More errors:
peptide = amide. Amine links (secondary/tertiary amines) are quite different as they are just R-N-R linkages.
Dehydration reaction is the release of a water from a molecule. The process of peptide formation involves fusion of two molecules into one with the ejection of a water. This is not dehydration. (It is just condensation.)
On hydrogen bonding, the role of hydrogen is equally as important as the role of N/O/F. The combination of a large electronegativity (N/O/F) and a large electropositivity (H and other alkali metals) gives rise to very 'efficient' charge separation across the X-H bond (where X=N/O/F), i.e. a larger-than-usual dipole is generated. The bonding profile is *almost* ionic. With such high charge separation, a X-H ... X-H ... X-H lattice structure is especially stable, and we call this hydrogen bonding.
In many cases, we only require X-H ... X-R to classify as 'hydrogen bonding'. The large electronegativity of X will almost always ensure it has a large and negative partial charge. The positive partial charge however is generally delocalized over the rest of the molecule. Here, the importance of H is signified, that if X is primarily taking electron density from H (which has no neighbouring atoms to help delocalize the loss of electron density), it must carry a large and positive charge, facilitating the strong dipole-dipole interaction.
On dispersion forces between R groups that make up the tertiary structure, in many cases it is the hydrophobicity of these groups that play an important part. i.e. These R groups do not have large affinities to each other, but because water does not liking anything non-polar, they are forced together by a strong solvent-induced force. A large part of protein folding comes from the hydrophobic effect, which is not completely understood by the scientific community. I have been working in a research group that has been looking into a new force between waters that may explain this interaction.
On denaturation, proteins aren't THAT sensitive to temperature. Covalent cleavage isn't too energy costly, it wouldn't require thousands of degree (that's to melt/boil salts). With the correct enzymes or pH, the amide bond can be quite easily broken. The secondary and tertiary structure aren't too sensitive to temperature either. From their optimal temperature, it takes quite a few degrees for them to denature. With understanding of the Boltzmann distribution (i.e. kinetic energy is a distribution), it won't be a sharp change either. You are correct in saying that in almost all cases the primary structure won't be affected (but your reason isn't entirely correct).
Not all catalysts weaken the bond. The classification of 'catalyst' is fairly loose, some catalysts may simply be support for an active species, some catalysts do not directly participate in the reaction but instead activates the reactant, some catalysts covalently bonds to the substrate in the process, and many do as you describe, which stabilizes and weaken the bonds in the substrate for other molecules to 'attack'.
That's really all. A lot of these you won't need to know (and most probably won't learn this from a teacher either), since many are advanced concepts. However, it is good to keep in mind that generalizations should be avoided in science. Don't make the statements too general, even when explaining things!
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