Q.What type of bonding helps in stabilising the $\alpha$-helix structure of proteins?

Concept and Intuition

Intramolecular hydrogen bonding forms a ring-like structure (often a favourable 5- or 6-membered ring) within a single molecule, and requires the donor and acceptor groups to be geometrically close — this is why it is common in ortho-substituted aromatic compounds. When the same functional groups are placed para to each other, the distance is too large for the H-bond geometry to form intramolecularly, so such molecules instead hydrogen-bond between molecules (intermolecular), which also explains their higher melting/boiling points and solubility behaviour compared to their ortho isomers.

Step-by-Step Solution

1. Salicylic acid (2-hydroxybenzoic acid): –OH and –COOH are ortho → forms a stable intramolecular H-bond (chelation).
2. Salicylaldehyde (2-hydroxybenzaldehyde): –OH and –CHO are ortho → forms an intramolecular H-bond.
3. Catechol (1,2-dihydroxybenzene): the two –OH groups are ortho → forms an intramolecular H-bond. …

Concept and Intuition

A protein's structure is described at several levels. The primary structure is simply the sequence of amino acids joined by covalent peptide bonds. The secondary structure describes local, regular folding patterns of this backbone — the α-helix and β-pleated sheet — and these patterns exist because of a repeating, geometrically regular pattern of hydrogen bonds between the backbone carbonyl oxygen (C=O) of one residue and the backbone amide hydrogen (N–H) of another residue further along the chain. This hydrogen-bonding pattern is the defining feature of secondary structure. In addition, weak van der Waals attractions between closely packed atoms of adjacent turns/strands contribute a secondary, reinforcing stabilisation to the folded shape.

Covalent bonds (other than the backbone peptide bonds themselves) and disulphide linkages (covalent S–S bridges between cysteine residues) come into play at the tertiary structure level, where they help lock the overall 3-D folded shape of the whole polypeptide, and at the quaternary level for holding multiple subunits together. They are not what defines or stabilises the secondary (helix/sheet) structure.

Step-by-Step Solution

1. Recall the four options: A. Hydrogen bonds, B. Covalent bonds, C. Disulphide linkages, D. Van der Waals forces.
2. Secondary structure = regular backbone folding (helix/sheet), stabilised primarily by hydrogen bonding between backbone C=O and N–H groups → A is relevant. …

Concept and Intuition

Intramolecular hydrogen bonding requires a donor (like -OH) and an acceptor (like the oxygen of a nearby -NO$_2$ group) positioned close together within the same molecule, typically forming a favourable 5- or 6-membered ring. HF and CH$_3$OH are small molecules with only one hydrogen-bonding group each, so they can only bond between molecules (intermolecular). o-Cresol has a methyl group ortho to -OH, but methyl has no lone pair/electronegative atom to accept a hydrogen bond, so it too only forms intermolecular H-bonds.

Step-by-Step Solution

1. HF: single -F per molecule, hydrogen bonds only intermolecularly (chains of HF).
2. CH$_3$OH: single -OH per molecule, hydrogen bonds only intermolecularly. …

Concept and Intuition

Nitrogen is electronegative enough, and N–H bonds polar enough, that ammonia molecules do form hydrogen bonds with each other (N–H···N), though these are weaker than the O–H···O bonds in water because nitrogen is less electronegative than oxygen and has only one lone pair available per molecule for accepting a hydrogen bond (compared to water's two). This hydrogen bonding is exactly why ammonia's melting point (-77.7°C) and boiling point (-33.3°C) are markedly higher than the simple trend set by the heavier, non-hydrogen-bonded hydrides of the same group ($PH_3$, $AsH_3$, $SbH_3$) would predict.

Step-by-Step Solution

1. Assertion: "Ammonia cannot exhibit hydrogen bonding in solid and liquid states" — this is factually wrong; ammonia does hydrogen bond (N–H···N) in both its solid and liquid forms. …

Concept and Intuition

Hydrogen bonding is strong specifically when hydrogen is attached to a small, highly electronegative atom — classically F, O, or N. The electronegativity pulls electron density away from hydrogen (making it highly positively polarized), and the small size of the electronegative atom allows a close approach with the lone pair of a neighbouring molecule, both of which strengthen the bond. Among the options here — Cl, O, I, S — oxygen is both more electronegative and smaller than sulfur, chlorine, and iodine, so water's O–H bonds hydrogen-bond far more strongly than the S–H, Cl–H, or I–H bonds in the others. This is the reason water has an unusually high boiling point, viscosity, and surface tension for such a light molecule.

Step-by-Step Solution

1. Compare electronegativities: O (3.44) > Cl (3.16) > > S (2.58) ≈ I is even less relevant since I–H barely hydrogen bonds at all (I is neither small nor very electronegative).
2. Compare atomic sizes: O is much smaller than S, Cl, and I, allowing closer, stronger hydrogen bonds. …

Concept and Intuition

Volatility of a liquid is inversely related to the strength of intermolecular forces holding its molecules together — stronger intermolecular attraction means molecules need more energy to escape into the vapour phase, so the substance is less volatile (higher boiling point). Sulphuric acid, $H_2SO_4$ (structurally $(HO)_2SO_2$), has two $-OH$ groups per molecule capable of donating and accepting hydrogen bonds, similar to (but even more extensive than) water's hydrogen-bonding network.

Step-by-Step Solution

1. Examine the structure of $H_2SO_4$: it is a tetrahedral sulphur centre with two $=O$ groups and two $-OH$ groups.
2. Each $-OH$ group can hydrogen bond to neighbouring $H_2SO_4$ molecules (donating through O–H and accepting through the S=O oxygens), creating an extensive three-dimensional hydrogen-bonded network in the liquid.
3. This strong, extensive hydrogen bonding requires significantly more thermal energy to break apart the liquid structure and release individual molecules into the vapour phase.
4. As a direct consequence, $H_2SO_4$ has an unusually high boiling point (~337 °C) and very low volatility for a molecule of its size compared to other common mineral acids. …

Concept and Intuition

Hydrogen bonding depends on electronegativity of the atom bonded to H and on how many effective H-bonds the molecule forms. Water has two O–H donors and two lone pairs, building an extensive network. HF forms a very strong individual H-bond but is limited to one H per molecule (zig-zag chains). H2O2 H-bonds via two O–H groups but less effectively than water overall. In H2S, sulphur's low electronegativity makes H-bonding negligible.

Step-by-Step Solution

1. H2S: sulphur is only weakly electronegative, so essentially no hydrogen bonding — placed last.
2. H2O2 vs H2S: H2O2 clearly H-bonds more than H2S, so H2O2 > H2S.
3. HF: strong single H-bond but one H per molecule — ranks above H2O2.
4. H2O: most extensive H-bond network — ranks first.
5. Overall: $H_2O > HF > H_2O_2 > H_2S$, matching option (D).

Common Mistakes …

Concept and Intuition

Vaporisation only requires overcoming intermolecular forces (covalent and coordinate bonds are intramolecular and stay intact). Methanol has an O–H bond, so hydrogen bonding — much stronger than ordinary dipole-dipole attraction — is the dominant intermolecular force determining its boiling point.

Step-by-Step Solution

1. Identify bond types present in liquid methanol: intramolecular C–H/O–H covalent bonds (not broken on vaporisation) and intermolecular hydrogen bonds + weaker dipole-dipole/dispersion forces.
2. The dominant (strongest) intermolecular attraction in an alcohol is hydrogen bonding, due to the highly polar O–H bond and small H atom. …