Question

The purpose of this experiment is to synthesize a polymer using a Gilch polymerization of 2,5- bis(bromomethyl) - 1 -methoxy-4-(2-ethylhexyloxy) benzene, to investigate the polymer's photophysical properties, and to understand the mechanism of the polymerization reaction. Introduction Poly(p-phenylene vinylenes), or PPVS, are highly luminescent polymers that are used for a variety of commercial applications. Their general structure is shown in Figure . PPVS can be synthesized via a number of routes, including a Gilch polymerization reaction, which will be used in this experiment. The Gilch polymerization has a number of advantages compared to other polymerization reactions, including: (a) tolerance of a wide range of substituents; (b) high atom- economy; and () relative insensitivity to air and moisture (compared to other methods that require a glovebox). The Gilch route involves the addition of a strong base such as potassium tert- butoxide to an a,a'-dihalogenated p-xylene monomer (compound 2, Scheme 1). In the first step, potassium tert-butoxide deprotonates the acidic benzylic protons to generate quinoidal intermediate 4. Dimerization of that intermediate is followed by radical polymerization to generate saturated intermediate 7. Elimination of that macromolecular intermediate through the base-catalyzed removal of HX yields the desired polymer, compound 1 (Scheme 1). R H -x -H C-H H H 3 2 R : radical dimerization resonance forms with R H R R 5 6 R H R polymerization elimination X HX R R 7 1 Scheme 1: Mechanism for the synthesis of polymer 1 starting from dihalogenated monomer 2

Answer 1

0 215-bis (bromo method )-1 methoxy-4-(2-61h Pheay boy) benzene Br 101 BY mut! 422:20 Potassium tex-butorsche In the reaction scheint step 6 elimatron step, you are used Base, I don't know to whether you using kotby or any other Base. In that elimination step if you using tatby then you have to calculate for 2- equivalences The immol was 0.946 and am kothy soktron indarily = nmol V(WIL) mot = 112.21 Starting material Amorent = 200 mg Br ) -CC944 20.946 V v=oi946 V 0.473 mL weight (mg) 200 = 0.473 moro 422-20 for this reaction we ruquins l equirlane of Base. Reaget I weight. I out mmol equrlan sm. I 200 ing | 422.20 0.473 I kotby | 0.236 in [cam in the 1.0.2 112.21 10.473 | mmol = weight (mg) mut a weight (g) 0.2 0.000472x1000=0.473 mut = 422.20 - (mol) mm molarity = mol ammol VCL) vom 2 v=volune V (m2) = mmol molarity 0.473 Kotbu inmol is 0.473 thun V(ml)=0 201236 ml We requiny 0.236 mL of kot by soletron in THE first step you have to use 0.236 unL and Step 6, elinimton you have to use 0.236 mL. According to my knowledge if we add a equinlane of base in step I leads to other undesined product[Dianinie intermediate?