Build each parent skeleton from the stated parent name, place –OH or –OR at the indicated locant, then add substituents at their locants. Cyclic names mean the functional group carbon is C-1 unless another locant is explicitly supplied.
(i) CH3C(OH)(CH3)CH2CH3. (ii) C6H5CH2CH(OH)CH3. (iii) HOCH2CH2C(OH)(CH3)CH2C(OH)(CH3)CH3. (iv) 2,3-diethylphenol. (v) CH3CH2CH2OCH2CH3. (vi) CH3CH(OCH2CH3)CH(CH3)CH2CH3. (vii) C6H11CH2OH. (viii) CH3CH2C(OH)(C6H11)CH2CH3. (ix) cyclopent-3-en-1-ol. (x) HOCH2CH(CH2CH3)CH(Cl)CH3.
List the carbon skeletons possible for five carbons and place one –OH group at non-equivalent positions. A primary alcohol has –OH on a carbon attached to one carbon, a secondary alcohol on a carbon attached to two carbons, and a tertiary alcohol on a carbon attached to three carbons.
Primary: pentan-1-ol, 2-methylbutan-1-ol, 3-methylbutan-1-ol, 2,2-dimethylpropan-1-ol. Secondary: pentan-2-ol, pentan-3-ol, 3-methylbutan-2-ol. Tertiary: 2-methylbutan-2-ol.
The –OH group in propanol makes the molecule polar and allows O–H...O hydrogen bonding between molecules. Considerable energy is needed to break these attractions during boiling. Butane is non-polar and its molecules attract each other only through weak dispersion forces, so it boils at a lower temperature.
Propanol forms intermolecular hydrogen bonds, while butane has only weak van der Waals forces.
The oxygen atom and O–H bond of an alcohol interact strongly with water molecules through hydrogen bonding. Hydrocarbons are non-polar and do not provide sites for hydrogen bonding, so water cannot solvate them effectively.
Alcohols can form hydrogen bonds with water; hydrocarbons cannot.
In the first step, borane adds across the C=C bond so that boron attaches to the less substituted carbon. Oxidation with alkaline hydrogen peroxide replaces B by –OH. Example: CH3CH=CH2 →(BH3) CH3CH2CH2BH2 →(H2O2/NaOH) CH3CH2CH2OH.
Hydroboration-oxidation is addition of BH3 to an alkene followed by oxidation with H2O2/NaOH to give an alcohol with anti-Markovnikov orientation.
C7H8O as a monohydric phenol is methylphenol, also called cresol. The methyl group may be ortho, meta or para to –OH on the benzene ring, giving 2-, 3- and 4-methylphenol respectively.
The three monohydric phenols are 2-methylphenol, 3-methylphenol and 4-methylphenol.
o-Nitrophenol forms intramolecular hydrogen bonding, so its molecules are less associated with one another and it has a lower boiling point. p-Nitrophenol forms intermolecular hydrogen bonding and remains more strongly associated, so it is less steam volatile.
o-Nitrophenol is steam volatile.
C6H5CH(CH3)2 + O2 → C6H5C(CH3)2OOH; then C6H5C(CH3)2OOH + H+ + H2O → C6H5OH + (CH3)2CO.
Cumene is oxidised to cumene hydroperoxide, which on acid hydrolysis gives phenol and acetone.
C6H5Cl + NaOH →(623 K, 300 atm) C6H5ONa + NaCl; C6H5ONa + HCl → C6H5OH + NaCl.
Chlorobenzene is fused with aqueous NaOH at high temperature and pressure to give sodium phenoxide, which on acidification gives phenol.
Step 1: CH2=CH2 + H3O+ → CH3CH2+ + H2O. Step 2: CH3CH2+ + H2O → CH3CH2OH2+. Step 3: CH3CH2OH2+ + H2O → CH3CH2OH + H3O+. The acid is regenerated, so it acts as a catalyst.
Acid-catalysed hydration of ethene proceeds by protonation, nucleophilic attack of water and deprotonation.
C6H6 + conc. H2SO4 → C6H5SO3H + H2O. C6H5SO3H + NaOH → C6H5SO3Na + H2O. C6H5SO3Na + 2NaOH → C6H5ONa + Na2SO3 + H2O. C6H5ONa + H+ → C6H5OH.
Benzene is sulphonated, the sulphonic acid is fused with NaOH to sodium phenoxide, and acidification gives phenol.
(i) C6H5CH=CH2 + H2O →(H+) C6H5CH(OH)CH3. (ii) C6H11CH2Br + OH- → C6H11CH2OH + Br- by SN2 attack on the primary halide. (iii) CH3CH2CH2CH2CH2Br + KOH(aq) → CH3CH2CH2CH2CH2OH + KBr.
(i) Hydrate styrene. (ii) Use bromomethylcyclohexane with aqueous NaOH. (iii) Use 1-bromopentane with aqueous KOH/NaOH.
2C6H5OH + 2Na → 2C6H5ONa + H2. C6H5OH + NaOH → C6H5ONa + H2O. Phenoxide ion is resonance stabilised, while ethoxide ion has no comparable resonance stabilisation and the ethyl group has a +I effect, so phenol is more acidic than ethanol.
Phenol reacts with sodium metal and sodium hydroxide. Phenol is more acidic than ethanol.
In o-nitrophenol, the –NO2 group has strong –I and –R effects, dispersing the negative charge in the conjugate base. In o-methoxyphenol, –OCH3 shows a +R electron-releasing effect toward the ring, increasing electron density and making loss of H+ less favourable.
The nitro group withdraws electron density and stabilises phenoxide ion; the methoxy group donates electron density and destabilises it.
The lone pair on oxygen overlaps with the pi system of benzene. Resonance structures place extra electron density mainly at the ortho and para carbons, making the ring more reactive toward electrophiles than benzene and directing substitution to ortho and para positions.
The –OH group donates electron density to the benzene ring by resonance and increases electron density at ortho and para positions.
(i) CH3CH2CH2OH + 2[O] → CH3CH2COOH + H2O. (ii) C6H5OH + Br2/CS2 → o-bromophenol + p-bromophenol + HBr. (iii) C6H5OH + dilute HNO3 → o-nitrophenol + p-nitrophenol + H2O. (iv) C6H5OH + CHCl3 + NaOH(aq) → o-hydroxybenzaldehyde after acidification.
(i) Propan-1-ol gives propanoic acid. (ii) Phenol gives o- and p-bromophenol. (iii) Phenol gives o- and p-nitrophenol. (iv) Phenol gives salicylaldehyde after acidification.
(i) C6H5ONa + CO2 →(pressure) o-HOC6H4COONa; acidification gives salicylic acid. (ii) C6H5OH + CHCl3 + NaOH → o-HOC6H4CHO after acidification. (iii) RONa + R'X → ROR' + NaX. Example: C2H5ONa + CH3Br → C2H5OCH3 + NaBr. (iv) CH3OC2H5 is unsymmetrical because methyl and ethyl groups are different.
Kolbe's reaction carboxylates sodium phenoxide; Reimer-Tiemann formylates phenol; Williamson synthesis forms ethers from alkoxides and alkyl halides; an unsymmetrical ether has two different groups bonded to oxygen.
Step 1: CH3CH2OH + H+ → CH3CH2OH2+. Step 2: the protonated alcohol loses water on heating to form CH3CH2+. Step 3: base removes a beta hydrogen: CH3CH2+ → CH2=CH2 + H+. Overall, CH3CH2OH →(conc. H2SO4, 443 K) CH2=CH2 + H2O.
Ethanol is protonated, loses water on heating, and deprotonation gives ethene.
(i) CH3CH=CH2 + H2O →(H+) CH3CH(OH)CH3. (ii) C6H5CH2Cl + KOH(aq) → C6H5CH2OH + KCl. (iii) C2H5MgCl + HCHO → C2H5CH2OMgCl →(H3O+) C2H5CH2OH. (iv) CH3MgBr + (CH3)2CO → (CH3)3COMgBr →(H3O+) (CH3)3COH.
(i) Acid-catalysed hydration. (ii) Hydrolysis with aqueous alkali. (iii) Reaction with formaldehyde followed by hydrolysis. (iv) Reaction with acetone followed by hydrolysis.
Strong oxidising agents convert primary alcohols to acids; PCC stops oxidation at aldehyde. Phenol reacts readily with bromine water to give 2,4,6-tribromophenol. Benzyl alcohol is oxidised to benzoic acid by acidified permanganate. Alcohol dehydration needs an acid and heat, and ketone reduction gives the secondary alcohol.
(i) Acidified KMnO4 or K2Cr2O7. (ii) PCC or controlled oxidation. (iii) Bromine water. (iv) Acidified KMnO4. (v) Conc. H2SO4, heat. (vi) NaBH4 or LiAlH4.
Ethanol has an O–H bond and can form O–H...O hydrogen bonds between molecules. Methoxymethane has oxygen lone pairs but no O–H bond, so its intermolecular attractions are weaker. More energy is therefore needed to boil ethanol.
Ethanol molecules form intermolecular hydrogen bonds; methoxymethane molecules do not hydrogen-bond with one another.
(i) CH3CH2CH2ONa + CH3CH2CH2Br → CH3CH2CH2OCH2CH2CH3 + NaBr. (ii) C6H5ONa + C2H5Br → C6H5OC2H5 + NaBr. (iii) (CH3)3CONa + CH3Br → (CH3)3COCH3 + NaBr. Use methyl halide rather than tert-butyl halide to avoid elimination. (iv) CH3ONa + C2H5Br → CH3OC2H5 + NaBr.
(i) Sodium propoxide and 1-bromopropane. (ii) Sodium phenoxide and ethyl bromide. (iii) Sodium tert-butoxide and methyl bromide. (iv) Sodium methoxide and ethyl bromide.
Alkoxide ions are strong bases as well as nucleophiles. With tertiary halides, E2 elimination predominates. For example, (CH3)3CBr + C2H5ONa → (CH3)2C=CH2 + C2H5OH + NaBr rather than tert-butyl ethyl ether. Therefore, ethers containing tertiary groups should be made using a tertiary alkoxide and a methyl or primary alkyl halide.
Williamson synthesis fails or gives poor yield with secondary and especially tertiary alkyl halides because elimination competes strongly.
Propan-1-ol is protonated: CH3CH2CH2OH + H+ → CH3CH2CH2OH2+. A second propan-1-ol molecule attacks the primary carbon and displaces water: CH3CH2CH2OH + CH3CH2CH2OH2+ → CH3CH2CH2O(H)+CH2CH2CH3 + H2O. Deprotonation gives CH3CH2CH2OCH2CH2CH3.
Propan-1-ol is dehydrated with acid at about 413 K to give 1-propoxypropane by an SN2 path.
Ether formation by acid dehydration needs nucleophilic attack by an alcohol molecule. In secondary and tertiary alcohols, steric hindrance and stable carbocation/alkene formation make elimination favourable, so alkenes are the major products instead of ethers.
Secondary and tertiary alcohols undergo elimination more readily than substitution under acid dehydration conditions.
In aryl alkyl ethers, the oxygen lone pair is conjugated with the benzene ring. Resonance donation forms structures in which the ortho and para carbons bear increased electron density. Electrophiles therefore attack faster than in benzene, and the major products are ortho and para substituted ethers.
The alkoxy group donates electron density to the aromatic ring by resonance and increases electron density mainly at ortho and para positions.
Step 1: CH3OCH3 + HI → CH3OH+CH3 + I-. Step 2: I- attacks methyl carbon and cleaves the C–O bond: CH3OH+CH3 + I- → CH3I + CH3OH. Step 3 with excess HI: CH3OH + HI → CH3I + H2O.
Methoxymethane is protonated and then iodide attacks a methyl group by SN2 to give methanol and methyl iodide; excess HI converts methanol to methyl iodide.
(i) C6H5OCH3 + CH3Cl/AlCl3 → o-CH3C6H4OCH3 + p-CH3C6H4OCH3. (ii) C6H5OCH3 + HNO3/H2SO4 → o-NO2C6H4OCH3 + p-NO2C6H4OCH3. (iii) C6H5OCH3 + Br2/CH3COOH → o-BrC6H4OCH3 + p-BrC6H4OCH3. (iv) C6H5OCH3 + CH3COCl/AlCl3 → o-CH3COC6H4OCH3 + p-CH3COC6H4OCH3.
Anisole gives mainly ortho and para substituted products, with para usually major because of steric effects.