Lab Report for the synthesis of MEH-PPV. I guess its asking how much do you need for all of the reactant to be used up?

https://www.chm.uri.edu/mlevine/chm292-2014/polymer.pdf (the lab procedure)

Answer 1

The mass of 2,5-bis(bromomethyl)-1-methoxy-4-(2-ethylhexyloxy)benzene = 200 mg, i.e. 0.2 g

(Note: 1 g = 1000 mg)

The molar mass of 2,5-bis(bromomethyl)-1-methoxy-4-(2-ethylhexyloxy)benzene = 422 g/mol

Therefore, the no. of moles of 2,5-bis(bromomethyl)-1-methoxy-4-(2-ethylhexyloxy)benzene

= 0.2/422 = 4.74*10^-4 mol

Since you have to take 4 equivalents of potassium tert-butoxide, mentioned in the procedure,

The required amount = 4*4.74*10^-4 mol = 1.896 mmol

(Note: 1 mol = 1000 mmol)

= 1.896 mmol / (2 mmol/mL)

(Note: 1 M = 1 mmol/mL)

= 0.948 mL