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I was looking into how the solubility of amides compares to other compounds e.g. acids and amines, and came across these two statements on the wikipedia page for amides:

https://en.wikipedia.org/wiki/Amide#Hydrogen_bonding_and_solubility

"The presence of a C=O dipole and, to a lesser extent a N–C dipole, allows amides to act as H-bond acceptors. In primary and secondary amides, the presence of N–H dipoles allows amides to function as H-bond donors as well."

"Typically amides are less soluble than comparable amines and carboxylic acids since these compounds can both donate and accept hydrogen bonds."

I understand that amides can act as both H bond donors and acceptors, so I'm wondering how the second statement can make sense?

Additionally, would it be correct to say amides have greater solubility in water since they can form more H bonds in total compared to amines and acids?

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  • $\begingroup$ Amides have electron delocalisation, so the net charge on an individual atom is less $\endgroup$ – Safdar Faisal Jul 10 '20 at 6:36
  • $\begingroup$ Amines can be better soluble because of their basicity - acid-base reactions are primary enhancer of solubility. $\endgroup$ – Mithoron Jul 10 '20 at 14:47
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Urea ($\ce{H2N-CO-NH2}$) is the quintessential amide. It is extremely soluble in water, and you could say it is a hydrogen donor and acceptor in water.

However, if you replace one of the $\ce{-NH2}$ groups with an alkane or other organic group, solubility is reduced, depending on the size of the group.

Amides generally have higher boiling points and melting points than related alcohols and acids - because of their unique ability to be internal hydrogen donors and acceptors. This donor-acceptor capability makes their solid state more stable and less able to be solvated to a greater degree in water. You could think of the internal bonds being like a dimerization, where the organic chain effect is doubled, while the double $\ce{-CO-NH2}$ interaction gains little benefit from solution in water.

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