"Organic Compounds Containing Oxygen" — Practice

Question 1

Predict the products when (a) ethanal and (b) $2,2$-dimethylpropanal are separately treated with concentrated aqueous sodium hydroxide.

Accepted Answer

**(a)** Ethanal has three $\alpha$-hydrogens, so an enolate forms and attacks a second molecule. The product is $3$-hydroxybutanal, which dehydrates on warming to but-$2$-enal.

**(b)** $2,2$-Dimethylpropanal has a fully substituted $\alpha$ carbon and therefore **no** $\alpha$-hydrogen.

It undergoes the Cannizzaro reaction, giving $2,2$-dimethylpropan-$1$-ol and sodium $2,2$-dimethylpropanoate in equal amounts.

The two aldehydes differ only in substitution at one carbon, yet give completely different reaction types.

Question 2

Explain why propanone reacts with hydrogen cyanide more slowly than ethanal, and why $2,2,4,4$-tetramethylpentan-$3$-one barely reacts at all.

Accepted Answer

Propanone has two methyl groups against ethanal's one. Each donates electron density inductively and by hyperconjugation, reducing the partial positive charge on the carbonyl carbon.

Each also crowds that carbon, obstructing the approach of cyanide.

In the tetramethyl ketone both groups are tertiary butyl, so the steric obstruction becomes overwhelming and cyanide cannot reach the carbon at all.

Electronics sets the trend and sterics makes it absolute at the extreme.

Question 3

Give the products of treating butan-$2$-one with iodine and sodium hydroxide, and explain why butan-$3$-one would give the same result.

Accepted Answer

The haloform reaction requires a methyl group directly attached to the carbonyl.

Butan-$2$-one has one, so all three of its methyl hydrogens are replaced by iodine and the resulting triiodomethyl group is displaced by hydroxide.

The products are **iodoform**, a yellow precipitate, and sodium propanoate.

Butan-$3$-one is the same compound named from the other end, since numbering must give the lowest locant. There is no separate substance.

Question 4

Give the major product of heating pentan-$2$-ol with concentrated sulphuric acid, and state whether rearrangement occurs.

Accepted Answer

Protonation and loss of water gives a secondary cation at carbon two.

A hydride shift from carbon three would give another secondary cation, which is no improvement, so **no rearrangement occurs**.

Loss of a proton follows Zaitsev, giving the more substituted alkene.

The major product is pent-$2$-ene, with pent-$1$-ene as the minor product.

Question 5

Distinguish between propan-$1$-ol, propan-$2$-ol and $2$-methylpropan-$2$-ol using the Lucas test, and explain the chemistry.

Accepted Answer

The reagent is concentrated hydrochloric acid with zinc chloride, a Lewis acid that assists ionisation.

$2$-Methylpropan-$2$-ol is tertiary and forms a stable cation immediately, so cloudiness appears at once.

Propan-$2$-ol is secondary and reacts within about five minutes.

Propan-$1$-ol is primary and would require an unstable primary cation, so no cloudiness appears at room temperature. The order follows carbocation stability exactly.

Question 6

Explain why phenol is brominated by bromine water without a catalyst, while benzene requires a Lewis acid and gives only monosubstitution.

Accepted Answer

Phenol's oxygen donates a lone pair strongly into the ring by resonance, greatly increasing the electron density at the ortho and para positions.

The ring is activated enough to attack molecular bromine directly, with no need to polarise it using a Lewis acid.

The activation is so strong that all three available positions react, giving $2,4,6$-tribromophenol as a white precipitate.

Benzene has no such donor, so bromine must first be polarised by iron bromide, and the resulting bromobenzene is deactivated, stopping the reaction at one substitution.

Question 7

Suggest a Williamson synthesis of tert-butyl methyl ether, and explain why the alternative combination fails.

Accepted Answer

Use **potassium tert-butoxide with iodomethane**. The methyl halide is primary and undergoes clean $	ext{SN}2$ displacement.

The alternative, methoxide with tert-butyl bromide, fails: methoxide is a strong base and the halide is tertiary, so **elimination** dominates and $2$-methylpropene is formed instead.

The general rule follows: the bulky group must come from the alkoxide and the simple one from the halide.

Both routes give the same ether on paper, and only the mechanism distinguishes them.

Question 8

Give the products of cleaving ethoxybenzene and $2$-ethoxy-$2$-methylpropane with hot hydrogen iodide.

Accepted Answer

**Ethoxybenzene:** the aryl-oxygen bond cannot break, so iodide attacks the ethyl group by $	ext{SN}2$. Products are **phenol** and iodoethane.

**$2$-Ethoxy-$2$-methylpropane:** one group is tertiary, so the reaction is $	ext{SN}1$. The tertiary cation forms and captures iodide.

Products are $2$-iodo-$2$-methylpropane and **ethanol**.

With excess hot acid the ethanol would go on to give iodoethane as well, which is why the conditions must be stated.

Question 9

Explain why an ester is hydrolysed irreversibly by alkali but reversibly by acid, and give one industrial consequence.

Accepted Answer

Acid-catalysed hydrolysis is the exact reverse of Fischer esterification, so both directions are accessible and an equilibrium is set up. Driving it forward requires a large excess of water.

In alkali the carboxylic acid formed is immediately deprotonated by the excess hydroxide.

The resulting carboxylate carries a negative charge and cannot be attacked by an alcohol, so the reverse reaction is impossible and the hydrolysis goes to completion.

Industrially this is **saponification**: fats are hydrolysed quantitatively with alkali to give soap, which no acid-catalysed route could achieve.

Question 10

Arrange ethanoyl chloride, ethanoic anhydride, ethyl ethanoate and ethanamide by reactivity towards ammonia, and state which conversions between them are possible in one step.

Accepted Answer

Reactivity: ethanoyl chloride $>$ ethanoic anhydride $>$ ethyl ethanoate $>$ ethanamide.

Chloride is an excellent leaving group and chlorine donates poorly by resonance, leaving the carbonyl strongly electrophilic. Nitrogen donates strongly and the amide ion is a very poor leaving group.

Any derivative can be converted directly into one **below** it: chloride to anhydride, ester or amide; ester to amide.

Going upwards is impossible directly and requires an activating reagent such as thionyl chloride acting on the parent acid.

Question 11

A compound $	ext{C}_3	ext{H}_6	ext{O}$ gives an orange precipitate with $2,4$-dinitrophenylhydrazine, a silver mirror with Tollens' reagent, and no yellow precipitate with iodine and alkali. Identify it.

Accepted Answer

The hydrazone shows a **carbonyl group**, so the compound is an aldehyde or a ketone.

The silver mirror shows it is an **aldehyde**, ruling out propanone.

The negative iodoform result confirms there is no methyl group attached to the carbonyl, which is consistent with an aldehyde of this size.

With three carbons, the compound is **propanal**.

Question 12

How would you distinguish between benzaldehyde and ethanal using two tests, and between phenol and benzyl alcohol using one?

Accepted Answer

Both aldehydes give a silver mirror with **Tollens' reagent**, so that alone does not distinguish them.

**Fehling's solution** responds only to aliphatic aldehydes: ethanal gives a red precipitate of copper(I) oxide, benzaldehyde gives nothing.

Ethanal also gives a positive **iodoform** test, since it carries a methyl group next to the carbonyl, while benzaldehyde does not.

For the second pair, **neutral iron(III) chloride** gives a violet colour with phenol and no colour with benzyl alcohol, since only a phenolic hydroxyl responds.

Practice — "Organic Compounds Containing Oxygen"

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