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Aramid fibers are strong synthetic fibers. Kevlar is called a para-aramid, and it's outstandingly strong. Why does the para-position give Kevlar its strength?

When the Kevlar monomers are in the para-position, will this make the chains closer (shorten the length of the bond), and make strong interactions between the chains?

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    $\begingroup$ This question is a bit unclear to me. What exactly would you call an "advantage"? Which aromatic molecule are you talking about? $\endgroup$ – M.A.R. Feb 26 '16 at 14:40
  • $\begingroup$ I read that their is para, ortho, and meta position in aromatic ring and I read one of good properties of aramid relates to it's para position of molecules in aromatic ring, so does the para position relates to good properties? Is it due to cloaing the chains togeter? Thanks $\endgroup$ – studentrr Feb 26 '16 at 14:45
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    $\begingroup$ This is a decent question. Had you chosen appropriate tags and given us enough context, this never would've been closed. Voted to reopen and +1. $\endgroup$ – M.A.R. Feb 26 '16 at 19:58
  • $\begingroup$ "...will this make the chains closer (shorten the length of the bond)..." Which bond do you think will be shortened? Do you mean the length of a chemical bond or the interactions between individual chains? $\endgroup$ – jerepierre Feb 26 '16 at 20:07
  • $\begingroup$ I ment if the length of the bond shorter, will it be strong? and will the para structure provide strength to material by closing the chains together, because the rings connected to chains by para position? Thanks $\endgroup$ – studentrr Feb 26 '16 at 21:05
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para isomers fits into the crystal lattice more perfectly than ortho and meta due to steric reasons and hence generally they have more melting point.

Also in certain aromatic compound ortho isomers form intra molecular hydrogen bonding which reduces its boiling point (generally) compared to ortho and para where in these isomers would be forming inter molecular hydrogen bonding.

In the case of para aramid you can see that the two peptide linkage formed are with two other para amid molecules (always).Had it been ortho aramid then the two bonds will be more likely to get bonded to a single other ortho amid molecule.And the polymeristaion ends there (at the most polymer of para aramid will be having less length and has unsymmetrical structure).But in para amid it can polymerise to any lengh.

These are the two possible polymers for ortho.

the molecule dimerises in case of ortho aramid

this the maximum polymerization that can happen (if possible)

there is no difference in the bond strength formed between two para amid molecule compared to two otho amid molecule after all both are peptide linkages only (i don't know if the ortho aramid exists or not but what will happen is this for sure).(sorry the figure looks completely distorted.I am not really good with the paint application)

finally its the capability to polymerize to a great extent that makes it strong.

Its also worth mentioning that the functional (or substituent) groups at C1 and C4 (C1& C4-carbon numbering in benzene) can't bond themselves,as the bonds in any polyatomic group,can't bend and elongate as they want (after all the distance between C1 and C4 even if we consider diagonally,is itself quite long.And secondly thinking like this insane,say a bond bending in way they want.It will never happen with para compounds.The least may be meta compounds can.)

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