7. Alcohols, Phenols and Ethers

What you will learn

• Classify and name alcohols, phenols and ethers.
• Connect bent oxygen geometry and lone pairs with chemical behaviour.
• Plan preparation and recognise carbonyl/Grignard product classes.
• Separate O–H reactions, C–O reactions and ring reactions.
• Explain acidity, hydrogen bonding and solvent trends.
• Choose valid ether syntheses and cleavage pathways.

Big idea: the oxygen connections decide the reaction map.

A hydroxyl group on a saturated carbon, a hydroxyl group on an aromatic ring and an oxygen between two carbon groups lead to distinct chemistry. The differences become manageable when you label bonds before selecting a reaction.

Use the paired oxygen models to separate a connectivity shorthand from a spatial arrangement. Then return to the reaction tables: identify the bond that changes and explain why that particular bond is affected under the stated conditions.

Williamson’s ether-forming approach made bond disconnection a practical planning idea: combine a suitable alkoxide with an accessible electrophilic carbon. Later named phenol transformations show the same broad lesson—control the functional group and reaction pathway to introduce a substituent at a predictable site.

The same oxygen atom can connect to a saturated carbon, an aromatic carbon or two carbon groups. Those connections define different functional classes.

An alcohol has a hydroxyl group attached directly to a saturated carbon. A primary, secondary or tertiary label counts carbon groups attached to that hydroxyl-bearing carbon. The label does not count hydroxyl groups. Mono-, di- and polyhydric describe how many hydroxyl groups are present.

A phenol has hydroxyl directly attached to an aromatic ring carbon. A ring elsewhere in the molecule is not enough: benzyl alcohol has its hydroxyl on a saturated side-chain carbon and is an alcohol rather than a phenol. Compounds can carry more than one phenolic hydroxyl group.

An ether oxygen is bonded to two carbon groups. The groups may be alkyl, aryl or one of each. A symmetrical ether has matching groups; an unsymmetrical ether has different groups. Ether classification does not mean the oxygen is a simple rigid straight bridge.

Classify $\mathrm{HOCH_2CH_2OH}$ in two independent ways.

There are two hydroxyl groups, so it is dihydric. Each hydroxyl-bearing carbon is bonded to one other carbon, so each hydroxyl site is primary.

Result: A dihydric alcohol with two primary alcohol sites. Functional-group count and substitution class are different descriptions.

1. Is C₆H₅CH₂OH a phenol?
2. What makes an ether symmetrical?

Answers

1. No. The OH group is attached to a saturated side-chain carbon.
2. The two carbon groups bonded to oxygen are the same.

A name must identify both the carbon skeleton and the position of the oxygen-containing group.

For a simple alcohol, choose a suitable parent containing the hydroxyl-bearing carbon and use an -ol suffix. Give the principal hydroxyl group an appropriate low locant under the naming rules. For multiple hydroxyl groups use endings such as -diol or -triol, retaining the necessary locants.

Phenol is an established parent name. Number additional substituents relative to its hydroxyl-bearing carbon as required. An ether can be named as an alkoxy substituent on an appropriate parent; this makes an unsymmetrical ether unambiguous without treating oxygen as part of an alkane carbon chain.

Name $\mathrm{CH_3CH(OH)CH_2CH_2CH_3}$.

The parent contains five carbons. Number from the end that gives the hydroxyl its lower locant.

Result: Pentan-2-ol; the hydroxyl locant is $2$.

Name $\mathrm{CH_3OCH_2CH_2CH_3}$.

Use the longer carbon group as the propane parent and the smaller oxygen-containing substituent as methoxy.

Result: 1-Methoxypropane.

When converting a name into a structure, draw the parent skeleton before placing substituents. Then check valence at the branch points. Oxygen ordinarily has two bonds in these neutral functional groups. A compound name should not silently change an alcohol into an ether by moving the H atom.

1. What does the ending -diol tell you?
2. What parent name directly indicates an aromatic OH connection?

Answers

1. There are two hydroxyl groups used in the named parent description.
2. Phenol.

The oxygen centre is bent, and its lone pairs help explain hydrogen bonding, basicity and resonance effects.

An alcohol oxygen has two sigma bonds and two lone-pair domains in a simple local model. Repulsions and the actual bonding environment make the bonded-atom arrangement bent. Ether oxygen has the same basic count of bonded and lone-pair domains. A displayed C–O–C formula written on one line is a connectivity shorthand, not proof of a linear angle.

Phenol differs because a lone pair can interact with the aromatic electronic system. This conjugation influences bond character and helps stabilise the phenoxide ion relative to a simple alkoxide comparison. Do not infer that every oxygen lone pair participates identically or that a resonance drawing represents atoms jumping between positions.

A student draws an ether as a straight C–O–C rod because its condensed formula is written left to right. Correct the drawing.

Keep the same two carbon–oxygen connections. Bend the two bonds around oxygen and show its lone pairs in a fuller Lewis sketch.

Result: Connectivity is preserved, but the local bonded geometry is bent. The 3D model explains why a line of symbols is not a bond-angle measurement.

For an exam, distinguish three drawing tasks. A condensed formula records connectivity quickly. A Lewis formula shows bonding pairs, lone pairs and charges. A spatial sketch uses wedges and dashed bonds when orientation matters. Each adds information for a different purpose.

1. How many lone pairs are placed on an ordinary neutral alcohol oxygen in a Lewis model?
2. Why is a separate structural explanation needed for phenol?

Answers

1. Two.
2. Its oxygen lone-pair system can interact with the aromatic ring.

Alcohol and phenol reactions become easier to organise when you ask whether the O–H bond or the C–O bond is being changed.

Acid-catalysed hydration of an alkene adds water in a pattern governed by the pathway, commonly the Markovnikov alcohol. Hydroboration followed by oxidation often supplies the complementary less-substituted alcohol under its specific reagent sequence. These are different routes, not a claim that adding water alone gives either product at will.

$$\mathrm{CH_{3}CH=CH_{2}} + \mathrm{H_{2}O} \xrightarrow{\mathrm{H^+}} \mathrm{CH_{3}CH(OH)CH_{3}}$$

Reduction of an aldehyde gives a primary alcohol; reduction of a ketone gives a secondary alcohol. The reagent supplies reducing equivalents. Catalytic hydrogenation is one useful overall representation.

$$\mathrm{CH_{3}COCH_{2}CH_{3}} + \mathrm{H_{2}} \xrightarrow{\mathrm{catalyst}} \mathrm{CH_{3}CH(OH)CH_{2}CH_{3}}$$

A Grignard reagent adds its carbon group to a carbonyl carbon. Hydrolytic work-up converts the resulting magnesium alkoxide into an alcohol. Formaldehyde, other aldehydes and ketones lead respectively to primary, secondary and tertiary alcohol classes after the addition. Follow the new C–C bond instead of only memorising product classes.

$$\mathrm{CH_{3}MgBr} + \mathrm{HCHO} \longrightarrow \mathrm{CH_{3}CH_{2}OMgBr}$$

$$\mathrm{CH_{3}CH_{2}OMgBr} + \mathrm{H_{2}O} \longrightarrow \mathrm{CH_{3}CH_{2}OH} + \mathrm{MgOHBr}$$

Phenol can be obtained from a haloarene under appropriate strong substitution conditions followed by acidification. Fusion of a benzenesulfonate with alkali supplies another classical phenoxide route. Hydrolysis of a suitable aromatic diazonium salt replaces its diazonium position with hydroxyl.

$$\mathrm{C_{6}H_{5}SO_{3}Na} + 2 \mathrm{NaOH} \xrightarrow{\Delta} \mathrm{C_{6}H_{5}ONa} + \mathrm{Na_{2}SO_{3}} + \mathrm{H_{2}O}$$

$$\mathrm{C_{6}H_{5}N_{2}}^{+} + \mathrm{H_{2}O} \longrightarrow \mathrm{C_{6}H_{5}OH} + \mathrm{N_{2}} + \mathrm{H}^{+}$$

The cumene process passes through a hydroperoxide and then gives phenol together with propanone. The two products are linked by the carbon skeleton. A useful overall atom-balanced summary is shown below; it does not replace the peroxide-intermediate mechanism.

$$\mathrm{C_{9}H_{12}} + \mathrm{O_{2}} \longrightarrow \mathrm{C_{6}H_{5}OH} + \mathrm{CH_{3}COCH_{3}}$$

Alcohols and phenols can hydrogen-bond through O–H. This commonly raises boiling points relative to appropriate hydrocarbons or ethers. As a nonpolar carbon portion grows, water solubility often decreases. Branching and multiple hydroxyl groups modify the comparison. State which structural feature changes in a series.

Alcohols react with sufficiently active metals to produce alkoxides and hydrogen. Phenols are more acidic than typical aliphatic alcohols because phenoxide is better stabilised than a simple alkoxide by its electronic environment. This is a comparison of conjugate-base stability, not a claim that the O–H bond simply falls apart without a partner.

$$2 \mathrm{CH_{3}CH_{2}OH} + 2 \mathrm{Na} \longrightarrow 2 \mathrm{CH_{3}CH_{2}ONa} + \mathrm{H_{2}}$$

$$\mathrm{C_{6}H_{5}OH} + \mathrm{NaOH} \longrightarrow \mathrm{C_{6}H_{5}ONa} + \mathrm{H_{2}O}$$

Electron-withdrawing ring substituents can further stabilise a phenoxide-type conjugate base and increase acidity. Electron-donating substituents often act in the opposite direction, with position and mechanism of the effect important. Phenol is much weaker than a typical carboxylic acid; it does not respond to bicarbonate in the same simple way.

Ester formation joins an alcohol with an acyl-derived group. Alcohols can undergo acid-catalysed esterification with carboxylic acids. Phenols often require a more reactive acylating reagent for efficient acylation; do not transfer every alcohol condition directly to phenol.

$$\mathrm{CH_{3}COOH} + \mathrm{CH_{3}CH_{2}OH} \xrightarrow{\mathrm{H^+}} \mathrm{CH_{3}COOCH_{2}CH_{3}} + \mathrm{H_{2}O}$$

Protonating the hydroxyl group can make water a better leaving group than hydroxide. Substitution can then give a haloalkane; elimination can give an alkene. Carbocation-forming pathways can rearrange. Reagent, solvent, substitution class and temperature affect which pathway is favoured.

The classical Lucas comparison uses a hydrochloric-acid/zinc-chloride system and observes formation of a separate haloalkane phase. Under comparable ordinary test conditions, tertiary alcohols respond more readily than secondary ones, while simple primary alcohols are much slower. This is a substrate- and condition-dependent observation, not an exception-free stopwatch rule.

$$\mathrm{CH_{3}CH_{2}CH_{2}OH} + \mathrm{HBr} \longrightarrow \mathrm{CH_{3}CH_{2}CH_{2}Br} + \mathrm{H_{2}O}$$

$$\mathrm{CH_{3}CH(OH)CH_{3}} \xrightarrow{\mathrm{acid},\Delta} \mathrm{CH_{3}CH=CH_{2}} + \mathrm{H_{2}O}$$

Controlled oxidation of a primary alcohol can stop at an aldehyde or proceed to a carboxylic acid depending on reagent and conditions. Secondary alcohols commonly give ketones. Tertiary alcohols lack the required hydrogen on the hydroxyl-bearing carbon for that same simple carbonyl-forming route; stronger oxidation can instead break carbon–carbon bonds.

$$\mathrm{CH_{3}CH_{2}OH} \xrightarrow{\mathrm{Cu},\Delta} \mathrm{CH_{3}CHO} + \mathrm{H_{2}}$$

The hydroxyl group activates the aromatic ring towards many electrophilic substitutions and favours ortho and para positions. Reagent strength and medium determine whether one or several substitutions occur. Bromine water commonly gives extensive ring bromination, whereas carefully chosen milder conditions can give a different distribution.

$$\mathrm{C_{6}H_{5}OH} + 3 \mathrm{Br_{2}} \longrightarrow \mathrm{C_{6}H_{2}Br_{3}OH} + 3 \mathrm{HBr}$$

Kolbe–Schmitt carboxylation uses a phenoxide and carbon dioxide to introduce a carboxyl group, commonly at an ortho position under its characteristic conditions. Acid work-up gives the hydroxy acid. The Reimer–Tiemann transformation introduces a formyl group, commonly ortho, using its chloroform/alkali sequence followed by work-up.

Dilute nitric-acid conditions commonly give an ortho/para nitrophenol mixture; stronger nitration can give extensive substitution. An ortho nitrophenol can form an internal hydrogen bond, while a para isomer more readily associates between molecules. This helps explain different volatility and separation behaviour: identify the donor–acceptor alignment, not only the position names.

$$\mathrm{C_{6}H_{5}ONa} + \mathrm{CO_{2}} \xrightarrow{\mathrm{specified\ pressure\ and\ temperature}} \mathrm{C_{6}H_{4}(OH)COONa}$$

$$\mathrm{C_{6}H_{5}OH} + \mathrm{CHCl_{3}} + 3 \mathrm{NaOH} \longrightarrow \mathrm{C_{6}H_{4}(OH)CHO} + 3 \mathrm{NaCl} + 2 \mathrm{H_{2}O}$$

Heating phenol with zinc dust can remove oxygen, giving benzene. Oxidation of a phenolic ring can produce quinone-type structures under suitable conditions. Use conditions and product connectivity to distinguish these transformations from substitution at saturated carbon.

$$\mathrm{C_{6}H_{5}OH} + \mathrm{Zn} \longrightarrow \mathrm{C_{6}H_{6}} + \mathrm{ZnO}$$

Plan a product-class prediction for methylmagnesium bromide adding to propanone, followed by work-up.

Propanone has two methyl groups already attached to its carbonyl carbon. Addition supplies another methyl group, and the oxygen becomes hydroxyl after work-up.

Result: The product is a tertiary alcohol, 2-methylpropan-2-ol. The carbonyl carbon becomes the hydroxyl-bearing carbon; no carbon is lost.

1. Why is phenol more acidic than a simple alcohol?
2. What alcohol class usually results from Grignard addition to a ketone followed by work-up?
3. Does tertiary alcohol resistance to simple carbonyl oxidation imply immunity to all oxidation?

Answers

1. Its conjugate base is stabilised by its aromatic electronic environment relative to a simple alkoxide comparison.
2. A tertiary alcohol.
3. No. Stronger conditions can cause other changes, including C–C bond cleavage.

Two small alcohols have large industrial importance, but their similar names do not make their biological effects or production routes interchangeable.

Methanol is an important chemical feedstock and solvent. A classical industrial synthesis combines carbon monoxide with hydrogen under appropriate catalytic conditions. Its uses do not make it suitable for ingestion; methanol is toxic.

$$\mathrm{CO} + 2 \mathrm{H_{2}} \xrightarrow{\mathrm{industrial\ catalyst}} \mathrm{CH_{3}OH}$$

Ethanol can be produced by biological conversion of suitable sugars or by industrial hydration of ethene. These routes give the same molecular compound but operate through different systems. Fermentation has biological constraints, while direct hydration uses a catalyst and suitable physical conditions.

$$\mathrm{C_{6}H_{12}O_{6}} \xrightarrow{\mathrm{biological\ conversion}} 2 \mathrm{C_{2}H_{5}OH} + 2 \mathrm{CO_{2}}$$

$$\mathrm{CH_{2}=CH_{2}} + \mathrm{H_{2}O} \xrightarrow{\mathrm{catalyst}} \mathrm{CH_{3}CH_{2}OH}$$

Concentrating ethanol from water requires attention to phase behaviour, including azeotropic limitations discussed in the solutions chapter. Industrial alcohol may contain deliberate additives or other contaminants. A chemistry label or apparent purity does not establish suitability for consumption.

Do fermentation and ethene hydration give different alcohol molecules?

Compare the product connectivity, not the production method. Each route can give the same ethanol structure. Impurities and process context may differ.

Result: The product molecule can be the same, while preparation, mixture composition and suitability for a use differ.

1. What carbon-containing feedstock appears in the classical methanol synthesis shown?
2. Why can ethanol–water separation connect back to the solutions chapter?

Answers

1. Carbon monoxide.
2. The mixture’s vapour–liquid behaviour and azeotrope affect ordinary distillation.

Ethers share oxygen with alcohols, but the absence of an O–H bond changes hydrogen bonding and several reaction pathways.

Acid-catalysed dehydration of suitable primary alcohols can give symmetrical ethers under appropriate conditions. Elimination competes, and crowded alcohols are poor choices for a simple clean intermolecular substitution route.

$$2 \mathrm{CH_{3}CH_{2}CH_{2}OH} \xrightarrow{\mathrm{acid,\ suitable\ conditions}} \mathrm{CH_{3}CH_{2}CH_{2}OCH_{2}CH_{2}CH_{3}} + \mathrm{H_{2}O}$$

Williamson synthesis uses an alkoxide and a suitable alkyl electrophile. The bond forms through nucleophilic substitution. Choose a less-crowded, commonly primary electrophile to favour substitution over elimination. A tertiary halide is generally a poor electrophile for this SN2-based preparation.

$$\mathrm{CH_{3}ONa} + \mathrm{CH_{3}CH_{2}CH_{2}Br} \longrightarrow \mathrm{CH_{3}OCH_{2}CH_{2}CH_{3}} + \mathrm{NaBr}$$

Choose a useful Williamson disconnection for 1-methoxypropane.

Break one C–O bond conceptually. Methoxide plus a primary propyl halide is a suitable substitution pairing. An alternative methyl-halide electrophile with propoxide can also be considered under suitable conditions.

Result: Use a primary or methyl electrophile rather than forcing the nucleophile to attack a tertiary carbon.

An ether cannot donate an O–H hydrogen bond to another identical ether molecule, but its oxygen can accept hydrogen bonds from water or another donor. This explains why “cannot hydrogen-bond to itself in the alcohol way” does not mean “cannot interact through hydrogen bonds at all”. Its carbon portion still affects water solubility.

Strong hydrogen halides can cleave suitable ethers after oxygen protonation. For many simple unsymmetrical alkyl ethers, attack at a less-hindered carbon supports an SN2 pathway. If an appropriate tertiary carbocation route is available, different selectivity can occur.

$$\mathrm{CH_{3}OCH_{2}CH_{2}CH_{3}} + \mathrm{HI} \longrightarrow \mathrm{CH_{3}I} + \mathrm{CH_{3}CH_{2}CH_{2}OH}$$

In an aryl alkyl ether, cleavage commonly breaks the alkyl–oxygen bond rather than the aromatic carbon–oxygen bond. The latter is not an ordinary accessible alkyl SN2 site. Thus methoxybenzene can give a methyl halide plus phenol in a familiar acid-cleavage comparison.

$$\mathrm{C_{6}H_{5}OCH_{3}} + \mathrm{HI} \longrightarrow \mathrm{C_{6}H_{5}OH} + \mathrm{CH_{3}I}$$

The alkoxy group activates an aromatic ring and directs many electrophilic substitutions towards ortho and para positions. Ether oxygen donation and ring attachment must be distinguished from a simple saturated ether’s substitution chemistry. Some ethers can also form hazardous peroxides on exposure to air during storage; apparent solvent stability is not a guarantee of inertness.

Methoxybenzene can undergo ring halogenation, nitration and suitable Friedel–Crafts alkylation or acylation, with strong ortho/para influence. These are aromatic substitutions, distinct from cleavage of the methyl–oxygen bond.

1. Why is a tertiary halide usually a poor Williamson electrophile?
2. Which bond is usually cleaved in an aryl methyl ether by HI?

Answers

1. Crowding inhibits SN2 and elimination can dominate.
2. The methyl–oxygen bond, producing methyl iodide and phenol in the familiar comparison.

Try these original questions without looking at the answers. Show equations, conditions and units when relevant.

7.1 Classify benzyl alcohol and phenol by their OH-bearing carbons.

7.2 Distinguish monohydric from primary.

7.3 Name $\mathrm{CH_3CH_2CH(OH)CH_2CH_3}$.

7.4 Draw butane-1,2-diol in condensed form.

7.5 Name $\mathrm{CH_3CH_2OCH_2CH_2CH_3}$.

7.6 Why should C–O–C not be drawn automatically as a linear local geometry?

7.7 What product class follows ordinary ketone reduction?

7.8 What product class follows ordinary aldehyde reduction?

7.9 Why can a tertiary alcohol form after Grignard addition to a ketone?

7.10 What happens if a Grignard reagent meets water before carbonyl addition?

7.11 Give one route from an aromatic diazonium group to phenol.

7.12 Name the companion organic product in the cumene-to-phenol process.

7.13 Why do comparable alcohols often boil higher than ethers?

7.14 Why does water solubility often decrease as an alcohol carbon skeleton grows?

7.15 Explain phenol’s acidity using its conjugate base.

7.16 Would an electron-withdrawing ring substituent commonly strengthen phenolic acidity? Explain.

7.17 What gas is released when a suitable alcohol reacts with sodium?

7.18 What structural change occurs in alcohol esterification?

7.19 Why can an acid-activated OH group leave more readily?

7.20 Give the main simple carbonyl oxidation product of a secondary alcohol.

7.21 Why is simple aldehyde/ketone formation by the same route unsuitable for a tertiary alcohol?

7.22 Explain why phenol commonly directs electrophiles ortho and para.

7.23 What functional group is introduced in Kolbe–Schmitt phenoxide carboxylation?

7.24 What group is introduced by the Reimer–Tiemann transformation?

7.25 What does heating phenol with zinc dust illustrate?

7.26 Why is a historical industrial use not proof that an alcohol is safe to consume?

7.27 Give an atom-balanced summary of glucose fermentation to ethanol.

7.28 Select a suitable electrophile class for a Williamson ether synthesis.

7.29 Why may acid dehydration of a tertiary alcohol fail as a clean symmetrical-ether route?

7.30 Can an ether accept a hydrogen bond from water?

7.31 Give the familiar products of HI cleavage of methoxybenzene.

7.32 Why is aryl–oxygen cleavage not predicted by ordinary methyl SN2 reasoning?

7.33 Plan a mechanism-based alcohol/ether reaction answer.

Answers

7.1 Benzyl alcohol is an alcohol on a saturated side-chain carbon; phenol has OH directly on an aromatic carbon.

7.2 Monohydric counts OH groups; primary counts carbon neighbours at the OH-bearing carbon.

7.3 Pentan-3-ol.

7.4 $\mathrm{HOCH_2CH(OH)CH_2CH_3}$.

7.5 1-Ethoxypropane.

7.6 The oxygen bonding and lone-pair domains produce a bent bonded-atom arrangement.

7.7 A secondary alcohol.

7.8 A primary alcohol.

7.9 The carbonyl carbon already has two carbon groups and gains a third before becoming the OH-bearing carbon.

7.10 Proton transfer consumes the reagent and produces its corresponding hydrocarbon.

7.11 Hydrolysis under suitable conditions replaces the diazonium group and releases nitrogen.

7.12 Propanone.

7.13 Alcohols can donate and accept an extended intermolecular O–H hydrogen-bond network.

7.14 The growing nonpolar portion becomes more influential relative to its hydroxyl interaction with water.

7.15 Phenoxide is stabilised by its aromatic electronic environment relative to an ordinary alkoxide comparison.

7.16 It can stabilise the conjugate base further, with the effect depending on position and the electronic pathway.

7.17 Hydrogen.

7.18 The alcohol oxygen becomes bonded to an acyl group, with the appropriate accompanying products for the chosen reagent.

7.19 Protonation permits departure as water rather than hydroxide.

7.20 A ketone.

7.21 There is no hydrogen on the OH-bearing carbon to remove in that straightforward transformation.

7.22 Oxygen donation stabilises relevant ring-substitution intermediates at those positions.

7.23 A carboxyl-derived group, commonly ortho after the full work-up.

7.24 A formyl group, commonly ortho in the principal product.

7.25 Removal of oxygen to give benzene in the familiar reduction/deoxygenation comparison.

7.26 Toxicity, impurities and intended use differ; a chemical use does not establish biological suitability.

7.27 $\mathrm{C_6H_{12}O_6\rightarrow2C_2H_5OH+2CO_2}$.

7.28 A methyl or suitable primary alkyl electrophile is commonly preferred for SN2 access.

7.29 Elimination and carbocation pathways compete strongly with the desired intermolecular substitution.

7.30 Yes, through an oxygen lone pair, even though the ether has no O–H donor.

7.31 Phenol and methyl iodide, $\mathrm{C_6H_5OH}$ and $\mathrm{CH_3I}$.

7.32 The aromatic carbon is not an accessible saturated SN2 centre, and its bond environment differs.

7.33 Locate the oxygen group, identify whether O–H or C–O changes, activate the relevant bond if needed, evaluate substitution versus elimination, then draw product connectivity with conditions.

Quick-check answers appear directly after each topic. The selected explanations below show how to check the reasoning, rather than memorise a final line.

Why may acid dehydration of a tertiary alcohol fail as a clean symmetrical-ether route?

Elimination and carbocation pathways compete strongly with the desired intermolecular substitution.

Can an ether accept a hydrogen bond from water?

Yes, through an oxygen lone pair, even though the ether has no O–H donor.

Give the familiar products of HI cleavage of methoxybenzene.

Phenol and methyl iodide, $\mathrm{C_6H_5OH}$ and $\mathrm{CH_3I}$.

Why is aryl–oxygen cleavage not predicted by ordinary methyl SN2 reasoning?

The aromatic carbon is not an accessible saturated SN2 centre, and its bond environment differs.

Plan a mechanism-based alcohol/ether reaction answer.

Locate the oxygen group, identify whether O–H or C–O changes, activate the relevant bond if needed, evaluate substitution versus elimination, then draw product connectivity with conditions.