The Student Room Group

List the functional groups in this polymer: -CO(CH2)4CONHC(CH3)2CH2O-

Okay so I thought it was amide and ketone.

But apparently not ketone?

Can someone expalin what it is please.

Also could we classify alcohol as a functional group and would we need to be specific i.e. tertiary alcohol etc.
Reply 1
The C on the left has four bonds - one to the (CH2)4 group, two to the O in the C=O, where does the bond dangling off the left hand of the polymer go to?

What kind of bond it that?
Original post by Pigster
The C on the left has four bonds - one to the (CH2)4 group, two to the O in the C=O, where does the bond dangling off the left hand of the polymer go to?

What kind of bond it that?


From what you said ester, but I don't get how you deduced that.
Reply 3
The section of polymer you gave us is itself made of two individual monomers. Can you identify those?
Original post by Pigster
The section of polymer you gave us is itself made of two individual monomers. Can you identify those?


carboxlyic acid and primary alchool?
Reply 5
Perhaps I should have been clearer.

The section of polymer is made up of two different monomers which have bonded together to make the repeat unit you gave us.

Can you identify (name) the two monomers?
Original post by Pigster
Perhaps I should have been clearer.

The section of polymer is made up of two different monomers which have bonded together to make the repeat unit you gave us.

Can you identify (name) the two monomers?


hexanoic acid and 2,2 methylaminopropanol?
Reply 7
Hexanedioic acid and 2-amino-2-methylpropanol.

The amide linkage is shown in the middle, the ester is split.

Perhaps two repeat units would show it more clearly:

-CO(CH2)4CONHC(CH3)2CH2OCO(CH2)4CONHC(CH3)2CH2O-
Original post by Pigster
Hexanedioic acid and 2-amino-2-methylpropanol.

The amide linkage is shown in the middle, the ester is split.

Perhaps two repeat units would show it more clearly:

-CO(CH2)4CONHC(CH3)2CH2OCO(CH2)4CONHC(CH3)2CH2O-


All I see is a ketone, surely for an ester there should be O-C=0
Reply 9
CH2OCO(
is right there in the middle

CH2 - O - C=O - (CH2)4
Original post by Pigster
CH2OCO(
is right there in the middle

CH2 - O - C=O - (CH2)4


thank you, missed that.

I now understand.

Can I jjust ask you, how did you know to draw out the two monomers??
It isn't a thing to 'know'.

It was just a train of conversation to try to get you to recognise what was there: a di-acid and an amino-alcohol, hence why you'd get amide and an ester linkages.

The best thing to do in these situations is to draw out the full displayed formula, ensuring that each C has four bonds.
Original post by Mihael_Keehl
thank you, missed that.

I now understand.

Can I jjust ask you, how did you know to draw out the two monomers??


Maybe this interactive on polyester formation will help ...
Original post by Pigster
It isn't a thing to 'know'.

It was just a train of conversation to try to get you to recognise what was there: a di-acid and an amino-alcohol, hence why you'd get amide and an ester linkages.

The best thing to do in these situations is to draw out the full displayed formula, ensuring that each C has four bonds.


Oh of course! It said polymer, I was reading monomer :/

Thank you.

Original post by charco
Maybe this interactive on polyester formation will help ...



Yes thank you, nice visual presentation :smile:

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