How to Identify Hydrogen Bond Donors and Acceptors

Two substances have exactly the same molecular formula, \(\ce{C4H10O}\). One boils at about \(35^\circ\mathrm{C}\), while the other boils at about \(118^\circ\mathrm{C}\). Both contain oxygen, so you might predict that both can form hydrogen bonds between their own molecules. But only one can.

The difference is where the hydrogen atoms are attached. Butan-1-ol, \(\ce{CH3CH2CH2CH2OH}\), has an O-H bond. Ethoxyethane, \(\ce{CH3CH2OCH2CH3}\), has an oxygen atom between two carbon atoms, but no O-H bond.

Here is the method you need: a hydrogen bond donor has a hydrogen directly bonded to N, O, or F. A hydrogen bond acceptor has a suitable electron-rich site, usually an available lone pair on N, O, or F. A molecule can be both, one, or neither.

Butan-1-ol is both a donor and an acceptor. Ethoxyethane is an acceptor only. That difference helps explain their boiling points, but it also raises an interesting question: if ethoxyethane cannot donate hydrogen bonds, why can it still hydrogen bond with water?

01What makes a hydrogen bond donor or acceptor?

Picture two neighbouring molecules.

One has an O-H bond. Oxygen attracts the shared electrons more strongly than hydrogen, leaving the oxygen partially negative and the hydrogen partially positive.

The partially positive hydrogen can be attracted to an electron-rich region on another molecule, such as an oxygen lone pair.

That attraction is a hydrogen bond.

The two roles are different.

Hydrogen bond donor: look for the hydrogen

For the conventional hydrogen bonds studied in HSC Chemistry, look for a hydrogen covalently attached to:

• Oxygen: O-H
• Nitrogen: N-H
• Fluorine: F-H

These bonds are strongly polar because the attached atom attracts bonding electrons more strongly than hydrogen.

That leaves the hydrogen with a partial positive charge, represented by \(\delta^+\). It can interact with an appropriate acceptor.

For example, methanol, \(\ce{CH3OH}\), can donate a hydrogen bond through its O-H hydrogen.

But methane, \(\ce{CH4}\), is not a conventional hydrogen bond donor. Its hydrogens are attached to carbon.

The important question is not whether a molecule contains hydrogen. Almost every organic molecule does.

It is what the hydrogen is bonded to.

Hydrogen bond acceptor: look for available electron density

A hydrogen bond acceptor provides an electron-rich region that can interact with the partially positive hydrogen of a donor.

For standard HSC organic examples, common acceptors include:

• The oxygen of an alcohol, ether, or carbonyl group.
• The nitrogen of an ordinary, unprotonated amine.
• Other suitable electronegative atoms with available lone pairs.

An oxygen atom in an ether has two lone pairs, even though it has no hydrogen attached.

That makes the ether an acceptor without making it a donor.

A useful way to remember the relationship is that the donor presents the hydrogen, while the acceptor provides the electron-rich site. However, the hydrogen is not handed over. It remains covalently attached to its original atom.

If the idea of partial charges is unfamiliar, the Bond Polarity and Electronegativity guide explains why some bonds share electrons unevenly.

Where the simple rule needs care

The N/O/F rule is a useful school-level model of conventional hydrogen bonding, not a complete definition of every possible hydrogen-bond interaction.

An atom having lone pairs does not automatically make it a strong acceptor.

For example, the nitrogen lone pair in an amide is delocalised into the adjacent carbonyl group. Amide nitrogen is therefore not normally counted as a hydrogen bond acceptor. The carbonyl oxygen remains an acceptor.

Similarly, a positively charged nitrogen with four covalent bonds, such as in \(\ce{NH4+}\), has no available lone pair and cannot act as a nitrogen acceptor.

Some weaker hydrogen bonds involve less familiar donors or acceptors, but these are not the starting point for ordinary HSC structure comparisons. In particular, a carbon-fluorine bond in an organic molecule does not automatically provide a strong hydrogen bond acceptor.

02Compare the complete molecular structure

A molecular formula tells you which atoms are present. It does not tell you which atoms are connected.

That is why \(\ce{C4H10O}\) alone cannot establish whether a molecule is a hydrogen bond donor.

You need the structural formula.

A reliable decision process

1. Find the O, N, and F atoms. Check how each is connected to the rest of the molecule.
2. Find any O-H, N-H, or F-H bonds. These identify conventional hydrogen bond donor sites.
3. Identify suitable lone-pair acceptor sites. Oxygen atoms in alcohols, ethers, and carbonyl groups, and unprotonated amine nitrogens, are common examples. Check for exceptions such as amide nitrogen or nitrogen with four bonds.
4. Compare the whole molecules. Decide whether each is donor only, acceptor only, both, or neither.
5. Check the other molecule involved. Hydrogen bonding requires a compatible donor and acceptor. Being an acceptor does not mean a substance can hydrogen bond with itself.

Here are some useful comparisons.

Notice that methylamine and ammonium both have N-H bonds, but they differ as acceptors.

In methylamine, nitrogen has three covalent bonds and one available lone pair. In ammonium, nitrogen has four covalent bonds and no lone pair.

You cannot determine acceptor ability just by recognising the element.

03Can a molecule hydrogen bond with water but not with itself?

Yes. This is the most useful consequence of separating donor and acceptor roles.

Consider ethoxyethane again.

In a pure sample, every ethoxyethane molecule has an acceptor oxygen but no conventional hydrogen bond donor.

Two acceptors cannot supply the two complementary roles needed for a conventional hydrogen bond.

However, add water and the situation changes.

Water has O-H donor groups, so its hydrogen can interact with the ether oxygen.

The ethoxyethane molecule has not changed chemically. It has simply encountered a suitable hydrogen bond donor.

This distinction leads to four possible combinations.

This does not mean every possible hydrogen bond forms all the time.

Molecules are moving, bonds are continually forming and breaking, and their orientations matter. Hydrogen bonding competes with other intermolecular interactions.

It also doesn’t guarantee complete solubility in water. The size of a molecule’s nonpolar region, the energy required to disrupt existing attractions, and the interactions available after mixing all influence solubility.

The donor-acceptor test tells you which interactions are possible. It does not, by itself, tell you their exact strengths or the final physical properties.

04What can a boiling-point experiment actually establish?

Return to butan-1-ol and ethoxyethane.

Both have the formula \(\ce{C4H10O}\) and the same molar mass, approximately \(74.1\ \mathrm{g\,mol^{-1}}\).

If molar mass alone controlled boiling point, their boiling temperatures should be similar.

They are not.

How could we investigate that difference experimentally?

First design: measure the boiling points

A suitable comparison would determine the boiling point of each pure substance under the same external pressure.

In a conventional distillation arrangement, the apparatus has the following spatial sequence.

At one end, a distillation flask contains the liquid sample. A controlled electric heating system supplies energy.

As the liquid boils, vapour rises towards the distillation head. A temperature sensor positioned in the vapour path measures the temperature of the vapour passing into the side arm.

The vapour then travels through a condenser, where cooling water flows through an outer jacket and removes heat. The vapour condenses and collects in a receiving vessel.

The temperature of the vapour becomes approximately steady while a pure liquid distils at constant pressure.

The thermometer measures temperature, not the strength of an individual hydrogen bond.

Pressure must be controlled because boiling occurs when the liquid’s vapour pressure equals the external pressure. Changing the external pressure changes the boiling temperature.

These particular liquids are flammable, especially ethoxyethane. This is an explanation of the measurement principle, not a suggested student practical.

What should the first experiment show?

From the structures, we predict that butan-1-ol can form an intermolecular hydrogen-bond network, while ethoxyethane cannot form conventional hydrogen bonds between its own molecules.

That suggests butan-1-ol should need a higher temperature to boil.

Published reference measurements agree.

Reference values: NIST Chemistry WebBook boiling-point compilations, rounded to one decimal place.

These are reference measurements, not values obtained from an experiment performed for this article.

The flaw in the first design

There is a second problem.

The boiling-point comparison helps explain why butan-1-ol associates more strongly with itself.

It does not directly establish whether ethoxyethane can accept hydrogen bonds from another substance.

A student might reason:

“Ethoxyethane has a much lower boiling point, so its oxygen cannot form hydrogen bonds.”

That conclusion is incorrect.

The experiment tests the combined interactions in two pure liquids. It does not independently test how the ether oxygen interacts with a separate donor.

To test acceptor ability, we need a different experimental arrangement.

Revised test: introduce a hydrogen bond donor

One approach is infrared spectroscopy.

Infrared radiation can be absorbed by molecules when its frequency matches a molecular vibration, such as the stretching vibration of an O-H bond.

Hydrogen bonding changes the electronic environment of an O-H bond. It commonly shifts its stretching absorption towards lower wavenumbers and can broaden the absorption band.

A wavenumber is measured in reciprocal centimetres, \(\mathrm{cm^{-1}}\).

In a conceptual experiment, we could investigate an alcohol vapour by itself and then in the presence of an ether vapour.

The apparatus would contain an infrared source, a gas sample cell, and a detector in that order.

Infrared radiation travels from the source through the gas cell to the detector. The detector measures transmitted radiation, allowing the instrument to determine which wavenumbers were absorbed.

First, measure the alcohol under controlled conditions without an ether present.

Next, introduce the ether while controlling temperature, the amount of alcohol, and other relevant experimental conditions.

If the ether oxygen accepts hydrogen bonds from alcohol O-H groups, we would expect changes in the O-H stretching absorption consistent with that interaction.

This provides a way to investigate the acceptor role independently of the pure ether’s ability to donate hydrogen bonds.

However, changing the gas mixture can also change concentration, collision behaviour, and other environmental conditions. Suitable control mixtures and comparison spectra are needed.

Infrared absorption is evidence about molecular interactions. It is not a photograph of a hydrogen bond, and a shifted peak alone does not establish a unique explanation.

The experimental logic is therefore:

Structure predicts the donor and acceptor sites. Boiling-point evidence tests the overall strength of attractions in pure substances. Spectroscopy can provide additional evidence about particular interactions.

These are related observations, but they answer different questions.

05Boiling breaks intermolecular attractions, not the molecule’s covalent bonds

One final distinction matters when explaining these boiling-point measurements.

When butan-1-ol boils, its molecules separate from one another and move into the gas phase.

The O-H covalent bond inside each butan-1-ol molecule remains intact.

Likewise, ethoxyethane still contains its C-O-C structure after vaporisation.

For ethoxyethane, the physical change is:

\[ \ce{CH3CH2OCH2CH3(l) -> CH3CH2OCH2CH3(g)} \]

The symbols \((l)\) and \((g)\) indicate liquid and gas.

Boiling does not remove the molecule’s donor or acceptor sites.

A butan-1-ol molecule in the gas phase still has an O-H bond and oxygen lone pairs.

Ethoxyethane vapour still contains an oxygen acceptor.

Both can participate in hydrogen bonding if suitable partners approach them under the right conditions.

If boiling routinely broke O-H or C-O covalent bonds, cooling the vapour would not simply recover the unchanged original substance.

This is why boiling-point explanations should focus on attractions between molecules, not covalent bonds within them.

06Questions and solutions

The first question checks the structural method. The later questions test whether you can apply it when experimental observations, chemical changes, or competing interactions complicate the picture.

Question 1

Consider two molecules:

• Butan-2-ol: \(\ce{CH3CH(OH)CH2CH3}\)
• Butan-2-one: \(\ce{CH3C(=O)CH2CH3}\)

For each molecule:

(a) Identify whether it can donate and accept conventional hydrogen bonds.

(b) State whether molecules of the pure substance can form conventional hydrogen bonds with one another.

(c) Explain whether the molecule can hydrogen bond with water.

Solution 1

Butan-2-ol is both a hydrogen bond donor and acceptor. Butan-2-one is an acceptor only. Both can hydrogen bond with water, but only butan-2-ol can form conventional hydrogen bonds between its own molecules.

For part (a), butan-2-ol contains an O-H bond. Its hydrogen can act as a donor, while the oxygen lone pairs provide an acceptor site.

Butan-2-one has a carbonyl oxygen with lone pairs, so it can accept hydrogen bonds. However, it has no hydrogen bonded to oxygen, nitrogen, or fluorine. It cannot donate a conventional hydrogen bond.

For part (b), one butan-2-ol molecule can donate through its O-H group to another butan-2-ol molecule’s oxygen.

Two butan-2-one molecules cannot form a conventional hydrogen bond with each other because neither has a suitable donor hydrogen.

For part (c), water can donate an O-H hydrogen bond to either molecule’s oxygen.

Butan-2-ol can also donate its O-H hydrogen to the oxygen of water.

The common mistake is to assume that the presence of oxygen automatically allows self-hydrogen-bonding. It provides an acceptor site, but a conventional hydrogen bond still needs a donor.

Question 2

A chemist investigates 3-hydroxypropanal, \(\ce{HOCH2CH2CHO}\).

The molecule contains an alcohol group at one end and an aldehyde group, \(\ce{-CHO}\), at the other.

The following infrared measurements are illustrative, not actual experimental data.

All measurements are made at the same temperature and 3-hydroxypropanal concentration, with appropriate background corrections.

(a) Identify all conventional donor and acceptor sites in 3-hydroxypropanal.

(b) Calculate the total carbonyl wavenumber change from the dry reference to the water-containing environment. Define your sign convention.

(c) Describe two different hydrogen-bond orientations possible between 3-hydroxypropanal and water.

(d) A student claims that the entire observed shift must be caused by hydrogen bonding and that subtracting the polar-solvent shift gives the exact hydrogen-bond contribution. Evaluate both claims.

Solution 2

3-Hydroxypropanal is both a hydrogen bond donor and acceptor. Its carbonyl stretching wavenumber decreases by \(14\ \mathrm{cm^{-1}}\) in the water-containing environment, which is consistent with hydrogen bonding but does not prove that hydrogen bonding accounts for the entire shift.

For part (a), the molecule has one O-H hydrogen that can donate a hydrogen bond.

It also has two oxygen acceptor sites: the alcohol oxygen and the aldehyde carbonyl oxygen.

For part (b), define the wavenumber change as final minus initial:

\[ \Delta\tilde{\nu} =\tilde{\nu}_{\mathrm f} -\tilde{\nu}_{\mathrm i} \]

Here, \(\tilde{\nu}\) represents wavenumber, in \(\mathrm{cm^{-1}}\), with subscripts \(\mathrm f\) and \(\mathrm i\) denoting final and initial.

\[ \begin{aligned} \Delta\tilde{\nu} &=1716-1730\\ &=-14\ \mathrm{cm^{-1}} \end{aligned} \]

The negative sign means the absorption has shifted towards a lower wavenumber.

For part (c), two possible orientations are:

Water as donor: A water O-H hydrogen interacts with the carbonyl oxygen of 3-hydroxypropanal.

3-Hydroxypropanal as donor: Its alcohol O-H hydrogen interacts with the oxygen of a water molecule.

Water could also donate to the alcohol oxygen, giving another possible interaction.

The carbonyl measurement is particularly relevant to hydrogen bonding involving the aldehyde oxygen. It does not rule out hydrogen bonding at the alcohol group.

For part (d), the student’s interpretation is too strong.

Even the polar solvent that cannot donate hydrogen bonds produces a lower carbonyl stretching wavenumber:

\[ \begin{aligned} \Delta\tilde{\nu}_{\mathrm{polar}} &=1724-1730\\ &=-6\ \mathrm{cm^{-1}} \end{aligned} \]

This shows that the carbonyl vibration can respond to the surrounding solvent without that solvent donating a hydrogen bond.

Water produces an additional change, which is consistent with an extra interaction such as hydrogen bonding.

However, the effects of solvent polarity and hydrogen bonding are not necessarily independent or additive. Subtracting the two shifts does not automatically isolate an exact hydrogen-bond contribution.

The precise trap is treating an infrared shift as a direct measurement of one intermolecular force, rather than an observation of a bond vibrating in a changed environment.

Question 3

Two separate samples each contain \(0.100\ \mathrm{mol}\) of a pure substance at its own normal boiling point.

Sample A contains butan-1-ol. Sample E contains ethoxyethane.

For this question, use the following illustrative molar enthalpies of vaporisation, rounded for calculation. They are supplied problem data, not reported experimental measurements:

Assume both samples vaporise completely at constant pressure, without chemical reactions or heat losses. The liquids begin at their own boiling points, so no additional heating to reach those temperatures is required.

(a) Calculate the energy absorbed by each sample during vaporisation.

(b) Calculate the difference in energy per mole.

(c) A student says that the difference is the energy of one hydrogen bond between butan-1-ol molecules. Explain why this conclusion is unjustified.

(d) After the vapours cool and condense separately, do the recovered substances retain their original donor and acceptor abilities? Explain.

Solution 3

The butan-1-ol sample absorbs \(4.4\ \mathrm{kJ}\), while the ethoxyethane sample absorbs \(2.7\ \mathrm{kJ}\). The additional \(17\ \mathrm{kJ\,mol^{-1}}\) associated with butan-1-ol’s vaporisation is not the energy of an individual hydrogen bond.

For part (a), use:

\[ q=n\Delta H_{\mathrm{vap}} \]

where \(q\) is heat absorbed in kilojoules, \(n\) is amount in moles, and \(\Delta H_{\mathrm{vap}}\) is molar enthalpy of vaporisation in kilojoules per mole.

For butan-1-ol:

\[ \begin{aligned} q_{\mathrm A} &=n\Delta H_{\mathrm{vap,A}}\\ &=(0.100)(44)\\ &=4.4\ \mathrm{kJ} \end{aligned} \]

For ethoxyethane:

\[ \begin{aligned} q_{\mathrm E} &=n\Delta H_{\mathrm{vap,E}}\\ &=(0.100)(27)\\ &=2.7\ \mathrm{kJ} \end{aligned} \]

In each substitution, moles multiplied by \(\mathrm{kJ\,mol^{-1}}\) gives kilojoules.

Both values are positive because vaporisation is endothermic: energy is absorbed by the substance.

For part (b), the additional energy required for the alcohol sample is:

\[ \begin{aligned} \Delta q&=4.4-2.7\\ &=1.7\ \mathrm{kJ} \end{aligned} \]

For the equal sample amounts, the additional energy per mole is:

\[ \begin{aligned} \frac{\Delta q}{n} &=\frac{1.7\ \mathrm{kJ}}{0.100\ \mathrm{mol}}\\ &=17\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

For part (c), that \(17\ \mathrm{kJ\,mol^{-1}}\) difference is not an individual hydrogen-bond energy.

The molar enthalpy of vaporisation describes the overall energy difference between liquid and vapour at the relevant boiling conditions.

It includes the effects of all intermolecular attractions and the different molecular arrangements in each phase.

The two substances also differ in molecular shape and in the strengths and arrangements of their other intermolecular interactions.

Furthermore, one mole of alcohol molecules does not correspond to exactly one mole of identical hydrogen bonds. Liquid hydrogen-bond networks are continually rearranging, and individual molecules can have different local environments.

The higher vaporisation enthalpy is consistent with stronger overall intermolecular attraction in the alcohol, but it does not isolate one hydrogen-bond strength.

For part (d), both substances retain their original donor and acceptor abilities after condensation.

Butan-1-ol remains a donor and acceptor because its O-H bond and oxygen lone pairs remain present.

Ethoxyethane remains an acceptor only because its ether oxygen still has lone pairs, but it still has no O-H or N-H bond.

Vaporisation and condensation change the physical state, not the covalent connectivity of these molecules.

Question 4

Methylamine and ethanoic acid undergo an acid-base reaction.

Before reaction, the neutral structures are:

• Methylamine: \(\ce{CH3NH2}\)
• Ethanoic acid: \(\ce{CH3COOH}\)

Assume proton transfer is effectively complete and the dominant products are:

• Methylammonium: \(\ce{CH3NH3+}\)
• Ethanoate: \(\ce{CH3COO-}\)

Ignore minor amounts of unreacted starting substances.

(a) Classify the donor and acceptor abilities of both neutral reactants.

(b) Classify the donor and acceptor abilities of both product ions.

(c) Can the product ions form hydrogen bonds with each other and with water?

(d) A student says that the reaction has destroyed hydrogen bonding because methylamine loses its nitrogen lone pair and ethanoic acid loses its O-H hydrogen. Explain the error, and distinguish the chemical reaction from boiling.

Solution 4

Both neutral reactants can donate and accept hydrogen bonds. After proton transfer, methylammonium is a donor but not an acceptor, while ethanoate is an acceptor but not a donor. Hydrogen bonding remains possible between the product ions and with water.

For part (a), methylamine contains N-H bonds that can donate hydrogen bonds, and nitrogen has a lone pair that can accept them.

Ethanoic acid contains an O-H group that can donate a hydrogen bond. Its carbonyl oxygen can accept one.

Both starting substances are therefore donors and acceptors.

For part (b), the reaction is:

\[ \ce{CH3NH2 + CH3COOH -> CH3NH3+ + CH3COO-} \]

Methylamine gains a proton, producing methylammonium.

The nitrogen now forms four covalent bonds and has no lone pair available for acceptance.

However, it still has N-H bonds, so it remains a conventional donor.

Ethanoic acid loses its acidic O-H proton, producing ethanoate.

The ethanoate ion has no O-H or N-H bond, so it cannot donate a conventional hydrogen bond.

Its negatively charged, oxygen-rich carboxylate group can accept hydrogen bonds. The negative charge is distributed across the two oxygens by resonance.

For part (c), methylammonium can donate a hydrogen bond through N-H to an ethanoate oxygen.

Water can also accept hydrogen bonds from methylammonium, and donate hydrogen bonds to ethanoate.

These interactions occur alongside strong electrostatic attractions involving the charged ions.

The presence of hydrogen bonds does not mean we should ignore ionic forces.

For part (d), the student has confused a change in the location of donor and acceptor sites with the elimination of all hydrogen-bonding possibilities.

Neither ion can perform both roles individually, but the two ions provide complementary roles when placed together.

There is also a deeper distinction.

The acid-base reaction transfers a proton, changing the covalent bonding: an O-H bond is broken and an N-H bond is formed.

Ordinary boiling does not do that. Boiling separates molecules without changing their covalent structures.

So the correct comparison is:

The acid-base reaction changes which species and bonding sites exist. Boiling changes the physical arrangement of existing molecules.

Question 5

Extension challenge: when a molecule can hydrogen bond with itself internally

Two positional isomers have the molecular formula \(\ce{C7H6O2}\).

Both contain a phenolic O-H group and an aldehyde group, \(\ce{-CHO}\), attached to a benzene ring.

Their structures differ only in the positions of the groups:

• Molecule A is 2-hydroxybenzaldehyde. Its O-H and CHO groups are attached to neighbouring ring carbons.
• Molecule B is 4-hydroxybenzaldehyde. Its O-H and CHO groups are on opposite sides of the ring.

An intramolecular hydrogen bond is a hydrogen bond formed between suitable sites within the same molecule, rather than between separate molecules.

The following infrared results are hypothetical illustrative observations, not measured spectra.

The compounds are examined at the same temperature in a solvent that neither donates nor accepts significant hydrogen bonds. Assume they remain chemically unchanged and that the observed O-H spectral changes are dominated by hydrogen-bond interactions.

(a) Classify both molecules as hydrogen bond donors, acceptors, both, or neither.

(b) Propose the most convincing explanation for the different concentration-dependent results.

(c) A student concludes that molecule A must have very strong intermolecular hydrogen bonding because its O-H absorption remains shifted even at low concentration. Evaluate this conclusion.

(d) Does the proposed interaction in molecule A prevent it from hydrogen bonding with water? Explain.

(e) Can you determine which isomer has the higher boiling point from the classifications and spectral data alone? Explain what makes this a more difficult comparison than butan-1-ol versus ethoxyethane.

Solution 5

Both molecules are donors and acceptors, but the observations suggest different dominant hydrogen-bonding arrangements. Molecule A is consistent with intramolecular hydrogen bonding, while molecule B shows concentration-dependent intermolecular association. The data do not establish which substance has the higher boiling point.

For part (a), both structures contain a phenolic O-H bond, which provides a hydrogen bond donor.

Both also contain oxygen lone pairs, including those on the aldehyde carbonyl oxygen, which provides a suitable acceptor site.

Their donor-acceptor classifications are identical.

That is precisely why simply counting functional groups is insufficient here.

For part (b), molecule A has its O-H and carbonyl groups on neighbouring positions of the benzene ring.

Its geometry allows the phenolic O-H hydrogen to interact with the carbonyl oxygen within the same molecule, forming a favourable six-membered intramolecular hydrogen-bonded arrangement.

This interaction does not require a second molecule.

Consequently, diluting A does not automatically remove that hydrogen bond.

This provides a reasonable explanation for its broadly similar O-H absorption at both concentrations.

Molecule B has its O-H and carbonyl groups on opposite sides of the ring.

Those groups are too far apart for the corresponding favourable internal interaction.

At low concentration, its molecules are relatively far apart, reducing opportunities for intermolecular hydrogen bonding.

The relatively sharp O-H absorption near \(3600\ \mathrm{cm^{-1}}\) is consistent with a larger proportion of unassociated O-H groups.

At higher concentration, intermolecular encounters and associations become more likely.

An O-H group on one molecule can donate to the carbonyl oxygen of another molecule, producing hydrogen-bonded associations that account for a lower, broader O-H absorption.

For part (c), the student’s conclusion is not justified.

The tempting argument is that a persistent hydrogen-bonded O-H signal must indicate an especially strong attraction between different molecules.

But dilution provides the crucial constraint.

Reducing concentration makes intermolecular encounters less likely. It does not similarly prevent suitable groups within the same molecule from interacting.

So a hydrogen-bonded signal that survives substantial dilution is evidence consistent with intramolecular hydrogen bonding, not necessarily unusually strong intermolecular bonding.

The observation is not absolute proof, but the molecular geometry and concentration dependence make the internal interpretation convincing under the stated assumptions.

For part (d), molecule A can still form hydrogen bonds with water.

An intramolecular hydrogen bond is not a permanent covalent connection that consumes the donor and acceptor sites.

Hydrogen-bonded arrangements are dynamic. When water approaches, its molecules can compete for A’s hydrogen bond donor and acceptor sites.

Water might donate to A’s carbonyl oxygen, or accept from A’s phenolic O-H group.

Whether those interactions dominate depends on the surrounding environment and the relative stability of the available arrangements.

The existence of an intramolecular hydrogen bond does not make a molecule incapable of intermolecular hydrogen bonding.

For part (e), the donor-acceptor classification and infrared measurements are not enough to determine the exact boiling-point order.

Both molecules have the same molecular formula, similar overall sizes, and the same types of conventional donor and acceptor sites.

But their geometries permit different balances between internal and external hydrogen bonding.

Intramolecular hydrogen bonding in A can compete with the intermolecular associations that help hold a liquid together.

This suggests that B may be better able to form an extended intermolecular hydrogen-bonding network, provided other factors are comparable.

However, the actual boiling points also depend on dispersion attractions, molecular geometry, the range of structures present in the liquid, and the thermodynamics of the liquid-to-vapour transition.

The observed infrared behaviour in dilute solutions does not provide all that information.

The crucial difference from butan-1-ol and ethoxyethane is that these two molecules have the same basic donor and acceptor classification.

Their behaviour cannot be explained by saying one possesses an O-H group and the other does not.

The difficult step is recognising that having a donor and an acceptor does not specify how those sites are actually used.

A site can participate in an interaction within its own molecule, between two identical molecules, or with a different species. Those possibilities compete.

That is why molecular connectivity and three-dimensional geometry matter even after donor and acceptor sites have been identified.

07What to do with this method in the next question

When an HSC Chemistry question asks whether a molecule can form hydrogen bonds, start by identifying the donor and acceptor sites from the complete structure.

Then ask which molecules are interacting.

For a pure substance, check whether identical molecules can supply complementary donor and acceptor roles.

For a mixture, check whether one substance can donate to an acceptor on the other, even if neither substance forms the same interactions in its pure state.

For boiling-point comparisons, identify the hydrogen-bonding possibilities, then consider the other intermolecular forces and structural differences. Do not treat a measured temperature difference as the energy of a single hydrogen bond.

Finally, keep the physical process separate from the chemical structure. Boiling rearranges and separates molecules without removing their donor or acceptor sites. Chemical reactions, such as proton transfer, can actually change those sites.

The next useful decision is whether those possible hydrogen bonds make a substance sufficiently compatible with water to dissolve appreciably. For that distinction, see Can Molecules Without OH Groups Hydrogen Bond with Water?.

The site test tells you where hydrogen bonding is possible. The surrounding molecules, geometry, and competing attractions determine how much that possibility influences what you observe.