Organic Compounds Containing Oxygen — Practice Questions with Answers

Alcohols possess an $-OH$ group with a hydrogen directly bonded to the highly electronegative oxygen atom, allowing extensive intermolecular hydrogen bonding between alcohol molecules.

Ethers, despite also containing oxygen, lack any O-H bond (oxygen is bonded to two carbon groups instead) — this means ether molecules cannot hydrogen bond with each other (though they can still accept hydrogen bonds from other O-H or N-H containing molecules). Since hydrogen bonding is a considerably stronger intermolecular force than the weaker dipole-dipole/dispersion forces present in ethers, alcohols require significantly more thermal energy to separate/vaporise, resulting in notably higher boiling points than comparable ethers.

Step 1: The cyanide ion ($CN^-$, generated from a small amount of base, or from HCN dissociation) attacks the electrophilic carbonyl carbon of the aldehyde, with the carbonyl pi bond electrons shifting fully onto the oxygen, forming a tetrahedral alkoxide intermediate.

Step 2: This negatively charged alkoxide oxygen is then protonated (by HCN or water present), yielding the final cyanohydrin product, which contains both a hydroxyl group and a nitrile ($-C\equiv N$) group on the same carbon.

Carboxylic acids possess both a hydrogen-bond donor (the acidic O-H) and two hydrogen-bond acceptor sites (the carbonyl oxygen and the hydroxyl oxygen), allowing two carboxylic acid molecules to form two simultaneous, mutually reinforcing hydrogen bonds with each other, creating a stable cyclic dimer structure.

This dual, cyclic hydrogen bonding arrangement is particularly strong and geometrically favourable, making the dimer significantly more stable than two separate, individually hydrogen-bonded carboxylic acid molecules — this is precisely why carboxylic acids are commonly found as stable dimers, especially in the vapour phase or when dissolved in non-polar solvents (where competing hydrogen bonding with solvent molecules is unavailable).

Both phenol and carboxylic acids are acidic due to resonance stabilisation of their respective conjugate bases, but carboxylic acids are considerably more acidic than phenol.

In a carboxylate ion, the negative charge is delocalised equally over two oxygen atoms (via two equivalent resonance structures), providing very effective, symmetric stabilisation. In a phenoxide ion, the negative charge is delocalised into the aromatic ring itself (onto specific carbon atoms, which are less electronegative and less able to comfortably stabilise negative charge compared to oxygen) — this comparatively less effective stabilisation (delocalisation onto carbon, rather than onto a second oxygen atom) is precisely why phenol is a considerably weaker acid than carboxylic acids.

Aldol addition refers to the base-catalysed reaction of two aldehyde (or ketone) molecules possessing alpha-hydrogens, where an enolate ion (formed by removing an alpha-hydrogen) attacks the carbonyl carbon of a second molecule, forming a beta-hydroxy aldehyde (or ketone) product — the "aldol" — without any loss of water.

Aldol condensation refers to the subsequent dehydration (loss of a water molecule) of this initially formed aldol product, typically upon further heating, producing an alpha,beta-unsaturated aldehyde (or ketone), with a newly formed carbon-carbon double bond conjugated to the remaining carbonyl group.

Acetic acid has an electron-donating methyl group ($-CH_3$) attached to the carboxyl carbon — this electron-donating group destabilises the resulting carboxylate anion (concentrating, rather than dispersing, negative charge density), making acetic acid's conjugate base comparatively less stable.

Formic acid, by contrast, has only a hydrogen atom (rather than an electron-donating alkyl group) attached to the carboxyl carbon — lacking this destabilising electron donation, formic acid's conjugate base (formate ion) is comparatively more stable than acetate — this greater conjugate base stability is precisely why formic acid is a measurably stronger acid than acetic acid.

The Cannizzaro reaction is a base-catalysed disproportionation reaction specifically occurring with aldehydes that lack any alpha-hydrogen atoms (and hence cannot undergo the typical aldol reaction) — for example, formaldehyde, benzaldehyde, or trimethylacetaldehyde.

In this reaction, two molecules of the same aldehyde react together in the presence of concentrated base — one molecule is oxidised to the corresponding carboxylate salt (losing a hydrogen), while the other molecule is simultaneously reduced to the corresponding primary alcohol (gaining a hydrogen), giving an equimolar mixture of a carboxylic acid salt and a primary alcohol as the two products.

Phenols form a coloured coordination complex specifically with iron(III) ions from ferric chloride — the phenolic oxygen (with its available lone pairs, and the acidic character of the O-H group) can coordinate to the iron centre in a manner facilitated by the adjacent aromatic ring's electronic structure, producing a characteristic, often violet/purple colouration.

Simple aliphatic alcohols, lacking the adjacent aromatic ring system (and its associated electronic effects that favour this specific type of coordination/complex formation with iron), do not typically form this same characteristic coloured complex with ferric chloride — this distinct colour reaction is therefore a useful, simple qualitative test to specifically distinguish phenols from ordinary alcohols.

Aldehydes have a hydrogen atom directly attached to the carbonyl carbon — this specific structural feature allows the carbonyl carbon to be relatively easily oxidised (the C-H bond can be broken as part of the oxidation mechanism, ultimately converting the aldehyde to a carboxylic acid).

Ketones, by contrast, have two carbon-containing groups attached to the carbonyl carbon, with no hydrogen directly present on that specific carbon — since oxidation of ketones would require breaking a carbon-carbon bond (a much higher-energy, generally impractical process under these mild oxidative testing conditions), ketones are essentially resistant to oxidation by mild oxidising agents like Tollen's reagent or Fehling's solution, unlike aldehydes, providing a useful basis for distinguishing these two carbonyl compound classes.

Carboxylic acids possess an acidic O-H group capable of extensive intermolecular hydrogen bonding, including forming particularly strong, stable cyclic dimers (as discussed regarding carboxylic acid structure) — this strong hydrogen bonding network requires substantial thermal energy to disrupt during boiling.

Esters, by contrast, lack any O-H group entirely (the acidic hydrogen has been replaced by an alkyl/aryl group) — ester molecules therefore cannot engage in this type of strong intermolecular hydrogen bonding with each other, relying instead only on weaker dipole-dipole and dispersion forces. This absence of hydrogen bonding capability is precisely why esters generally have considerably lower boiling points than carboxylic acids of comparable molecular weight.

Step 1 (protonation): The hydroxyl oxygen of 2-butanol is protonated by the acid catalyst, converting the poor leaving group (-OH) into a good leaving group (water): $CH_3CH(OH)CH_2CH_3 + H^+ \rightarrow CH_3CH(OH_2^+)CH_2CH_3$

Step 2 (ionisation): Water leaves, forming a secondary carbocation intermediate: $CH_3CH(OH_2^+)CH_2CH_3 \rightarrow CH_3CH^+CH_2CH_3 + H_2O$

Step 3 (deprotonation/elimination): A base (often another alcohol molecule, or water) removes a beta-hydrogen, forming the alkene. Two products are possible: but-1-ene or but-2-ene.

Major product (by Zaitsev's rule, favouring the more substituted alkene): but-2-ene ($CH_3-CH=CH-CH_3$)

Step 1 (enolate formation): Base removes an alpha-hydrogen from one molecule of propanal ($CH_3CH_2CHO$), forming a resonance-stabilised enolate ion.

Step 2 (nucleophilic addition): This enolate ion's alpha-carbon attacks the carbonyl carbon of a second propanal molecule, forming a new carbon-carbon bond and a tetrahedral alkoxide intermediate.

Step 3 (protonation): The alkoxide is protonated, giving the aldol addition product: $CH_3CH_2CH(OH)CH(CH_3)CHO$ (3-hydroxy-2-methylpentanal)

Step 4 (condensation, dehydration on heating): Loss of water from this aldol product gives the final conjugated, alpha,beta-unsaturated aldehyde: 2-methylpent-2-enal ($CH_3CH_2CH=C(CH_3)CHO$)

Moles of ethanol $= \dfrac{5}{46} \approx 0.1087\text{ mol}$

Since oxidation of ethanol to acetic acid proceeds 1:1 (no carbon atoms gained or lost): moles of acetic acid formed $= 0.1087\text{ mol}$

Mass of acetic acid $= 0.1087\times60 \approx \mathbf{6.52\text{ g}}$

The hydroxyl group's oxygen lone pair can be donated (via resonance) directly into the aromatic ring, significantly increasing electron density specifically at the ortho and para positions relative to the -OH group.

This substantially increased electron density at these positions makes phenol's ring considerably more reactive/nucleophilic towards attacking an electrophile compared to benzene's unactivated ring, dramatically accelerating the rate of electrophilic aromatic substitution.

Product with excess bromine water: Since phenol is so strongly activated (and bromine water is a mild enough electrophile source), substitution readily occurs at all three activated positions simultaneously, giving 2,4,6-tribromophenol as a white precipitate, without requiring a Lewis acid catalyst (unlike the bromination of benzene itself, which requires $FeBr_3$ and gives only mono-substitution).

Reaction: $CH_3COOH + C_2H_5OH \rightarrow CH_3COOC_2H_5 + H_2O$ (1:1 stoichiometry)

Since acetic acid (3 mol) is in excess and ethanol (2 mol) is the limiting reagent, moles of ester formed = moles of limiting reagent (ethanol) = 2 mol.

Mass of ethyl acetate formed $= 2\times88 = \mathbf{176\text{ g}}$

Step 1: The Grignard reagent ($CH_3MgBr$) attacks the electrophilic carbonyl carbon of methyl acetate ($CH_3COOCH_3$), displacing the methoxide leaving group and forming an intermediate ketone (acetone, $CH_3COCH_3$), releasing $CH_3OMgBr$.

Step 2: Since excess Grignard reagent is present, a second equivalent of $CH_3MgBr$ immediately attacks this newly formed, highly reactive ketone intermediate, adding across the carbonyl to form a magnesium alkoxide.

Step 3 (acidic workup): Aqueous acid protonates this alkoxide, giving the final tertiary alcohol product.

Final product: tert-Butyl alcohol ($(CH_3)_3COH$) — esters characteristically react with two equivalents of Grignard reagent (via this ketone intermediate) to ultimately produce a tertiary alcohol.

Step 1 (repeated alpha-halogenation): The base-catalysed enolisation and subsequent iodination of acetone's methyl groups occurs repeatedly (three times) at one methyl group, converting $CH_3-CO-CH_3$ into $CI_3-CO-CH_3$ (triiodoacetone) — this repeated halogenation occurs preferentially because each successive iodine substitution makes the remaining alpha-hydrogens on that same carbon increasingly acidic (more easily removed for further enolisation).

Step 2 (cleavage by hydroxide): Hydroxide ion (present from the sodium hypoiodite/hydroxide system) attacks the carbonyl carbon of the triiodoacetone intermediate, ultimately cleaving the molecule and displacing the highly stabilised triiodomethyl carbanion, which is immediately protonated to form iodoform.

Products: Iodoform ($CHI_3$, a characteristic yellow solid with a distinctive smell) and sodium acetate ($CH_3COONa$).

Boiling point order: n-Pentane < Diethyl ether < n-Butanol

n-Pentane (purely non-polar hydrocarbon) has only weak London dispersion forces between molecules, giving it the lowest boiling point of the three.

Diethyl ether has a polar C-O-C linkage, giving it modest dipole-dipole interactions in addition to dispersion forces (though it cannot hydrogen bond with itself, lacking an O-H or N-H group) — this gives it a somewhat higher boiling point than pentane, but still considerably lower than an alcohol.

n-Butanol has a full -OH group, allowing extensive intermolecular hydrogen bonding between molecules — this considerably stronger intermolecular force gives butanol by far the highest boiling point among these three comparably-sized molecules.

Step 1 (protonation): The pi electrons of propene attack a proton from the acid catalyst ($H_3O^+$), forming the more stable secondary carbocation (rather than the less stable primary carbocation), consistent with Markovnikov's rule: $CH_3-CH=CH_2 + H^+ \rightarrow CH_3-CH^+-CH_3$

Step 2 (nucleophilic attack by water): A water molecule (acting as a nucleophile) attacks this secondary carbocation, forming a protonated alcohol (oxonium) intermediate: $CH_3-CH^+-CH_3 + H_2O \rightarrow CH_3-CH(OH_2^+)-CH_3$

Step 3 (deprotonation): Another water molecule removes the proton, regenerating the acid catalyst and yielding the final product: 2-propanol (isopropanol), $CH_3-CH(OH)-CH_3$.

A lower $pK_a$ value corresponds to a stronger acid — so, based on the given values, acidic strength increases in the order: acetic acid < chloroacetic acid < dichloroacetic acid.

Each successive chlorine atom substituted onto the alpha-carbon exerts an additional electron-withdrawing inductive ($-I$) effect, due to chlorine's high electronegativity.

This progressively stronger electron withdrawal (with increasing chlorine substitution) increasingly stabilises the resulting carboxylate conjugate base, by helping to further disperse/delocalise its negative charge — since a more stabilised conjugate base directly corresponds to a stronger parent acid, this explains why acidic strength (and hence $-I$ effect magnitude) increases progressively with each additional chlorine substituent: acetic acid (no Cl) < chloroacetic acid (1 Cl) < dichloroacetic acid (2 Cl).