Alcohols, Phenols and Ethers · Lesson 12 of 12
Chapter Summary and Practice
“Connect structure, physical properties, acid–base equilibria and reaction pathways to solve integrated synthesis and identification problems.”
• Use local bonding to connect classification, properties and reaction choice. • Construct multistep routes with valid conditions and work-up sequences. • Diagnose product errors using atom balance, charge and carbon tracking. • Solve mixed chapter-wide problems with explicit chemical reasoning.
One oxygen, three distinct structural settings
Alcohols, phenols and ethers share oxygen chemistry but differ in connectivity. A saturated C–OH group can be protonated, deprotonated, substituted, dehydrated or oxidised according to the substrate. An aromatic C–OH group has a more acidic proton, a conjugated C–O bond and an activated ring. An ether has no OH hydrogen but retains oxygen lone pairs, making protonation, Lewis-acid coordination and selected acid cleavage possible. Reaction predictions should start with these distinctions, not with a reagent list in isolation.
| Decision | Alcohol | Phenol | Ether |
|---|---|---|---|
| Local linkage | sp³ C–OH | Aromatic C–OH | C–O–C |
| Ordinary self OH hydrogen bonding | Yes | Yes | No OH donor |
| Hydrogen bonding with water | Donor and acceptor | Donor and acceptor, with hydrophobic ring | Acceptor |
| Acid–base behaviour | Weak acid; oxygen accepts H⁺ | More acidic; NaOH gives phenoxide | Oxygen accepts H⁺; no OH proton |
| Ordinary HX substitution | Possible at suitable saturated carbon | Not an ordinary route to aryl halide | Cleavage after oxygen protonation |
| Aromatic substitution | Only if an appropriate aromatic part exists | OH activates ortho/para | Ar–OR activates ortho/para |
Classification is local. Benzyl alcohol is a primary alcohol, not a phenol; vinyl alcohol is an enol, not an allylic alcohol. Carbon degree counts neighbours of the OH-bearing saturated carbon. Hydroxyl count is independent of degree. A common name must be translated to a structural formula before reaction planning: tert-butyl alcohol and isobutyl alcohol have the same molecular formula but different reaction centres.
Use physical properties as structural evidence
For similar sizes, alcohol self hydrogen bonding raises boiling points above corresponding ethers and hydrocarbons. Increasing chain length strengthens dispersion, while branching often lowers boiling point within a related isomer set. Water solubility involves replacement of old interactions by new water–solute interactions: an ether can accept hydrogen bonds from water even though it cannot form an OH donor network with itself. A growing nonpolar skeleton generally reduces solubility for a fixed hydroxyl count.
2-Nitrophenol can satisfy an OH hydrogen bond internally, reducing extended association and making it more steam volatile than 4-nitrophenol. The same nitro substitution also influences acidity through electron withdrawal and, at ortho/para positions, additional resonance stabilisation of the phenoxide. Physical behaviour and acid–base behaviour are connected to structure, but they are not interchangeable measurements.
A compact quantitative reference
| Relationship | Meaning | Conditions and limits |
|---|---|---|
| Phenol pKa≈10; ethanol≈15.9 | Ka ratio≈10⁵·⁹≈7.94×10⁵ | Representative aqueous comparison; not a direct ratio of H₃O⁺ concentrations |
| n=m/M | Mass m divided by molar mass M gives amount n | Use g with g mol⁻¹ to obtain mol |
| Product amount = reactant amount × coefficient ratio | Translate a balanced equation into a theoretical amount | Requires the limiting reactant and the stated reaction extent |
| Actual amount = theoretical amount × fractional yield | Apply measured conversion/recovery as appropriate | A yield percentage is divided by 100; do not invent actual efficiency |
| 1 glucose → 2 ethanol + 2 CO₂ | Ideal fermentation stoichiometry | Actual biochemical yield can be lower |
| 1 cumene hydroperoxide → 1 phenol + 1 acetone | Theoretical cleavage stoichiometry | Separate isolated yields may differ |
A pKa difference is logarithmic. A 2-unit difference means a hundredfold Ka difference, not twice the acidity on a linear scale. A balanced chemical equation provides mole ratios, not equal masses. A hydride shift moves a bonding pair and H together; deprotonation removes H⁺ while retaining the pair on the original framework. These are distinct bookkeeping operations and should be checked separately.
Reaction choice follows the required bond change
| Target transformation | Reagents/conditions | What must be checked |
|---|---|---|
| Alkene → Markovnikov alcohol | Dilute aqueous acid | Carbocation stability, rearrangement and symmetry |
| Alkene → anti-Markovnikov alcohol | Borane, then H₂O₂/aqueous NaOH | Less substituted C receives OH; staged reagents |
| Aldehyde/ketone → alcohol | NaBH₄ or suitable reduction | Primary/secondary product; selective functional groups |
| Acid/ester → primary alcohol | LiAlH₄ in dry ether, then water | Ester gives an alcohol from each fragment |
| Carbonyl + carbon group → alcohol | RMgX in dry ether, then protonation | New C–C bond and carbon count |
| Phenol → phenoxide | NaOH | OH proton removed; C–O retained |
| Alcohol → ester | Acyl reagent and suitable acid/base conditions | Acylation at oxygen; phenol needs an activated acyl reagent |
| Alcohol → halide | HX, PBr₃ or SOCl₂ as suitable | Degree, leaving-group activation and rearrangement |
| Alcohol → alkene | Suitable acid and heating | β-H, E1/E2 distinction and possible isomerisation |
| Primary alcohol → aldehyde/acid | Controlled dry oxidant / suitable excess aqueous oxidant | Oxidation endpoint and salt before acid work-up |
| Phenol → ortho COOH | Sodium phenoxide + heated pressurised CO₂, then acid | Kolbe–Schmitt, extra carbon from CO₂ |
| Phenol → ortho CHO | CHCl₃/aqueous NaOH, heat, then acid | Reimer–Tiemann, extra carbon from CHCl₃ |
| Alcohol/phenol → ether | Alkoxide/phenoxide + suitable methyl/primary halide | Williamson SN2 versus E2 |
| Ether → halide + alcohol/phenol | Concentrated HI/HBr and heat | Attacked carbon, cation exceptions, excess acid |
| Anisole → substituted aromatic ether | Br₂, nitrating mixture or Friedel–Crafts system | Ring substitution under specified conditions; OCH₃ remains |
Connect synthesis routes without inventing a shortcut
Problem
Prepare propan-1-ol, propan-2-ol and methyl propyl ether using propene as the common carbon skeleton.
- 1.For the primary alcohol use hydroboration–oxidation: propene → organoborane → propan-1-ol. For the secondary alcohol use dilute aqueous acid hydration: propene → propan-2-ol.
- 2.To obtain CH₃OCH₂CH₂CH₃ from the primary alcohol, make sodium propoxide under dry conditions, then react with methyl iodide. Oxygen attacks methyl carbon; the three-carbon propyl skeleton stays intact.
- 3.Alternatively convert propan-1-ol to a primary propyl halide and react with methoxide. Both are legitimate ordinary Williamson disconnections, but using methyl iodide avoids β elimination on the electrophile.
- 4.Do not use propan-2-ol when the target requires an n-propyl group: it would retain the wrong oxygen-bearing connectivity unless another transformation were deliberately added.
Problem
Explain a conceptual separation of phenol and anisole followed by recovery of phenol.
- 1.Aqueous NaOH converts phenol into ionic sodium phenoxide while anisole, lacking an OH proton, remains neutral. Ionic phenoxide is favoured in the aqueous phase; a suitable immiscible organic phase can retain neutral anisole.
- 2.Acidify the separated phenoxide-containing aqueous fraction to return C₆H₅O⁻ to C₆H₅OH. Depending on concentration, neutral phenol can then be recovered by appropriate separation.
- 3.This uses a reversible change in protonation and solubility. It does not require HI cleavage of anisole or destruction of either aromatic ring; actual recovery depends on partitioning and experimental conditions.
Problem
Connect phenol to salicylic acid and then to acetylsalicylic acid.
- 1.Phenol + NaOH gives sodium phenoxide. Heat with pressurised CO₂ and then acidify to give o-HOC₆H₄COOH; the new carboxyl carbon comes from CO₂.
- 2.Use ethanoic anhydride with an acid catalyst to acetylate the phenolic OH, giving o-CH₃COOC₆H₄COOH plus ethanoic acid. The carboxylic acid group remains.
- 3.Check the oxygen attachments: one ring substituent is COOH through carbon, and the other is OC(=O)CH₃ through oxygen. Confusing either connection changes the compound.
- 4.The two stages demonstrate ring carbon–carbon bond formation followed by oxygen acylation. Naming both stages explains the route more clearly than a single arrow labelled “conversion”.
Mechanism and notation checks that prevent common errors
Curved arrows start at electron pairs or bonds and point to the place receiving the pair. A positive charge is an electron deficiency, not an arrow source. An SN2 arrow pair shows bond formation and leaving-group departure in the same step. An E2 elimination needs β-H removal, formation of C=C and departure of the leaving group. An E1 sequence includes a genuinely plausible carbocation and can rearrange. Primary ethanol dehydration should be represented without requiring a free ethyl cation.
Resonance arrows connect contributors of one species, while reaction arrows connect reactants and products. A resonance picture must conserve atom positions and total charge. Phenoxide ring contributors place negative charge at ortho/para sites; they do not mean that Na⁺ is covalently attached to one of those carbons. Quinones must be drawn with their actual conjugated diketone bonds, rather than adding carbonyls to an unchanged aromatic circle.
For any reaction, check atom count, net charge, oxygen connectivity, essential conditions and protonation state. An equation balanced with [O] is a formal oxygen-equivalent relation; a complete ionic equation needs the actual oxidant and reduction product. A first-stage ether-cleavage answer may retain an alcohol, whereas excess hot acid can remove it by another reaction. Aryl phenol remains an important limit. These checks are part of chemical reasoning, not just presentation.
Quick Quiz
Quiz
Which route adds a new carbon–carbon bond before producing an alcohol?
Explanation: The carbon reagent supplies a new group to the carbonyl carbon.
Which pair requires different aromatic-versus-side-chain reaction reasoning?
Explanation: Phenol has aryl C–O; benzyl alcohol has OH on a saturated side-chain carbon.
What should be checked before reporting only an alcohol from ether cleavage?
Explanation: Reaction extent changes the final product set.
How large is Ka(A)/Ka(B) when pKa(A)=7 and pKa(B)=10?
Explanation: 10^(10−7)=1000, with A the stronger acid.
Which route preserves a phenolic oxygen while adding a carboxyl carbon to the ring?
Explanation: CO₂ provides the new carbon and the original ring oxygen remains phenolic.
Practice Problems
- Identify and name CH₃CH(CH₃)CH(OH)C(CH₃)₂CH₃, CH₃OCH₂CH(CH₃)₂ and p-CH₃C₆H₄OH.
- Construct 1-phenylpropan-2-ol, 2-ethoxy-3-methylpentane and 3-cyclohexylpentan-3-ol, then classify the hydroxyl compounds.
- Explain how the eight C₅H₁₂O alcohol isomers can be generated without duplication and distinguish the three C₇H₈O phenols.
- Arrange n-butane, ethoxyethane, pentanal and pentan-1-ol by boiling point, and explain why an ether still dissolves partly in water.
- Prepare both propanols from propene and explain hydration of styrene to 1-phenylethanol.
- Design Grignard preparations of propan-1-ol and 2-methylpropan-2-ol, and explain the reagent order.
- Give two routes to phenol, including one from benzene and one from cumene.
- Compare phenol, 4-methylphenol, 3-nitrophenol and ethanol by acidity and calculate Ka(phenol)/Ka(ethanol).
- Predict products of butan-1-ol with HBr, hot concentrated H₂SO₄ and PCC. Explain how extended acid exposure affects the alkene mixture.
- Explain the rearranged bromide from 3-methylbutan-2-ol and HBr.
- Give phenol products with bromine water, cold Br₂/CS₂, dilute nitric acid and chromic acid.
- Prepare salicylic acid and salicylaldehyde from phenol and identify the source of each new carbon.
- Plan ethoxybenzene and tert-butyl methyl ether syntheses, and identify a failed reversed Williamson pairing.
- Predict first-stage HI cleavage of anisole, ethyl isopropyl ether and benzyl ethyl ether.
- Compare phenol oxygen acetylation with anisole Friedel–Crafts acetylation.
- Calculate ethanol from 90.0 g glucose at 70% yield and phenol from 0.80 mol cumene hydroperoxide at 85% yield.
- Explain why 2-nitrophenol is steam volatile and why this does not contradict its increased acidity relative to phenol.
- Choose an alkene for 4-methylheptan-4-ol and for 2-cyclohexylbutan-2-ol.
For the first choose a five-carbon parent and number to keep OH at C3, then compare substituent locants: 2,2,4-trimethylpentan-3-ol, a secondary alcohol. The ether is 1-methoxy-2-methylpropane. The ring compound is 4-methylphenol, a monohydric phenol. Naming requires the correct parent and local functional class before locant comparison.
Key Takeaways
• Local oxygen connectivity determines the relevant functional class and reaction centre. • Intermolecular forces explain properties only when size and molecular shape are also considered. • Reaction conditions and work-up stages are part of a complete transformation. • Carbon tracking distinguishes reduction, addition, substitution and rearrangement. • Phenoxide resonance connects acidity with activated ring chemistry. • Williamson planning requires an accessible electrophilic carbon and attention to base-induced elimination. • Ether cleavage depends on structure, acid amount and possible further reactions. • Correct mechanisms conserve atoms, charge and plausible valence at every stage. • Integrated synthesis succeeds by connecting verified individual steps rather than assuming shortcuts.
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Cleavage and Aromatic Reactions of Ethers
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