$\alpha$-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 $\alpha$-helix structure stable?

> [!TLDR] > The $\alpha$-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 $\alpha$-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 $\alpha$-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. 1. **The backbone has built-in hydrogen bond donors and acceptors.** Every amino acid in a polypeptide chain has a carbonyl group ($\ce{C=O}$) and an amide group ($\ce{N-H}$) in its backbone. In an unfolded chain, these groups are free to hydrogen bond with water. But in the $\alpha$-helix, they pair with each other instead. 2. **The hydrogen bonding pattern is regular and predictable.** In a right-handed $\alpha$-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. 3. **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. 4. **Why not other interactions?** - **Disulfide bonds** ($\ce{-S-S-}$) are covalent and form only between cysteine residues — they are rare in $\alpha$-helices and not required for helix formation. - **Hydrophobic interactions** and

α-Helix is a secondary structure of proteins formed by twisting of polypeptide chain into right handed screw l

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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.

Note

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:

\text{X—H} \cdots \text{Y}

where X and Y are N, O, or F. The dotted line ( $\cdots$ ) 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 ( $H_2O$ ) boils at $100^\circ C$ , while hydrogen sulfide ( $H_2S$ ) boils at $-60^\circ C$ . Both are similar molecules, but water forms hydrogen bonds; $H_2S$ 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 ( $\delta^+$ ).

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 ( $\delta^+$ ) is attracted to a lone pair of electrons on another electronegative atom (the acceptor).

This is electrostatic — Coulomb's law governs it:

F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1 q_2}{r^2}

$q_1$ = partial positive charge on H
$q_2$ = 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 $\delta^+$
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:

E_{\text{H-bond}} \propto \frac{\delta^+ \cdot \delta^-}{r^2}

Where:

$\delta^+$ = partial charge on H (depends on electronegativity of donor atom)
$\delta^-$ = 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 …

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The $\alpha$ -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 $\alpha$ -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 $\alpha$ -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.

The backbone has built-in hydrogen bond donors and acceptors. Every amino acid in a polypeptide chain has a carbonyl group ( $\ce{C=O}$ ) and an amide group ( $\ce{N-H}$ ) in its backbone. In an unfolded chain, these groups are free to hydrogen bond with water. But in the $\alpha$ -helix, they pair with each other instead.
The hydrogen bonding pattern is regular and predictable. In a right-handed $\alpha$ -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.
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.
Why not other interactions?
- Disulfide bonds ( $\ce{-S-S-}$ ) are covalent and form only between cysteine residues — they are rare in $\alpha$ -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. …

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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 $\alpha$ -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., $-\text{NH}_3^+$ and $-\text{COO}^-$ ). These are also part of tertiary structure stabilization.

How to Avoid:

Memorize the "backbone rule": The $\alpha$ -helix is stabilized by interactions between the backbone atoms (the $-\text{NH}$ and $-\text{CO}$ 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 $\alpha$ -helix.

How to Avoid:

Learn the exact pattern: The hydrogen bond forms between the carbonyl oxygen ( $\text{C=O}$ ) of one amino acid and the amide hydrogen ( $\text{N-H}$ ) of the amino acid four residues later.
Write it precisely: "Hydrogen bonds between the $-\text{C=O}$ group of the $n^{\text{th}}$ amino acid and the $-\text{N-H}$ group of the $(n+4)^{\text{th}}$ 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 $\alpha$ -helix is a single-chain structure. The bonds are intra-molecular (within the same chain). Inter-chain hydrogen bonds are found in $\beta$ -pleated sheets (between strands).

How to Avoid:

Use the keyword "intra-chain" or "within the same polypeptide chain" in your answer.
Compare with $\beta$ -sheet: Remember that $\beta$ -sheets can be inter-chain (between chains) or intra-chain, but the $\alpha$ -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 $\alpha$ -helix is left-handed. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$Q. In which of the following intramolecular H-bonding is absent? (A) Salicylic acid (B) Salicylaldehyde (C) Quinol (D) Catechol$
›Reveal solutionSolution
Intramolecular hydrogen bonding requires the two interacting groups to be close together (typically ortho-positioned); quinol's para –OH groups are too far apart, so it lacks intramolecular H-bonding.

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

Salicylic acid (2-hydroxybenzoic acid): –OH and –COOH are ortho → forms a stable intramolecular H-bond (chelation).

Salicylaldehyde (2-hydroxybenzaldehyde): –OH and –CHO are ortho → forms an intramolecular H-bond.

Catechol (1,2-dihydroxybenzene): the two –OH groups are ortho → forms an intramolecular H-bond. …
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AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
$Q. Which of the following forces stabilise secondary structure of proteins? A. Hydrogen bonds B. Covalent bonds C. Disulphide linkages D. Vander Waals forces Correct answer is (A) A, B only (B) A, C, D only (C) A, D only (D) B, C only$
›Reveal solutionSolution
Secondary structure of proteins (helices/sheets) is held together by hydrogen bonds between backbone C=O and N–H groups, with weak van der Waals interactions adding extra stability — not covalent bonds or disulphide linkages, which belong to primary and tertiary/quaternary structure respectively. Answer: (C).

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

Recall the four options: A. Hydrogen bonds, B. Covalent bonds, C. Disulphide linkages, D. Van der Waals forces.

Secondary structure = regular backbone folding (helix/sheet), stabilised primarily by hydrogen bonding between backbone C=O and N–H groups → A is relevant. …
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AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$\mathrm{CH_3OH}$
›Reveal solutionSolution
Only o-nitrophenol among the choices has two suitably positioned groups (ortho -OH and -NO $_2$ ) close enough to hydrogen-bond within the same molecule, forming a stable 6-membered ring — the others hydrogen-bond only between molecules.

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

HF: single -F per molecule, hydrogen bonds only intermolecularly (chains of HF).

CH $_3$ OH: single -OH per molecule, hydrogen bonds only intermolecularly. …
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AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
$Q. Assertion (A): Ammonia cannot exhibit hydrogen bonding in solid and liquid states. Reason (R): Ammonia has higher melting and boiling points than predicted. (A) Both (A) and (R) are correct and (R) is the correct explanation of (A). (B) Both (A) and (R) are correct but (R) is not the correct explanation of (A). (C) (A) Is correct but (R) is incorrect. (D) (A) Is incorrect but (R) is correct$
›Reveal solutionSolution
Ammonia does hydrogen-bond in condensed phases (the Assertion is false), but it genuinely has anomalously high melting/boiling points as a consequence (the Reason is true).

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

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. …
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AP EAPCET 2021Set ap-2021-09-03-FN1 markMCQ
$HCl$
›Reveal solutionSolution
This tests which of HCl, H2O, HI, H2S forms the strongest hydrogen bonds, based on the electronegativity and atomic size of the atom attached to hydrogen. The answer is $H_2O$ .

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

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).

Compare atomic sizes: O is much smaller than S, Cl, and I, allowing closer, stronger hydrogen bonds. …
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AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
$H_2SO_4$
›Reveal solutionSolution
$H_2SO_4$ has multiple $-OH$ groups that engage in extensive intermolecular hydrogen bonding, which is why it is a viscous, high-boiling, low-volatility liquid compared to other mineral acids.

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

Examine the structure of $H_2SO_4$ : it is a tetrahedral sulphur centre with two $=O$ groups and two $-OH$ groups.

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.

This strong, extensive hydrogen bonding requires significantly more thermal energy to break apart the liquid structure and release individual molecules into the vapour phase.

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. …
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AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
$H_2O_2$
›Reveal solutionSolution
The overall hydrogen-bonding strength order is $H_2O > HF > H_2O_2 > H_2S$ — option (D).

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

H2S: sulphur is only weakly electronegative, so essentially no hydrogen bonding — placed last.

H2O2 vs H2S: H2O2 clearly H-bonds more than H2S, so H2O2 > H2S.

HF: strong single H-bond but one H per molecule — ranks above H2O2.

H2O: most extensive H-bond network — ranks first.

Overall: $H_2O > HF > H_2O_2 > H_2S$ , matching option (D).

Common Mistakes …
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AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
$Q. The dominant intermolecular force that must be overcome to convert liquid methanol to its vapour is ______ (A) Covalent Bonds (B) Dipole-Dipole Interactions (C) Hydrogen Bonds (D) Coordinate Bonds$
›Reveal solutionSolution
Methanol molecules are held together mainly by O–H···O hydrogen bonds; vaporisation must overcome these. Answer: Hydrogen Bonds.

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

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.

The dominant (strongest) intermolecular attraction in an alcohol is hydrogen bonding, due to the highly polar O–H bond and small H atom. …
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Q. $\alpha$ -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 $\alpha$ -helix structure stable?

Concept understanding — Hydrogen Bonding

Hydrogen Bonding: From Intuition to Precision

The Precise Definition

Why Only N, O, and F?

What Makes Hydrogen Bonding Special?

Real-World Consequences

Why this formula?

Hydrogen Bonding: Why It Happens — The Reasoning, Not Just the Rule

1. The Core Requirement: A "Naked" Proton

2. The Electrostatic Attraction (The "Why")

3. Why Only F, O, N?

4. The "Formula" for Hydrogen Bond Strength

5. Directionality — The "Linear" Preference

Concept: Secondary Structure of Proteins — The α-Helix

Method: Hydrogen Bonding Analysis in α-Helix

Steps:

1. Confusing the Type of Bond (Hydrogen vs. Disulfide vs. Ionic)

2. Writing "Hydrogen Bonds" Without Specifying the Atoms

3. Forgetting the "Intra-chain" Nature of the Bonds

4. Mentioning "Van der Waals Forces" as the Primary Stabilizer

5. Confusing the Direction of the Helix (Right-Handed vs. Left-Handed)

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

Q.α\alphaα-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\alphaα-helix structure stable?

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Q. $\alpha$ -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 $\alpha$ -helix structure stable?

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