"Organic Compounds Containing Oxygen" — Practice

Question 1

How would you distinguish between butan-1-ol, butan-2-ol and 2-methylpropan-2-ol?

Accepted Answer

Use the Lucas test, adding concentrated hydrochloric acid with zinc chloride to each. The tertiary alcohol 2-methylpropan-2-ol gives turbidity immediately, since a tertiary carbocation forms at once. Butan-2-ol is secondary and gives turbidity in about five minutes, while butan-1-ol is primary and gives none without heating. Oxidation offers an alternative: the primary alcohol gives an aldehyde then an acid, the secondary gives a ketone, and the tertiary resists. A third route is the iodoform test, which butan-2-ol alone gives.

Question 2

Arrange in increasing order of acid strength: phenol, 4-nitrophenol, 4-methylphenol, ethanol, acetic acid, with reasons.

Accepted Answer

Ethanol is weakest, because ethoxide has no way to delocalise its negative charge. 4-Methylphenol comes next, as resonance stabilisation applies but the donating methyl group pushes density into the ring and partly destabilises the anion. Phenol follows, then 4-nitrophenol, whose nitro group withdraws density by both induction and resonance and stabilises the phenoxide considerably. Acetic acid is strongest, since carboxylate spreads its charge over two equivalent electronegative oxygens rather than one oxygen and some ring carbons.

Question 3

Explain why phenol reacts with bromine water at room temperature without a catalyst while benzene does not react at all.

Accepted Answer

The hydroxyl group donates a lone pair into the aromatic ring by resonance, raising the electron density at the ortho and para positions substantially. That activation is strong enough that bromine itself, without any Lewis acid catalyst to polarise it, is attacked by the ring. All three activated positions react, so 2,4,6-tribromophenol precipitates immediately as a white solid. Benzene has no such activating group, so its pi cloud is not electron-rich enough to attack unpolarised bromine, and a catalyst such as iron(III) bromide is essential.

Question 4

Give the correct reagents for preparing 2-methoxy-2-methylpropane by the Williamson synthesis, and explain why the alternative pairing fails.

Accepted Answer

Use sodium tert-butoxide with methyl iodide. The Williamson synthesis proceeds by an SN2 mechanism requiring backside attack on the halide carbon, and a methyl halide is the least hindered substrate possible. The alternative, sodium methoxide with tert-butyl chloride, fails because a tertiary carbon is far too crowded for backside attack, so the strongly basic methoxide removes a beta hydrogen instead and eliminates, giving 2-methylpropene. The general rule is to put the bulk on the alkoxide and use the least hindered halide available.

Question 5

Predict the product when concentrated sodium hydroxide acts on ethanal and on benzaldehyde, and state the deciding factor.

Accepted Answer

The deciding factor is whether an alpha hydrogen is present. Ethanal has three, so it undergoes the aldol reaction: a carbanion forms and attacks a second molecule's carbonyl, giving 3-hydroxybutanal, which dehydrates on warming to but-2-enal. Benzaldehyde has no alpha hydrogen, since the carbon next to its carbonyl is part of the ring and carries none available for removal, so it undergoes the Cannizzaro reaction and disproportionates into benzoic acid and benzyl alcohol. The two reactions are mutually exclusive.

Question 6

Explain why ethanal reacts faster than propanone with hydrogen cyanide, and why benzaldehyde is slower than ethanal.

Accepted Answer

Nucleophilic addition requires a nucleophile to reach an electron-poor carbon, and two factors control how easily. Propanone carries two methyl groups, each donating electron density into the carbonyl carbon and reducing its positive character, and each physically blocking one approach path. Ethanal has one methyl group and one hydrogen, so it is both more electrophilic and less crowded. Benzaldehyde is slower than ethanal for a different reason: the aromatic ring donates electron density into the carbonyl by resonance, spreading the positive charge and making the carbon less attractive to a nucleophile.

Question 7

A compound of formula $\mathrm{C_3H_6O}$ gives a positive iodoform test but no silver mirror with Tollens' reagent. Identify it and justify.

Accepted Answer

The formula fits both propanal and propanone. Tollens reagent oxidises aldehydes to acids, depositing metallic silver, so the absence of a silver mirror rules out propanal and identifies the compound as propanone. The positive iodoform test confirms this, since propanone has a methyl group attached directly to the carbonyl carbon, which is exactly what that test detects. Propanal would give the opposite pattern of results, a positive Tollens test and a negative iodoform test.

Question 8

How would you distinguish benzaldehyde from ethanal using two tests?

Accepted Answer

Both are aldehydes, so both give a silver mirror with Tollens reagent and that test alone cannot separate them. Fehling's solution does: it responds to aliphatic aldehydes only, so ethanal gives a red precipitate of copper(I) oxide while benzaldehyde gives nothing. The iodoform test provides a second discrimination, since ethanal has a methyl group adjacent to its carbonyl and gives a yellow precipitate, while benzaldehyde has a ring there and does not. Either test settles it, and together they are conclusive.

Question 9

Name a reagent that converts an acyl chloride to an aldehyde without over-reduction, and explain how it is controlled.

Accepted Answer

Rosenmund reduction, using $\mathrm{H_2}$ over palladium supported on $\mathrm{BaSO_4}$. Ordinary palladium would reduce the aldehyde further to the primary alcohol, so the catalyst is deliberately poisoned, usually with sulphur or quinoline, to lower its activity to the point where reduction stops after one step. This is the same principle as Lindlar's catalyst in alkyne chemistry: the metal is intentionally weakened so a reaction that would otherwise run to completion halts at a useful intermediate.

Question 10

A molecule contains both a ketone and a carboxylic acid. How would you reduce the ketone to a secondary alcohol while leaving the acid untouched?

Accepted Answer

Use $\mathrm{NaBH_4}$. It is a mild hydride source, reactive enough to add to a ketone but not to a carboxylic acid, whose carbonyl is already deactivated by the lone pair donated from the adjacent hydroxyl and is therefore much less electrophilic. $\mathrm{LiAlH_4}$ would be wrong, since it is powerful enough to reduce the acid to a primary alcohol as well and would destroy the selectivity required. If both groups needed reducing, $\mathrm{LiAlH_4}$ would achieve it in a single step. A question naming the reagent has usually told you half the answer.

Question 11

Suggest two routes for converting bromoethane into propanoic acid.

Accepted Answer

Both routes add one carbon, since the target has three carbons and the starting material two. The first uses the nitrile: treat bromoethane with alcoholic potassium cyanide to give propanenitrile, then hydrolyse it with dilute acid to propanoic acid. The second uses a Grignard reagent: form ethylmagnesium bromide in dry ether, bubble carbon dioxide through it, and work up with dilute acid. Both are standard chain-extension methods, and either would be accepted.

Question 12

Why does hot alkaline potassium permanganate oxidise the side chain of ethylbenzene to benzoic acid but leave tert-butylbenzene untouched?

Accepted Answer

Because the oxidation begins by attacking a benzylic hydrogen, meaning a hydrogen on the carbon directly attached to the ring, and proceeds through a benzylic radical or cation that the ring stabilises by resonance. Ethylbenzene has two such hydrogens, so the reaction proceeds and continues until the whole side chain has been cut back to a single carboxyl carbon, giving benzoic acid regardless of the original chain length. Tert-butylbenzene has no benzylic hydrogen at all, its attached carbon carrying three methyl groups, so there is nothing to abstract and it survives unchanged.

Practice — "Organic Compounds Containing Oxygen"

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