Organic Compounds Containing Oxygen

Four carbonyl compounds. Rank them by how readily a nucleophile attacks the carbonyl carbon.

$\mathrm{C}{\mathrm{H}}_3\mathrm{CHO},\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3,\mathrm{C}{\mathrm{H}}_3\mathrm{COOH},\mathrm{C}{\mathrm{H}}_3\mathrm{COCl}$

The obvious reasoning: the carbonyl carbon is electrophilic because oxygen withdraws density from it, so the compound with the most electron-withdrawing atoms attached should be the most electrophilic. Ethanoic acid has two oxygens. It should win.

The actual order is

$\mathrm{C}{\mathrm{H}}_3\mathrm{COCl}\gg \mathrm{C}{\mathrm{H}}_3\mathrm{CHO}>\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3\gg \mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$

Ethanoic acid comes last, and not narrowly. A carboxylic acid does not undergo nucleophilic addition at all under conditions where an aldehyde reacts instantly.

The mistake was counting only what the attached group takes and ignoring what it gives back.

Every group attached to a carbonyl carbon does two things: it withdraws through the sigma bond ($-I$) and, if it has a lone pair, it donates into the carbonyl through the pi system ($+R$). The second effect is the one that matters, because it feeds density directly into the carbon under attack.

Chlorine's failure to donate is the same size-matching failure met in the p-block, where $\mathrm{B}{\mathrm{F}}_3$ turned out to be the weakest Lewis acid of the boron halides for exactly the opposite reason. A 3p lone pair cannot reach a carbon 2p orbital properly; an oxygen 2p lone pair can.

That comparison sets up the whole chapter, because the same question orders everything in it.

Oxygen is more electronegative than carbon and hydrogen, so every compound here has an electron-poor carbon and an electron-rich oxygen. That polarisation makes the carbon electrophilic, makes any attached hydrogen acidic, and leaves lone pairs so the oxygen is also a nucleophile and a base.

How strongly each shows up depends entirely on what else can share the load.

1. Alcohols, Phenols and Ethers

Alcohols carry OH on an $s{p}^3$ carbon, classified primary, secondary or tertiary by how many carbons that carbon bears. Phenols carry OH directly on an aromatic ring. Ethers have oxygen between two carbon groups.

Preparation

Alcohols come from hydration of alkenes, reduction of aldehydes, ketones or acids, hydrolysis of haloalkanes, and Grignard reagents with carbonyl compounds.

The Grignard route is worth knowing as a pattern: methanal gives a primary alcohol, any other aldehyde gives a secondary, and a ketone gives a tertiary.

Phenol is made industrially by the cumene process, and in the laboratory from chlorobenzene, benzenesulphonic acid, or a diazonium salt with warm water. Ethers are made by the Williamson synthesis, an alkoxide displacing a halide.

Illustration 1

The cumene process makes phenol from benzene and propene. Follow the atoms, and say why this route displaced every alternative.

${\mathrm{C}}_6{\mathrm{H}}_6+\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3{)}_2\text{(Friedel-Crafts alkylation)}$

${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3{)}_2+{\mathrm{O}}_2\to \text{cumene hydroperoxide}\stackrel{{\mathrm{H}}^{+}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}+\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3$

Count what came out. Every molecule of phenol arrives with one molecule of propanone alongside it, and both are large-volume industrial chemicals.

That is why the route won. A process producing one saleable product and one waste stream must charge the customer for disposing of the waste. A process producing two saleable products splits its costs across both, and the world happens to want propanone in roughly the quantity that phenol demand generates it.

Oxidation is also the cheapest possible reagent here, since the oxidant is air. Compare the laboratory routes, which need chlorobenzene at high temperature and pressure, or a diazonium salt made from aniline in three steps.

Why the boiling points differ so much

Alcohols and phenols hydrogen bond to each other; ethers cannot, having no O-H bond.

Ethanol boils at 351 K and its isomer dimethyl ether at 249 K, a gap of over a hundred kelvin from the same molecular formula. Ethers do accept hydrogen bonds from water, which is why they are appreciably soluble in it while unable to bond to themselves.

2. Acidity: The Central Comparison

$\text{carboxylic acid}>\text{phenol}>\text{water}>\text{alcohol}$

Phenol is about a million times more acidic than ethanol, and the reason is entirely in the anion. Phenoxide delocalises its charge into the ring; ethoxide has nowhere to put it.

A second contribution: the carbon bearing OH in phenol is $s{p}^2$ and therefore more electronegative than an alcohol's $s{p}^3$ carbon, pulling density away even before the proton leaves.

Carboxylic acids beat phenols because carboxylate spreads its charge over two equivalent oxygens, both far more electronegative than carbon, giving two identical contributing structures and a genuinely symmetrical ion.

Substituent effects

Anything that stabilises the anion increases acidity. Electron-withdrawing groups increase acidity: 4-nitrophenol beats phenol, and 2,4,6-trinitrophenol (picric acid) beats acetic acid. Electron-donating groups decrease it: 4-methylphenol is weaker than phenol.

Illustration 2

Rank ethanol, p-cresol, phenol, p-nitrophenol and ethanoic acid by acidity, and give one argument that covers all five.

The measured $\mathrm{p}{K}_a$ values run 16, 10.3, 10.0, 7.2 and 4.8 in that order, so acidity increases down the list. Every step is explained by the same question: how well is the conjugate base stabilised?

Ethoxide carries its charge on one oxygen with nowhere to put it, and the alkyl group pushes electrons toward that charge, making things worse. Worst anion, weakest acid.

Phenoxide spreads the charge around the ring by resonance, which is worth six orders of magnitude against ethanol.

p-Cresolate is phenoxide with a methyl group donating electrons into a ring already carrying negative charge, so it is slightly destabilised and p-cresol is slightly the weaker acid.

p-Nitrophenoxide is the reverse: the nitro group withdraws by resonance and can place the negative charge directly on its own oxygens, so the anion is far better off and the acid gains nearly three $\mathrm{p}{K}_a$ units.

Acetate shares the charge equally between two oxygens, and both are fully equivalent.

Now the part worth carrying. Phenoxide and acetate are both described as resonance-stabilised, yet the carboxylic acid beats the phenol by more than five $\mathrm{p}{K}_a$ units. The difference is where the charge lands: phenoxide's resonance pushes charge onto ring carbons, and carbon holds negative charge badly, while acetate keeps it entirely on oxygen. Counting resonance structures is not enough — the electronegativity of the atoms sharing the charge decides the outcome.

3. Reactions of Alcohols

With hydrogen halides, and the Lucas test

Alcohols react with HX to give haloalkanes, and the rate depends sharply on class because the reaction goes through a carbocation.

Illustration 3

Benzyl alcohol and allyl alcohol are both primary, so the Lucas test should show nothing. Both give turbidity immediately, faster than most tertiary alcohols. Explain.

The classification into primary, secondary and tertiary is a proxy, and here the proxy fails.

What the Lucas test really measures is how easily a carbocation forms, since that is the slow step. Counting attached carbons is a reasonable stand-in for that when hyperconjugation and induction are the only stabilising mechanisms available. They are not the only ones.

So the test reports exactly what it was designed to report. It was the substitution-count shorthand that was approximate, not the test.

Trap. Any question offering benzyl or allyl alongside ordinary alcohols is checking whether you classify by structure or reason by carbocation stability. The same warning applies to every ${\mathrm{S}}_{\mathrm{N}}1$ reaction, to dehydration and to the HI cleavage of ethers.

Dehydration

Concentrated sulphuric acid at high temperature gives an alkene, with ease running $3^{\circ }>2^{\circ }>1^{\circ }$, again through the carbocation. Saytzeff's rule applies, so the more substituted alkene dominates.

Oxidation, and how it distinguishes the classes

A primary alcohol oxidises to an aldehyde and onward to an acid; stopping at the aldehyde needs a mild reagent such as PCC, or immediate distillation of the volatile aldehyde. A secondary alcohol gives a ketone and stops, no hydrogen remaining on the carbinol carbon. A tertiary alcohol resists oxidation entirely.

Understand this through the hydrogen count rather than memorising it, and the exceptions never surprise you.

4. Phenols

The ring makes phenol behave quite differently from an alcohol. The OH donates a lone pair into the ring, activating it so strongly that phenol reacts with bromine water at room temperature without any catalyst, giving 2,4,6-tribromophenol immediately as a white precipitate. Benzene under the same conditions does nothing.

Illustration 4

Convert phenol into aspirin, and say why the ring's activation is essential at the first step and irrelevant at the second.

Step 1, Kolbe. Sodium phenoxide with carbon dioxide under pressure gives sodium salicylate, which on acidification gives salicylic acid, that is 2-hydroxybenzoic acid.

Carbon dioxide is a feeble electrophile, far too weak to attack benzene at all. It works here only because phenoxide is one of the most activated rings in ordinary chemistry: it carries a full negative charge that resonance places directly on the ortho and para carbons. The activation is what makes the step possible.

Step 2, acetylation. Salicylic acid with ethanoic anhydride gives aspirin, acetylsalicylic acid.

That step is an ordinary esterification of the phenolic OH and involves the ring not at all. What it achieves is medicinal rather than electronic: free salicylic acid is corrosive to the stomach lining, and capping the phenol as an ester makes it tolerable while the body hydrolyses it back afterwards.

One synthesis, two steps, and the ring's electronics matter enormously in one and not at all in the other. Ask what each step needs before deciding which property is doing the work.

5. Ethers

Williamson synthesis works by ${\mathrm{S}}_{\mathrm{N}}2$, and that constrains what can be made. The halide must be primary, because a tertiary halide meets the strongly basic alkoxide and eliminates instead.

So to make tert-butyl methyl ether, use tert-butoxide with methyl iodide, not methoxide with tert-butyl chloride. Choosing the wrong pairing gives an alkene, and this is a favourite trap.

Illustration 5

Ethers are famously unreactive, which is why they are used as solvents. Yet hydrogen iodide cleaves them. Predict the products from ethyl methyl ether and from tert-butyl methyl ether, and explain the difference.

The oxygen is protonated first in both cases, turning a terrible leaving group, $\mathrm{R}{\mathrm{O}}^{-}$, into a good one, ROH. What happens next depends on the carbons.

Ethyl methyl ether. Both groups are ordinary alkyls, so no cation is stable enough to form and the mechanism is ${\mathrm{S}}_{\mathrm{N}}2$. Iodide attacks the less hindered carbon, which is the methyl.

$\mathrm{C}{\mathrm{H}}_3-\mathrm{O}-{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{HI}\to \mathrm{C}{\mathrm{H}}_3\mathrm{I}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$

tert-Butyl methyl ether. Now one carbon can support a stable tertiary cation, so the mechanism switches to ${\mathrm{S}}_{\mathrm{N}}1$: the C-O bond breaks on its own and iodide captures the cation.

$(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}-\mathrm{O}-\mathrm{C}{\mathrm{H}}_3+\mathrm{HI}\to (\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{CI}+\mathrm{C}{\mathrm{H}}_3\mathrm{OH}$

The iodide went to the more substituted carbon, exactly opposite to the first case.

Note that neither answer needed memorising. Ask which mechanism operates, and the mechanism names the carbon: ${\mathrm{S}}_{\mathrm{N}}2$ picks the accessible one, ${\mathrm{S}}_{\mathrm{N}}1$ picks the one that makes the better cation.

6. Aldehydes and Ketones

The carbonyl carbon is $s{p}^2$, planar and strongly electrophilic. The characteristic reaction is nucleophilic addition, the opposite of the electrophilic addition alkenes undergo, and the difference comes entirely from the polarity of the bond.

Preparation

Both come from oxidation of alcohols, ozonolysis of alkenes and hydration of alkynes. Three routes are specific to aldehydes: Rosenmund reduction (acyl chloride over poisoned Pd on $\mathrm{BaS}{\mathrm{O}}_4$), Stephen's reaction (nitrile with $\mathrm{SnC}{\mathrm{l}}_2$ and HCl, then hydrolysis), and the Etard reaction (toluene with chromyl chloride).

Ketones come from Friedel-Crafts acylation for aryl ketones, and from a Grignard reagent with a nitrile or acyl chloride. Gattermann-Koch formylates benzene with CO and HCl over $\mathrm{AlC}{\mathrm{l}}_3$ and CuCl, solving the problem that formyl chloride is too unstable to use directly.

Why aldehydes are more reactive than ketones

Electronic: a ketone has two alkyl groups donating into the carbonyl carbon; an aldehyde has one. Steric: a ketone has two bulky groups shielding the carbon; an aldehyde has one group and a hydrogen.

Aromatic aldehydes are less reactive than aliphatic ones, because the ring donates by resonance in addition to whatever the substituent does.

Named reactions worth knowing cold

Aldol condensation requires an alpha hydrogen. Cannizzaro reaction requires none. The two are mutually exclusive, and asking whether an alpha hydrogen exists answers a large share of the questions set from this chapter.

Clemmensen reduces the carbonyl to $\mathrm{C}{\mathrm{H}}_2$ with zinc amalgam and HCl; Wolff-Kishner does the same with hydrazine and base. Use Clemmensen when the molecule tolerates acid and Wolff-Kishner when it tolerates base.

Illustration 6

Mixing ethanal and propanal with dilute base gives a mess. Explain, then give a version of the reaction that gives one product cleanly.

Both aldehydes have alpha hydrogens, so both form carbanions and both offer carbonyls to be attacked. Four combinations are possible and all four occur.

Four products in comparable amounts, none of them separable easily. A crossed aldol between two similar partners is a preparative dead end.

The fix is to make one partner incapable of being the nucleophile. Use benzaldehyde, which has no alpha hydrogen, so it can only be attacked and never attack.

${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}+\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}\stackrel{\mathrm{dilute}\mathrm{NaOH}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CH}=\mathrm{CHCHO}$

One nucleophile and one electrophile means one product, cinnamaldehyde. This is the Claisen-Schmidt reaction, and the same principle governs every crossed condensation: remove one of the two roles from one of the two partners.

Distinguishing tests

Illustration 7

Benzaldehyde gives a silver mirror with Tollens' reagent and nothing at all with Fehling's. Both are mild oxidising agents in alkaline solution. Why does one work and the other not?

Because they are not equally mild, and benzaldehyde is not equally easy to oxidise.

Fehling's copper(II), held in a tartrate complex, is a weaker oxidant than Tollens' silver(I). It has just enough power for an aliphatic aldehyde and not quite enough for an aromatic one.

Benzaldehyde is harder to oxidise than ethanal because its carbonyl is conjugated with the ring, which delocalises the carbonyl pi system and lowers its energy. A stabilised starting material is a reluctant one.

So the pair of tests is more informative than either alone.

Two tests separate three classes, and the discriminating step is the one where the reagents differ in strength rather than in kind.

Illustration 8

Predict what concentrated sodium hydroxide does to each of these: (a) ethanal, (b) benzaldehyde, (c) 2,2-dimethylpropanal, $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}-\mathrm{CHO}$.

One question decides all three: does the carbonyl compound have an α-hydrogen?

(a) Ethanal has three. Hydroxide removes one to give an enolate, which attacks a second molecule of ethanal. The aldol product dehydrates on warming to but-2-enal. This is aldol condensation.

(b) Benzaldehyde has none — the carbon next to the carbonyl is part of the ring and carries no hydrogen. With no enolate available, hydroxide attacks the carbonyl carbon directly and a hydride is transferred to a second molecule. One aldehyde is reduced to benzyl alcohol and the other oxidised to benzoate. This is the Cannizzaro reaction.

(c) 2,2-Dimethylpropanal also has none, since the neighbouring carbon carries three methyl groups and no hydrogen. Cannizzaro again, giving neopentyl alcohol and the carboxylate.

Cannizzaro is a disproportionation: the same compound is both oxidised and reduced, which is why it consumes two molecules and gives two different products in equal amounts. That also makes it wasteful, since half your aldehyde becomes acid.

The crossed version fixes that. Run the reaction with excess formaldehyde and formaldehyde is preferentially oxidised, because it is the least hindered and most readily attacked, so the aldehyde you actually care about is reduced cleanly to its alcohol in nearly full yield.

7. Carboxylic Acids

Preparation

Oxidation of a primary alcohol or aldehyde; hydrolysis of a nitrile, ester or amide; and a Grignard reagent with carbon dioxide. The nitrile and Grignard routes both add a carbon to the chain, which is what makes them useful for building up.

Oxidation of an alkylbenzene side chain with hot alkaline permanganate gives benzoic acid regardless of side-chain length, provided there is at least one benzylic hydrogen. Tert-butylbenzene is therefore untouched, which is a neat diagnostic.

Characteristic reactions

Esterification with an alcohol and acid catalyst is reversible, driven forward by removing water. Decarboxylation of the sodium salt with soda lime gives an alkane with one carbon fewer. Hell-Volhard-Zelinsky uses bromine with red phosphorus to substitute at the alpha carbon.

Reduction with $\mathrm{LiAl}{\mathrm{H}}_4$ gives a primary alcohol. $\mathrm{NaB}{\mathrm{H}}_4$ is too mild to touch a carboxylic acid, which is a useful selectivity.

Illustration 9

Esterification of ethanoic acid with ethanol removes a molecule of water. Which oxygen ends up in the water, the acid's or the alcohol's?

The equation cannot tell you, because both are oxygens. Label one and find out.

Run the reaction with ethanol enriched in ${}^{18}\mathrm{O}$. If the alcohol's oxygen leaves in the water, the label appears in the water and the ester is ordinary. If the acid's oxygen leaves, the label stays in the ester.

The result: the ${}^{18}\mathrm{O}$ is found in the ester.

$\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5{}^{18}\mathrm{OH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{CO}{-}^{18}\mathrm{O}-{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$

So the alcohol keeps its oxygen and contributes it whole to the ester, while the water's oxygen came from the acid's OH group.

That settles the mechanism. The alcohol attacks the carbonyl carbon as a nucleophile, giving a tetrahedral intermediate, and it is the acid's OH that is protonated and expelled as water. The bond broken is the acyl-oxygen bond, not the alkyl-oxygen bond, which one experiment established and no amount of arrow-pushing could have proved.

Illustration 10

Use the Hell-Volhard-Zelinsky reaction to make glycine from ethanoic acid.

HVZ substitutes bromine at the alpha carbon, using bromine with a little red phosphorus.

$\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}\stackrel{\mathrm{B}{\mathrm{r}}_2/red\mathrm{P}}{\to }\mathrm{BrC}{\mathrm{H}}_2\mathrm{COOH}$

The alpha carbon now carries a good leaving group, so an ordinary ${\mathrm{S}}_{\mathrm{N}}2$ displacement with excess ammonia installs the amino group.

$\mathrm{BrC}{\mathrm{H}}_2\mathrm{COOH}+2N{\mathrm{H}}_3\to {\mathrm{H}}_2\mathrm{NC}{\mathrm{H}}_2\mathrm{COOH}+\mathrm{N}{\mathrm{H}}_4\mathrm{Br}$

Glycine, the simplest amino acid, in two steps from vinegar.

Note why HVZ was needed rather than direct halogenation. Free radical bromination would attack indiscriminately and would not favour the alpha position at all. HVZ works through the enol of the acyl bromide, which exists only at the alpha carbon, so the selectivity is built into the mechanism.

This is the general route to alpha-amino and alpha-hydroxy acids, and it is where this chapter hands over to biomolecules.

Illustration 11

A molecule contains an ester, a ketone and a nitro group. Which reagent reduces which, and in what order would you use them?

Reducing agents differ in strength, and that difference is a tool rather than a limitation.

To reduce only the ketone, use $\mathrm{NaB}{\mathrm{H}}_4$: the ester and the nitro group survive untouched. To reduce only the nitro group, use catalytic hydrogenation or $Sn/HCl$. To reduce everything reducible, use $\mathrm{LiAl}{\mathrm{H}}_4$.

Trap. A question naming $\mathrm{NaB}{\mathrm{H}}_4$ has already told you the answer to half of itself. Selectivity questions are usually solved by reading the reagent rather than the substrate, and the standard trap is applying $\mathrm{LiAl}{\mathrm{H}}_4$ where $\mathrm{NaB}{\mathrm{H}}_4$ was specified and reducing groups the question wanted preserved.

Note finally that the carbonyl of a carboxylic acid does not undergo nucleophilic addition the way an aldehyde does, because the OH donates a lone pair into it. That is the point this chapter opened on, and it is why acids need the more forceful reagent.

Summary

The reactivity order of carbonyl compounds towards nucleophiles is acyl chloride, aldehyde, ketone, ester, amide, carboxylate, and it is set by how strongly the attached group donates back into the carbonyl, not by how much it withdraws. Chlorine barely donates, because a 3p lone pair cannot reach a carbon 2p orbital, so an acyl chloride's carbon is left naked.

The same question, what shares the load, orders acidity in the other direction: carboxylic acid, phenol, water, alcohol, at $\mathrm{p}{K}_a$ 4.8, 10, 15.7 and 16.

Alcohols hydrogen bond and ethers cannot, so ethanol boils 102 K above dimethyl ether. The cumene process makes phenol and propanone together from benzene, propene and air, which is why it displaced every alternative.

The Lucas test measures carbocation stability, not substitution count, so benzyl and allyl alcohols give instant turbidity despite being primary. Alcohol oxidation is a count of hydrogens on the carbinol carbon: two gives an aldehyde then an acid, one gives a ketone, none gives nothing.

Phenoxide is activated enough for carbon dioxide to attack it, which is the Kolbe route to salicylic acid and then aspirin. Williamson synthesis needs the primary halide, and HI cleavage picks the accessible carbon by ${\mathrm{S}}_{\mathrm{N}}2$ or the better cation by ${\mathrm{S}}_{\mathrm{N}}1$.

Aldehydes beat ketones on both electronic and steric grounds. Aldol and Cannizzaro are decided by one question about alpha hydrogens, and a crossed aldol works only when one partner cannot act as the nucleophile.

Tollens detects all aldehydes and Fehling only aliphatic ones, because Fehling's copper(II) is the weaker oxidant and benzaldehyde is stabilised by conjugation with the ring.

Isotopic labelling shows that esterification breaks the acyl-oxygen bond, since ${}^{18}\mathrm{O}$ from the alcohol ends up in the ester. HVZ gives selective alpha-bromination through the enol, and ammonia then converts it to an alpha-amino acid.

Choose reducing agents by strength: $\mathrm{NaB}{\mathrm{H}}_4$ for aldehydes and ketones alone, $\mathrm{LiAl}{\mathrm{H}}_4$ for esters and acids as well, and catalytic hydrogenation for nitro groups and multiple bonds.