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Monomer question (1 Viewer)


Well-Known Member
Feb 25, 2015
Uni Grad

Since there is a C-O-C bond its most likely a condensation polymer since most of ur addition polymers are C=C i.e. ethene based.

Essentially you need separate the chain u are given into the repeating units shown in the red lines in the pic attached and then break apart the repeating unit into two pieces shown in the blue line. You break apart the two pieces in the one repeating unit by essentially looking for a strange bond like C-O-C or sometimes C-N-C. note: it doesn't matter if u break it apart on one side where u give one molecule the C-O and the other just the C

From here redraw the two separate sides between the red and blue lines, then simplify add on atoms to form an OH (you see now that on one side we have have an O so u just add one H, and the other side with just the C we need to add a full OH, as can be seen one OH and H added is basically the water molecule that was lost during the condensation reaction being added back in). These added in oxygens and hydrogens are shown in the pink.

You can use this same type of approach for all these types of condensation Qs and addition Qs. So summarising the steps
1. Find the repeating molecule
2. Break repeating molecule into two molecules that repeat for condensation (addition is just like one thing repeating over and over so don't need to do this bit)
3. Draw out the molecules from part 2
4. For condensation add in oxygen and hydrogen atoms to form OHs on each end and for addition just close two adjacent bonds to make a double bond
Last edited:


New Member
Oct 11, 2018
Polyester is a condensation polymer. Ester reactions are in "The Acidic Environment" section 9.3.5 (page 56 of the syllabus). The two monomers can be figured out by breaking the C-O-C bond in the C-O-C=O and inserting a water molecule. The C in the C=O gets an OH added to it, turning it into a COOH. The C-O gets a H, turning it into a C-O-H. The two monomers are HOOC-CH2-COOH (propanedioic acid) and HOCH2CH2CH2OH (1,3-propanediol).

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