Corrected exercises: carboxylic acids and their derivatives (202-SNU-RE)

The two resonance forms of ethanoate place the negative charge alternately on one oxygen and on the other, the double bond each time occupying the other position. The decisive feature is that the two forms are EXACTLY EQUIVALENT: they differ only in how the oxygens are labelled, so they contribute equally and the stabilization is at its maximum. The reality of the ion is neither one nor the other but their superposition, in which the charge is shared equally, one half on each oxygen, and the two bonds have exactly the same character, halfway between single and double. That is why both measure $127$ pm, a value between the $123$ pm of the double bond and the $136$ pm of the single bond in the non-ionized acid. This measured equality is the experimental proof of delocalization, and you should know how to cite it: it is far more convincing than a drawing of arrows.

The ratio is $10^{16.0-4.76}=10^{11.24}$, about $1.7\times 10^{11}$: ethanoic acid is one hundred and seventy billion times more acidic than ethanol. The reason lies entirely in the conjugate bases. The ethoxide ion $CH_{3}CH_{2}O^{-}$ carries its charge on an oxygen that has no neighbour able to share it: there is no resonance form, and the charge stays whole on a single atom. The ethanoate ion, for its part, has the neighbouring carbonyl to delocalize it. This shows what a simple resonance over two atoms is worth: more than eleven orders of magnitude, far more than the effect seen for the phenoxide ion in the previous chapter, with its six orders of magnitude, the difference being that the two forms of the carboxylate are equivalent whereas those of the phenoxide are not.

Chlorine is strongly ELECTRON-WITHDRAWING by the inductive effect: it pulls electrons toward itself, so it helps to disperse the negative charge of the carboxylate and STABILIZES the conjugate base. Each additional chlorine strengthens the effect, hence the steady drop from $4.76$ to $0.66$, four orders of magnitude for three chlorines. Note that trichloroethanoic acid, at $0.66$, is in practice a STRONG acid, comparable to phosphoric acid, which is remarkable for a carboxylic acid. The second series measures something else: DISTANCE. There is always a single chlorine, but it moves away from the carboxyl group. At position $2$ it is an immediate neighbour and the effect is at its maximum, $2.86$. At position $3$, almost nothing is left, $4.05$. At position $4$, with $4.52$, we are almost back to the $4.82$ of unsubstituted butanoic acid. The inductive effect is transmitted through the chain of $\sigma$ bonds and fades very quickly, by a factor close to ten for each intervening bond, and it is gone after three. This is the fundamental difference from a RESONANCE effect, which is transmitted without attenuation along a conjugated system, however long. Be careful not to mix up the two series: the first counts substituents, the second counts bonds.

The figure shows that two acid molecules pair head to tail through TWO simultaneous hydrogen bonds, closing an eight-membered ring. Each molecule is both donor, through its $O-H$, and acceptor, through its $C=O$, and the geometry of the carboxyl group puts these two groups exactly where they need to be for the pairing to be perfect. The result is a DIMER so stable that it persists in the vapour phase, which can be measured: the apparent molar mass of gaseous ethanoic acid is close to $120$, not $60$ g/mol. Boiling then means vaporizing an entity of double the mass, which costs much more energy. Propan-1-ol also forms hydrogen bonds, but it has only ONE donor site and its acceptor is the same oxygen: it forms disordered chains, never a closed pairing with two bonds. Hence the $21$ degrees of difference. The general conclusion is worth noting: it is not only the NUMBER of hydrogen bonds that counts, it is their GEOMETRY.

With $K_{a}=10^{-4.76}=1.74\times 10^{-5}$ and $C=0.10$ mol/L, the usual approximation gives $[H_{3}O^{+}]=\sqrt{K_{a}C}=\sqrt{1.74\times 10^{-6}}=1.32\times 10^{-3}$ mol/L, so $pH=2.88$. The percent ionization is $\frac{1.32\times 10^{-3}}{0.10}=1.3$ percent, well below $5$ percent: the approximation is justified. The comparison with phenol is instructive. Phenol, with $pK_{a}=10.0$, gave a pH of $5.50$ under the same conditions, a hydronium concentration about FOUR HUNDRED times lower. The structural reason is that of part a): the carboxylate has two EQUIVALENT resonance forms, both on very electronegative oxygens, whereas the phenoxide has only one on the oxygen and three on carbons, atoms much less able to carry a negative charge. A carboxylic acid and a phenol are therefore two weak acids separated by five orders of magnitude, which is precisely what the hydrogencarbonate test of the previous chapter detects with the naked eye.

First method, the VIGOROUS oxidation of a primary alcohol or an aldehyde, with dichromate in aqueous acid or with permanganate. The chain does not get longer. Second method, the hydrolysis of a NITRILE in hot aqueous acid or base; since the nitrile itself generally comes from a halide by the action of $CN^{-}$, the chain gets longer by one carbon. Third method, the CARBONATION of a Grignard reagent with carbon dioxide, followed by hydrolysis; the chain also gets longer by one carbon. For aromatic acids one can add the oxidation of a side chain by hot permanganate, which always gives benzoic acid whatever the length of the chain, provided it carries a benzylic hydrogen. The two chain-lengthening methods are complementary: the nitrile starts from a primary halide and copes poorly with hindered substrates, while the Grignard reagent accepts aryl and tertiary halides but tolerates no acidic proton anywhere in the molecule.

Step $1$: $C_{6}H_{5}MgBr+CO_{2}\rightarrow C_{6}H_{5}COOMgBr$. The phenyl carbanion attacks the carbon of carbon dioxide, which is electrophilic for the same reason as a carbonyl, giving a magnesium carboxylate. Step $2$: $C_{6}H_{5}COOMgBr+H_{3}O^{+}\rightarrow C_{6}H_{5}COOH$. Acid hydrolysis protonates the carboxylate and releases the acid. The carbon dioxide must be dry because the Grignard reagent is destroyed instantly by the slightest trace of water: the carbanion, the conjugate base of a hydrocarbon with $pK_{a}$ above $50$, pulls a proton off water, with $pK_{a}$ $15.7$, without hesitation, and you recover benzene instead of the desired product. In practice dry ice, solid $CO_{2}$, is used, which has the advantage of being at once the reagent, the drying agent and the coolant.

The amount of bromobenzene is $\frac{15.7}{157.01}=0.1000$ mol. The stoichiometry is one to one throughout the sequence, so the maximum is $0.1000$ mol of benzoic acid, that is $0.1000\times 122.12=12.2$ g. With an overall yield of $72$ percent, you obtain $0.72\times 12.2=8.79$ g. It is worth noting that the product is LIGHTER than the reagent despite the addition of a carbon and two oxygens: bromine, at $79.90$ g/mol, weighs more by itself than the $COOH$ group that replaces it. A plausibility check based on the idea that atoms are added so the mass must increase would therefore be misleading here.

Step $1$: shake the ether solution with an aqueous solution of $NaHCO_{3}$. Only benzoic acid, with $pK_{a}$ $4.20$, is stronger than carbonic acid at $6.35$: it is therefore the only one deprotonated, and it moves into the aqueous phase as sodium benzoate, with effervescence of $CO_{2}$. Phenol, at $10.0$, and naphthalene, which is neutral, stay in the ether. Separate the phases, then acidify the aqueous phase with $HCl$: benzoic acid, neutral again, precipitates and is filtered off. Step $2$: shake the remaining ether phase with a $NaOH$ solution. This time phenol, stronger than water at $15.7$, is deprotonated and moves into the aqueous phase as phenoxide; naphthalene, which has no acidic hydrogen, stays in the ether. Separate, then acidify the aqueous phase to recover the phenol. Step $3$: evaporate the ether to recover the naphthalene. The procedure thus exploits exactly the discrimination test of the previous chapter, but turns it into a PREPARATIVE method: this is the classic acid-base separation, and it regularly turns up in laboratory exams.

3-Oxobutanoic acid has a ketone group in the BETA position relative to the carboxyl group. This arrangement allows a remarkable concerted mechanism: the electrons move in a cycle through six atoms, the carbon-carbon bond breaks, the $CO_{2}$ leaves, and an ENOL forms which then tautomerizes to propanone, as in the exercise of the previous chapter. The six-membered cyclic transition state is hardly strained, so the barrier is low, and the loss of a gas makes the process irreversible. Butanoic acid, for its part, has no carbonyl in the beta position: no six-membered cycle can form, a carbanion would have to be torn off directly, which is prohibitive, hence its stability up to more than $300$ degrees. The rule to remember is that only BETA-keto acids and beta-dicarboxylic acids decarboxylate easily, which will be exploited in the malonic ester synthesis of exercise $8$. The physiological consequence is diabetic KETOACIDOSIS: lacking insulin, the body oxidizes fatty acids massively and accumulates ketone bodies, including 3-oxobutanoic acid, which decarboxylates spontaneously to propanone. This very volatile compound is eliminated through the lungs and gives the patient's breath a characteristic fruity smell, a clinical emergency sign that the chemistry of this paragraph fully explains.

The acyl chloride is $R-CO-Cl$, and the leaving group is $Cl^{-}$. The anhydride is $R-CO-O-CO-R$, and the leaving group is a CARBOXYLATE ion $R-COO^{-}$. The ester is $R-CO-OR'$, and the leaving group is an alkoxide ion $R'O^{-}$. The amide is $R-CO-NR'_{2}$, and the leaving group would be an amide anion $R'_{2}N^{-}$. The carboxylate is $R-COO^{-}$, and the leaving group would be $O^{2-}$, which never happens. Note that in every case what leaves is the conjugate base of something, which explains why the table of $pK_{a}$ values is enough to order everything.

The decreasing order is: acyl chloride, anhydride, ester, amide, carboxylate. It follows exactly the order of the $pK_{a}$ values in the table: $-7$ for $HCl$, $4.8$ for the carboxylic acid, $16$ for the alcohol, $36$ for the amine. The rule fits in one sentence: a derivative is the more reactive the WEAKER the BASE that its leaving group is, that is, the stronger the conjugate acid of that group. This is exactly the leaving-group criterion met in the chapter on alcohols, except that here it governs not an $S_{N}1$ or an $S_{N}2$ but the elimination step of an acyl substitution. The range is considerable: between $HCl$ and an amine there are $43$ units of $pK_{a}$, that is $43$ orders of magnitude in basicity, which explains why the two ends of the scale behave in practice like compounds from different families.

The heteroatom bonded to the carbonyl has a lone pair that it can donate by resonance, which creates a resonance form with a $C=X$ double bond and a positive charge on $X$. This donation reduces the positive charge on the carbon, so it makes it LESS electrophilic, and it stabilizes the whole molecule. Its efficiency depends on two things: the availability of the lone pair and the quality of the orbital overlap. Nitrogen, not very electronegative and of a size comparable to carbon, donates very well: the amide is strongly stabilized. Oxygen, more electronegative, holds its lone pair more tightly: the ester is moderately stabilized. Chlorine, very electronegative and above all equipped with diffuse $3p$ orbitals that overlap poorly with the $2p$ orbitals of carbon, donates almost nothing: the acyl chloride is practically not stabilized. The resulting order, acyl chloride more reactive than ester, itself more reactive than amide, is the same as in b). The two factors never contradict each other because they are two readings of the same property: an atom that holds its electrons poorly is also an atom that bears a negative charge well, hence both a poor resonance donor and a good leaving group.

The practical rule is that you can go DOWN the scale freely, but you never go back up in one direct step. Going down means replacing a leaving group with a nucleophile whose conjugate base is weaker, which is thermodynamically favourable. Application. Acyl chloride to ester: possible, you go down, just add the alcohol. Ester to acyl chloride: IMPOSSIBLE directly, you would have to go up; you must go through the acid, obtained by hydrolysis, then use $SOCl_{2}$, which is not a simple substitution but an activation. Anhydride to amide: possible, you go down. Amide to ester: impossible directly, you would be going up; you must hydrolyse the amide to the acid, which requires harsh conditions, then esterify. Ester to amide: possible, you go down, though slowly, and this is the ammonolysis of esters, used industrially. This single rule advantageously replaces a table of reactions to memorize.

Ethanoyl chloride expels a chloride ion, the conjugate base of an acid with $pK_{a}$ $-7$, hence an extraordinarily weak base that leaves with no difficulty; moreover, chlorine hardly stabilizes the carbonyl by resonance, so the carbon remains very electrophilic. The two factors add up and water, though a mediocre nucleophile, is enough to attack it at once. The fumes observed are in fact the gaseous $HCl$ released, which hydrates on contact with the moisture of the air. Ethyl ethanoate would have to expel an ethoxide ion, the conjugate base of an acid with $pK_{a}$ $16$, hence a STRONG base: the gap is $23$ units of $pK_{a}$, that is $23$ orders of magnitude in the basicity of the leaving group. On top of that comes the stabilization by donation from the oxygen. The activation barrier is therefore so high that spontaneous hydrolysis at room temperature is imperceptible, and it has to be catalyzed by an acid or, better, by a base that makes the reaction complete, which is the subject of exercise $5$.

Step $1$, the ADDITION. The nucleophile attacks the carbonyl carbon perpendicular to the plane, exactly as on a ketone. The $\pi$ electrons shift onto the oxygen and the carbon, which was $sp^{2}$ and flat, becomes $sp^{3}$ and TETRAHEDRAL. The species obtained is called the tetrahedral intermediate: it carries both the incoming nucleophile and the leaving group, and it is the tipping point of the whole reaction. Step $2$, the ELIMINATION. A lone pair on the charged oxygen drops back down to re-form the $C=O$ double bond, which forces one of the two groups to leave. The carbon becomes $sp^{2}$ and flat again. This is where the whole chapter is decided: the intermediate expels whichever of the two groups is the WEAKER BASE, so if the incoming nucleophile is a stronger base than the leaving group, the latter departs and the reaction goes through; otherwise the intermediate reverts and nothing happens. That is the difference from a ketone: a ketone has no group that can leave, so its tetrahedral intermediate is simply protonated, and the addition is final.

An $S_{N}2$ requires the nucleophile to attack from the BACK of the bond being broken, in line with it. But the carbon of an ester is $sp^{2}$: its three substituents and the carbon itself lie in one plane, and the back of the $C-O$ bond lies in that same plane, occupied by the two other bonds. There is simply no room, and on top of that the $\pi$ cloud of the carbonyl, located on either side of the plane, repels the nucleophile. This is exactly the argument that made aryl and vinyl halides inert. An $S_{N}1$ is no more possible: it would require forming an ACYLIUM cation $R-CO^{+}$ by spontaneous departure of the alkoxide, but the alkoxide is a strong base that does not leave on its own, and the acylium cation, although it exists for chlorides activated by $AlCl_{3}$ as in Friedel-Crafts acylation, is not accessible here. Only the addition-elimination route remains, which is why all acid derivatives, without exception, follow it.

It proves that the bond that breaks is that of the ACID, between its carbon and its $-OH$ group, and not that of the ALCOHOL, between its oxygen and its carbon. The reasoning is direct: since the oxygen-18 supplied by the alcohol is found in the ester, the alcohol has kept its $C-O$ bond and supplied only its oxygen and its carbon to the product; and since the water formed is not labelled, the oxygen it carries away must come from the acid. Without the labelling, the two hypotheses would be strictly indistinguishable, because the overall equation is identical in both cases. This is an excellent example, worth being able to cite, of what an isotopic experiment contributes: it measures no rate and no yield, it TRACKS the atoms and decides between two mechanisms that the stoichiometry confuses.

Step $1$: protonation of the carbonyl oxygen of the acid, which activates the carbon. Step $2$: addition of the alcohol, through its nucleophilic oxygen, to this carbon. Step $3$: deprotonation of the oxygen that came from the alcohol; this gives the neutral tetrahedral intermediate, which carries three oxygens on the same carbon. Step $4$: protonation of one of the TWO $-OH$ groups that came from the acid, turning it into $-OH_{2}^{+}$. This is where everything is decided, and it is what the labelling of c) reveals: it is an oxygen from the ACID that is protonated and then expelled, never the one from the alcohol. Step $5$: departure of water, with re-formation of the $C=O$ double bond. Step $6$: final deprotonation, which releases the ester and regenerates the acid catalyst. Note that all the steps are reversible, which is the very definition of an equilibrium, and that the proton is never consumed, which confirms the catalysis.

The reaction goes through the tetrahedral intermediate, in which four groups crowd around one carbon where there were only three at the start. Any crowding on the alcohol side or the acid side therefore raises the energy of this state, and the rate drops. Methanol brings only a methyl: the reaction is fast. Propan-2-ol brings two methyls on the carbon bearing the oxygen: it is slow. 2-Methylpropan-2-ol brings three: the intermediate is unreachable and, worse, this tertiary alcohol prefers to dehydrate or ionize in the acidic medium, so you obtain alkene instead of ester. The method to use then is to change the acid partner rather than insist: first convert the acid into the ACYL CHLORIDE with $SOCl_{2}$, then react the tertiary alcohol with it in the presence of pyridine. Since the acyl chloride is infinitely more reactive, as exercise $3$ established, the reaction takes place at low temperature, it is complete rather than an equilibrium since the $HCl$ is neutralized, and the crowding is no longer prohibitive. This is the clearest illustration of the practical value of the hierarchy: when the gentle route fails, you go up a rung on the scale.

Let $x$ be the amount of ester formed at equilibrium. The reaction table gives $1-x$ for the acid and for the alcohol, and $x$ for the ester as for the water. Write $K=\frac{x^{2}}{(1-x)^{2}}=4.0$. Since everything is positive, you can take the square root of both sides, which avoids the quadratic: $\frac{x}{1-x}=2.0$, so $x=2-2x$ and then $x=\frac{2}{3}=0.667$ mol. At equilibrium there is therefore $0.333$ mol of acid and as much alcohol left, and $0.667$ mol of ester and as much water have formed. The yield is $66.7$ percent. Note that this figure depends neither on the temperature, only weakly, nor on the amount of catalyst: it is a THERMODYNAMIC limit, and no technical improvement will get past it for a fixed starting composition.

With $4.00$ mol of alcohol, the relation becomes $K=\frac{x^{2}}{(1-x)(4-x)}=4.0$, that is $x^{2}=4(4-5x+x^{2})$, then $3x^{2}-20x+16=0$. The discriminant is $400-192=208$, so $x=\frac{20-\sqrt{208}}{6}=\frac{20-14.42}{6}=0.930$ mol, the root with the plus sign being rejected since it would exceed the initial amount of acid. The yield relative to the acid therefore rises from $66.7$ to $93.0$ percent. The expected comment is twofold. On the one hand, Le Chatelier's principle works exactly as announced: adding a reactant shifts the equilibrium. On the other hand, the yield is calculated here relative to the LIMITING reactant, the acid, and it would be absurd to calculate it relative to the alcohol, of which more than three quarters remains unused. In practice the cheaper reactant is used in excess, and the surplus is recovered by distillation.

First way, remove the water as it forms, with a Dean-Stark apparatus, a drying agent or a molecular sieve. Second way, distil off the ester if it is the most volatile component of the mixture, which is often the case since it can no longer form hydrogen bonds with itself. As for concentrated sulfuric acid, it is usually invoked as a catalyst, which is true but affects only the RATE and never the position of the equilibrium. Its real action on the yield is entirely different: concentrated, it is strongly DEHYDRATING and binds the water formed. It then acts like the first way, by removing a product. This distinction is regularly tested and is worth remembering: a catalyst never shifts an equilibrium, and if adding sulfuric acid improves the yield, it is not as a catalyst.

The equation is $CH_{3}COOC_{2}H_{5}+HO^{-}\rightarrow CH_{3}COO^{-}+C_{2}H_{5}OH$. The mechanism has three steps. First, the hydroxide ion adds to the carbonyl carbon and forms the tetrahedral intermediate. Second, this intermediate eliminates the ethoxide ion and regenerates the carbonyl: we then have, side by side in the medium, an ethanoic acid and an ethoxide ion. Third, and this is the decisive step, the ethoxide, a strong base whose conjugate acid has $pK_{a}$ $16$, instantly pulls the proton off the ethanoic acid, with $pK_{a}$ $4.8$. This acid-base reaction is favoured by $10^{11.2}$, so it is completely shifted. It makes the saponification IRREVERSIBLE for a simple reason: the final product is not the acid but the CARBOXYLATE, a negatively charged species that no alcohol will attack, since a nucleophile would have to approach a centre that is already electron-rich. The product is therefore trapped out of reach of the reverse reaction. Note that the base is CONSUMED here in stoichiometric amount and is not a catalyst, unlike the acid in the Fischer esterification: this is the fundamental difference between the two reactions.

A triglyceride is the triester of propane-1,2,3-triol, glycerol, with three long-chain fatty acids. Its saponification therefore gives GLYCEROL and three long-chain CARBOXYLATE ions, whose sodium salts make up the soap. A soap is a molecule with two personalities: a long nonpolar, hydrophobic hydrocarbon tail, and a charged, hydrophilic carboxylate head. In solution, above a certain concentration, these molecules assemble into spherical MICELLES, the tails gathered at the centre away from the water and the charged heads facing outward. A greasy soil dissolves in the nonpolar core of the micelle, while the charged exterior ensures dispersion in water and the electrostatic repulsion between micelles, which keeps them from coalescing. In hard water, rich in $Ca^{2+}$ and $Mg^{2+}$ ions, the carboxylates form INSOLUBLE salts with these divalent cations, which precipitate as a grey deposit, the familiar bathtub ring: the soap is used up without cleaning anything. Synthetic detergents replace the carboxylate head with a sulfonate or sulfate head, whose calcium and magnesium salts remain SOLUBLE, so they keep their full effectiveness in hard water.

The first resonance form is the usual drawing, $CH_{3}-CO-NH_{2}$, with a $C=O$ double bond and a $C-N$ single bond. The second results from the donation of the lone pair of the NITROGEN toward the carbon: the $C-N$ bond becomes double, the $C=O$ bond becomes single, the oxygen carries a negative charge and the nitrogen a positive charge. This second form is far from negligible, it is given a weight of the order of $40$ percent, because it places the negative charge on the oxygen, a very electronegative atom that welcomes it readily. The real molecule is the superposition of the two, and its $C-N$ bond therefore has PARTIAL DOUBLE-BOND CHARACTER. Now a double bond is shorter than a single one: $133$ pm is a value between the $147$ pm of a single $C-N$ bond in an amine and the $127$ pm of a true $C=N$ double bond. Symmetrically, the $C=O$ bond of the amide, at $124$ pm, is slightly LONGER than the $122$ pm of a ketone, since it has lost a little of its double-bond character. The two measurements confirm each other and are the experimental proof of the resonance.

The frame contains SIX atoms: the two alpha carbons, the carbonyl carbon, the oxygen, the nitrogen and the hydrogen carried by the nitrogen. The property they share is being COPLANAR. The reason follows directly from a): since the $C-N$ bond has partial double-bond character, it cannot rotate freely, and the nitrogen adopts a hybridization close to $sp^{2}$ instead of the usual $sp^{3}$ of amines, with its lone pair in a pure $p$ orbital involved in the conjugation. The three substituents of the nitrogen and the three of the carbon therefore end up in the same plane. The consequence for proteins is major and you must be able to state it: the polypeptide chain is not a flexible chain with total freedom. Of the three bonds of each residue, the one in the peptide link is LOCKED, and only the other two, around the alpha carbon, can rotate. The folding of a protein is therefore described by only two angles per amino acid, and it is this local rigidity that makes regular motifs possible, the alpha helix and the beta sheet. Add that the conformation adopted is almost always TRANS, more than $99$ percent of the time, proline being the exception.

Apply $k=A\exp\left(-\frac{E_{a}}{RT}\right)$ with $RT=8.314\times 298=2478$ J/mol. For the amide, $\frac{E_{a}}{RT}=\frac{80000}{2478}=32.3$, so $k=1.0\times 10^{13}\times \mathrm{e}^{-32.3}=9.5\times 10^{-2}$ s$^{-1}$. For ethane, $\frac{12000}{2478}=4.84$, so $k=1.0\times 10^{13}\times \mathrm{e}^{-4.84}=7.9\times 10^{10}$ s$^{-1}$. The ratio is $\frac{7.9\times 10^{10}}{9.5\times 10^{-2}}\approx 8.3\times 10^{11}$, nearly a thousand billion. The mean lifetime of a conformation, equal to $\frac{1}{k}$, is $10$ SECONDS for the amide against $13$ picoseconds for ethane. The order of magnitude is what matters: ten seconds is an eternity on the molecular scale and is perfectly observable by nuclear magnetic resonance, whereas thirteen picoseconds means that ethane rotates billions of times per second and no conformation has an individual existence. An amide is therefore RIGID in the literal sense, and the usual phrase “free rotation around a single bond” stops being true here.

First fact, the weakness of the basicity. The lone pair on the nitrogen of an amine is FREE and available: it picks up a proton without difficulty, hence a conjugate $pK_{a}$ of $10.6$. The lone pair on the nitrogen of an amide is, on the contrary, ENGAGED in the conjugation with the carbonyl, as a) established; mobilizing it to bind a proton would cost the loss of all the resonance stabilization. Second fact, the site of protonation. Since the second resonance form places a NEGATIVE charge on the oxygen, that is the most electron-rich site in the molecule, and so it is the one that picks up the proton; moreover, protonating the oxygen PRESERVES the conjugation, whereas protonating the nitrogen would destroy it. The difference in basicity is quantified by the difference in conjugate $pK_{a}$ values, $10.6-(-0.5)=11.1$ units, that is ELEVEN orders of magnitude. In concrete terms, an amine is protonated from a slightly acidic pH, whereas concentrated sulfuric acid is needed to protonate an amide, which explains in passing why proteins do not denature at the first change of pH through attack on their peptide bonds.

In aqueous acid, the hydrolysis gives ethanoic acid and the ammonium ion; in basic medium, it gives the ethanoate ion and ammonia. In both cases prolonged reflux is needed, often several hours, for two cumulative reasons. The first is REACTIVITY: the amide is the least electrophilic derivative on the scale, as exercise $3$ showed, both because the nitrogen donates strongly by resonance and because the leaving group would be an amide anion, an extremely strong base. The second concerns the overall balance: a simple hydrolysis would be an equilibrium, and it is only the final acid-base step, protonation of the amine in acidic medium or deprotonation of the acid in basic medium, that makes the reaction complete. At least one EQUIVALENT of acid or base is therefore needed, not a catalytic trace. With $LiAlH_{4}$, the result is quite different and often poorly remembered: ethanamide gives ETHANAMINE, that is, an amine, because the intermediate loses its oxygen as an aluminium alkoxide and passes through an iminium ion that the hydride then reduces; it is therefore the $C-O$ bond that is broken, not the $C-N$ bond. Ethanenitrile also gives ethanamine, by double addition of hydride across the triple bond. So remember the rule by family: ester and acid give an ALCOHOL, amide and nitrile give an AMINE.

The equation is $CH_{3}COOH+SOCl_{2}\rightarrow CH_{3}COCl+SO_{2}+HCl$. The mechanism is not a direct substitution: thionyl chloride first converts the poor leaving group $-OH$ into a chlorosulfite, an excellent leaving group, exactly as it did on an alcohol in the previous chapter. The practical reason for preferring it is decisive in the laboratory: BOTH by-products, $SO_{2}$ and $HCl$, are GASES at room temperature. They therefore leave the medium by themselves, which on the one hand shifts the equilibrium toward the product according to Le Chatelier's principle, and on the other hand makes purification trivial, since all that is needed is to distil the acyl chloride without having to separate anything. With $PCl_{3}$ or $PCl_{5}$, you obtain phosphorous or phosphoric acids and oxychlorides, all liquids or solids, which must be separated from the product.

The amount of acid is $\frac{12.0}{60.05}=0.1998$ mol, and the stoichiometry is one to one. The maximum mass of ethanoyl chloride is therefore $0.1998\times 78.50=15.69$ g. The gases released are $0.1998$ mol of $SO_{2}$, that is $12.80$ g, and as much $HCl$, that is $7.29$ g, so $20.09$ g in all. Check: the reactants consumed weigh $12.0$ g of acid plus $0.1998\times 118.96=23.77$ g of thionyl chloride, that is $35.77$ g; the products weigh $15.69+20.09=35.78$ g. The agreement to within rounding confirms the calculation, and it can be checked even faster on the molar masses themselves: $60.05+118.96=179.01$ on one side, $78.50+64.06+36.46=179.02$ on the other. Note that more than half of the mass involved leaves as gas, which is the price of the activation.

Toward the ANHYDRIDE: react the ethanoyl chloride with the ethanoate ion, supplied as sodium ethanoate. Toward the ESTER: add ethanol, in the presence of pyridine to capture the $HCl$ formed. Toward the AMIDE: add methanamine. In all three cases you go down the scale, so the reaction is fast, complete and not an equilibrium. For the amide, two equivalents of amine are needed because the reaction releases one molecule of $HCl$ per molecule of amide formed. Now the amine is a much better base than the amide, and this $HCl$ protonates it instantly to the methylammonium ion, which no longer has a lone pair and is no longer nucleophilic at all. The first equivalent therefore does the substitution, the second neutralizes the acid. This amine, often expensive, can of course be saved by using a cheap tertiary base such as pyridine or triethylamine, which is always done in practice.

The ionic solid obtained is methylammonium ethanoate. What happened is a competition of rates: the carboxylic acid is an acid with $pK_{a}$ $4.8$ and the amine a base with conjugate $pK_{a}$ $10.6$, so the acid-base reaction between them is favoured by nearly six orders of magnitude and above all is INSTANTANEOUS, whereas nucleophilic acyl substitution is slow. The acid-base reaction therefore wins the race, and the damage is twofold: the acid, now a carboxylate, finds itself on the very last rung of the reactivity scale, and the amine, now an ammonium ion, has lost its lone pair. Neither can do anything any more. First way out: heat strongly, above $150$ or even $200$ degrees Celsius, which shifts the acid-base equilibrium and drives off the water formed; this is brutal but it is what industry does for nylon. Second way: ACTIVATE the acid beforehand, either by converting it to the acyl chloride with $SOCl_{2}$ as in a), or by using a coupling agent such as dicyclohexylcarbodiimide, which converts the $-OH$ into a good leaving group in situ. It is this second route, workable at room temperature, that makes peptide synthesis possible.

The product is 2-methylpropan-2-ol, a TERTIARY alcohol, with release of ethanol. The sequence is as follows: a first equivalent of Grignard reagent adds to the ester and the tetrahedral intermediate expels the ethoxide ion, giving propanone. But this ketone is MUCH more electrophilic than the ester it comes from, for the two reasons of exercise $3$: it has no donor heteroatom to stabilize its carbonyl by resonance, and it is less hindered. It therefore reacts with a second equivalent of Grignard reagent FASTER than ester remains to be consumed, and the addition is this time final since a ketone has no leaving group. It is therefore impossible to stop at the ketone: even with a single equivalent, you would get a mixture of tertiary alcohol and unconverted ester, never the pure ketone. This is the general rule for sequences in which the intermediate is more reactive than the substrate. To obtain the ketone nevertheless, you must use a deliberately milder reagent, a lithium dialkylcuprate on the acyl chloride, or go through a Weinreb amide whose tetrahedral intermediate is stabilized by chelation and collapses only on the final hydrolysis.

The equation is $2\,CH_{3}COOC_{2}H_{5}\rightarrow CH_{3}COCH_{2}COOC_{2}H_{5}+C_{2}H_{5}OH$, in the presence of one equivalent of sodium ethoxide followed by a final acidification. The product is ethyl 3-oxobutanoate, universally called ethyl acetoacetate. The mechanism is that of a nucleophilic acyl substitution in which the nucleophile is an enolate: ethoxide removes an alpha hydrogen from a first ester molecule, the enolate formed attacks the carbonyl of a second molecule, and the tetrahedral intermediate expels an ethoxide ion. This is the addition-elimination pattern of exercise $4$, except that the nucleophile is a carbon, which creates a carbon-carbon bond.

The simple ester has a $pK_{a}$ of $25.0$ and ethanol $16.0$: the deprotonation is unfavourable by $25.0-16.0=9.0$ units, that is a constant of $10^{-9}$. At any moment there is therefore one enolate molecule for a billion ester molecules, which is nevertheless enough since this enolate is consumed and immediately regenerated. The reaction goes through thanks to a final step that is often forgotten: the product formed carries, between its two carbonyl groups, doubly activated hydrogens whose $pK_{a}$ is only $10.7$. Ethoxide therefore removes them quantitatively, with a constant of $10^{16.0-10.7}=10^{5.3}$, that is two hundred thousand. This very favourable deprotonation takes the product out of the game and PULLS the whole sequence of unfavourable equilibria that precedes it, according to Le Chatelier's principle. That is exactly why a WHOLE equivalent of base is needed: it is not a catalyst but a reagent, consumed stoichiometrically to form the anion of the product, which is only reprotonated at the final acidification. A trace of base would not trap the product and the reaction would go nowhere.

The medium contains an alkoxide and an ester, and an alkoxide attacks an ester: this is a TRANSESTERIFICATION. If methoxide were used with an ethyl ester, a mixture of methyl and ethyl esters would gradually form, hence a mixture of Claisen products. By matching the alkoxide to the alkoxy group of the ester, the transesterification does take place but is INVISIBLE, since it gives back exactly the same compound. This is a characteristic refinement of organic chemistry: the side reaction is not prevented, it is made harmless. For the crossed version, the condition is the same as for the crossed aldol reaction of the previous chapter: one of the two esters must have NO alpha hydrogen, so that it can play only the role of electrophile. Classic examples are ethyl benzoate, ethyl methanoate, diethyl oxalate and diethyl carbonate. Thus, ethyl benzoate and ethyl ethanoate cleanly give ethyl benzoylacetate. The intramolecular variant should also be cited, the DIECKMANN condensation, which closes a diester into a five- or six-membered ring for the entropic reasons already seen.

Step $1$: treat diethyl propanedioate with sodium ethoxide. Since its $pK_{a}$ is only $13.3$, the deprotonation is this time FAVOURABLE by $16.0-13.3=2.7$ units, hence quantitative, which is the whole point of the reagent. Step $2$: add 1-bromopropane; the malonate anion, a carbon nucleophile, carries out an $S_{N}2$ and ends up bearing a propyl group. Step $3$: hydrolyse the two esters in hot aqueous acid, which gives the corresponding diacid, then heat. This gives pentanoic acid, $CH_{3}CH_{2}CH_{2}CH_{2}COOH$. The decarboxylation step is justified exactly by exercise $2$: the diacid obtained has its two $COOH$ groups on the SAME carbon, so each is in the beta position relative to the other. The six-membered cyclic transition state is available, one $CO_{2}$ leaves and brings irreversibility, and an enol forms which tautomerizes to the acid. The net result of the sequence is worth remembering as a formula: the malonic ester synthesis turns $R-X$ into $R-CH_{2}-COOH$, that is, it adds TWO carbons and installs an acid group.

The sequence is identical, with ethyl 3-oxobutanoate in place of the malonate. Step $1$: sodium ethoxide, deprotonation that is even easier since the $pK_{a}$ is only $10.7$. Step $2$: bromoethane, alkylation by $S_{N}2$. Step $3$: acid hydrolysis then heating; the acid obtained is a BETA-KETO acid, so it too decarboxylates and you obtain pentan-2-one, $CH_{3}COCH_{2}CH_{2}CH_{3}$. The structural distinction between the two syntheses is clear and it is what you must be able to state: the malonic ester synthesis gives a substituted CARBOXYLIC ACID, $R-CH_{2}-COOH$, adding two carbons; the acetoacetic ester synthesis gives a METHYL KETONE, $CH_{3}-CO-CH_{2}-R$, adding three carbons. In both cases the limitation is the same and comes from step $2$: it is an $S_{N}2$, so the halide must be primary, or methyl, or at a stretch allylic or benzylic. A tertiary halide would give only alkene by $E2$, the malonate anion also being a base.

The degree of unsaturation is $\frac{2\times 4+2-8}{2}=\frac{10-8}{2}=1$, the two oxygens being ignored. Only one unsaturation, and two oxygens to place. The compatible families are therefore: carboxylic ACIDS, which use the unsaturation in their $C=O$ and their two oxygens in the carboxyl group; ESTERS, likewise; compounds carrying both a $C=O$ and a separate $-OH$, such as hydroxyaldehydes and hydroxyketones; and finally, more exotic, unsaturated or cyclic diols and cyclic ethers containing oxygen. So you must not conclude right away that it is an acid just because there are two oxygens.

Effervescence with $NaHCO_{3}$ proves a carboxylic ACID, since only an acid stronger than carbonic acid, with $pK_{a}$ $6.35$, releases $CO_{2}$ from it; a phenol or an alcohol would do nothing, as the chapter on alcohols showed. Two acids fit $C_{4}H_{8}O_{2}$: butanoic acid and 2-methylpropanoic acid. What follows tells them apart. $LiAlH_{4}$ reduces the acid to a PRIMARY alcohol, dehydration gives the corresponding alkene, and ozonolysis cuts this alkene into two carbonyl fragments. If A were butanoic acid, you would get butan-1-ol, then but-1-ene, and ozonolysis would give methanal and PROPANAL, an aldehyde that reduces Tollens' reagent. If A is 2-methylpropanoic acid, you get 2-methylpropan-1-ol, then 2-methylpropene, and ozonolysis gives methanal and PROPANONE, a ketone that does not reduce Tollens' reagent. It is this second result that is observed: A is 2-METHYLPROPANOIC ACID. Note the elegance of the set-up: three successive reactions serve only to carry the information about the branching through to a colour test.

The absence of reaction with $NaHCO_{3}$ rules out the acid, and the absence of reaction with sodium rules out any $O-H$, hence any alcohol. The fact that a basic reflux provides an acid AND an alcohol is the signature of a SAPONIFICATION, so B is an ESTER. There are four of them with formula $C_{4}H_{8}O_{2}$: propyl methanoate, 1-methylethyl methanoate, ethyl ethanoate and methyl propanoate. First piece of information: the alcohol gives a yellow precipitate of triiodomethane, so it carries the $CH_{3}CH(OH)-$ unit or is ethanol itself. That leaves only ethyl ethanoate, whose alcohol is ethanol, and 1-methylethyl methanoate, whose alcohol is propan-2-ol. Second piece of information: the acid does not reduce Tollens' reagent. Now methanoic acid is the only carboxylic acid with a HYDROGEN on its functional carbon: it therefore also behaves like an aldehyde and does reduce Tollens' reagent. The methanoate is thus ruled out, and B is ETHYL ETHANOATE. This trap of reducing methanoic acid comes up regularly in exams.

C is neither an acid nor an ester, but it has an $O-H$ since it reacts with sodium, and a carbonyl that is an ALDEHYDE since 2,4-DNPH and Tollens' reagent are both positive. It is therefore a hydroxyaldehyde of formula $C_{4}H_{8}O_{2}$. The last clue settles it: the loss of water in basic medium with formation of a CONJUGATED compound is the signature of an alcohol in the BETA position relative to the carbonyl, the only position that allows the $E1cb$ elimination which yields an alpha,beta-unsaturated aldehyde. C is therefore 3-HYDROXYBUTANAL, and the conjugated compound obtained is but-2-enal. It is exactly the aldol of ethanal, the product of the aldol reaction studied in the chapter on aldehydes and ketones, and the dehydration observed is its second phase.

Synthesis of A. Step $1$: magnesium in anhydrous ether, which gives 1-methylethylmagnesium bromide. Step $2$: dry ice, that is solid, dry $CO_{2}$, then acid hydrolysis. The carbanion attacks the carbon dioxide and you obtain 2-methylpropanoic acid, with lengthening by one carbon. This route is preferred to the nitrile route because the substrate is SECONDARY: the cyanide ion, which is also a base, would cause a competing $E2$ elimination there and the yield would collapse. Synthesis of B. The obvious route, oxidizing part of the ethanol to ethanoic acid then esterifying with the rest, is precisely the one to avoid: Fischer esterification is an EQUILIBRIUM capped at $66.7$ percent, as exercise $5$ calculated. The route asked for is therefore: step $1$, oxidation of part of the ethanol with $K_{2}Cr_{2}O_{7}$ in aqueous acid, which gives ethanoic acid; step $2$, action of $SOCl_{2}$, which gives ethanoyl chloride; step $3$, addition of the remaining ethanol, in the presence of pyridine. This last reaction goes down the reactivity scale, so it is fast, COMPLETE and not an equilibrium, all the more so since the pyridine captures the $HCl$ and rules out any reverse reaction. An extra step was paid for to remove a thermodynamic limit, which is the central reasoning of the chapter.

The equation is $HOOC-C_{6}H_{4}-COOH+HO-CH_{2}CH_{2}-OH\rightarrow \left[-OC-C_{6}H_{4}-CO-O-CH_{2}CH_{2}-O-\right]+2\,H_{2}O$. Each monomer is BIFUNCTIONAL, which is the very condition for a chain to grow from both ends, and each bond created is an ESTER bond, formed by esterification with loss of water. Check: $166.13+62.07=228.20$ g/mol for the two monomers, minus $2\times 18.02=36.03$ g/mol for the two molecules of water eliminated, which gives $192.17$ g/mol, exactly the value given for the repeat unit. This check is indispensable: the most frequent error is to add up the monomers without subtracting the water, which then throws off every calculation of the degree of polymerization.

The amount of repeat units is $\frac{25.0}{192.17}=0.1301$ mol, that is about $7.8\times 10^{22}$ repeat units. The average degree of polymerization is the number of repeat units per chain: $\overline{DP}=\frac{30000}{192.17}=156$. A bottle therefore contains about $0.130/156=8.3\times 10^{-4}$ mol of chains, nearly $5\times 10^{20}$ macromolecules. The order of magnitude deserves a comment: $156$ repeat units per chain may seem modest for a material that must withstand the pressure of a carbonated drink, but it is already enough for the chains to entangle and partially crystallize. Note that this is an AVERAGE: a polymer is never a pure substance, it is described by a distribution of masses, which sets it apart from all the other species in the course.

The equation is $H_{2}N-\left(CH_{2}\right)_{6}-NH_{2}+HOOC-\left(CH_{2}\right)_{4}-COOH\rightarrow \left[-NH-\left(CH_{2}\right)_{6}-NH-CO-\left(CH_{2}\right)_{4}-CO-\right]+2\,H_{2}O$. Check: $116.21+146.14=262.35$ g/mol, minus $36.03$ g/mol of water, equals $226.32$ g/mol, the value given. The names in fact tell the structure: the first $6$ counts the carbons of the diamine, the second those of the diacid. The chains hold together better than those of polyethylene for a very specific reason: each repeat unit contains two AMIDE groups, hence two $N-H$ donor groups and two $C=O$ acceptor groups. Neighbouring chains pair up through a dense network of HYDROGEN BONDS, regularly spaced along the backbone. Polyethylene has only London forces between its chains. As a result nylon melts at about $265$ degrees Celsius against $130$ for polyethylene, and it can be spun into fibres that resist tension, whereas polyethylene is used as flexible packaging. The planarity and rigidity of the amide link, established in exercise $6$, also contribute by imposing a regular extended conformation on the chains, which favours crystallization.

The lower phase, aqueous and denser, contains hexane-1,6-diamine, soluble in water, usually with sodium hydroxide added; the upper phase, hexane, contains hexanedioyl chloride, soluble in nonpolar solvents. The acyl chloride is used rather than the acid for the reason demonstrated in exercise $7$: brought together, a carboxylic acid and an amine merely undergo an instantaneous acid-base reaction and form an inert salt, so you would have to heat above $200$ degrees Celsius to obtain the amide. The acyl chloride, at the very top of the reactivity scale, reacts at room temperature and completely, the $HCl$ released being captured by the sodium hydroxide in the aqueous phase. The thread forms exactly at the INTERFACE because it is the only place where the two reagents, each confined to its own solvent, can meet. The reaction there is so fast that it immediately produces a solid film of polymer; by grasping it with tweezers and pulling, you remove it and expose a fresh interface where polymerization resumes, hence a continuous thread that can be wound up over several metres. This is the famous nylon rope experiment.

The target degree of polymerization is $\overline{DP}=\frac{20000}{226.32}=88.4$. The Carothers equation then gives $p=1-\frac{1}{\overline{DP}}=1-\frac{1}{88.4}=0.9887$, that is $98.9$ percent conversion. The comment is the heart of the question. In an ordinary synthesis, a yield of $99$ percent is a rare luxury and a yield of $90$ percent is already excellent. Here, $90$ percent conversion would give a $\overline{DP}$ of only $10$, that is a viscous oil with no mechanical properties, not a material. Step-growth polymerization is therefore quite exceptionally demanding: a NEAR-QUANTITATIVE conversion is needed to obtain a usable polymer. Three very real industrial constraints follow: a rigorously equal stoichiometry between the two monomers, extreme purity of the reagents since any monofunctional impurity permanently blocks one end of a chain, and very thorough removal of the water formed, which explains the high-vacuum, high-temperature reactors of nylon and polyester plants.

Glycolysis consists of heating the polymer with a large excess of ethane-1,2-diol in the presence of a catalyst. The principle is a TRANSESTERIFICATION: the diol attacks the carbonyls of the ester bonds in the chain and cuts them. Since esterification is an equilibrium, as exercise $5$ showed, a massive excess of diol shifts this equilibrium toward depolymerization, according to Le Chatelier's principle. The monomer is recovered, bis(2-hydroxyethyl) terephthalate, which is purified and repolymerized: the resulting material is of virgin quality, unlike mechanical recycling by grinding and remelting, which degrades the chains at each cycle. Polyethylene cannot be treated this way because its backbone is made only of SINGLE carbon-carbon bonds. There is no electrophilic carbon, no leaving group, hence no possible substitution reaction: the only way to cut it would be to break $C-C$ bonds by pyrolysis at very high temperature, which gives a mixture of hydrocarbons and not a clean monomer. This is a general conclusion of the chapter: a CONDENSATION polymer, built by a reversible reaction, can be chemically depolymerized; an ADDITION polymer, built by an irreversible reaction on alkenes, cannot.