Q.α-Helix is a secondary structure of proteins formed by twisting of polypeptide chain into right handed screw like structures. Which type of interactions are responsible for making the α-helix structure stable?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Hydrogen Bonding
Hydrogen Bonding: From Intuition to Precision
Imagine you're holding two magnets. If you bring the north pole of one close to the south pole of another, they snap together. Now imagine a much weaker version of that — a tiny tug, not a full lock. That's the spirit of hydrogen bonding.
In chemistry, atoms in a molecule share electrons through covalent bonds. But electrons aren't shared equally in all cases. Some atoms are greedy — they pull the shared electrons closer to themselves. Oxygen, nitrogen, and fluorine are the biggest electron-hoarders. When one of these atoms bonds with hydrogen, the hydrogen ends up with a slight positive charge (because its electron has been pulled away), and the other atom gets a slight negative charge.
Now here's the key: that slightly positive hydrogen is attracted to any nearby slightly negative atom (like oxygen, nitrogen, or fluorine) on another molecule. This attraction is a hydrogen bond.
A hydrogen bond is not a true chemical bond like a covalent or ionic bond. It's an intermolecular force — a strong dipole-dipole attraction — but weaker than covalent bonds (about 1/10th to 1/20th the strength).
The Precise Definition
A hydrogen bond is an attractive interaction between a hydrogen atom covalently bonded to a highly electronegative atom (N, O, or F) and another electronegative atom (N, O, or F) that has a lone pair of electrons.
We can write it as:
X—H⋯Y
where X and Y are N, O, or F. The dotted line (⋯) represents the hydrogen bond. X—H is the donor (the molecule that provides the hydrogen), and Y is the acceptor (the molecule that provides the lone pair).
Why Only N, O, and F?
Three things make these three elements special:
- High electronegativity — They pull electrons hard, creating a large partial positive charge on hydrogen.
- Small size — The lone pair on Y is compact, allowing the hydrogen to get very close. Closer distance means stronger attraction.
- Lone pairs — They have unshared electron pairs that can act as the acceptor.
Chlorine is electronegative, but it's too large — the hydrogen can't get close enough for a strong bond. Carbon is not electronegative enough.
What Makes Hydrogen Bonding Special?
Unlike other dipole-dipole interactions, hydrogen bonds are directional and stronger. They're about 5–30 kJ/mol, compared to 0.5–2 kJ/mol for ordinary van der Waals forces. This strength has dramatic consequences.
Real-World Consequences
Water's high boiling point — Water (H2O) boils at 100∘C, while hydrogen sulfide (H2S) boils at −60∘C. Both are similar molecules, but water forms hydrogen bonds; H2S does not (sulfur is not electronegative enough). Those bonds must be broken to boil water, requiring much more energy.
Ice floats — In liquid water, molecules jostle and form temporary hydrogen bonds. When water freezes, the molecules arrange into a hexagonal lattice held open by hydrogen bonds. This structure is less dense than liquid water — hence ice floats. Without hydrogen bonding, ice would sink, and lakes would freeze from the bottom up, killing aquatic life.
DNA double helix — The two strands of DNA are held together by hydrogen bonds between base pairs (adenine-thymine and guanine-cytosine). These bonds are strong enough to keep the strands together, but weak enough to be unzipped during replication. …
Why this formula?
Hydrogen Bonding: Why It Happens — The Reasoning, Not Just the Rule
Hydrogen bonding is not a full covalent bond — it's a special type of intermolecular attraction. To understand why it occurs, we must look at the electronic structure of the atoms involved.
1. The Core Requirement: A "Naked" Proton
A hydrogen bond forms when a hydrogen atom is covalently bonded to a highly electronegative atom (like F, O, or N). Why?
- Electronegativity difference pulls the bonding electron pair strongly toward the electronegative atom.
- The hydrogen atom is left with almost no electron cloud — it becomes a partially positive proton (δ+).
Key idea: The hydrogen is now a small, dense positive charge — it can get very close to a lone pair on another electronegative atom.
2. The Electrostatic Attraction (The "Why")
The partially positive hydrogen (δ+) is attracted to a lone pair of electrons on another electronegative atom (the acceptor).
This is electrostatic — Coulomb's law governs it:
F=4πε01⋅r2q1q2
- q1 = partial positive charge on H
- q2 = partial negative charge on lone pair
- r = distance between them
Because the hydrogen is so small, r is very small → force is strong (stronger than van der Waals, weaker than covalent).
3. Why Only F, O, N?
Not all electronegative atoms work. The atom must have:
| Property | Why it matters |
|---|---|
| High electronegativity | Pulls electron density away from H, creating δ+ |
| Small atomic size | Allows close approach of the H to the lone pair |
| At least one lone pair | Provides the negative site for attraction |
F, O, and N satisfy all three. Cl is electronegative but too large — the H cannot get close enough for a strong bond.
4. The "Formula" for Hydrogen Bond Strength
There is no single formula for hydrogen bond energy, but the strength depends on:
EH-bond∝r2δ+⋅δ−
Where:
- δ+ = partial charge on H (depends on electronegativity of donor atom)
- δ− = partial charge on acceptor lone pair
- r = distance between H and acceptor atom
Typical strengths (for context):
- Covalent bond: ~400 kJ/mol
- Hydrogen bond: 10–40 kJ/mol
- van der Waals: ~1–5 kJ/mol
5. Directionality — The "Linear" Preference
Hydrogen bonds are directional: the strongest interaction occurs when the donor H–X bond and the acceptor lone pair are collinear (180° angle).
Why? Because:
- The positive charge on H is concentrated along the bond axis …
The stability of the α-helix comes from hydrogen bonding between the backbone amide groups.
Reasoning:
- In an α-helix, the polypeptide chain coils such that the carbonyl oxygen (C=O) of one amino acid residue is positioned close to the amide hydrogen (N−H) of the residue four places further along the chain.
- A hydrogen bond forms between this C=O (acceptor) and N−H (donor), linking each turn of the helix to the next. …
The α-helix is stabilized primarily by intramolecular hydrogen bonds between the carbonyl oxygen of one amino acid and the amide hydrogen of the amino acid four residues ahead, forming a regular helical pattern.
The α-helix is one of the most elegant examples of how a long, floppy polypeptide chain can fold into a precise, repeating shape. The key question is: what holds this spiral together? It’s not covalent bonds between side chains — those come later in tertiary structure. Instead, the stability of the α-helix comes from a very specific pattern of hydrogen bonds that form within the backbone itself.
Let’s break down why hydrogen bonding is the correct answer, and why other interactions (like disulfide bonds or hydrophobic forces) are not the primary stabilizers here.
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The backbone has built-in hydrogen bond donors and acceptors. Every amino acid in a polypeptide chain has a carbonyl group (C=O) and an amide group (N−H) in its backbone. In an unfolded chain, these groups are free to hydrogen bond with water. But in the α-helix, they pair with each other instead.
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The hydrogen bonding pattern is regular and predictable. In a right-handed α-helix, the carbonyl oxygen of residue n forms a hydrogen bond with the amide hydrogen of residue n+4. This means every turn of the helix (about 3.6 amino acids) is locked in place by these bonds. The bonds run parallel to the helix axis and are nearly linear, which makes them strong.
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These are intramolecular, not intermolecular, bonds. The hydrogen bonds form within the same polypeptide chain, not between different chains. This is what distinguishes secondary structure from quaternary structure.
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Why not other interactions?
- Disulfide bonds (−S−SX−) are covalent and form only between cysteine residues — they are rare in α-helices and not required for helix formation.
- Hydrophobic interactions and ionic bonds involve side chains and are more important for tertiary folding, not the regular backbone pattern of secondary structure.
- Van der Waals forces contribute to packing but are not the primary stabilizing force. …
Concept: Secondary Structure of Proteins — The α-Helix
The α-helix is a common secondary structure in proteins, where the polypeptide chain coils into a right-handed spiral. Its stability comes from specific interactions between amino acid residues.
Method: Hydrogen Bonding Analysis in α-Helix
Why this method?
The α-helix is stabilized primarily by hydrogen bonds that form between the backbone atoms — not the side chains. This method identifies the pattern and location of these bonds.
Steps:
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Identify the backbone atoms involved
- The carbonyl oxygen (C=O) of one amino acid residue.
- The amide hydrogen (N–H) of another residue.
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Locate the bonding pattern
- In an α-helix, the hydrogen bond forms between the C=O of residue n and the N–H of residue n+4.
- This means each turn of the helix involves about 3.6 amino acid residues.
-
Check directionality
- All hydrogen bonds are oriented parallel to the helix axis. …
Here are the common mistakes students make when answering this question, along with how to avoid each.
1. Confusing the Type of Bond (Hydrogen vs. Disulfide vs. Ionic)
The Mistake: Students often write that disulfide bonds (−S−S−) or ionic bonds (salt bridges) stabilize the α-helix. This is incorrect.
Why it’s wrong:
- Disulfide bonds are covalent bonds formed between cysteine residues. They stabilize tertiary (3D folding) and quaternary structures, not the local helical twist.
- Ionic bonds occur between charged side chains (e.g., −NH3+ and −COO−). These are also part of tertiary structure stabilization.
How to Avoid:
- Memorize the "backbone rule": The α-helix is stabilized by interactions between the backbone atoms (the −NH and −CO groups of the peptide bonds), not the side chains (R-groups).
- Visualize the helix: The side chains point outward away from the helix core. They do not participate in holding the helix together.
2. Writing "Hydrogen Bonds" Without Specifying the Atoms
The Mistake: Simply writing "Hydrogen bonds" is too vague. In exams, you must specify which atoms are involved.
Why it’s wrong: There are many hydrogen bonds in proteins (e.g., between side chains and water). The examiner wants to know you understand the specific pattern in the α-helix.
How to Avoid:
- Learn the exact pattern: The hydrogen bond forms between the carbonyl oxygen (C=O) of one amino acid and the amide hydrogen (N-H) of the amino acid four residues later.
- Write it precisely: "Hydrogen bonds between the −C=O group of the nth amino acid and the −N-H group of the (n+4)th amino acid."
3. Forgetting the "Intra-chain" Nature of the Bonds
The Mistake: Students say the helix is stabilized by bonds between different polypeptide chains (inter-chain).
Why it’s wrong: The α-helix is a single-chain structure. The bonds are intra-molecular (within the same chain). Inter-chain hydrogen bonds are found in β-pleated sheets (between strands).
How to Avoid:
- Use the keyword "intra-chain" or "within the same polypeptide chain" in your answer.
- Compare with β-sheet: Remember that β-sheets can be inter-chain (between chains) or intra-chain, but the α-helix is always intra-chain.
4. Mentioning "Van der Waals Forces" as the Primary Stabilizer
The Mistake: Listing van der Waals forces as the main reason for stability.
Why it’s wrong: While van der Waals forces do exist between the tightly packed atoms of the helix, they are secondary to hydrogen bonds. The primary stabilizing force is the hydrogen bond network.
How to Avoid:
- Rank the interactions: State clearly: "The primary stabilizing force is hydrogen bonding. Van der Waals interactions provide additional, but minor, stability."
- Don't lead with van der Waals: Always mention hydrogen bonds first and most prominently.
5. Confusing the Direction of the Helix (Right-Handed vs. Left-Handed)
The Mistake: Stating that the α-helix is left-handed. …
- COMEDK 2026Set 2026-A1 markMCQQ.Choose the correct reason: o-hydroxybenzaldehyde is a liquid at room temperature while p-hydroxy-benzaldehyde is a high melting solid, because (A) o-hydroxybenzaldehyde shows only weak Vanderwal's forces of attraction while in p hydroxy benzaldehyde there is strong Vanderwal's forces of attraction in solid state (B) o-hydroxybenzaldehyde shows association of the molecules while in p-hydroxy benzaldehyde there is no association among the molecules (C) o-hydroxybenzaldehyde shows intramolecular hydrogen bonding while in p-hydroxy benzaldehyde there is no intramolecular hydrogen bonding (D) p-hydroxybenzaldehyde, shows intramolecular hydrogen bonding while in o-hydroxy benzaldehyde there is stronger intermolecular hydrogen bonding
›Reveal solutionSolution
The key difference is that ortho-hydroxybenzaldehyde forms a stable intramolecular hydrogen bond, which prevents intermolecular association and keeps it a liquid; para-hydroxybenzaldehyde cannot do this, so it forms intermolecular hydrogen bonds, leading to a high-melting solid. The correct option is (C).
The Concept & Intuition
Why do some small organic molecules melt at room temperature while others, with the same molecular formula, are solids? The answer lies in intermolecular forces — the attractions between separate molecules. For a substance to be a solid at room temperature, its molecules must stick together strongly enough to overcome thermal motion. The most powerful of these "sticky" forces is often hydrogen bonding.
Now, here’s the twist: a molecule can form a hydrogen bond either between two different molecules (intermolecular) or within the same molecule (intramolecular). If a molecule can satisfy its hydrogen-bonding "needs" internally, it has no reason to stick strongly to its neighbors. That molecule will behave like a small, non-polar molecule — likely a liquid. But if it cannot bond internally, it must reach out to neighbors, creating a network of strong intermolecular bonds — and that network makes a solid.
This is exactly the case with the two isomers of hydroxybenzaldehyde.
Step-by-Step Reasoning
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Identify the structural difference.
In ortho-hydroxybenzaldehyde, the –OH group and the –CHO group are adjacent (positions 1 and 2 on the benzene ring). In para-hydroxybenzaldehyde, they are opposite (positions 1 and 4).
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Consider the possibility of intramolecular hydrogen bonding.
For an intramolecular hydrogen bond to form, the donor (–OH) and the acceptor (the carbonyl oxygen of –CHO) must be close enough in space to form a stable six-membered ring.
- In the ortho isomer, the two groups are right next to each other. A hydrogen bond can easily form between the H of –OH and the O of –C=O, creating a stable chelate ring (a six-membered ring including the hydrogen).
- In the para isomer, the groups are far apart. The –OH and –CHO cannot bend enough to touch each other; no intramolecular hydrogen bond is possible.
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Analyze the consequences for intermolecular forces.
- Ortho isomer: The –OH hydrogen is already "tied up" in an intramolecular bond. It is no longer available to form strong hydrogen bonds with neighboring molecules. The only forces between molecules are weak van der Waals (dispersion) forces. Weak forces mean a low melting point — a liquid at room temperature.
- Para isomer: The –OH hydrogen is free. It can form strong intermolecular hydrogen bonds with the carbonyl oxygen of a neighboring molecule. This creates a network of molecules held together by many hydrogen bonds. Breaking this network requires a lot of energy, so the melting point is high — it is a solid at room temperature. …
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- COMEDK 2024Set 2024-A1 markMCQQ.Arrange the following amines in the decreasing order of their solubilities in water. (A) III > II > I (B) II > III > I (C) I > II > III (D) I > III > II
›Reveal solutionSolution
Water solubility of amines depends on hydrogen bonding with water and the hydrophobic bulk of the alkyl/aryl groups. The order is ethylamine (II) > diethylamine (III) > aniline (I), so the correct option is (B).
Concept & Intuition
Solubility in water for amines is governed by two competing factors:
- Hydrogen bonding: The N–H bonds in primary and secondary amines can donate hydrogen bonds to water, and the lone pair on nitrogen can accept hydrogen bonds. More N–H groups mean more potential for strong interactions.
- Hydrophobic character: Larger hydrocarbon groups (alkyl chains or aromatic rings) disrupt water’s structure, reducing solubility. Aromatic rings are especially bulky and nonpolar.
Here, we compare:
- I (aniline): Aromatic ring + one –NH₂ group. The ring is large, flat, and hydrophobic; the lone pair on nitrogen is partially delocalized into the ring, making it less available for hydrogen bonding.
- II (ethylamine): Small alkyl chain (two carbons) + a primary amine (–NH₂). Two N–H bonds for donation, small hydrophobic part.
- III (diethylamine): Two ethyl groups on nitrogen, secondary amine (one N–H bond). More hydrophobic than ethylamine, but still aliphatic.
Step-by-step reasoning
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Identify the type of amine and number of N–H bonds
- I: Primary aromatic amine (aniline) — one –NH₂ group, two N–H bonds.
- II: Primary aliphatic amine (ethylamine) — one –NH₂ group, two N–H bonds.
- III: Secondary aliphatic amine (diethylamine) — one N–H bond.
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Assess hydrogen-bonding capacity
- Both I and II have two N–H donors, but the lone pair on nitrogen in aniline is less basic (delocalized into the ring), so it is a weaker hydrogen-bond acceptor.
- III has only one N–H donor, so it can form fewer hydrogen bonds with water than a primary amine of similar size.
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Consider the hydrophobic contribution
- II: Only two carbons — very small hydrophobic part, so water molecules easily surround it.
- III: Four carbons total (two ethyl groups) — more hydrophobic, so solubility decreases relative to II. …
- COMEDK 2022Set 20221 markMCQQ.Which of the following is a false statement? (A) Methoxymethane has higher boiling point than ethanol. (B) Alcohols are more soluble is water than hydrocarbons of comparable molecular mass. (C) ortho-and para-nitrophenols are more acidic than phenols. (D) Phenol is more acidic than alcohol.
›Reveal solutionSolution
The false statement is (A).
Concept: hydrogen bonding and its effect on boiling point; acidity of phenols.
(A) 'Methoxymethane has higher boiling point than ethanol.' - FALSE. Both are C2H6O (Mr = 46). Ethanol has an O-H and forms intermolecular HYDROGEN BONDS, so it boils at 351 K (78 C), whereas dimethyl ether (methoxymethane) has no O-H, only weak dipole-dipole forces, and boils at 248 K (-25 C). Ethanol's boiling point is far HIGHER.
(B) TRUE - alcohols hydrogen-bond with water. …
- COMEDK 2021Set 20211 markMCQQ.Why, ketones and carboxylic acids have higher boiling point as compared to aldehydes? (A) Due to −CHO group (B) Due to Fazan's rule (C) Due to inter molecular hydrogen bonding (D) Due to intramolecular hydrogen bonding
›Reveal solutionSolution
[!TLDR]
Higher boiling points come from stronger intermolecular attraction, and hydrogen bonding between molecules is the strongest of these forces here, so the answer is intermolecular hydrogen bonding.
Concept
Boiling point depends on how strongly neighbouring molecules attract each other. This CBSE Class 12 topic (aldehydes, ketones and carboxylic acids) ranks the forces as: hydrogen bonding > dipole-dipole > van der Waals. A carboxylic acid has an −OH group that lets molecules hydrogen-bond to one another, typically forming a dimer.
Solution
- Aldehydes (−CHO) are polar and attract by dipole-dipole forces, but the aldehyde hydrogen is not acidic enough to form true intermolecular hydrogen bonds among aldehyde molecules.
- Carboxylic acids (−COOH) hydrogen-bond strongly through their −OH, often as cyclic dimers, giving very high boiling points.
- These extra intermolecular attractions require more energy to overcome, raising the boiling point. …
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