Organic Compounds Containing Oxygen

Acetaldehyde with concentrated alkali gives an aldol. Benzaldehyde under the very same conditions gives a Cannizzaro product instead — an alcohol and a carboxylate. What decides which reaction happens?

One structural feature: whether the molecule has an $\alpha$-hydrogen.

Hydroxide's first move is to look for an acidic hydrogen next to the carbonyl. Acetaldehyde has three, so hydroxide removes one, and the resulting enolate attacks another molecule of aldehyde. That is the aldol.

Benzaldehyde has none — the carbon next to its carbonyl is part of the ring and carries no hydrogen. With nothing to deprotonate, hydroxide instead adds directly to the carbonyl carbon, and the resulting alkoxide transfers a hydride to a second molecule. One molecule is oxidised and the other reduced.

Reading a structure for its $\alpha$-hydrogens is the single most useful habit in this chapter, and it also decides the haloform test, the Hell-Volhard-Zelinsky reaction and the Claisen condensation.

1. Aldol and Cannizzaro in detail

The aldol gives a $\beta$-hydroxy carbonyl compound, which dehydrates on warming to an $\alpha ,\beta$-unsaturated one. A crossed aldol between two different partners bearing $\alpha$-hydrogens gives four products and is synthetically useless; it becomes useful only when one partner has no $\alpha$-hydrogen and cannot enolise.

The Cannizzaro reaction needs no $\alpha$-hydrogen at all. In a crossed Cannizzaro with methanal, methanal is always the one oxidised, because its carbonyl is the most electrophilic and accepts hydroxide first, and it is the best hydride donor.

Illustration 1

Predict the products when a mixture of benzaldehyde and methanal is treated with concentrated sodium hydroxide.

Neither has an $\alpha$-hydrogen, so this is a crossed Cannizzaro.

Methanal is oxidised to sodium formate; benzaldehyde is reduced to benzyl alcohol.

Methanal always loses. It is both the most electrophilic carbonyl, so hydroxide adds there first, and the best hydride donor, so it hands the hydride on — two reasons pointing the same way.

2. Carbonyl reactivity

Aldehydes are more reactive than ketones towards nucleophilic addition for two reasons that reinforce each other. Alkyl groups donate electron density, reducing the partial positive charge on the carbonyl carbon; and they crowd that carbon, obstructing the incoming nucleophile. A ketone has two alkyl groups where an aldehyde has one.

Aromatic carbonyls are less reactive still, because the ring donates by resonance into the carbonyl.

Illustration 2

Arrange in order of decreasing reactivity towards hydrogen cyanide: methanal, ethanal, propanone, benzaldehyde.

Methanal has no alkyl groups at all: most reactive.

Ethanal has one, propanone has two.

Benzaldehyde has one substituent, but it is a ring that donates by resonance into the carbonyl and also hinders approach.

Order: methanal $>$ ethanal $>$ propanone $>$ benzaldehyde.

Resonance donation is worth more than an alkyl group here, which is why benzaldehyde sits below propanone despite having only one substituent.

3. Alcohols: substitution, dehydration and oxidation

The Lucas test exploits the first row: with concentrated hydrochloric acid and zinc chloride, a tertiary alcohol turns cloudy immediately, a secondary within five minutes, and a primary not at all at room temperature.

Oxidation levels matter. A primary alcohol gives an aldehyde with a mild reagent such as pyridinium chlorochromate and a carboxylic acid with permanganate or dichromate. A secondary alcohol gives a ketone with either. A tertiary alcohol resists oxidation entirely, because there is no hydrogen on the carbon bearing the hydroxyl.

Illustration 3

Give the major product of dehydrating $3,3$-dimethylbutan-$2$-ol with hot concentrated sulphuric acid.

Protonation and loss of water gives a secondary cation.

A methyl group migrates from the adjacent quaternary carbon, giving a tertiary cation.

Loss of a proton then gives the more substituted alkene, $2,3$-dimethylbut-$2$-ene.

The skeleton has rearranged and Zaitsev has been obeyed. Both consequences follow from the carbocation, which is why dehydration is never used when the skeleton must be preserved.

4. Phenols: an activated ring and an acidic hydroxyl

Phenol's oxygen donates strongly into the ring, which has two consequences. The ring is very activated, so bromination in water gives the $2,4,6$-tribromide immediately without any catalyst. And the hydroxyl is far more acidic than an alcohol's, at $p{K}_a$ $10$ against $16$.

Three named reactions exploit the activated ring:

Phenol also fails to react in ways alcohols do: it will not undergo substitution of its hydroxyl by halide, because the $\text{C}-\text{O}$ bond has partial double-bond character from the resonance donation.

Illustration 4

Explain why phenol does not react with hydrogen bromide to give bromobenzene, although ethanol readily gives bromoethane.

In phenol the oxygen lone pair is delocalised into the ring, giving the carbon-oxygen bond partial double-bond character and shortening it.

Breaking that bond would also destroy the conjugation, so the substitution is energetically prohibitive.

The carbon involved is also $s{p}^2$ and aromatic, and neither an $\text{SN}1$ nor an $\text{SN}2$ pathway operates at such a centre.

Aryl halides cannot be made this way at all. The route to bromobenzene is direct bromination of benzene, or a diazonium reaction, never substitution on phenol.

5. Ethers: making them and breaking them

Williamson synthesis couples an alkoxide with a halide, and it is an $\text{SN}2$ reaction. The halide must therefore be primary; a tertiary halide gives elimination instead. When an ether has one bulky and one simple group, the bulky group must come from the alkoxide and the simple one from the halide.

Cleavage by hydrogen iodide splits the ether, and which fragment keeps the iodide depends on the mechanism. With primary and secondary groups the reaction is $\text{SN}2$ and iodide attacks the less hindered carbon. With a tertiary group the reaction is $\text{SN}1$ and iodide goes to the tertiary carbon. With an aryl group the aryl-oxygen bond never breaks, so the product is always a phenol plus an alkyl iodide.

Illustration 5

Give the products of cleaving (a) $2$-methoxy-$2$-methylpropane and (b) methoxybenzene with hot hydrogen iodide.

(a) One group is tertiary, so cleavage is $\text{SN}1$. The tertiary carbocation forms and captures iodide, giving $2$-iodo-$2$-methylpropane and methanol.

(b) The aryl-oxygen bond cannot be broken, so iodide attacks the methyl group. The products are phenol and iodomethane.

In each case identify the mechanism first. Which fragment takes the iodide is entirely a consequence of whether the pathway is $\text{SN}1$ or $\text{SN}2$.

6. Carboxylic acid derivatives: the reactivity ladder

$\text{acid chloride}>\text{anhydride}>\text{ester}>\text{amide}$

Two factors run together. The leaving group gets worse down the list, from chloride through carboxylate and alkoxide to amide. And resonance donation into the carbonyl gets stronger, from chlorine's poor overlap through to nitrogen's excellent donation, which reduces the electrophilicity of the carbon.

Because the order is fixed, any derivative can be converted into one below it but not above it by direct reaction with the appropriate nucleophile.

Ester hydrolysis runs by two quite different mechanisms. In acid it is the exact reverse of esterification and is therefore reversible, needing excess water to drive it. In base it is irreversible, because the carboxylic acid formed is immediately deprotonated to the carboxylate, which no alcohol can attack. That irreversibility is why saponification goes to completion.

Illustration 6

Explain why an amide can be made from an acid chloride but an acid chloride cannot be made from an amide by direct reaction with hydrogen chloride.

An acid chloride sits at the top of the reactivity ladder, so ammonia attacks it readily and chloride, an excellent leaving group, departs.

Going the other way would require chloride, a poor nucleophile, to attack the least electrophilic derivative, and would require the amide ion, a very poor leaving group, to depart.

Both requirements fail, so the reaction does not occur.

The ladder is a one-way street. Making a more reactive derivative always requires an activating reagent such as thionyl chloride rather than a direct exchange.

Illustration 7

Arrange ethanoyl chloride, ethanoic anhydride, ethyl ethanoate and ethanamide by their rate of hydrolysis, and explain using both factors.

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

Leaving group: chloride is excellent, carboxylate good, alkoxide poor and the amide ion very poor. Hydrolysis requires that group to depart, so the order follows directly.

Resonance: chlorine's $3p$ orbital overlaps the carbon $2p$ poorly, so almost no donation occurs and the carbonyl stays strongly electrophilic. Nitrogen's $2p$ overlaps well, donating heavily and reducing the electrophilicity of the amide carbon.

The two arguments give the same order, which is why the ladder is so reliable and why an amide requires prolonged heating with acid or alkali to hydrolyse at all.

7. Distinguishing tests

The Fehling exception is examined constantly: aromatic aldehydes give no reaction with Fehling's solution but do give a silver mirror with Tollens', so the pair together distinguishes benzaldehyde from ethanal.

Illustration 8

How would you distinguish between propan-$2$-ol, propan-$1$-ol, propanone and propanal using simple tests?

$2,4$-dinitrophenylhydrazine: propanone and propanal give precipitates; the alcohols do not.

Tollens' or Fehling's: propanal responds, propanone does not. That separates the two carbonyl compounds.

Iodoform: propanone and propan-$2$-ol both give a yellow precipitate, since both carry the methyl carbinol or methyl ketone pattern. Propan-$1$-ol does not.

Lucas: propan-$2$-ol turns cloudy within minutes; propan-$1$-ol does not react at room temperature.

Two tests separate all four. Choosing tests that split the set in half at each step is the efficient strategy, rather than testing for each compound in turn.

Illustration 9

A compound ${\text{C}}_4{\text{H}}_8\text{O}$ gives a $2,4$-dinitrophenylhydrazone but no silver mirror, and gives a yellow precipitate with iodine and alkali. Identify it.

The hydrazone shows a carbonyl group; the absence of a silver mirror rules out an aldehyde, so it is a ketone.

The iodoform test requires a methyl group attached to the carbonyl.

With four carbons, a methyl ketone must be butan-$2$-one.

Three tests, one structure. Each removes a possibility, and the molecular formula fixes the rest.

Illustration 10

Explain why the Hell-Volhard-Zelinsky reaction works on ethanoic acid but not on benzoic acid or on $2,2$-dimethylpropanoic acid.

The reaction brominates the $\alpha$ carbon and proceeds through an enol of the acid bromide, which requires an $\alpha$-hydrogen.

Ethanoic acid has three.

Benzoic acid has none, since the carbon next to its carbonyl is aromatic. $2,2$-Dimethylpropanoic acid has none either, its $\alpha$ carbon being fully substituted.

The same structural test as the aldol and the haloform. Reading a structure for $\alpha$-hydrogens answers all three questions at once.

Illustration 11

Give the product of a Claisen condensation of ethyl ethanoate, and explain why ethyl methanoate cannot undergo the same reaction with itself.

Ethoxide removes an $\alpha$-hydrogen from ethyl ethanoate; the resulting enolate attacks a second ester molecule and ethoxide leaves.

The product is ethyl $3$-oxobutanoate, acetoacetic ester.

Ethyl methanoate has no $\alpha$-hydrogen, so no enolate can form and it cannot act as the nucleophilic partner.

It can still act as the electrophile, which makes it useful in a crossed Claisen where the other partner supplies the enolate.

Illustration 12

Explain why saponification of an ester goes to completion while acid-catalysed hydrolysis does not.

Acid-catalysed hydrolysis is the exact reverse of Fischer esterification, so both directions are accessible and an equilibrium is established.

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

The resulting carboxylate is negatively charged and cannot be attacked by the alcohol, so the reverse reaction is impossible.

Removing the product drives the reaction, which is why saponification is quantitative and is used industrially for soap.

Illustration 13

Suggest how to convert ethanoic acid into ethanamide, and explain why heating the acid directly with ammonia is unsatisfactory.

Treat the acid with thionyl chloride to give ethanoyl chloride, then add ammonia. The chloride is at the top of the ladder, so ammonia attacks readily and the amide forms at once, in the cold.

Heating the acid with ammonia directly first gives ammonium ethanoate, an ionic salt.

Converting that salt to the amide requires driving off water at around $200^{\circ }$C, and the equilibrium is unfavourable, so the yield is poor.

Activating the acid first is the general strategy, and the same reasoning applies to making esters of hindered alcohols, where the direct Fischer route also fails.

Summary

• The presence of an $\alpha$-hydrogen decides between aldol and Cannizzaro, and governs the haloform, Hell-Volhard-Zelinsky and Claisen reactions too.
• In a crossed Cannizzaro with methanal, methanal is always oxidised and the other aldehyde reduced.
• Aldehydes beat ketones towards nucleophiles on both electronic and steric grounds; aromatic carbonyls are less reactive still.
• Alcohol reactivity with $\text{HX}$ and towards dehydration runs $3^{\circ }>2^{\circ }>1^{\circ }$, and both may rearrange.
• Oxidation: $1^{\circ }$ gives an aldehyde with mild reagents and an acid with strong ones; $3^{\circ }$ resists entirely.
• Phenol's ring is strongly activated, so bromine water gives the tribromide with no catalyst.
• Phenol's $\text{C}-\text{O}$ bond has partial double-bond character, so its hydroxyl cannot be replaced by halide.
• Reimer-Tiemann gives salicylaldehyde; Kolbe-Schmitt gives salicylic acid; diazonium coupling gives an azo dye.
• Williamson needs a primary halide; the bulky group must come from the alkoxide.
• Ether cleavage: $\text{SN}2$ sends iodide to the less hindered carbon, $\text{SN}1$ to the tertiary one, and the aryl-oxygen bond never breaks.
• Two reasons give the same ladder: leaving group ability falls and resonance donation rises going down it.
• Derivative reactivity: acid chloride $>$ anhydride $>$ ester $>$ amide, and the ladder is a one-way street.
• Saponification is irreversible because the carboxylate cannot be attacked by an alcohol.
• Aromatic aldehydes give no Fehling reaction but do reduce Tollens' reagent — the standard distinguishing pair.