Like $NH_3$, nitrogen atom of amine is trivalent and carries an unshared pair of electrons. Nitrogen orbitals in amines are therefore $sp^3$ hybridised and the geometry of amines is pyramidal. Lower aliphatic amines are soluble in water due to the formation of hydrogen bond with water molecules. The solubility decreases as the molar mass of amines increases due to increase in size of hydrophobic part. Higher amines are insoluble in water. However amines are less soluble in water than alcohols because of low electronegativity of nitrogen as compared to oxygen. Boiling points of isomeric amines follow the order 1º > 2º > 3º. It is due to the fact the amines are held together due to hydrogen bonding. Extent of hydrogen bonding is more in primary amines than in secondary amines as two hydrogen atoms are available for hydrogen bond formation. Tertiary amines do not show hydrogen bonding because of the absence of hydrogen atom attached to nitrogen. Amines can be prepared from, nitro compounds, nitriles, amides etc. (a) Complete the following equations : (i) $CH_3CONH_2 \xrightarrow[\text{(ii) } H_2O]{\text{(i) } LiAlH_4}$ (ii) 4-Nitrotoluene ($4\text{-}CH_3C_6H_4NO_2$, drawn: benzene ring with $NO_2$ and $CH_3$ para to each other) $\xrightarrow{Fe + HCl}$ (b) Why primary amines have higher boiling points than tertiary amines ? (c) Classify the following amines as primary, secondary or tertiary : (i) N,N-dimethylnaphthalen-2-amine (drawn: naphthalene bearing $-N(CH_3)_2$) (ii) Naphthalen-2-amine (drawn: naphthalene bearing $-NH_2$) OR (c) Out of Butan-1-amine and Butan-1-ol, which is more soluble in water ?

> [!TLDR] > Part (a): amide → ethanamine (LiAlH$_4$); nitrotoluene → p-toluidine (Fe/HCl); 1° amines boil higher than 3° (H-bonding); the naphthalene amines are tertiary and primary respectively. > Part (b): butan-1-ol is more soluble in water than butan-1-amine because O is more electronegative than N (stronger H-bonds). --- ## Part (a) ### (i) Completing the equations 1. LiAlH$_4$ reduces the amide fully to a primary amine: $$CH_3CONH_2 \xrightarrow[\text{(ii) }H_2O]{\text{(i) }LiAlH_4} CH_3CH_2NH_2\ \text{(ethanamine)}$$ 2. Fe/HCl reduces the nitro group to an amino group (methyl unaffected): $$4\text{-}CH_3C_6H_4NO_2 \xrightarrow{Fe+HCl} 4\text{-}CH_3C_6H_4NH_2\ \text{(4-methylaniline)}$$ ### (b) Why 1° amines have higher boiling points than 3° amines Primary amines have **two N–H bonds**, so they form extensive intermolecular hydrogen bonds (N–H···N). Tertiary amines have **no N–H bond** and cannot hydrogen-bond with one another (only weak dipole/London forces). For similar molar mass, boiling point order is 1° > 2° > 3°, so primary amines boil higher. ### (c) Classification - **(i) N,N-dimethylnaphthalen-2-amine:** nitrogen bears two methyl groups and the naphthyl group, no N–H → **tertiary amine**. …

Like NH₃, nitrogen atom of amine is trivalent and carries an unshared pair of electrons. Nitrogen orbitals in

Question

NH_3

(a) Complete the following equations :

(i)

CH_3CONH_2 \xrightarrow[\text{(ii) } H_2O]{\text{(i) } LiAlH_4}

(ii) 4-Nitrotoluene (

4\text{-}CH_3C_6H_4NO_2

, drawn: benzene ring with

NO_2

and

CH_3

para to each other)

\xrightarrow{Fe + HCl}

(b) Why primary amines have higher boiling points than tertiary amines ?

(i) N,N-dimethylnaphthalen-2-amine (drawn: naphthalene bearing

-N(CH_3)_2

)

(ii) Naphthalen-2-amine (drawn: naphthalene bearing

-NH_2

)

(OR)

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Part (a)Concept understanding — Amine Boiling Point Trend

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 ( $NH_3$ ) 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.

Note

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 ( $RNH_2$ )	2	Strong, extensive network	Highest
Secondary ( $R_2NH$ )	1	Moderate	Middle
Tertiary ( $R_3N$ )	0	None (only van der Waals)	Lowest

Watch out

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 …

Part (b)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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Part (a): amide → ethanamine (LiAlH $_4$ ); nitrotoluene → p-toluidine (Fe/HCl); 1° amines boil higher than 3° (H-bonding); the naphthalene amines are tertiary and primary respectively.

Part (b): butan-1-ol is more soluble in water than butan-1-amine because O is more electronegative than N (stronger H-bonds).

Part (a)

(i) Completing the equations

LiAlH $_4$ reduces the amide fully to a primary amine:

CH_3CONH_2 \xrightarrow[\text{(ii) }H_2O]{\text{(i) }LiAlH_4} CH_3CH_2NH_2\ \text{(ethanamine)}

Fe/HCl reduces the nitro group to an amino group (methyl unaffected):

4\text{-}CH_3C_6H_4NO_2 \xrightarrow{Fe+HCl} 4\text{-}CH_3C_6H_4NH_2\ \text{(4-methylaniline)}

(b) Why 1° amines have higher boiling points than 3° amines

Primary amines have two N–H bonds, so they form extensive intermolecular hydrogen bonds (N–H···N). Tertiary amines have no N–H bond and cannot hydrogen-bond with one another (only weak dipole/London forces). For similar molar mass, boiling point order is 1° > 2° > 3°, so primary amines boil higher.

(c) Classification

(i) N,N-dimethylnaphthalen-2-amine: nitrogen bears two methyl groups and the naphthyl group, no N–H → tertiary amine. …

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

Showing the 12 most recent of 26 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. Alcohols are more soluble in water than hydrocarbons of comparable molecular masses. Why?$
›Reveal solutionSolution
Solubility in water depends on the ability to form hydrogen bonds with water; alcohols have a polar -OH group that hydrogen-bonds with water, whereas hydrocarbons are non-polar and cannot.

Water is a highly polar, hydrogen-bonding solvent. A solute dissolves well in water when it can interact with water molecules through similarly strong intermolecular forces (like dissolves like).

Alcohols contain a polar -OH group. The oxygen's lone pairs and the O-H hydrogen can both participate in hydrogen bonding with surrounding water molecules, allowing alcohol molecules to be effectively surrounded and stabilised by the solvent (solvation). This makes alcohols appreciably soluble in water, especially the shorter-chain ones. …
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CBSE 2026Set ANNUAL1 mark
$Q. Write true or false: Intermolecular hydrogen bond is formed between two different molecules of the same or different compounds.$
›Reveal solutionSolution
True. Intermolecular hydrogen bonding occurs between two distinct molecules, while intramolecular hydrogen bonding occurs within one molecule.

A hydrogen bond forms when a hydrogen atom, covalently bonded to a highly electronegative atom (like F, O, or N), is attracted to a lone pair on another electronegative atom. This attraction can occur:

Intermolecularly: between the hydrogen of one molecule and the electronegative atom of a DIFFERENT molecule (which can be of the same compound, e.g., water-water hydrogen bonding, or a different compound, e.g., water-ethanol hydrogen bonding). …
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CBSE 2025Set 56/6/11 markMCQ
$(CH_3)_3N$
›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:

$(CH_3)_3N$ (trimethylamine) is a tertiary amine with no N–H bonds

$CH_3CH_2CH_2NH_2$ (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 $-NH_2$ 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 ( $(CH_3)_3N$ : 59 g/mol; $CH_3CH_2CH_2NH_2$ : 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. …
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CBSE 2025Set A1 mark
$Q. Fill in the blank: The lower aliphatic amines are gases with ______ odour.$
›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 …

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CBSE 2025Set ANNUAL1 mark
$Q. Solubility of water in alcohol is due to ______ bond.$
›Reveal solutionSolution

Water dissolves in alcohol because both form intermolecular hydrogen bonds through their -OH groups.

Both water (H-O-H) and alcohols (R-O-H) have a highly polar O-H bond, with oxygen able to act as a hydrogen-bond acceptor and the O-H hydrogen able to act as a hydrogen-bond donor. When water and alcohol are mixed, new O-H...O hydrogen bonds form between the water and alcohol molecules, similar in strength to the hydrogen bonds broken w …

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CBSE 2024Set 56/3/11 markMCQ
$Q. For the following question, two statements are given - one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : The boiling point of ethanol is higher than that of dimethyl ether. Reason (R) : Ethanol molecules are associated through hydrogen bonding whereas in dimethyl ether, it is not possible.$
›Reveal solutionSolution
The key idea is that hydrogen bonding between ethanol molecules creates stronger intermolecular forces than the dipole-dipole forces in dimethyl ether, raising ethanol’s boiling point. The Assertion is true, the Reason is true, and the Reason correctly explains the Assertion — so the answer is (A).

Let’s unpack this step by step.

What the question is really asking

You have two organic compounds with the same molecular formula, $\ce{C2H6O}$ : ethanol ( $\ce{CH3CH2OH}$ ) and dimethyl ether ( $\ce{CH3OCH3}$ ). Despite having identical molar masses, their boiling points differ sharply — ethanol boils at $78.4^\circ \text{C}$ , dimethyl ether at $-24.8^\circ \text{C}$ . The Assertion states this fact; the Reason claims the cause is hydrogen bonding in ethanol but not in ether. We need to check both statements and their logical link.

Why boiling point depends on intermolecular forces

Boiling point is the temperature at which the vapour pressure of a liquid equals atmospheric pressure. To boil, molecules must overcome the attractive forces holding them together in the liquid phase. Stronger intermolecular forces → more energy needed → higher boiling point. The three main types (in increasing strength) are: London dispersion forces (present in all molecules), dipole-dipole interactions (in polar molecules), and hydrogen bonding (a special, very strong dipole-dipole interaction involving H bonded to N, O, or F).

Hydrogen bonding in ethanol — the critical difference

In ethanol, the —OH group has a hydrogen atom covalently bonded to a highly electronegative oxygen. This creates a large partial positive charge on the H and a large partial negative charge on the O. The H of one ethanol molecule is strongly attracted to the O of another, forming a hydrogen bond. These bonds link ethanol molecules into extended networks, so you need a lot of thermal energy to break them apart and vaporise the liquid.

Why dimethyl ether cannot hydrogen-bond

Dimethyl ether has an oxygen atom, but every hydrogen is bonded to carbon — not to oxygen. The C–H bond is not polar enough to create the strong partial positive charge needed for hydrogen bonding. The only intermolecular forces in dimethyl ether are weaker dipole-dipole interactions (from the C–O–C dipole) and London forces. That’s why it boils at a much lower temperature. …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. The reason for high boiling points of alcohols as compared to isomeric ethers is :$
(a) Solubility of alcohols in water
(b) High reactivity
(c) Association of molecules due to hydrogen-bonding
(d) van der Waals forces
›Reveal solutionSolution

Alcohols have much higher boiling points than isomeric ethers because alcohol molecules are held together by intermolecular hydrogen bonding, while ether molecules are not.

An alcohol (R-OH) has a polar O-H bond, and the hydrogen (attached to the highly electronegative oxygen) can form a hydrogen bond with the lone pair of oxygen on a neighbouring alcohol molecule. This causes alcohol molecules to associate into chains/clusters through extensive intermolecular hydrogen bonding.

An ether (R-O-R'), though isomeric in molecular formula and having a similar dipole moment, has no O-H bond, so its molecules cannot hydrogen-bond to each other (only weaker dipole-dipole and van der Waals forces act between ether molecules).

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Ethanol and dimethyl ether form a pair of functional isomers. The boiling point of ethanol is higher than that of dimethyl ether, due to the presence of -$
(a) H-bonding in dimethyl ether
(b) H-bonding in ethanol
(c) $CH_3CH_2$ group of ethanol
(d) $CH_3$ group of dimethyl ether
›Reveal solutionSolution

Ethanol's -OH group lets it hydrogen-bond with other ethanol molecules; dimethyl ether has no O-H bond and so cannot hydrogen-bond at all.

Ethanol ( $CH_3CH_2OH$ ) has an $-OH$ group, allowing intermolecular hydrogen bonding between its molecules — a strong dipole-dipole interaction that raises the boiling point significantly. Dimethyl ether ( $CH_3-O-CH_3$ ) has its oxygen bonded to two carbon atoms, with no O–H bond at all, so it cannot hydrogen-bond (option a is therefore wrong — dimethyl ether has no H-bonding). Its inter …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. In which of the following is hydrogen bond not formed?$
(a) CH3COOH
(b) NH3
(c) C2H5OH
(d) CH3COCH3
›Reveal solutionSolution
Acetone lacks an O–H/N–H/F–H bond, so it cannot form (self-associating) hydrogen bonds, unlike the other three.

CH3COOH (acetic acid): has O–H → forms strong hydrogen bonds (even dimerizes).

NH3: has N–H → forms hydrogen bonds.

C2H5OH (ethanol): has O–H → forms hydrogen bonds. …
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CBSE 2024Set sz1 markMCQ
$Q. Select the correct one: In which of the following compounds does hydrogen bond occur?$
(a) SiH4
(b) LiH
(c) HI
(d) NH3
›Reveal solutionSolution

Hydrogen bonding occurs in NH3 because nitrogen is small and electronegative enough to polarize the N-H bond strongly, unlike Si in SiH4, Li in LiH, or I in HI.

A hydrogen bond forms when a hydrogen atom, covalently bonded to a small and highly electronegative atom (N, O, or F), is attracted to a lone pair on a similar electronegative atom of a neighbouring molecule.

Checking each option: SiH4 -- silicon has low electronegativity (comparable to hydrogen), so Si-H bonds are not polar enough. LiH -- this is an ionic hydride (Li+ H-), not a case of hydrogen bonding between molecules. HI -- although iodine is electronegative, it is too large and its electronegativity too low for the H-I bond to support hydrogen bonding (HI is in fact one of the weakest hydrogen halide acids in this regard). NH3 -- nitrogen is small and sufficiently electronegative, so the N-H bond is polar en …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Alcohol boils at higher temperature than the corresponding hydrocarbon due to$
(a) intermolecular hydrogen bond
(b) intra molecular hydrogen bond
(c) van der waal's force of attraction
(d) dipole-dipole interaction
›Reveal solutionSolution

The –OH group lets alcohol molecules hydrogen-bond to one another, an attraction hydrocarbons of similar mass simply cannot form.

Alcohol molecules associate with each other through intermolecular hydrogen bonding, since the O–H bond is highly polar (O is electronegative) and each molecule has both an H-bond donor (O–H) and acceptor (lone pairs on O). Considerable extra thermal energy is required to break these hydrogen bonds before the molecules can escape into the vapour ph …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Alcohol are soluble in water because -$
(a) they have higher molecular weight than water.
(b) they form Hydrogen Bond with water.
(c) they displace water.
(d) None of these.
›Reveal solutionSolution

Alcohols dissolve in water because their -OH group can hydrogen-bond with water molecules.

An alcohol molecule, R-OH, carries a polar -OH group in which the O-H bond is highly polarised. This -OH group can form hydrogen bonds with the -OH of water molecules (both as a hydrogen-bond donor and acceptor), just as water molecules hydrogen-bond with each other. This intermolecular hydrogen bonding lets the alcohol molecules mix intimately with water, so lower alcohols (methanol, ethanol, propanol) are miscible with water in all proportions. As the hydrocarbon (R) part of the alcohol becomes larger, the non-polar part dominates, hydrogen bonding contribution becomes relatively less significant, and solubility in w …

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Part (a)

(i) Completing the equations

(b) Why 1° amines have higher boiling points than 3° amines

(c) Classification

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