Alcohols, Phenols and Ethers

The chapter opens with classification (7.1) — mono/di/tri/polyhydric alcohols and phenols, and primary/secondary/tertiary/allylic/benzylic/vinylic alcohols sorted by the hybridisation of the carbon bearing $-OH$ — before fixing IUPAC and common nomenclature (7.2) for all three families, and comparing bond angles and bond lengths across methanol, phenol and methoxymethane (7.3). Section 7.4 is the heart of the chapter: preparation of alcohols from alkenes, carbonyl compounds and Grignard reagents (7.4.1), preparation of phenols from haloarenes, benzenesulphonic acid, diazonium salts and cumene (7.4.2), the physical properties that hydrogen bonding controls (7.4.3), and the full chemistry (7.4.4) split cleanly into reactions that break the $O-H$ bond (acidity, esterification — shared by both alcohols and phenols), reactions that break the $C-O$ bond (substitution, dehydration, oxidation — alcohols only), and reactions unique to the aromatic ring of phenols (nitration, halogenation, Kolbe's reaction, Reimer-Tiemann reaction, reduction with zinc, oxidation to quinones). Section 7.5 profiles the two industrially vital alcohols, methanol and ethanol. Section 7.6 repeats the preparation-properties-reactions arc for ethers: dehydration and Williamson synthesis, boiling point and solubility trends, and the cleavage and electrophilic substitution chemistry of the $C-O-C$ linkage. One question should run through every section: does this reaction break the $O-H$ bond, the $C-O$ bond, or does it happen on the ring — because that single fork sorts almost the whole chapter.

Classifying alcohols, phenols and ethers

An alcohol has one or more hydroxyl groups attached to an $sp^3$ carbon of an aliphatic system, such as $CH_3OH$; a phenol has the hydroxyl group attached directly to an $sp^2$ carbon of an aromatic ring, such as $C_6H_5OH$. Both are classified by hydroxyl count as monohydric (one $-OH$), dihydric (two), trihydric (three), or polyhydric (many) — ethane-1,2-diol and propane-1,2,3-triol are the dihydric and trihydric alcohols everyone meets first, as glycol and glycerol. Monohydric alcohols split further by the hybridisation of the carbon carrying $-OH$. Where that carbon is $sp^3$, the alcohol is primary, secondary or tertiary according to whether the hydroxyl-bearing carbon is itself attached to one, two or three other carbons; an allylic alcohol has $-OH$ on an $sp^3$ carbon immediately next to a carbon-carbon double bond, and a benzylic alcohol has $-OH$ on an $sp^3$ carbon immediately next to an aromatic ring — both allylic and benzylic alcohols can independently be primary, secondary or tertiary. Where the hydroxyl-bearing carbon is $sp^2$ instead, the alcohol is a vinylic alcohol ($CH_2=CH-OH$, $-OH$ directly on a double-bonded carbon) or a phenol ($-OH$ directly on an aromatic ring carbon); vinylic alcohols are unstable and rapidly tautomerise to the corresponding carbonyl compound, which is why they are rarely isolated. Ethers are classified by whether the two groups flanking oxygen match: a simple or symmetrical ether has identical alkyl or aryl groups on both sides of oxygen (diethyl ether, $C_2H_5-O-C_2H_5$), while a mixed or unsymmetrical ether has two different groups ($C_2H_5-O-CH_3$ or $C_2H_5-O-C_6H_5$).

Nomenclature of alcohols, phenols and ethers

Common names build on the alkyl group plus the word "alcohol" (methyl alcohol for $CH_3OH$); IUPAC names replace the terminal "-e" of the parent alkane with "-ol", numbering the longest chain containing $-OH$ from whichever end gives the hydroxyl group the lower locant — methanol, propan-1-ol, propan-2-ol (isopropyl alcohol), butan-2-ol (sec-butyl alcohol), 2-methylpropan-2-ol (tert-butyl alcohol). Polyhydric alcohols keep the "-e" of the alkane and add "-diol", "-triol" and so on with locants for every $-OH$: $HOCH_2CH_2OH$ is ethane-1,2-diol (glycol), and glycerol is propane-1,2,3-triol. Cyclic alcohols use the prefix "cyclo-" with the hydroxyl carbon numbered C-1, as in cyclohexanol. Phenol itself is both the common and the accepted IUPAC name for $C_6H_5OH$; substituted phenols still use ortho (1,2-), meta (1,3-) and para (1,4-) informally alongside systematic locants — o-cresol is 2-methylphenol, m-cresol is 3-methylphenol. The three dihydroxybenzenes carry historic common names worth knowing alongside their systematic ones: catechol (benzene-1,2-diol), resorcinol (benzene-1,3-diol), and hydroquinone or quinol (benzene-1,4-diol). Common ether names list both alkyl/aryl groups alphabetically before the word "ether" ($CH_3OC_2H_5$ is ethylmethyl ether; identical groups take the prefix "di-", as in diethyl ether); IUPAC treats an ether as a hydrocarbon in which one hydrogen is replaced by an alkoxy or aryloxy group, $-OR$ or $-OAr$, choosing the larger group as the parent chain and naming the smaller as the substituent — $CH_3OCH_3$ is methoxymethane, $C_2H_5OC_2H_5$ is ethoxyethane, and $C_6H_5OCH_3$ (anisole) is methoxybenzene.

Structure and bonding: bond angles and bond lengths

In alcohols the $C-O-H$ angle is close to, but slightly less than, the tetrahedral $109.5\degree$ because the two lone pairs on oxygen repel the bonding pairs more than the bonding pairs repel each other. In phenol the $C-O$ bond length is $136\ \text{pm}$, shorter than the $C-O$ bond in methanol, for two combined reasons: the hydroxyl carbon is $sp^2$ hybridised (a shorter, more s-character-rich bond than an $sp^3$ carbon gives) and the oxygen's lone pair conjugates into the aromatic ring, lending the $C-O$ bond partial double-bond character — the same conjugation that will explain phenol's acidity below. In ethers the $C-O-C$ bond angle is slightly greater than tetrahedral, since the two bulky alkyl groups repel each other more than two hydrogens would; the $C-O$ bond length, $141\ \text{pm}$, is close to that of an alcohol's $C-O$ bond.

where: $n$ is the number of carbons in a saturated, monohydric, acyclic alcohol; the general formula mirrors the parent alkane $C_nH_{2n+2}$ with one hydrogen replaced by $-OH$, and the same substitution pattern, one hydrogen for one $-OH$, underlies the general formulas of both mono-substituted phenols and simple ethers.

Preparation of alcohols from alkenes

Alkenes give alcohols by two routes with opposite regiochemistry. Acid-catalysed hydration adds water across the double bond in the presence of dilute acid, following Markovnikov's rule for unsymmetrical alkenes — the hydroxyl group ends up on the more substituted carbon, because the mechanism proceeds through the more stable carbocation (worked in full in the first derivation below). Hydroboration-oxidation, discovered by H. C. Brown (who shared the 1979 Nobel Prize in Chemistry with G. Wittig for this and related boron chemistry), adds diborane, $(BH_3)_2$, across the double bond so that boron attaches to the less substituted, less hindered carbon; oxidative workup with $H_2O_2$ and aqueous $NaOH$ then replaces boron with $-OH$ with retention of that regiochemistry, giving the anti-Markovnikov alcohol in excellent yield. Reading the two routes together: hydration puts $-OH$ where the carbocation would be most stable (more substituted carbon), while hydroboration-oxidation puts $-OH$ exactly opposite (less substituted carbon) — the single most exam-tested contrast in this section.

Preparation of alcohols from carbonyl compounds and Grignard reagents

Aldehydes and ketones are reduced to alcohols by catalytic hydrogenation over finely divided platinum, palladium or nickel, or by the hydride reagents $NaBH_4$ or $LiAlH_4$: aldehydes give primary alcohols, ketones give secondary alcohols. Carboxylic acids are reduced to primary alcohols by $LiAlH_4$, a strong but expensive reducing agent reserved for special syntheses; commercially, acids are instead first esterified and the ester then catalytically hydrogenated to the alcohol.

Grignard reagents add to the carbonyl carbon of an aldehyde or ketone to form a magnesium alkoxide adduct, which aqueous acid hydrolysis converts to the alcohol. Because the identity of the carbonyl compound fixes how substituted the resulting carbon is, methanal ($HCHO$) gives a primary alcohol, any other aldehyde gives a secondary alcohol, and any ketone gives a tertiary alcohol — one reagent, three possible outcomes, sorted entirely by which carbonyl compound is chosen.

Preparation of phenols

Four routes convert benzene derivatives to phenol. From haloarenes, chlorobenzene is fused with $NaOH$ at $623\ \text{K}$ and $320\ \text{atm}$, and acidifying the resulting sodium phenoxide liberates phenol. From benzenesulphonic acid, benzene is sulphonated with oleum, and the sulphonic acid is converted to sodium phenoxide by fusion with molten $NaOH$, then acidified. From diazonium salts, an aromatic primary amine treated with nitrous acid ($NaNO_2 + HCl$) at $273-278\ \text{K}$ gives a diazonium salt, which hydrolyses to phenol on warming with water or dilute acid. From cumene, the industrial route that supplies most of the world's phenol: cumene (isopropylbenzene) is air-oxidised to cumene hydroperoxide, which dilute acid converts to phenol with acetone as a valuable by-product — a single process that supplies two major industrial chemicals from one hydrocarbon feedstock.

Physical properties: boiling point and solubility

Boiling points of alcohols and phenols rise with chain length (stronger van der Waals forces) and fall with branching (less surface area for neighbouring molecules to grip), but the dominant effect is intermolecular hydrogen bonding through the $-OH$ group, which gives alcohols and phenols markedly higher boiling points than hydrocarbons, ethers, haloalkanes or haloarenes of comparable molecular mass. Ethanol (46 g/mol) boils at $351\ \text{K}$ while propane (44 g/mol), with no hydrogen bonding at all, boils at only $231\ \text{K}$; methoxymethane (46 g/mol), whose oxygen can accept but not donate a hydrogen bond to another ether molecule, boils at an intermediate $248\ \text{K}$. Solubility in water follows the same logic: alcohols and phenols hydrogen-bond to water molecules, and the lower members are miscible with water in all proportions, but solubility falls as the hydrophobic alkyl or aryl portion grows larger relative to the single hydrogen-bonding $-OH$ group.

Acidity of alcohols and phenols

Alcohols and phenols react with active metals (Na, K, Al) to liberate hydrogen and form alkoxides or phenoxides, and phenols additionally dissolve in aqueous $NaOH$ to form sodium phenoxide — a reaction alcohols do not give, since alcohols are too weakly acidic for hydroxide ion to deprotonate them. Both are Bronsted acids, donating a proton to a sufficiently strong base, but their acid strengths differ enormously.

where: $K_a$ is the acid dissociation constant and $pK_a$ its negative logarithm, so a lower $pK_a$ means a stronger acid. Among simple alcohols, electron-releasing alkyl groups push electron density onto oxygen, which strengthens (shortens and stiffens) the $O-H$ bond and destabilises the alkoxide's negative charge, so acid strength falls as more alkyl substitution accumulates on the hydroxyl carbon: methanol is the most acidic simple alcohol, tertiary alcohols the least. All simple alcohols are, however, weaker acids than water itself — an alkoxide ion is therefore a stronger base than hydroxide, which is why sodium ethoxide, not sodium hydroxide, is used when a stronger base is needed in organic synthesis. Phenol breaks this pattern dramatically: its $pK_a$ is $10.0$, versus $15.9$ for ethanol, meaning phenol is roughly a million times more acidic. Table 7.3's data sharpens the trend further — electron-withdrawing nitro groups lower $pK_a$ (raise acidity) most strongly from the ortho and para positions (o-nitrophenol $pK_a = 7.2$, p-nitrophenol $7.1$, meta only $8.3$), while electron-releasing methyl groups raise $pK_a$ (lower acidity) modestly, as in o-, m- and p-cresol, all around $pK_a = 10.1-10.2$. The full resonance argument behind this million-fold jump, and behind the ortho/para-versus-meta substituent pattern, is the subject of the second derivation below.

Esterification of alcohols and phenols

Both alcohols and phenols react with carboxylic acids, acid chlorides and acid anhydrides to form esters, the introduction of an acetyl group being called acetylation when the acylating agent is ethanoic acid, its chloride or its anhydride. Esterification with a carboxylic acid or anhydride needs a trace of concentrated $H_2SO_4$ as catalyst and is reversible, so water is removed as it forms to drive the equilibrium toward ester; esterification with an acid chloride is carried out in the presence of a base such as pyridine, which neutralises the $HCl$ by-product and pulls the equilibrium the same way. Acetylation of salicylic acid by this route produces aspirin, giving this seemingly abstract reaction a familiar medicine-cabinet outcome.

where: the reaction is an equilibrium favouring ester formation only when water is continually removed, and the acid-chloride variant simply replaces $R'COOH$ with $R'COCl$ and pyridine for the acid catalyst, releasing $HCl$ instead of $H_2O$.

Substitution and dehydration reactions of alcohols

Reactions that break the $C-O$ bond happen only in alcohols, never (except with zinc dust) in phenols — the aryl $C-O$ bond's partial double-bond character, from the same ring conjugation that boosts phenol's acidity, makes it far too strong for ordinary nucleophilic substitution. Alcohols react with hydrogen halides to give alkyl halides, $ROH + HX \rightarrow RX + H_2O$; the differing reactivity of primary, secondary and tertiary alcohols with concentrated $HCl$/anhydrous $ZnCl_2$ (the Lucas reagent) is the basis of the Lucas test — tertiary alcohols turn turbid almost immediately as the halide separates out, secondary alcohols after some minutes, and primary alcohols not at all at room temperature. Alcohols also react with phosphorus trihalides ($PX_3$) to give alkyl halides directly.

Dehydration — loss of a water molecule to give an alkene — occurs on heating with concentrated $H_2SO_4$ or $H_3PO_4$, or over anhydrous $ZnCl_2$ or alumina; ethanol needs $443\ \text{K}$ with concentrated $H_2SO_4$, while secondary and tertiary alcohols dehydrate under noticeably milder conditions, giving the reactivity order tertiary greater than secondary greater than primary — exactly the order of increasing carbocation stability, which the third derivation below traces mechanistically.

Oxidation of alcohols

Oxidation forms a carbon-oxygen double bond by cleaving an $O-H$ and a $C-H$ bond together, so it is also called dehydrogenation. A primary alcohol is oxidised first to an aldehyde and, with a stronger oxidant, on to a carboxylic acid; strong agents such as acidified $KMnO_4$ take a primary alcohol straight to the acid, while anhydrous $CrO_3$, or better still pyridinium chlorochromate (PCC) — a complex of $CrO_3$ with pyridine and $HCl$ — stops cleanly at the aldehyde. A secondary alcohol is oxidised by $CrO_3$ to a ketone and, under ordinary conditions, no further. A tertiary alcohol has no $C-H$ bond on its hydroxyl carbon and so resists oxidation entirely; only under harsh conditions (strong oxidants, elevated temperature) do various $C-C$ bonds cleave, yielding a mixture of smaller carboxylic acids. Passing alcohol vapour over heated copper at $573\ \text{K}$ achieves the same carbon-type-dependent outcome by catalytic dehydrogenation: primary alcohols give aldehydes, secondary alcohols give ketones, and tertiary alcohols, lacking the needed $C-H$ bond, dehydrate to an alkene instead. In the body, this same chemistry turns ingested methanol into methanal and then the highly toxic methanoic acid, which is why methanol poisoning causes blindness and is treated with intravenous ethanol to outcompete methanol for the oxidising enzyme.

Reactions unique to phenols: electrophilic substitution and named reactions

The $-OH$ group on a benzene ring donates electron density by resonance, activating the ring toward electrophilic aromatic substitution and directing incoming electrophiles ortho and para, where the resonance-donated electron density concentrates. Nitration with dilute $HNO_3$ at $298\ \text{K}$ gives a mixture of ortho- and para-nitrophenol, separable by steam distillation because the ortho isomer's intramolecular hydrogen bond (chelation with the adjacent $-OH$) makes it more volatile, while the para isomer's intermolecular hydrogen bonding (linking molecules together) makes it less so; concentrated $HNO_3$ instead gives 2,4,6-trinitrophenol, picric acid, though in poor yield directly (an industrial route first sulphonates phenol, then displaces the sulphonic groups with nitro groups under concentrated $HNO_3$). Halogenation with bromine in a low-polarity solvent (CHCl₃ or CS₂) at low temperature gives monobromophenol without any Lewis-acid catalyst — the ring is so strongly activated by $-OH$ that bromine polarises spontaneously — while bromine water gives white, crystalline 2,4,6-tribromophenol directly. Two further reactions belong to phenol alone. Kolbe's reaction: sodium phenoxide, generated by treating phenol with $NaOH$, is more electron-rich than phenol itself and reactive enough to undergo electrophilic substitution even with the weak electrophile $CO_2$, giving, after acidification, ortho-hydroxybenzoic acid (salicylic acid) as the main product. Reimer-Tiemann reaction: phenol treated with chloroform and $NaOH$ introduces a $-CHO$ group at the ortho position via a dichlorocarbene intermediate that alkylates the ring, and alkaline hydrolysis of the resulting benzal chloride liberates salicylaldehyde. Phenol is also reduced straight to benzene by zinc dust — the one $C-O$ cleavage reaction phenols do undergo — and is slowly oxidised in air, or more rapidly by chromic acid, to conjugated diketones called quinones (benzoquinone), the dark discolouration familiar from phenol left standing.

Commercially important alcohols: methanol and ethanol

Methanol, once called "wood spirit" from its origin in the destructive distillation of wood, is now made by catalytic hydrogenation of carbon monoxide over a $ZnO-Cr_2O_3$ catalyst at high pressure and temperature. It is a colourless liquid, boiling point $337\ \text{K}$, highly poisonous — even small ingested quantities can cause blindness, larger ones death — and is used chiefly as a paint/varnish solvent and in manufacturing formaldehyde. Ethanol is obtained commercially by the fermentation of sugars: glucose and fructose (both $C_6H_{12}O_6$), produced from sucrose by the enzyme invertase, are converted to ethanol and $CO_2$ by the enzyme zymase, found in yeast, under anaerobic conditions; fermentation self-limits once the alcohol content exceeds about 14%, because zymase activity is inhibited beyond that point. Ethanol boils at $351\ \text{K}$, is used widely as a solvent and industrial feedstock, and is also made in bulk by direct hydration of ethene. Commercial (non-beverage) ethanol is rendered undrinkable — denatured — by adding copper sulphate (colour) and pyridine (foul odour).

Ethers: preparation, physical properties and reactions

Ethers are made two ways. Dehydration of alcohols under a protic acid gives ether rather than alkene at a lower temperature than alkene formation requires — ethanol gives ethoxyethane at $413\ \text{K}$ under $H_2SO_4$ versus ethene at $443\ \text{K}$ — through a bimolecular ($S_N2$) attack of a neutral alcohol molecule on a protonated alcohol; this route works well only for unhindered primary alcohols, since secondary and tertiary alcohols favour elimination (to the alkene) over substitution and give poor ether yields. Williamson synthesis, the general laboratory method for both symmetrical and unsymmetrical ethers, reacts an alkyl halide with a sodium alkoxide or aryloxide, $R-X + R'-O^-Na^+ \rightarrow R-O-R' + NaX$, by an $S_N2$ mechanism that works best with a primary alkyl halide; with a secondary or, especially, tertiary alkyl halide the strongly basic alkoxide favours elimination instead, so a tertiary halide such as $(CH_3)_3C-Br$ with $CH_3O^-Na^+$ gives only 2-methylpropene, no ether at all. Phenols, too, form aryl ethers by this method, reacting as the phenoxide. Ether $C-O$ bonds are weakly polar, giving ethers a small net dipole moment, but with no $-OH$ to hydrogen-bond to another ether molecule, their boiling points closely resemble alkanes of similar molecular mass and sit well below the corresponding alcohol's boiling point (ethoxyethane and n-pentane both boil near $308\ \text{K}$, versus $390\ \text{K}$ for butan-1-ol). Their water solubility, however, resembles alcohols of similar mass, since the ether oxygen can still accept a hydrogen bond from a water molecule even though it cannot donate one. Chemically, ethers are the least reactive functional group covered in this chapter; their $C-O$ bond cleaves only under drastic conditions with excess concentrated hydrogen halide, in the reactivity order $HI$ greater than $HBr$ greater than $HCl$. A dialkyl ether gives two molecules of alkyl halide; an alkyl aryl ether (anisole-type) cleaves selectively at the alkyl-oxygen bond, since the aryl-oxygen bond is strengthened by the same partial double-bond character seen in phenol, giving phenol plus alkyl halide — phenol itself does not react further, because its $sp^2$ ring carbon cannot undergo nucleophilic substitution. The alkoxy group $-OR$ activates an aromatic ring and directs ortho/para exactly like phenol's $-OH$: anisole brominates in ethanoic acid without any catalyst (90% para yield), undergoes Friedel-Crafts alkylation and acylation with $AlCl_3$, and nitrates with the mixed-acid combination to give ortho- and para-nitroanisole.

Board level - Ionisation of phenol at neutral pH using the Henderson-Hasselbalch equation

Phenol has $pK_a = 10.0$ (Table 7.3). Using the Henderson-Hasselbalch equation, $pH = pK_a + \log_{10}\dfrac{[\text{C}_6\text{H}_5\text{O}^-]}{[\text{C}_6\text{H}_5\text{OH}]}$, find the ratio of ionised phenoxide to un-ionised phenol in a solution held at pH 7.0, and state roughly what fraction of the phenol is ionised.

Step 1. Substitute the given values into the Henderson-Hasselbalch equation: $7.0 = 10.0 + \log_{10}\dfrac{[\text{C}_6\text{H}_5\text{O}^-]}{[\text{C}_6\text{H}_5\text{OH}]}$.

Step 2. Isolate the logarithm term: $\log_{10}\dfrac{[\text{C}_6\text{H}_5\text{O}^-]}{[\text{C}_6\text{H}_5\text{OH}]} = 7.0 - 10.0 = -3.0$.

Step 3. Undo the logarithm: $\dfrac{[\text{C}_6\text{H}_5\text{O}^-]}{[\text{C}_6\text{H}_5\text{OH}]} = 10^{-3.0} = 0.001$, so the ratio is $1:1000$. Out of every $1001$ phenol species present, only about $1$ is ionised, roughly $0.1\%$. Answer: the phenoxide-to-phenol ratio is 1:1000, so about 0.1 percent of phenol is ionised at pH 7 — phenol behaves as an essentially un-ionised, neutral molecule at physiological or neutral pH, unlike a typical carboxylic acid ($pK_a \approx 4-5$), which would be almost completely ionised at the same pH; this is also why phenol, unlike a carboxylic acid, does not liberate $CO_2$ from aqueous $NaHCO_3$.

Board level - Ranking boiling points across alcohols, an alkane and an aldehyde (NCERT Example 7.3 extended)

Arrange the following in increasing order of boiling point, and justify the order: pentan-1-ol, n-butane, pentanal, ethoxyethane.

Step 1. Identify which compounds can hydrogen-bond to each other. Pentan-1-ol has an $-OH$ group and can both donate and accept hydrogen bonds between its own molecules; n-butane has no polar group at all; ethoxyethane's oxygen can only accept a hydrogen bond (from something else) but has no $O-H$ to donate, so it cannot hydrogen-bond to another ether molecule; pentanal's carbonyl oxygen is similarly an acceptor only, but the $C=O$ dipole is somewhat stronger than an ether's $C-O-C$ dipole.

Step 2. Rank by intermolecular force strength at comparable molecular mass: n-butane (only weak van der Waals forces) is lowest-boiling, followed by ethoxyethane (weak dipole, no self-hydrogen-bonding), then pentanal (a stronger permanent dipole from the polarised carbonyl but still no self-hydrogen-bonding), with pentan-1-ol highest of the four because only it can form a full hydrogen-bonded network with its own molecules.

Step 3. Confirm this matches the general rule stated in the core ideas: alcohols outrank ethers, aldehydes and hydrocarbons of similar mass specifically because of $O-H\cdots O$ hydrogen bonding, which none of the other three functional groups here can supply between identical molecules. Answer: n-butane less than ethoxyethane less than pentanal less than pentan-1-ol, the order rising exactly with each compound's ability to hydrogen-bond.

Competitive level (JEE / NEET) - Predicting the outcome of a flawed Williamson synthesis and an ether cleavage (NCERT Examples 7.6 and 7.7 extended)

(a) Predict the major product when sodium ethoxide, $C_2H_5O^-Na^+$, is reacted with tert-butyl bromide, $(CH_3)_3C-Br$, in an attempt to make t-butyl ethyl ether, and explain the failure. (b) Suggest a correct route to t-butyl ethyl ether. (c) Predict the product(s) when methoxybenzene (anisole) is heated with excess concentrated HI.

Step 1. For part (a), recognise that sodium ethoxide is both a nucleophile and a strong base, and that the alkyl halide is tertiary. Tertiary alkyl halides are highly hindered at the carbon bearing the leaving group, so the bulky, basic ethoxide ion cannot approach that carbon for $S_N2$ substitution; instead it removes a $\beta$-hydrogen, giving elimination. Result: the major product is 2-methylprop-1-ene (an alkene), not the desired ether, because elimination dominates over substitution whenever a strong, hindered base meets a tertiary substrate.

Step 2. For part (b), invert the combination: keep the bulky group on the alkoxide (where steric hindrance does not block an $S_N2$ attack) and put the leaving group on a primary, unhindered alkyl halide instead. Reacting sodium tert-butoxide, $(CH_3)_3CO^-Na^+$, with ethyl bromide (a primary, unhindered halide) allows a clean $S_N2$ attack of the alkoxide oxygen on the primary carbon. Answer (b): $(CH_3)_3CO^-Na^+ + C_2H_5Br \rightarrow (CH_3)_3C-O-C_2H_5 + NaBr$ — always route Williamson synthesis through the primary alkyl halide, whichever side of the target ether it happens to sit on.

Step 3. For part (c), recall that alkyl aryl ethers cleave selectively at the alkyl-oxygen bond, because the aryl-oxygen bond is reinforced by partial double-bond character from ring conjugation and the $sp^2$ aryl carbon resists nucleophilic attack. Protonation of anisole's oxygen by HI gives an oxonium ion, and iodide ion then attacks the methyl carbon (the only alkyl carbon present, and unhindered) by $S_N2$, displacing phenol.

Step 4. With HI in excess and at high temperature, the liberated phenol cannot be further attacked (its $sp^2$ carbon blocks substitution), but the methyl iodide product is stable and accumulates; the phenol itself remains as phenol. Answer (c): $C_6H_5-O-CH_3 + HI \rightarrow C_6H_5OH + CH_3I$, phenol plus methyl iodide, with no further reaction of the phenol even in excess HI. Timing note: aim for under three minutes on the recognition step alone. The fastest diagnostic for any Williamson-type question is to ask "which carbon does the nucleophile have to attack?" — if that carbon is tertiary (or often even secondary), expect elimination and flip the synthesis around to use a primary halide instead. For ether cleavage with HX, the fastest diagnostic is to ask "is either side of the ether oxygen an aryl carbon?" — if yes, the aryl-oxygen bond survives and only the alkyl-oxygen bond breaks, regardless of how much HX is used or how hot the reaction runs.

The single most common slip is applying alcohol reactions to phenol: students routinely try to convert phenol to chlorobenzene using $HCl/ZnCl_2$ (the Lucas reagent), $PCl_3$ or $SOCl_2$, forgetting that phenol's aryl $C-O$ bond has partial double-bond character and does not undergo nucleophilic substitution — the only reagent that cleaves phenol's $C-O$ bond in this chapter is zinc dust, and that removes the oxygen entirely rather than replacing it with a halogen. A second frequent error is swapping the regiochemistry of alkene-to-alcohol reactions: acid-catalysed hydration is Markovnikov (hydroxyl on the more substituted carbon), hydroboration-oxidation is anti-Markovnikov (hydroxyl on the less substituted carbon), and mixing the two up under exam pressure is common enough to specifically check for the word "hydroboration" or "diborane" before assigning regiochemistry. Third, students often forget that the Lucas test relies on relative rate, not an absolute yes/no result — primary alcohols do eventually react with the Lucas reagent, just far too slowly to produce turbidity at room temperature within the timescale of the test. Fourth, Williamson synthesis questions are frequently answered without checking whether the alkyl halide side is primary; secondary and especially tertiary alkyl halides give elimination instead of the intended ether, and the correct fix is always to redesign the synthesis around a primary alkyl halide, keeping any bulky group on the alkoxide side. Fifth, many students assume phenol behaves like a carboxylic acid because both are "acidic," but phenol does not liberate $CO_2$ from aqueous sodium bicarbonate ($NaHCO_3$) — it is too weak an acid for that reaction, and this distinction is a standard qualitative test to tell phenols and carboxylic acids apart. Finally, in Kolbe's reaction and the Reimer-Tiemann reaction, remember both require the more nucleophilic phenoxide ion, generated first with $NaOH$ — attempting either reaction directly on neutral phenol without base gives little or no product.

• Classify any alcohol, phenol or ether from its structure — monohydric versus polyhydric, primary/secondary/tertiary/allylic/benzylic for alcohols, and symmetrical versus unsymmetrical for ethers.
• Assign IUPAC and common names to alcohols, phenols and ethers, and reverse the process to draw a structure from a given name.
• Predict which alcohol an alkene gives under acid-catalysed hydration versus hydroboration-oxidation, and explain the mechanistic reason the two give opposite regiochemistry.
• Predict the class of alcohol (primary, secondary, tertiary) formed when a Grignard reagent reacts with methanal, another aldehyde, or a ketone.
• State all four laboratory/industrial routes to phenol (haloarenes, benzenesulphonic acid, diazonium salts, cumene) and explain why the cumene route dominates industrially.
• Explain, using resonance structures, why phenol is about a million times more acidic than ethanol, and predict how ortho/para versus meta electron-withdrawing or electron-releasing substituents shift that acidity.
• Sort any given reaction of alcohols or phenols into O-H bond cleavage (acidity, esterification, shared by both), C-O bond cleavage (substitution, dehydration, oxidation — alcohols only), or ring substitution (nitration, halogenation, Kolbe's, Reimer-Tiemann — phenols only).
• Write the mechanisms for acid-catalysed alkene hydration, acid-catalysed alcohol dehydration, and explain why they are the same carbocation pathway run in opposite directions.
• Design a correct Williamson ether synthesis for a given target ether, choosing which side to make the alkyl halide and which the alkoxide to avoid elimination.
• Predict the products of ether cleavage by concentrated HX for both dialkyl and alkyl aryl ethers, including which C-O bond breaks and why.

Ask the tutor: Want to work through the resonance structures behind phenol's acidity in more detail, or practise designing a Williamson synthesis for a specific target ether?