Q.Why methyl ammine has less boiling point than methanol?
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Why Do Amines Boil Where They Do?
Imagine you are holding a handful of magnets. Some have one exposed pole, some have two, and some have none. The ones with exposed poles will stick to each other more strongly — you have to pull harder to separate them. That is exactly what happens inside a pot of boiling amine.
Amines are ammonia (NH3) with one, two, or three of its hydrogens replaced by carbon groups. The key detail: only N–H bonds can form strong hydrogen bonds between molecules. A tertiary amine has no N–H bond left — all three hydrogens are replaced. So it cannot hydrogen-bond with its neighbours. A primary amine has two N–H bonds; a secondary amine has one.
Hydrogen bonding here means the N–H group of one molecule attracts the lone pair on the nitrogen of another molecule. This is a strong intermolecular force, far stronger than the van der Waals forces that all molecules experience.
The Precise Trend
Boiling point order: Primary > Secondary > Tertiary (for similar molecular mass).
| Amine type | N–H bonds | Hydrogen bonding | Boiling point (relative) |
|---|---|---|---|
| Primary (RNH2) | 2 | Strong, extensive network | Highest |
| Secondary (R2NH) | 1 | Moderate | Middle |
| Tertiary (R3N) | 0 | None (only van der Waals) | Lowest |
This trend holds only when the molecular masses are roughly equal. If you compare a tiny primary amine (methylamine, M=31) with a large tertiary amine (triethylamine, M=101), the tertiary amine will boil higher — because its sheer size creates many more van der Waals contacts. The hydrogen-bonding advantage of the primary amine is overwhelmed by the mass difference. Always compare apples to apples.
Why Secondary Is Not the Highest …
Both methanol and methylamine form intermolecular hydrogen bonds, so their boiling points depend on how strong those bonds are, and oxygen is more electronegative than nitrogen. …
Methanol's O-H hydrogen bonds are stronger than methylamine's N-H hydrogen bonds (O is more electronegative than N), so methanol has the higher boiling point.
Both methanol (CH3OH, b.p. 65°C) and methylamine (CH3NH2, b.p. −6°C) are capable of intermolecular hydrogen bonding, since both have an electronegative atom bonded to hydrogen. However, oxygen (electronegativity ≈3.5) is significantly more electronegative than nitrogen (electronegativity ≈3.0). This makes the O-H bond in methanol more polar than the N-H bond in methylamine, so the hydrogen bonds formed between methanol molecules (O−H⋯O) are stronger than those between methylamine molecules (N−H⋯N). Stronger inter …
- CBSE 2025Set 56/6/11 markMCQQ.For the following question, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Boiling point of (CH3)3N is higher than that of CH3CH2CH2NH2. Reason (R) : Hydrogen bonding is more extensive in CH3CH2CH2NH2.
›Reveal solutionSolution
Boiling points are governed by intermolecular forces; primary amines form extensive hydrogen bonds while tertiary amines cannot. The assertion is false (trimethylamine boils lower than propylamine), but the reason is true (primary amines do hydrogen-bond more extensively). The correct answer is (D).
Understanding Boiling Points and Hydrogen Bonding in Amines
Boiling point reflects the energy needed to overcome intermolecular forces. For amines, the dominant force is either hydrogen bonding (when N–H bonds are present) or weaker dipole-dipole and van der Waals interactions (when they are not).
The key structural difference here:
- (CH3)3N (trimethylamine) is a tertiary amine with no N–H bonds
- CH3CH2CH2NH2 (propylamine) is a primary amine with two N–H bonds
Let's examine each statement systematically.
Step-by-Step Analysis
1. Hydrogen bonding capability
Hydrogen bonding requires a hydrogen atom bonded to an electronegative atom (N, O, or F). In propylamine, the −NH2 group has two hydrogens attached to nitrogen, allowing it to act as both a hydrogen bond donor and acceptor. Multiple molecules can link together in an extended network.
Trimethylamine has nitrogen bonded only to carbon atoms. It can accept hydrogen bonds (the lone pair on nitrogen can interact with H-bond donors), but it cannot donate hydrogen bonds. This severely limits intermolecular association.
2. Comparing intermolecular forces
For propylamine:
- Strong N–H···N hydrogen bonds between molecules
- Each molecule can form multiple hydrogen bonds
- Creates a cohesive liquid structure requiring significant energy to vaporize
For trimethylamine:
- Only dipole-dipole interactions and London dispersion forces
- No hydrogen bonding between trimethylamine molecules themselves
- Much weaker intermolecular attraction
3. Molecular mass consideration
Both compounds have similar molecular masses ((CH3)3N: 59 g/mol; CH3CH2CH2NH2: 59 g/mol), so van der Waals forces contribute similarly. The decisive factor is hydrogen bonding.
4. Experimental boiling points
The actual boiling points confirm our analysis:
- Trimethylamine: 2.9°C (just above freezing!)
- Propylamine: 47–49°C
Propylamine boils nearly 45°C higher due to extensive hydrogen bonding. …
- CBSE 2025Set A1 markQ.Fill in the blank: The lower aliphatic amines are gases with ______ odour.
›Reveal solutionSolution
The lowest members of the aliphatic amine family are gases at room temperature with a sharp, fishy smell resembling ammonia.
Simple aliphatic amines such as methylamine, dimethylamine, and trimethylamine are gases under ordinary conditions (their boiling points are low, close to or below room temperature) and have a characteristic pungent, fishy odour, similar to (but distinguishable from) ammonia — this smell is also noticeable in decaying fish, w …
- CBSE 2022Set ANNUAL1 markQ.Write the following in increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2
›Reveal solutionSolution
Boiling point depends on the strength and extent of intermolecular hydrogen bonding: O–H bonds hydrogen-bond more strongly than N–H bonds, and fewer N–H bonds mean weaker association.
All three compounds have similar molecular mass, so the difference in boiling point comes from hydrogen bonding:
- C2H5OH (ethanol): the O–H bond is highly polar (O is more electronegative than N), so alcohols form the strongest intermolecular hydrogen bonds among the three. Boiling point ≈ 78 °C.
- C2H5NH2 (ethylamine, a primary amine): has two N–H bonds available for hydrogen bonding, so molecules associate reasonably well, but N–H hydrogen bonds are weaker than O–H. Boiling point ≈ 17 °C. …
- CBSE 2020Set HE8221 markQ.Write True or False: Ethyl amine is soluble in water while aniline is not.
›Reveal solutionSolution
Lower aliphatic amines like ethylamine dissolve readily in water via hydrogen bonding, but aromatic amines like aniline are only sparingly soluble because the bulky, hydrophobic benzene ring dominates their behaviour.
The statement is True.
Amines can form hydrogen bonds with water through their N–H bonds and the lone pair on nitrogen. Solubility in water depends on the balance between this hydrogen-bonding ability and the hydrophobic bulk of the rest of the molecule.
- Ethylamine (C2H5NH2) has a small ethyl group, so the polar –NH2 group dominates; it is completely miscible with water. …
- CBSE 2018Set ANNUAL1 markQ.Primary amines have higher boiling points than tertiary amines. Why?
›Reveal solutionSolution
Boiling point depends on the strength of intermolecular forces; only amines with an N–H bond can hydrogen-bond with each other, and primary amines have two such N–H bonds versus none in tertiary amines.
A primary amine, R−NH2, has two N–H bonds. Because nitrogen is electronegative, these N–H bonds are polarised, and one molecule's N–H can hydrogen-bond to the lone pair on another molecule's nitrogen. This intermolecular hydrogen bonding is a relatively strong force that must be overcome for the liquid to boil, raising the boiling point.
A tertiary amine, R3N, has no N–H bond at all — all three positions on nitrogen are occupied by carbon substituents. Tertiary amine molecules therefore cannot hydrogen-bond with each other (there is no acidic H to donate), and can only interact through weaker dipole-dipole interactions and van der Waals forces.
As a result, for amines of comparable molecular mass, the order of boiling points is: …
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