Steigt die Schmelztemperatur je nach form des Polyamid?
Ich muss bestimmen, welche der beiden kunststoffe die höchste Schmelztemperatur hat. Das obere hat die höchste Schmelztemperatur, da es länger ist. Aber stimmt es, dass wenn alle Elemente so wie bei dem oberen Polyamid nach oben zeigen, dass die Schmelztemperatur steigt?
None of these two polyamides is branched, and you need correct side groups. A simple carbonyl group is not enough.
And whether it shows up or down, it doesn’t matter. Since they are linked to the rest of the chain by single bonds, they can be simply rotated. So you could also let the carbonyl group show down at the upper molecule (or at the lower to the top). This would make no difference, it remains the same molecule with the same properties.
What this is about is simply the length of the chain. Since both polymers have equal numbers of repeating units (everything n pieces), a repeating unit at the upper but is significantly longer than at the lower, the upper also has the higher melting temperature. The strength of the intermolecular forces (here vdW forces) increases with the chain length.
Well, I can understand that. However, it is also possible that the lower polymer has the higher melting point because it has a higher proportion of the more polar amide infections in the total molecule per monomer? As is known, the polar-polar interaction prevails in comparison with the V.d.W. forces.
Yeah, get the thoughts. But if we say that both have n=10, then the number of amide functions is also the same (i.e. 20) with both this should also be similar to the contribution of the dipole WW. In addition, however, you still have the increased vdW WW in the upper polymer.
Yes, the stronger the branching is, the better the molecule can arrange and thus the melting temperature rises.
Both molecules are equally branched (4 branches) the branches show only in different correct. At the top, the branches are only in one direction, so it can be more easily arranged, right?