Why are hydrogen bonds in an antiparallel beta sheet stronger than those in parallel beta sheets?

Question

Beta sheets are illustrated as such in most diagrams, where:

1. In an antiparallel β-sheet, the polypeptide strands are arranged such that a $\ce{C=O}$ and an $\ce{NH}$ from adjacent strands face each other, and the $\ce{H}$ forms an H-bond with the $\ce{O}$, with $\ce{C=O**HN}$ all lying on the same plane.
2. In a parallel β-sheet, the strands are literally parallel, and so the H-bond formed has to form a sort of angle, where the $\ce{C=O}$ and $\ce{NH}$ no longer lie on the same plane (when viewed from the perspective of the diagram).

Most explanations explain the stronger bond strength by stating the obvious from the diagram: that the "hydrogen bonds are aligned directly opposite each other, making for stronger and more stable bonds." (taken from chem libretexts, https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Protein_Structure/Secondary_Structure%3A_-Pleated_Sheet)

This seems intuitive, but my issue arises when considering some orbital geometry: in a peptide bond, the $\ce{C}$, $\ce{N}$ and $\ce{O}$ have orbitals with ${sp^2}$ hybridization, meaning that bond pairs/lone pairs are organized in a trigonal planar orientation around each of those atoms. As a consequence of this, one can also observe that all the atoms illustrated in the diagram should lie on the same plane, making things rather easy to visualize.

However, if one considers that the 2 lone pairs and one bond pair in the $\ce{O}$ atoms are orientated in a trigonal planar fashion, then shouldn't the ideal, strongest, H-bond is orientated at an angle (specifically $\pu{{60}^\circ}$) relative to the $\ce{C=O}$ bond? Now, if I observe the diagram again, it seems this $\pu{{60}^\circ}$ H-bond angle is better fulfilled in the parallel strand than in the antiparallel one. So shouldn't the H-bond in the parallel strand, theoretically, be stronger?

A possible explanation would be that the bond length in parallel strands would be longer, thereby weakening it. But that's not how most textbooks explain it, so what am I missing out here?

Answer 1

Why are hydrogen bonds in an antiparallel beta sheet stronger than those in parallel beta sheets?

They are probably not. The difference is small, and depends on sequence context. Also, the diagrams do not reflect the typical conformation of the backbone in beta sheets. To complicate matters, most sheets are mixed rather than purely parallel or anti-parallel, and many are irregular (different length of strands, bulges, one strand part of two sheets etc).

"hydrogen bonds are aligned directly opposite each other, making for stronger and more stable bonds."

No, the geometry around the hydrogen bond is fairly similar. Here is a depiction of structures from an ab initio study showing the geometry:

[OP ..] shouldn't the ideal, strongest, H-bond is orientated at an angle (specifically 60∘) relative to the C=O bond?

Based on analyzing known structures with a single hydrogen bond for a given acceptor, the hydrogen lies in the plane of the carbonyl group but has a fairly smooth distribution with many examples having a 180 degree angle (carbonyl carbon, carbonyl oxygen, hydrogen).

A possible explanation would be that the bond length in parallel strands would be longer, thereby weakening it. But that's not how most textbooks explain it, so what am I missing out here?

The ab initio study talks about the primary effect (having a hydrogen bond), secondary effects (repulsion between partial charges of oxygen atoms, repulsion between partial charge of hydrogen atoms, C-H...O hydrogen bond in the parallel sheet) and tertiary effects (induced dipoles due to formation of hydrogen bonds). You have to consider all of these; as the differences in hydrogen bonding geometry are subtle, the secondary and tertiary effects will become significant.