Exercise practice: Alcohols, Phenols and Ethers

4.1 Exercise 7.3: isomers of C5H12O

The molecular formula C5H12O fits a saturated alcohol. Build the carbon skeletons of pentane, then replace a hydrogen with OH at each distinct position. The straight chain gives pentan-1-ol, pentan-2-ol and pentan-3-ol. The branched skeleton 2-methylbutane gives 2-methylbutan-1-ol, 2-methylbutan-2-ol and 3-methylbutan-2-ol. The 2,2-dimethylpropane skeleton gives only 2,2-dimethylpropan-1-ol, because every other position is equivalent to one already counted. That is eight alcohols in total.

Classify them by the carbon that carries OH. Pentan-1-ol, 2-methylbutan-1-ol and 2,2-dimethylpropan-1-ol are primary, since that carbon touches one other carbon. Pentan-2-ol, pentan-3-ol, 3-methylbutan-2-ol are secondary. 2-Methylbutan-2-ol is the only tertiary alcohol, because its OH carbon touches three carbons.

4.2 Exercise 7.4 and 7.5: boiling point and solubility

Why propanol boils above butane. Propan-1-ol and butane have almost the same molecular mass, but propanol has an OH group and its molecules hydrogen bond to each other. Butane can only use weak van der Waals forces, so much less energy is needed to separate its molecules and its boiling point is far lower.

Why alcohols dissolve in water better than hydrocarbons do. The OH group forms hydrogen bonds with water molecules, and those attractions pay for much of the energy needed to break water's own hydrogen bonds. A hydrocarbon cannot form hydrogen bonds, so it disrupts the water network without giving anything back. As the carbon chain of an alcohol grows, the part that cannot hydrogen bond becomes larger and the solubility falls.

4.3 Exercise 7.6: hydroboration oxidation

Diborane adds across the double bond of propene so that boron goes to the terminal carbon, the one with more hydrogen atoms. Oxidation with hydrogen peroxide in aqueous sodium hydroxide then replaces the boron by an OH group with the same orientation. The product is propan-1-ol, the anti-Markovnikov alcohol, in good yield. Acid catalysed hydration of the same alkene would give propan-2-ol instead.

4.4 Exercise 7.7 and 7.8: phenols and steam distillation

The formula C7H8O allows three monohydric phenols, differing in where the methyl group sits on the ring: 2-methylphenol, 3-methylphenol and 4-methylphenol. It also allows benzyl alcohol, which is an alcohol rather than a phenol, and anisole, which is an ether. Only the three methylphenols are phenols.

When ortho and para nitrophenol are separated by steam distillation, the ortho isomer is the one that distils. Its OH group hydrogen bonds to the nitro group within the same molecule, so adjacent molecules cannot hold on to each other strongly and it becomes volatile. The para isomer hydrogen bonds between molecules, forming a network that keeps it in the flask.

4.5 Exercise 7.9 and 7.10: preparing phenol

From cumene. Oxidise cumene with air to cumene hydroperoxide, then treat it with dilute acid. The hydroperoxide breaks into phenol and acetone, so this route delivers two useful products at once.

From chlorobenzene. Fuse chlorobenzene with sodium hydroxide at 623 K and about 320 atmospheres, then acidify the sodium phenoxide that forms. The chlorine is replaced by OH even though an aryl halide normally resists nucleophilic substitution.

4.6 Exercise 7.14 and 7.15: acidity comparisons

Two reactions that show phenol is acidic. Phenol reacts with sodium metal to release hydrogen and form sodium phenoxide, and it reacts with aqueous sodium hydroxide to form the same salt and water. Ethanol does the first but not the second, which places phenol well above ethanol in acid strength: pKa 10.0 against 15.9.

Why o-nitrophenol is more acidic than o-methoxyphenol. The nitro group withdraws electron density through both the inductive and the resonance effect, which stabilises the phenoxide ion and makes the acid stronger. A methoxy group can donate electron density by resonance, which destabilises the phenoxide ion and weakens the acid. The nitro compound therefore loses its proton more easily.

4.7 Exercise 7.17: four equations

(i) Propan-1-ol with alkaline KMnO4. The primary alcohol is oxidised through the aldehyde to propanoic acid, CH3CH2COOH.

(ii) Phenol with bromine in CS2 at low temperature. Monobromination dominates and the main product is p-bromophenol, with some o-bromophenol.

(iii) Phenol with dilute nitric acid at 298 K. Nitration gives a mixture of o-nitrophenol and p-nitrophenol.

(iv) Phenol with chloroform and aqueous NaOH, then hydrolysis. This is the Reimer Tiemann reaction and the product is salicylaldehyde, 2-hydroxybenzaldehyde.

4.8 Exercise 7.19: mechanism of dehydration

Protonate the OH group of ethanol with concentrated sulphuric acid to give the oxonium ion. In the slow step, water leaves and a carbocation forms; this step sets the rate. Then a proton is removed from a neighbouring carbon and the carbon carbon double bond forms. The acid is released again in the last step, so sulphuric acid is only a catalyst. The reaction is driven forward by removing ethene as it appears.

4.9 Exercise 7.20 and 7.24: conversions

Propene to propan-2-ol. Acid catalysed hydration with dilute sulphuric acid gives the Markovnikov alcohol.

Benzyl chloride to benzyl alcohol. Warm with aqueous sodium hydroxide; the primary benzylic chloride is replaced by OH.

Ethylmagnesium chloride to propan-1-ol. React it with methanal, then hydrolyse the adduct, which adds one carbon and gives the primary alcohol.

Methylmagnesium bromide to 2-methylpropan-2-ol. React it with propanone and hydrolyse; the ketone becomes a tertiary alcohol with two methyl groups already in place.

Williamson routes. 1-Propoxypropane comes from sodium propoxide and 1-bromopropane. Ethoxybenzene comes from sodium phenoxide and bromoethane. 1-Methoxyethane comes from sodium methoxide and bromoethane. 2-Methoxy-2-methylpropane comes from sodium tert-butoxide and bromomethane, because putting the tertiary group in the alkoxide avoids the elimination that would destroy a tertiary halide.

4.10 Exercise 7.28 and 7.30: HI cleavage

1-Propoxypropane with HI. Protonation followed by iodide attack on the less crowded carbon gives propan-1-ol and 1-iodopropane; with excess HI and heat the alcohol becomes the iodide too.

Methoxybenzene with HI. The methyl oxygen bond breaks and not the aryl oxygen bond, so the products are phenol and iodomethane.

Mechanism of HI with methoxymethane. The oxygen is protonated to give an oxonium ion. Iodide attacks one methyl carbon from the back in an SN2 step, displacing methanol, and the product is iodomethane. With excess HI the methanol is converted to iodomethane as well.