Hello dear, once finish the reaction, you can make the compararison of the spectra of the copolymer with those of the monomers. You will find that the signal atribbuted to -C=C- that appear in the insatured monomer spectra has disappeared. The copolymer has to show the signal assigned to ester groups (peak aroun 1730 1/cm) and the symetric and antysimetric stretching of CH3 and CH2 around 1920 and 1840 1/cm). the signal around 1030 cm of C-O stretching of ester groups also must appear. Furthermore the signals of C=C stretching of aromatic, must appear at 3030 or 3000 1/cm, the same manner the signal corresponding to bending of aromatic have to appear arong 1456, 1590 1/cm this signal have low intensity.
I've done an analysis of the copolymer and the monomer , but the results only reveal FTIR peak of PMMA at ~ 1730 1 / cm , for styrene peak at ~ 1600 1 / cm (aromatic ring) did not appear. signal at 3030 or 3000 does not appear but the signals that appear are 3437 , 2993 , and 2948 1/cm.
(b) you must be careful if you are making the copolymer to not mis-identify a physical blend of some homo polystyrene (pSTY) and homo polymethylmethacrylate (pMMA) with the signature of a true copolymer of STY/MMA The FTIR experiment will not easily distinguish a blend from a true copolymer. For that you need some differential solubility or some chromatography.
thanks for the reply Mr. Mike Bender , greatly helped my research
i have done retest of FTIR my copolymer sample, it shows a peak at ~1730 as C=O stretc carboxyl (methyl methacrylate) and ~1600 as C=C stretc aromatic (styrene)
the intensity of styrene's peak is too low, but FTIR's machine shows poly(methyl methacrylate:styrene) at my copolymer, can it be considered of success?
my composition (PMMA = 25%wt, MMA = 52.5%wt and Styrene =22.5%wt)