CLASS 12 CHEMISTRY · CHAPTER 7 · STUDY NOTES
Alcohols, Phenols and Ethers
A comprehensive study of the classification, nomenclature, preparation, and chemical properties of alcohols, phenols, and ethers, highlighting their industrial importance and structural characteristics.
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Classification of Alcohols and Phenols
Identify the carbon–oxygen attachment before assigning a class.Where does the oxygen attach?
- Alcohol
- A compound with hydroxyl attached to a saturated sp³ carbon.
Alcohol has C(sp³)–OH; phenol has OH directly on aromatic carbon. Ether has oxygen between two carbon groups and no O–H bond. Benzyl alcohol remains an alcohol because CH₂ separates OH from the aromatic ring. A vinylic OH attachment differs from the alcohol definition used here.
Mono-, di- and trihydric counts hydroxyl groups. Primary, secondary and tertiary instead counts the carbon neighbours of the OH-bearing carbon: one, two and three. Methanol is the zero-carbon-neighbour special case. These classifications answer different questions.
Let’s look a little closer.
Go deeper: Count locally rather than judging the whole chain
An alcohol molecule with two OH groups can contain OH-bearing carbons of different degrees. Inspect each carbon separately; branching elsewhere does not automatically make an alcohol tertiary. Allylic and benzylic describe the position next to a double bond or aromatic ring.
Carbon hybridisation and connectivity influence chemical behaviour. An aromatic ring elsewhere in a molecule does not confer phenol acidity on a remote OH group. The structural drawings show connectivity; their angles are not measured tetrahedral geometry.
Take a moment. What do you think?
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Question
HOCH₂CH₂CH₂OH has two OH groups. Does that make each alcohol carbon secondary?
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Nomenclature and Structural Aspects
Use the name to reconstruct a specific molecule.Read names through molecular shape
- Hydroxyl locant
- The numbered parent-chain position at which OH is attached.
Select a suitable parent containing OH, number to give the hydroxyl appropriate low locants and use the suffix -ol. A four-carbon chain with OH on the second carbon is butan-2-ol. Multiple hydroxyls use -diol or -triol with their locants. In cyclic alcohol names, the OH-bearing carbon is position one.
For suitable phenol derivatives, 1,2-, 1,3- and 1,4- ring arrangements correspond to ortho, meta and para. Ethers can be named as alkoxy-substituted parents, such as methoxyethane. Oxygen is bent, with two bonds and two lone-pair regions; ether is not a straight C–O–C rod.
Go deeper: Names and diagrams must encode the same bonds
Common names and systematic names can describe the same connectivity. A name without positions may not uniquely distinguish isomers. Check the reconstructed structure before comparing reactivity or molecular formula.
The chapter geometry figure gives methanol C–O–H about 108.9°, phenol about 109° and methoxymethane C–O–C about 111.7°. Its phenol bond-length label and prose differ, so these notes flag that inconsistency rather than select one as exact. Lone-pair lobes are schematic probability regions.
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Question
A drawing labels a four-carbon chain butan-3-ol. How should it be renamed?
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Preparation of Alcohols
Track both the original carbon chain and any newly added carbon.Build an alcohol without losing the carbon story
- Carbonyl reduction
- Conversion of a carbon–oxygen double bond into an alcohol through reduction and appropriate work-up.
Acid-catalysed hydration of an alkene generally gives Markovnikov attachment through protonation, water attack and deprotonation. Hydroboration followed by alkaline peroxide oxidation instead gives the less substituted OH attachment without a free carbocation rearrangement. These routes share a net alcohol product class but have different mechanisms.
Reduction of an aldehyde generally gives a primary alcohol; a ketone gives a secondary alcohol. LiAlH₄ also reduces carboxylic acids and esters; an ester gives a primary alcohol from its acyl portion and an alcohol from its alkoxy portion. NaBH₄ is not a universal replacement for these stronger reductions.
Go deeper: Grignard addition creates a new carbon–carbon bond
Grignard addition to methanal gives a primary alcohol, another aldehyde a secondary alcohol and a ketone a tertiary alcohol after protonating work-up. The organic group of RMgX contributes a new carbon substituent at the former carbonyl carbon. The initial oxygen-bearing intermediate has a counterion, omitted in the simplified drawing.
Water is introduced after the addition because it would consume Grignard reagent beforehand. Aldehyde reduction preserves the original carbon count, whereas Grignard addition increases it. Hydride addition and protonating work-up are separate operations; a stage selector is a teaching comparison, not the complete microscopic pathway.
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Question
Propanal reacts with methylmagnesium bromide and then work-up. Identify the alcohol and its degree.
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Preparation of Phenols
Different routes can converge on the same ring–OH attachment.Different routes to a ring–OH bond
- Phenoxide
- The negatively charged conjugate base obtained when phenol loses its O–H proton.
Chlorobenzene under strong aqueous NaOH conditions gives phenoxide, followed by acidification to phenol. The chapter describes 623 K but gives conflicting pressure labels, 320 atm in prose and 300 atm in the figure. These notes retain the strong-pressure qualification without presenting the conflict as a settled single value.
A benzene sulphonic derivative can also give phenoxide with molten alkali and then phenol after acidification. Diazonium hydrolysis on warming replaces ring N₂⁺ by OH with nitrogen release. The diazonium preparation at 273–278 K is a separate cold stage, not the hydrolysis condition.
Go deeper: Cumene links a phenol product to an acetone coproduct
Cumene oxidation first produces a hydroperoxide. Acid cleavage gives phenol and acetone; the six aromatic carbons remain in phenol and the three side-chain carbons form acetone. Carbon ownership explains why acetone is a coproduct rather than a carbon added to the ring.
Phenoxide acidification changes protonation without breaking the aryl–oxygen bond. Simple alcohol-to-halide replacement does not extend unchanged to phenol because the aryl bond has conjugative/sp² character. Net route drawings omit some reagents, counterions and coproducts; read the caption before interpreting them as complete equations.
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Question
One mole of cumene yields phenol and acetone by the net route. How are its nine carbons distributed?
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Physical Properties of Alcohols and Phenols
Separate intramolecular structure from attractions between molecules.Follow hydrogen bonds between molecules
- Hydrogen bond
- An attractive contact involving a donor-bound hydrogen and a suitable acceptor, distinct from their covalent bonds.
Alcohols and phenols have O–H donors and oxygen acceptors, supporting intermolecular hydrogen bonds. Ethers can accept a hydrogen bond from water but lack an O–H donor for the same self-network. The source comparison gives pentane 309.1 K, ethoxyethane 307.6 K and butan-1-ol 390 K boiling points.
For comparable alcohols, increasing size raises dispersion attractions while branching can lower effective intermolecular contact and boiling point. Increasing nonpolar chain extent relative to OH generally reduces water solubility. Hydrogen bonding near OH does not make every long-chain alcohol freely water-soluble.
Go deeper: Use the right comparison for each property
Phenol can hydrogen-bond with water, but its aromatic nonpolar portion limits solubility compared with small alcohols. Temperature, molecular identity and solvent matter. A cartoon tail envelope communicates relative hydrophobic extent, not a calculated concentration or a quantitative solubility prediction.
A hydrogen bond does not permanently join two molecules covalently. Dashed contacts depict intermolecular attraction; solid bonds depict connectivity. Boiling and melting also depend on different physical circumstances, so a boiling-point argument is not automatically a complete crystal-packing explanation.
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Question
An ether oxygen accepts a hydrogen bond from water. Why does that not establish an ether self-network identical to ethanol?
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Chemical Reactions: Acidity
Compare how well the conjugate base is stabilised.Stabilise the conjugate base
- pKa
- The negative base-ten logarithm of the acid dissociation constant, for a specified medium and conditions.
Alcohols react with active metals such as sodium to form alkoxides and hydrogen: 2ROH + 2Na → 2RONa + H₂. Phenol also forms phenoxide and can react with aqueous NaOH; ordinary simple alcohols are not deprotonated by NaOH in the same useful way. Conjugate-base stabilisation is essential to the comparison.
Phenoxide delocalises negative charge through the ring, whereas simple alkoxide lacks that pathway. NO₂ at ortho/para supplies additional resonance stabilisation; meta retains an inductive contribution without the same conjugated pathway. Electron-donating methyl tends to reduce phenol acidity.
Go deeper: Logarithmic numbers are not percentage ionisation
Source pKa values are o-, m-, p-nitrophenol 7.2, 8.3, 7.1; phenol 10; cresols about 10.1–10.2; ethanol 15.9. Lower pKa means stronger acidity in a matched medium. Phenol/ethanol Ka ratio is 10^(15.9−10)=about 794,328, often loosely rounded to a million.
Charge contributors are alternative resonance descriptions, not separate molecules or an electron moving in a loop. Compare stabilisation of both acid and conjugate base rather than counting a decorative number of resonance arrows. Do not generalise one order to every solvent or every substituted alcohol.
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Question
Use pKa 7.1 and 8.3 to compare para- and meta-nitrophenol acidity.
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Chemical Reactions of Alcohols
Choose oxidation, substitution or elimination from the actual structure.Choose which bonds change
- Carbinol hydrogen
- A hydrogen attached to the carbon that also bears the alcohol OH group.
Oxidation of primary alcohol can stop at aldehyde or proceed to acid; secondary gives ketone. Tertiary lacks a carbinol hydrogen and resists this simple carbonyl-forming pathway, though harsh oxidation can cleave C–C bonds. Lucas reagent HCl/ZnCl₂ gives qualitative turbidity: tertiary immediate, secondary slower, primary usually clear at room temperature.
Alcohol esterification with a carboxylic acid is reversible and gives water; the alcohol oxygen is retained in the ester alkoxy link in the ordinary Fischer account. Acyl chlorides and anhydrides provide other ester routes. Alcohol substitution by hydrogen halides changes the C–O attachment and depends on substrate and conditions.
Dehydration pathways depend on substrate and conditions; the ethanol product comparison does not establish one universal carbocation mechanism.
Go deeper: Conditions change the pathway, not just the colour of a sketch
The ethanol comparison with acid favours ether around 413 K and ethene around 443 K. This is not a universal temperature switch for every alcohol. Primary dehydration should not be drawn as necessarily forming a stable free primary carbocation; a concerted elimination from protonated alcohol is possible.
Commercial methanol comes from catalytic CO hydrogenation. Sugar fermentation gives C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ without air in the chapter account; alcohol accumulation inhibits yeast activity. Ethanol can also come from ethene hydration. Methanol is toxic and denatured alcohol is deliberately unsuitable for drinking; industrial use is not evidence of edibility.
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Question
In glucose fermentation, what fraction of the six carbon atoms remains in ethanol?
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Chemical Reactions of Phenols
Ring position and reaction conditions determine different phenol behaviour.An activated ring has several pathways
- Intramolecular hydrogen bond
- A donor–acceptor hydrogen-bond contact within one molecule.
Phenol activates electrophilic substitution mainly at ortho/para sites. Dilute nitration gives mononitro products; concentrated nitration can give 2,4,6-trinitrophenol. Bromine water gives white 2,4,6-tribromophenol, whereas suitable low-polarity, low-temperature conditions favour monobromination. These are representative pathways rather than asserted isomer yields.
Ortho nitrophenol can form an internal O–H···O contact to a nitro oxygen. Para more readily forms intermolecular contacts; this helps explain their steam-distillation comparison. A hydrogen-bond drawing must connect a donor hydrogen to an acceptor oxygen, not vaguely link two functional-group names.
Go deeper: Carbon addition and redox are different transformations
Kolbe–Schmitt introduces carboxyl carbon from CO₂ at an ortho position after work-up. Reimer–Tiemann uses chloroform/alkali and hydrolysis to introduce an ortho aldehyde. COOH and CHO are distinct groups, even though both introduce one new carbon.
Oxidation can convert phenol to p-benzoquinone with two carbonyls and changed ring bonding. Zinc-dust heating instead replaces phenolic OH with H, with ZnO in net accounting. Quinone is not an unchanged aromatic ring with two extra complete bonds.
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Question
A diagram links phenolic O directly to the nitro N with a dashed hydrogen bond. What should be repaired?
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Preparation and Properties of Ethers
Assign the bulky group to the correct reaction partner.Choose the correct Williamson partners
- Williamson synthesis
- Ether preparation by nucleophilic substitution of a suitable alkyl halide by an alkoxide or phenoxide.
Williamson ether formation generally uses methyl or primary alkyl halide for SN2. A bulky alkoxide can still attack a suitable primary halide, while a tertiary halide with strong alkoxide base favours elimination. Swapping the partners can therefore change the outcome even if the proposed ether formula is unchanged.
Phenoxide with a suitable methyl halide makes an aryl methyl ether through attack at methyl carbon. An ordinary aryl halide is not an interchangeable SN2 electrophile. Acid-catalysed dehydration of two primary alcohol molecules can produce ether; different alcohols may give mixtures.
Go deeper: The net formula does not reveal mechanistic suitability
The oxygen atom of the nucleophile remains between the two groups in the product. Backside access belongs to the carbon bearing the leaving group, not to the carbon behind oxygen. A formula-only partner list can conceal a tertiary electrophile and a dominant elimination pathway.
For ethanol the chapter compares ether formation at 413 K with elimination at 443 K. Secondary/tertiary alcohols compete more strongly through alkene formation. Reagent identity, solvent, temperature and substrate all belong to a useful route; no animation should imply every ether is obtained selectively by one universal method.
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Question
For propyl methyl ether, which pairing is more suitable: propoxide + methyl bromide or methoxide + an ordinary bromobenzene?
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Chemical Reactions of Ethers
Follow cleavage at alkyl carbon separately from substitution on an aromatic ring.Protonate first, then choose the bond
- Oxonium ion
- An oxygen-centred positively charged species formed here by protonating an ether oxygen.
Strong HI or HBr protonates ether oxygen. For a suitable unsymmetrical dialkyl ether, nucleophilic attack at the less hindered alkyl carbon can give halide plus alcohol; excess reagent and heat may convert that alcohol further. A tertiary group permits an ionisation pathway that can change which bond cleaves.
Anisole cleavage retains aryl–O and breaks methyl–O, giving methyl iodide and phenol. Ordinary SN2 does not attack aryl carbon as if it were methyl carbon. Anisole also undergoes electrophilic ring bromination, nitration and acylation, with methoxy activating ortho/para positions.
Go deeper: Keep two reaction maps separate
Protonation produces an oxygen with three bonds and positive charge; the drawing must account for the added hydrogen. Product alcohol again has two oxygen bonds. Stage comparisons describe a plausible selected pathway, not all mechanisms for every ether.
In aromatic substitution, a ring hydrogen is replaced by an electrophile while methoxy remains attached. In HI cleavage, the methyl–O bond is broken while ring–O remains. The same starting anisole therefore has two distinct reaction maps; conditions determine which is relevant.
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Question
Why is iodobenzene + methanol an incorrect ordinary HI-cleavage drawing for anisole?
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