Corrected exercises: aldehydes and ketones (202-SNU-RE)

The carbonyl carbon forms three $\sigma$ bonds, to the oxygen and to the two groups $R$ and $R'$: it is therefore $sp^{2}$, and its three neighbours sit at the corners of a flat triangle, with angles close to $120$ degrees as the arc in the figure shows. Its fourth orbital, a pure $p$ orbital, is PERPENDICULAR to this plane and forms the $\pi$ bond with the oxygen. A nucleophile arriving in the plane of the figure would meet only filled $\sigma$ bonds and the two $R$ groups: it would have nothing to attack. It must instead arrive PERPENDICULAR to the plane, from above or from below, where the empty lobe of the antibonding $\pi^{*}$ orbital lies. Experiment even shows that the optimal path is not exactly at $90$ degrees but tilted slightly toward the oxygen, at about $107$ degrees from the $C=O$ bond: this is the Bürgi-Dunitz angle, which minimizes repulsion with the oxygen lone pairs. The important structural consequence is that as the attack proceeds, the carbon goes from flat $sp^{2}$ to tetrahedral $sp^{3}$, and that if the two faces are different, two stereochemically distinct products are obtained.

Oxygen has an electronegativity of $3.44$ against $2.55$ for carbon: it pulls the electrons of both bonds toward itself, and above all those of the more mobile $\pi$ bond. The carbon therefore carries a partial POSITIVE charge and the oxygen a partial NEGATIVE charge, which the high dipole moment of propanone, $2.88$ D, confirms, to be compared with the $1.15$ D of the ether and the $0.08$ D of propane. One can even write an ionic resonance form $C^{+}-O^{-}$, a minor contributor but a real one, that sums up the situation. The bond is SHORTER because it is double: two electron pairs between the same nuclei draw them closer than one pair does, hence $122$ pm against $143$ pm. There is no contradiction with its reactivity, because what the nucleophile breaks is not the whole bond but only its $\pi$ component, which is lateral and therefore poorly overlapped, leaving the $\sigma$ bond intact. And above all, the $\pi$ electrons do not go off into the void: they take refuge on the oxygen, a very electronegative atom that readily accepts a negative charge. That is the whole difference with the alkene, whose $\pi$ bond cleavage would create a positive charge on a carbon.

Butane has only London forces, which are very weak: it boils below zero. Propanone has a large permanent dipole, hence dipole-dipole interactions much stronger than mere London forces: it boils at $56$ degrees. Propan-2-ol also has a hydrogen on its oxygen, so it can form HYDROGEN BONDS with its neighbours, an even stronger interaction: $82$ degrees. The scale of the three values is therefore exactly the scale of the three types of intermolecular forces. As for the miscibility of propanone in water, there is nothing contradictory about it: a hydrogen bond needs a donor AND an acceptor. Propanone has no donor, so it cannot associate with itself; but its oxygen, rich in lone pairs, is an excellent ACCEPTOR, and water supplies the donor it lacks. It is the same reasoning as for the ether in the previous chapter, and the conclusion to remember is that a molecule can be very soluble in water without boiling high.

The order of increasing reactivity is: propanone < benzaldehyde < ethanal < methanal. Two reasons add up. The STERIC reason: the carbon must go from a flat, open environment to a much more crowded tetrahedral one, so the more bulky groups it carries, the more this transition state costs; methanal, with two tiny hydrogens, is the most accessible. The ELECTRONIC reason: an alkyl group is donating by the inductive effect, so each alkyl group lowers the partial positive charge on the carbon and makes it less attractive to a nucleophile; propanone has two, ethanal only one, methanal none. Benzaldehyde has only one substituent, like ethanal: the steric reason alone would place it next to ethanal. It is the second reason, pushed further, that makes it fall back. The aromatic ring is CONJUGATED with the carbonyl, so the electron density of the ring delocalizes the positive charge of the carbon by resonance, which is far more effective than a simple inductive effect: benzaldehyde is clearly less reactive than ethanal. It still stays ahead of propanone, which combines two donating groups and a double crowding; the hydration constants confirm it, close to $10^{-2}$ for benzaldehyde against $2\times 10^{-3}$ for propanone.

The three chlorine atoms are strongly ELECTRON-WITHDRAWING by the inductive effect, and they sit on the carbon directly next to the carbonyl. They therefore pull electron density away and INTENSIFY the positive charge of the carbonyl carbon, which becomes markedly more electrophilic than that of ethanal. This is the exact opposite of the effect of the alkyl groups in the previous question. The consequence for the hydrate is spectacular. Hydration is an equilibrium between the carbonyl and the geminal diol; by making the carbonyl very unstable and very reactive, we shift this equilibrium massively toward the hydrate. Whereas propanone is hydrated only to $0.2$ percent, chloral is almost completely hydrated, and its hydrate is a crystalline SOLID, stable, isolable and weighable, which is quite exceptional for a geminal diol. Chloral hydrate was in fact the first synthetic hypnotic, in use from $1869$. The same reasoning explains why ninhydrin, whose central carbonyl is flanked by two other electron-withdrawing carbonyls, is also stable in its hydrated form.

First step: the nucleophile attacks the carbonyl carbon perpendicular to the plane. The electrons of the $\pi$ bond shift onto the oxygen and the carbon becomes $sp^{3}$: the result is a tetrahedral ALKOXIDE, an intermediate carrying a negative charge on the oxygen. Second step: this alkoxide is protonated, by the water of the medium or by the final addition of dilute acid, which delivers the alcohol. It is the FIRST step that is rate-determining: it breaks a $\pi$ bond, creates a charge and turns a flat carbon into a crowded tetrahedral carbon, whereas the second is only a proton transfer, always very fast. This is why all the factors discussed in exercise $1$, the crowding and the charge of the carbon, do govern the overall rate.

In acidic medium, the order of the steps is REVERSED. The acid first protonates the OXYGEN, which has lone pairs, giving a cation for which one can write a resonance form with the positive charge on the CARBON. This carbon is then far more electrophilic than before, to the point that a weak, neutral nucleophile such as water or an alcohol is enough to attack it. The final loss of a proton gives back the catalyst. What the protonation brings can therefore be summed up in one sentence: it ACTIVATES the electrophile instead of activating the nucleophile, which is the only option when the nucleophile cannot be deprotonated. But there is a limit, and it is an acid-base one: if the medium becomes too acidic, the nucleophile itself, which almost always has a lone pair, is protonated in its turn and stops being nucleophilic. There must be enough acid to activate the carbonyl and not enough to kill the nucleophile, a constraint that returns in spectacular fashion in exercise $4$ with the bell-shaped curve for imines.

The percentage hydrated is $\frac{K}{1+K}$. For methanal, $\frac{2200}{2201}=99.95$ percent. For ethanal, $\frac{1.4}{2.4}=58.3$ percent. For propanone, $\frac{0.0020}{1.0020}=0.20$ percent. The gap between the extremes is a factor of $1.1\times 10^{6}$ on the constants, and it is entirely explained by the two reasons of exercise $1$: the two methyl groups of propanone crowd the tetrahedral carbon to be formed and at the same time lower its positive charge, whereas methanal has neither handicap. One can also read the equilibrium from the other end: the more stable the carbonyl, the less it hydrates. Finally, the hydrate is almost never isolable because the reaction is an EQUILIBRIUM in aqueous solution: as soon as the water is evaporated to recover it, Le Chatelier's principle shifts the equilibrium toward the carbonyl and the hydrate decomposes between your fingers. The only exceptions are those where the equilibrium is crushed on the hydrate side, like the chloral of exercise $1$.

The product is 2-hydroxypropanenitrile, $CH_{3}CH(OH)CN$. $HCN$ alone is not used because $HCN$ is a very weak acid, with a $pK_{a}$ close to $9.2$: it is therefore essentially in molecular form, and the $HCN$ molecule is a poor nucleophile. It is the cyanide ION $CN^{-}$ that is the real nucleophile, and a trace of $KCN$ is enough to supply enough of it to start and sustain the reaction, each cycle regenerating the cyanide ion during the final protonation by $HCN$. It is therefore a base CATALYSIS. The nitrile group is then a very useful starting point: hydrolyzed in aqueous acid, it gives a carboxylic acid, which turns the cyanohydrin into an $\alpha$-hydroxy acid, lactic acid in the present case; reduced by $LiAlH_{4}$, it gives a primary amine, hence a $\beta$-amino alcohol. The cyanohydrin is thus one of the few simple ways to LENGTHEN a carbon chain by one atom, like the Grignard reagent on methanal.

In the cyanohydrin of butan-2-one, $CH_{3}CH_{2}C(OH)(CN)CH_{3}$, the carbon bears four different groups, a methyl, an ethyl, $-OH$ and $-CN$: it is asymmetric. The carbonyl carbon of the starting material is $sp^{2}$ and FLAT, as the figure of exercise $1$ shows, so both of its faces are accessible to the nucleophile. The ketone is achiral and nothing in the medium distinguishes one face from the other: the cyanide ion arrives just as often from above as from below. The two paths lead to the two enantiomers, formed in rigorously equal amounts: the product is a RACEMIC mixture, optically inactive. Propanone, for its part, gives a cyanohydrin whose carbon bears two identical methyl groups: it is not asymmetric, and the product is not chiral at all. This is a general result: from achiral reagents in an achiral medium, one never obtains an enantiomeric excess. To steer the attack, a source of chirality is needed, a chiral catalyst or an enzyme, and that is exactly what the hydroxynitrile lyase of cassava does, producing only one of the two cyanohydrins.

The equation is $CH_{3}CH_{2}CHO+2\,CH_{3}OH\rightleftharpoons CH_{3}CH_{2}CH(OCH_{3})_{2}+H_{2}O$, with catalysis by an anhydrous strong acid. The product is 1,1-dimethoxypropane. Check. Propanal $C_{3}H_{6}O$ weighs $58.08$ g/mol and methanol $32.04$, that is $58.08+2\times 32.04=122.16$ g/mol on the reactant side. The acetal $C_{5}H_{12}O_{2}$ weighs $104.15$ and water $18.02$, that is $122.16$ g/mol on the product side. The two totals coincide exactly, which confirms that the stoichiometry is indeed TWO alcohols for one carbonyl and not one, a very frequent exam error.

The acid first protonates the carbonyl oxygen, which activates the carbon as in exercise $2$. Methanol attacks it and, after loss of a proton, a HEMIACETAL is obtained, a molecule that carries on the SAME carbon an $-OH$ group and an $-OCH_{3}$ group. This is the intermediate to name. To go further, the $-OH$ group must be driven out and replaced by a second $-OCH_{3}$. Now $HO^{-}$ is a poor leaving group, as the previous chapter established: it must therefore be PROTONATED to turn it into water, and this is possible only in acidic medium. Once the water has left, a cation remains, stabilized by the lone pair of the methoxy oxygen, called an oxocarbenium ion, which the second methanol attacks before a last proton is lost. This is why the acidic medium is not a mere accelerator but an absolute NECESSITY: in basic medium the reaction stops irretrievably at the hemiacetal.

To shift toward the acetal, one can use a large excess of alcohol, which then also serves as solvent, or remove the water as it forms, with a Dean-Stark apparatus or a drying agent such as anhydrous magnesium sulfate. Both means rely on Le Chatelier's principle, one by adding a reactant, the other by removing a product. To go back, it is enough to treat the acetal with water IN EXCESS and a trace of acid: the same equilibrium, pushed the other way by the mass of water, gives back the carbonyl and the alcohol quantitatively. It is this complete reversibility, controlled by a single easily manipulated parameter, that makes the acetal an ideal protecting group.

The difficulty is that $LiAlH_{4}$, the only reducing agent able to convert an ester to a primary alcohol, would also reduce the ketone, and even faster since the ketone is clearly more electrophilic than an ester. The strategy is therefore to MASK the ketone. Step $1$: treat with an excess of ethane-1,2-diol and a trace of para-toluenesulfonic acid, removing the water; this gives the cyclic acetal of the ketone, a dioxolane, while the ester, which is not electrophilic in this way, is untouched. The diol is preferred to two molecules of methanol because closing the five-membered ring is entropically favourable and makes the protection more efficient. Step $2$: $LiAlH_{4}$ in anhydrous ether, then hydrolysis; the ester is reduced to a primary alcohol, and the acetal, completely inert under these conditions, protects the ketone. Step $3$: mild acid hydrolysis, for instance dilute $H_{3}O^{+}$ at room temperature; the acetal opens again and the ketone is restored intact. The result is $CH_{3}CO-CH_{2}CH_{2}-CH_{2}OH$. This protect, transform, deprotect sequence is one of the most frequently tested lines of reasoning in the course.

An acetal no longer contains any carbonyl: its two oxygens are engaged in single $C-O$ bonds, exactly as in an ether. There is therefore no electrophilic carbon left to attack, no $\pi$ bond to open, and no acidic hydrogen. A nucleophile, a base or a Grignard reagent literally has nothing to do on this molecule: it is as inert as an ether, which is exactly the property sought. The difference with the HEMIACETAL is essential and often misunderstood. The hemiacetal keeps an $-OH$ group: in basic medium, this hydrogen is removed, the alkoxide thus formed expels the alkoxy group and the carbonyl is regenerated. The hemiacetal is therefore in permanent equilibrium with the starting carbonyl, in acidic medium AS WELL AS in basic medium, and it protects nothing. Only the complete acetal, which has no available hydrogen left, is stable in basic medium. This distinction is not academic: exercise $10$ will show that sugars are cyclic hemiacetals, which explains in one sentence why glucose reduces the Tollens reagent whereas sucrose, which is an acetal, does not.

With methanamine, a PRIMARY amine, the imine $\left(CH_{3}\right)_{2}C=N-CH_{3}$ is obtained, also called a Schiff base. With $N$-methylmethanamine, a SECONDARY amine, the enamine $CH_{2}=C(CH_{3})-N(CH_{3})_{2}$ is obtained. The difference comes from the number of hydrogens carried by the nitrogen. The mechanism is identical to the end: addition of the nitrogen, then loss of water, which leaves in both cases an iminium ion $C=N^{+}$. To stabilize itself, this ion must lose a proton. With a primary amine, the nitrogen still carries one: it loses it and the $C=N$ double bond remains, hence the imine. With a secondary amine, the nitrogen carries none; the only available proton is then on the ALPHA carbon, and its loss shifts the double bond toward the carbon, hence the enamine with its $C=C$ bond. The nature of the product is therefore entirely decided by the class of the amine, and not by the conditions.

The nitrogen, rich in its lone pair, attacks the carbonyl carbon. After proton transfer, a molecule is obtained that bears an $-OH$ and an $-NHR$ on the same carbon: this is the CARBINOLAMINE, sometimes called a hemiaminal, strictly analogous to the hemiacetal of exercise $3$. The water must then be eliminated, and this is where acid catalysis comes in: it PROTONATES the $-OH$ group to make it a good leaving group, exactly as in acetal formation. The departure of water produces the iminium ion, stabilized by the lone pair of the nitrogen, and the final loss of a proton gives the imine. The step speeded up by the acid is therefore the DEHYDRATION of the carbinolamine, the second phase of the reaction, and not the initial attack of the amine, which does not need it since the amine is already a good nucleophile.

Right-hand branch, at too high a pH: there is not enough acid to protonate the $-OH$ group of the carbinolamine, so dehydration does not occur and the reaction stays stuck at the intermediate. The rate decreases as the pH rises. Left-hand branch, at too low a pH: the amine is a base, with a conjugate $pK_{a}$ close to $10.6$, so in a clearly acidic medium it is entirely protonated to the ammonium ion. Its lone pair is then engaged in a bond with a proton, it is no longer nucleophilic at all, and nothing is left to attack the carbonyl. The rate collapses as the pH falls. The maximum therefore lies at the compromise, near $pH=4.5$: acidic enough for dehydration to proceed, little enough acidic for a sufficient fraction of the amine to remain in free form. This is the clearest illustration of the principle stated in exercise $2$: in acid catalysis, the acid activates the electrophile but always threatens the nucleophile. Note that with an aniline, a much weaker base since its anilinium ion has a $pK_{a}$ of $4.6$, the optimum shifts to a lower pH, the parasitic protonation being much less to be feared.

The problem is that the imine must be reduced in the presence of the starting carbonyl, with which it is in equilibrium. Now $NaBH_{4}$ reduces ketones very well: it would destroy the reagent instead of reducing the imine. Sodium cyanoborohydride $NaBH_{3}CN$ solves exactly this difficulty. The strongly electron-withdrawing cyano group depletes the boron and makes the hydrides much less reactive: at slightly acidic pH, this reagent is too mild to attack a neutral ketone, but it readily reduces the IMINIUM ion, which carries a full positive charge and is therefore far more electrophilic. A selective reduction is thus obtained, and the reagent is in addition stable in acidic aqueous medium, which $NaBH_{4}$ is not. From cyclohexanone and methanamine, the product is $N$-methylcyclohexanamine. Reductive amination is the method of choice for making a clean secondary amine, because it avoids the polyalkylations that a direct substitution on a halide would give.

First role, VISION. Retinal, an aldehyde derived from vitamin A, is bound to a lysine residue of opsin by an imine bond, forming rhodopsin. A photon isomerizes the 11-cis double bond to trans, which deforms the protein and triggers the nerve impulse; hydrolysis of the imine then allows the retinal to be released, reisomerized and reattached. Without reversibility, the eye would work only once. Second role, the TRANSAMINASES. Pyridoxal phosphate, a derivative of vitamin $B_{6}$, forms an imine with the amino acid substrate, which allows the amino group to be transferred from one carbon skeleton to another, a central reaction of nitrogen metabolism. Here again reversibility is the key point: the bond must form to hold the substrate, then break to release the product and regenerate the cofactor. One can also cite the glycation of hemoglobin, the subject of exercise $10$, where it is on the contrary the SECONDARY irreversibility, after rearrangement of the initial imine, that makes the phenomenon measurable in the clinic.

Solvent: $NaBH_{4}$ tolerates water, methanol and ethanol, in which it decomposes slowly enough to be usable; $LiAlH_{4}$ reacts violently with the slightest acidic proton and requires strictly dry anhydrous ether or tetrahydrofuran. Functions reduced: $NaBH_{4}$ reduces ONLY aldehydes and ketones, which is its value; $LiAlH_{4}$ also reduces esters, carboxylic acids, amides and nitriles, hence everything that carries a carbon linked to a heteroatom by a multiple bond. Reactivity: aluminium being less electronegative than boron, its hydrides are far more available and more nucleophilic. Handling: $NaBH_{4}$ is a stable powder that can be weighed in the open air, $LiAlH_{4}$ ignites on contact with moisture and is handled under an inert atmosphere, with a final hydrolysis carried out dropwise and under cooling. Butan-2-one gives butan-2-ol in both cases, as a racemic mixture for the reason demonstrated in exercise $2$.

Each $BH_{4}^{-}$ ion contains four hydrides, and each ketone consumes only one: $\frac{0.100}{4}=0.0250$ mol of $NaBH_{4}$ is therefore needed, that is $0.0250\times 37.83=0.946$ g. In practice clearly more is used, often double or more, for two reasons. First, the reagent is slowly DESTROYED by the protic solvent itself, methanol or water, which hydrolyzes it with evolution of dihydrogen: part of the charge never reaches the substrate. Second, the four hydrides are not equivalent: as the boron takes on alkoxy groups from the previous reductions, it becomes more crowded and its remaining hydrides are less reactive, so that the fourth is often unusable in practice. A chemist therefore reasons on the THEORETICAL hydrides for the exam calculation, and takes a margin for the real handling.

The product is $CH_{3}CH(OH)-CH_{2}-COOCH_{2}CH_{3}$: only the ketone is reduced, the ester stays intact. Two reasons explain this selectivity. The first comes from the reagent: $NaBH_{4}$ is too mild for an ester, as question a) established. The second, deeper, comes from the substrate: in an ester, the oxygen of the $-OR$ group donates its lone pair by resonance to the carbonyl carbon, which clearly lowers its positive charge and makes it far less electrophilic than a ketone carbon, which has no neighbouring resonance donor. It is the same hierarchy as that of the acid derivatives. The point of method to remember is therefore that a selectivity is always explained from both sides, by the reagent chosen AND by the intrinsic reactivity of the functional groups present, and that a good exam answer mentions both.

With benzaldehyde, the phenyl Grignard reagent gives, after hydrolysis, diphenylmethanol, a secondary alcohol. With propanone, it gives 2-phenylpropan-2-ol, a tertiary alcohol. The difference with reduction is structural: the nucleophile supplied by $NaBH_{4}$ is a HYDRIDE ION, hence a simple charged hydrogen atom, which merely completes the existing molecule; that of a Grignard reagent is a CARBANION, that is a whole carbon fragment, which creates a new carbon-carbon bond. This is what makes the Grignard reagent a tool for BUILDING a skeleton and the hydride a mere tool for changing a functional group. The counting rule seen in the chapter on alcohols still holds: methanal gives a primary alcohol, an aldehyde a secondary one, a ketone a tertiary one.

The two methods are the CLEMMENSEN reduction, with zinc-mercury amalgam in concentrated $HCl$, and the WOLFF-KISHNER reaction, with hydrazine then hot potassium hydroxide in a high-boiling solvent. Both convert $C=O$ to $CH_{2}$ and give propylbenzene here. They are complementary because their media are opposite. If the molecule carries an ACETAL, Wolff-Kishner must be chosen: the Clemmensen medium is strongly acidic and aqueous, it would immediately hydrolyze the acetal and deprotect the carbonyl that was to be kept, as exercise $3$ showed. If the molecule carries an ESTER, Clemmensen must on the contrary be chosen: the hot concentrated potash of Wolff-Kishner would saponify the ester irreversibly. The reasoning to remember goes beyond this particular case and can be put this way: when two methods give the same product, they are not told apart on yield but on COMPATIBILITY with the rest of the molecule.

Tautomerism is a CHEMICAL EQUILIBRIUM between two distinct molecules, the keto form $CH_{3}-CO-CH_{3}$ and the enol form $CH_{3}-C(OH)=CH_{2}$, which differ by the position of an ATOM, here a hydrogen, and correlatively by that of a double bond. The difference with resonance is total and must be stated plainly: two resonance forms are not two molecules, they are two imperfect drawings of a SINGLE molecule whose reality is intermediate, and a double-headed arrow is written between them. Two tautomers are on the contrary two real compounds, isolable in principle, separated by an activation barrier, present simultaneously in proportions fixed by an equilibrium constant, and two opposite half-arrows are written between them. Confusing the two is the heaviest error of the chapter, because it leads to writing impossible mechanisms.

It is never the acid that must be looked at but its CONJUGATE BASE. Removing a hydrogen from an alkane leaves a carbanion whose negative charge stays entirely localized on a carbon, an atom of low electronegativity with no way of spreading the charge: the anion is very unstable, so the acid is very weak. Removing an alpha hydrogen from a carbonyl gives on the contrary an ENOLATE, whose charge is delocalized by resonance on the carbon AND on the OXYGEN of the neighbouring carbonyl. The resonance form that puts the charge on the oxygen is by far the more stable, since oxygen is very electronegative, and it does all the work. The gap is $50-19.3=30.7$ $pK_{a}$ units, that is more than THIRTY orders of magnitude on the acidity constant: a factor of $10^{30}$, one of the largest resonance effects met in the course. It is also understood that this effect requires a precise geometry: the $C-H$ bond being broken must be parallel to the $p$ orbital of the carbonyl for overlap to be possible.

In pentane-2,4-dione, the hydrogens of the central carbon are alpha to TWO carbonyls at once. The anion formed therefore delocalizes its charge over three atoms instead of two, with this time TWO resonance forms carrying the charge on an oxygen. The stabilization is much stronger, and the resulting anion is a symmetric conjugated system of particularly low energy. The result is spectacular: with $pK_{a}=8.9$, this compound is more acidic than the ammonium ion, more acidic than phenol, and seven orders of magnitude more acidic than ethanol. An ordinary base such as sodium hydroxide therefore deprotonates it quantitatively, which makes 1,3-dicarbonyl compounds extremely convenient in synthesis: no exotic base is needed to make the enolate. The ratio of the acidity constants with propanone is $10^{19.3-8.9}=10^{10.4}$, about $2.5\times 10^{10}$, that is twenty-five billion.

The dashed line is an INTRAMOLECULAR HYDROGEN BOND, between the hydrogen of the enol and the oxygen of the carbonyl that stayed intact. It closes a pseudo-ring of six atoms, an ideal geometry that allows an almost perfect alignment, and it adds to the conjugation between the $C=C$ double bond of the enol and the neighbouring carbonyl. These two effects stabilize the enol form so much that it becomes the MAJOR form, which is exceptional. Going to water destroys part of this advantage, for a reason that must be stated carefully: water is a protic solvent, able to form its own hydrogen bonds with the two oxygens of the KETO form, which it therefore solvates very well. The enol form has already used up its donor in its internal bond and has nothing more to offer the solvent. Water thus preferentially stabilizes the keto form and the enol proportion falls from $76$ to $15$ percent. Hexane, nonpolar and aprotic, stabilizes neither form by hydrogen bonding and so lets the intramolecular advantage express itself fully. It is a fine example, to remember, of an equilibrium shifted by the choice of SOLVENT alone.

Racemization. The asymmetric carbon of $(R)$-3-phenylbutan-2-one is precisely an alpha carbon of the carbonyl, and it carries a hydrogen. The base removes it and forms the enolate, in which this carbon becomes $sp^{2}$ and therefore FLAT: all the stereochemical information is lost at that instant. Reprotonation can then occur indifferently from above or from below, with the same probability, and regenerates the two enantiomers in equal amounts. The mixture becomes racemic and the optical rotation vanishes. Note that the molecule has undergone no net transformation: it is chemically identical to itself, only its stereochemistry has been erased. Isotopic exchange. The same mechanism applies, but in $D_{2}O$ the proton that comes back is a DEUTERIUM. Each enolization-reprotonation cycle therefore replaces an alpha hydrogen by a deuterium, and as the process repeats indefinitely as long as the base is there, it continues until ALL the exchangeable hydrogens have been exchanged. Propanone has six, three on each methyl, and these are exactly the six observed. The selectivity of the experiment is what makes it convincing: hydrogens not located alpha are never exchanged, which shows that the mechanism does go through the enol and not through a general agitation of the molecule.

Step $1$: the hydroxide ion removes an alpha hydrogen from ethanal and forms the ENOLATE $^{-}CH_{2}CHO$, stabilized by resonance as in exercise $6$. This step is an equilibrium and very unfavourable, since the $pK_{a}$ of ethanal is $16.7$ against $15.7$ for water: only a tiny fraction of the aldehyde is deprotonated at any instant, but that is enough. Step $2$: this enolate, a carbon nucleophile, attacks the carbonyl carbon of a SECOND molecule of ethanal, which has stayed intact. This is the step that creates the carbon-carbon bond, and it is the heart of the reaction. Step $3$: the alkoxide formed removes a proton from the water of the medium, which delivers the product and regenerates the hydroxide ion, confirming that this is indeed a CATALYSIS. The product is 3-hydroxybutanal, the classic aldol, that is a compound carrying both the ALDehyde group and the alcOHol group, hence the name of the reaction.

On heating, but-2-enal is obtained, according to $2\,CH_{3}CHO\rightarrow CH_{3}CH=CHCHO+H_{2}O$. This dehydration is much easier than that of an ordinary alcohol for two distinct reasons, both of which must be given. First the MECHANISM: the carbon between the alcohol and the aldehyde carries a hydrogen alpha to the carbonyl, hence acidic; the base removes it, and the enolate thus formed directly expels the hydroxide ion. The $-OH$ therefore never needs to be protonated and no carbocation has to be formed, which explains why the reaction takes place in BASIC medium and around $80$ degrees, where an alcohol would require concentrated sulfuric acid at $180$ degrees. This is a mechanism called $E1cb$, for unimolecular elimination from the conjugate base. Then the THERMODYNAMICS: the product is an $\alpha,\beta$-unsaturated aldehyde, in which the $C=C$ double bond is CONJUGATED with the carbonyl. This conjugation strongly stabilizes the product and pulls the equilibrium. Without it, the dehydration would have no reason to be so favourable.

FOUR products are obtained. Each of the two aldehydes has alpha hydrogens, so each can supply the enolate; and each has a carbonyl, so each can play the electrophile. The four enolate-electrophile combinations are possible and give four different aldols in similar proportions, a mixture of no preparative interest. First condition to make the reaction useful: one of the two partners must NOT HAVE any alpha hydrogen, so that it cannot give an enolate and is doomed to the role of electrophile. This is the case of benzaldehyde or methanal, and the condensation of benzaldehyde with propanone cleanly gives benzylideneacetone. Second condition: add the enolizable partner SLOWLY to an excess of the other, kept in the presence of base. The concentration of the enolizable partner then stays so low that it statistically never meets another molecule of itself, and can react only with the electrophile in excess. This is called slow addition or kinetic dilution conditions. A third route exists in practice, the prior and quantitative formation of a specific enolate by a very strong, bulky base, but it goes beyond the syllabus.

The product is 3-methylcyclopent-2-en-1-one, obtained by intramolecular aldol reaction followed by dehydration (condensation). The enolate formed on a terminal methyl attacks the carbonyl at the other end of the molecule, which closes a ring of FIVE carbon atoms, and then conjugated dehydration follows as in b). The preference for the intramolecular path is explained by two arguments. Entropy argument: the intramolecular reaction does not bring two molecules together into one, so it loses no translational degrees of freedom; the entropy change is much less unfavourable than for the intermolecular version. Effective concentration argument: the two groups are attached by the same chain, so they meet permanently, which is equivalent to an enormous local concentration, out of all proportion with the real concentration in solution. These two arguments hold, however, only if the ring size is reasonable: five- and six-membered rings are strongly favoured, three- and four-membered rings are too strained, and rings of seven or more become improbable because the two ends meet too rarely. Here the chain imposes precisely a five-membered ring, hence the excellent yield.

The retro-aldol reaction is the reverse reaction: a base removes the hydrogen of the alcohol of an aldol, and the alkoxide thus formed splits by expelling an enolate, which CLEAVES the molecule into two carbonyl fragments. The bond broken is exactly the one that had been created, between the alpha carbon and the carbon that bore the carbonyl. The major biological example is the fourth step of GLYCOLYSIS, catalyzed by aldolase: fructose-1,6-bisphosphate, with six carbons, is cut there into two trioses, dihydroxyacetone phosphate and glyceraldehyde-3-phosphate. What the enzyme gains here is precisely reversibility. The same aldolase, used in the forward direction, ensures gluconeogenesis, that is the rebuilding of glucose from trioses when the body runs short. A single enzyme thus serves the two opposite metabolic pathways, and it is the relative concentration of the metabolites that decides the direction, exactly as Le Chatelier's principle predicts. We find again here the fact, stressed in exercise $3$, that a well-placed equilibrium is a tool and not a weakness.

2,4-Dinitrophenylhydrazine detects the presence of a CARBONYL group, aldehyde or ketone indifferently, by formation of an orange to red precipitate of hydrazone, which is incidentally a substituted imine of the type seen in exercise $4$. The Tollens reagent detects specifically an ALDEHYDE, by formation of a metallic silver mirror. The Fehling or Benedict reagent also detects aldehydes, by a brick-red precipitate of copper(I) oxide $Cu_{2}O$; it is slightly less sensitive than Tollens and often fails on aromatic aldehydes. The iodoform test detects an entirely different pattern: the presence of a $CH_{3}CO-$ or $CH_{3}CH(OH)-$ group, hence METHYL KETONES and methyl carbinols, as well as ethanal and ethanol. Results. Propanal: DNPH positive, Tollens positive, Fehling positive, iodoform negative, since its methyl is not carried by the carbonyl. Propanone: DNPH positive, Tollens negative, Fehling negative, iodoform POSITIVE, it is a methyl ketone. Propan-2-ol: DNPH negative since there is no carbonyl, Tollens negative, Fehling negative, iodoform POSITIVE since it is a methyl carbinol, which the diiodine of the test first oxidizes to propanone. Benzaldehyde: DNPH positive, Tollens positive, Fehling negative or doubtful, iodoform negative. These four rows show that no test taken alone identifies, but that the four together separate the four compounds.

Oxidizing a carbonyl requires inserting an oxygen between the carbon and one of its neighbours, or, what amounts to the same, replacing a hydrogen of this carbon by an $-OH$. Now an aldehyde carries a HYDROGEN directly on the carbonyl carbon: this hydrogen can be removed, and the aldehyde becomes a carboxylic acid without any carbon-carbon bond having been broken. A ketone has on this carbon only two other CARBONS: oxidizing it would require breaking a carbon-carbon bond, which demands violent conditions and destroys the molecule, so it does not happen under the mild Tollens conditions. All the selectivity of the test therefore rests on the presence or absence of this single hydrogen. The equation with propanal, in ammoniacal medium, is $CH_{3}CH_{2}CHO+2\,Ag(NH_{3})_{2}^{+}+3\,HO^{-}\rightarrow CH_{3}CH_{2}COO^{-}+2\,Ag+4\,NH_{3}+2\,H_{2}O$. The electron balance is easily checked: the aldehyde gives up two electrons, and each of the two silver ions captures one.

Phase $1$, the halogenations. In basic medium, the base removes a hydrogen from the methyl carried by the carbonyl, forming the enolate, which attacks the diiodine: a hydrogen is replaced by an iodine. The operation is repeated three times. These three halogenations are faster and faster, and the reason is electronic: each iodine already installed is strongly electron-withdrawing, so it stabilizes the next enolate by spreading its negative charge, which lowers the $pK_{a}$ of the remaining hydrogens and speeds up their removal. The practical consequence is that one can never stop at a mono-iodination by this route: the reaction always goes to $CI_{3}$. Phase $2$, the cleavage. The hydroxide ion adds to the carbonyl and the tetrahedral intermediate expels the carbanion $^{-}CI_{3}$. This expulsion, unthinkable with a $^{-}CH_{3}$, becomes possible because the three electron-withdrawing iodines stabilize the negative charge: $CHI_{3}$ has a $pK_{a}$ close to $14$, so $^{-}CI_{3}$ is a base weak enough to act as a leaving group. Phase $3$, the final proton transfer between the carboxylic acid formed and the carbanion gives the carboxylate and triiodomethane. The latter is a YELLOW solid with a melting point of $119$ degrees Celsius, with the characteristic hospital smell, and it is its precipitation that is observed.

The reaction is the Cannizzaro DISPROPORTIONATION: $2\,C_{6}H_{5}CHO+KOH\rightarrow C_{6}H_{5}CH_{2}OH+C_{6}H_{5}COOK$. One molecule is reduced to benzyl alcohol, the other is oxidized to benzoate: the same compound plays both roles, hence the name disproportionation. The mechanism is revealing: the hydroxide ion adds to the carbonyl, the alkoxide formed then transfers its HYDRIDE, that is its hydrogen with the electron pair, directly to the carbonyl carbon of a second molecule. The restriction to aldehydes with no alpha hydrogen is not due to a mechanistic impossibility but to a COMPETITION: if the aldehyde has an alpha hydrogen, the base removes it and the aldol reaction of exercise $7$, much faster, takes over all the substrate before the slow hydride transfer has time to occur. Cannizzaro is therefore observable only when the aldol reaction is structurally impossible, which is the case for benzaldehyde, methanal and 2,2-dimethylpropanal. The general lesson is worth remembering: two possible mechanisms almost never coexist, the faster one sweeps up the substrate.

The amount of benzaldehyde is $\frac{10.6}{106.12}=9.99\times 10^{-2}$ mol. With a stoichiometry of two aldehydes for one product of each kind, $4.99\times 10^{-2}$ mol of each is obtained, that is $4.99\times 10^{-2}\times 108.14\approx 5.40$ g of benzyl alcohol and $4.99\times 10^{-2}\times 160.21\approx 8.00$ g of potassium benzoate. The maximum molar yield of alcohol is therefore only $50$ percent, which is the structural weakness of the method: half of the substrate is sacrificed to reduce the other half. Adding excess methanal solves exactly this problem. A CROSSED Cannizzaro is then carried out, in which methanal systematically plays the role of hydride donor and is converted to methanoate, while all of the benzaldehyde is reduced to benzyl alcohol. The yield thus goes from $50$ to nearly $100$ percent. Methanal always wins as a donor for two cumulative reasons, exactly those of exercise $1$: it is the most electrophilic carbonyl, so the hydroxide ion adds to it fastest and this is where the donor alkoxide forms; and its hydride, carried by a carbon with no substituent at all, is the most accessible. This process is used industrially to make pentaerythritol, from ethanal and a large excess of methanal.

The degree of unsaturation is $\frac{2\times 5+2-10}{2}=\frac{12-10}{2}=1$, the oxygen being ignored. One unsaturation allows either a double bond, which can be a $C=O$ or a $C=C$, or a RING. All three possibilities must be kept open until the tests have decided, because the most common error is to assume a carbonyl straight away because the formula contains an oxygen.

The precipitate with 2,4-DNPH proves a carbonyl, and the silver mirror proves that it is an ALDEHYDE. The negative iodoform is no surprise: an aldehyde cannot carry the $CH_{3}CO-$ pattern. The decisive piece of information is the last: with concentrated $KOH$ in the hot, X gives TWO different products, which is the signature of the Cannizzaro disproportionation of exercise $8$. Now this reaction takes place only if the aldehyde has NO alpha hydrogen, otherwise the aldol reaction would win. The carbon next to the carbonyl must therefore carry only carbons. Among the aldehydes of formula $C_{5}H_{10}O$, only one meets this condition: 2,2-dimethylpropanal, $\left(CH_{3}\right)_{3}C-CHO$, whose alpha carbon is quaternary. X is therefore 2,2-dimethylpropanal, and the two products observed are 2,2-dimethylpropan-1-ol and potassium 2,2-dimethylpropanoate.

For Y, the positive DNPH and the negative Tollens require a KETONE, and the positive iodoform further requires the $CH_{3}CO-$ pattern, hence a methyl ketone. Two candidates exist with formula $C_{5}H_{10}O$: pentan-2-one and 3-methylbutan-2-one. The iodoform test cuts the molecule between the methyl and the rest, and releases the acid corresponding to the remaining fragment: this would be butanoic acid, unbranched, in the first case, and 2-methylpropanoic acid, branched, in the second. The statement specifies that the chain obtained is not branched: Y is therefore PENTAN-2-ONE. For Z, the absence of reaction with DNPH rules out any carbonyl, and the effervescence with sodium proves an $O-H$, hence an alcohol. But the degree of unsaturation is $1$ and it has to be placed somewhere: since it is not a $C=O$, it is either a $C=C$ or a ring. This is where bromine water comes in: an alkene decolorizes it instantly by addition, and Z does not decolorize it. There is therefore no carbon-carbon double bond, and the unsaturation is necessarily a RING. Z is CYCLOPENTANOL. The reasoning deserves to be underlined: it is a NEGATIVE test that allowed a positive conclusion, by eliminating the last alternative.

We must go from $C_{4}$ to $C_{5}$ and create a methyl ketone, hence add a methyl on the carbonyl carbon. Step $1$: methylmagnesium bromide in anhydrous ether, then acid hydrolysis; the methyl carbanion adds to butanal and pentan-2-ol is obtained, a secondary alcohol with five carbons. Step $2$: oxidation by PCC in dichloromethane, or by dichromate in acidic medium, which converts the secondary alcohol to a ketone. Pentan-2-one is obtained. Note that the choice of oxidant is immaterial here, since a ketone cannot be overoxidized, unlike the case of a primary alcohol where PCC would be compulsory.

Step $1$: gentle oxidation of ethanol by PCC, which gives ethanal. PCC is ESSENTIAL here, because an oxidant in aqueous medium would lead to ethanoic acid and the rest would be impossible. Step $2$: dilute sodium hydroxide, then heating, which carries out the aldol reaction then the dehydration of exercise $7$ and gives but-2-enal. The overall balance is $2\,CH_{3}CH_{2}OH\rightarrow CH_{3}CH=CHCHO+2\,H_{2}+H_{2}O$ in terms of atom counting, two molecules of ethanol being needed for one molecule of product. Theoretical mass yield: two moles of ethanol weigh $2\times 46.07=92.14$ g and give at best one mole of but-2-enal, that is $70.09$ g. The maximum mass yield is therefore $\frac{70.09}{92.14}=76.1$ percent. This figure is a stoichiometric ceiling, independent of any technical consideration, exactly like the $51$ percent of fermentation seen in the previous chapter: the loss corresponds to the hydrogen atoms removed in the oxidation and to the water molecule lost in the dehydration.

The functional group created is a HEMIACETAL, and more precisely a CYCLIC hemiacetal since the alcohol and the aldehyde belong to the same molecule. It is exactly the reaction of exercise $3$, in its intramolecular version: the lone pair of the oxygen on carbon $5$ attacks the carbonyl carbon, exactly as an external methanol would have done. The six-membered ring is preferred for the same reason as the five-membered ring of the intramolecular aldol reaction of exercise $7$: five- and six-membered rings are the only ones free of notable strain, with angles close to $109.5$ degrees, and they close quickly because the two ends meet often. The six-membered ring, called pyranose, wins here over the five-membered ring, called furanose, for reasons of conformational stability: the chair form allows almost all the hydroxyl groups to be placed in the equatorial position. Note the remarkable consequence: a molecule that biology describes as an aldehyde is practically never one.

Let $x$ be the fraction of the $\alpha$ form. Since the optical rotation of a mixture is additive, we write $112.2\,x+18.7\,(1-x)=52.7$, that is $93.5\,x=34.0$ and $x=0.364$. At equilibrium there is therefore about $36.4$ percent of the $\alpha$ form and $63.6$ percent of the $\beta$ form. The $\beta$ form dominates because it places its anomeric hydroxyl in the EQUATORIAL position, less crowded. Mutarotation is the following phenomenon: if pure $\alpha$-D-glucopyranose is dissolved, the optical rotation measured is at first $+112.2$ degrees, then drifts slowly and settles at $+52.7$. The reason is that the cyclic hemiacetal is in equilibrium with the open form: the ring opens, giving carbon $1$ back its flat aldehyde geometry, then closes again indifferently on one side or the other. Each opening-closing is an opportunity to change anomer, and the system converges to the equilibrium mixture. It is the same reasoning as in exercise $6$ for racemization through the enol: as soon as a FLAT intermediate exists, the stereochemical information of the carbon concerned is erased.

Glucose is a HEMIacetal, so it keeps an $-OH$ group on the anomeric carbon and stays in equilibrium with its open aldehyde form, which reduces Tollens; sucrose is a complete ACETAL, its glycosidic bond engaging the anomeric carbons of its TWO units, so that no open form is accessible and no aldehyde can appear. This is the hemiacetal versus acetal distinction of exercise $3$, transposed as it stands. It follows immediately that maltose and lactose, in which only one anomeric carbon is engaged, are REDUCING sugars, which experiment confirms.

The amount of glucose is $\frac{1.00}{180.16}=5.55\times 10^{-3}$ mol. The Tollens reaction consumes two silver ions per aldehyde group, so $1.11\times 10^{-2}$ mol of silver is deposited, that is $1.11\times 10^{-2}\times 107.87\approx 1.20$ g. One gram of sugar therefore deposits more than one gram of silver, which is the principle of silvering mirrors. The apparent paradox is resolved by Le Chatelier's principle. The open form is not a fixed stock of $0.01$ percent that would be used up: it is an EQUILIBRIUM. As soon as the Tollens reagent consumes an open molecule, the equilibrium is shifted and a cyclic molecule at once opens to restore the proportion. The cyclic form is therefore a reservoir that continuously feeds the reaction, and the transformation is complete; only its RATE is reduced by the low proportion of active form. A clear distinction must thus be made between what limits the YIELD, here nothing, and what limits the RATE, here the position of the cyclization equilibrium. This distinction is one of the most useful in the course and it holds for all minor intermediates, including the enol of exercise $6$.

Step $1$: the amino group of a hemoglobin residue, the N-terminal valine of the beta chain or a lysine, attacks the open form of glucose and forms an IMINE, that is a Schiff base, exactly as in exercise $4$. This bond is reversible, hence labile, and the imine formed can come undone if the blood glucose falls back. Step $2$: this imine slowly undergoes the AMADORI rearrangement, which converts it into a stable KETOAMINE, by a mechanism related to a tautomerism of the type of exercise $6$. This second form is no longer reversible under physiological conditions, and it is the one that is assayed. The kinetics is the important point: step $1$ is fast and reversible, step $2$ is slow and practically irreversible, so that the amount accumulated integrates the exposure to glucose instead of reflecting its instantaneous value.

The reason is purely kinetic and biological. Since glycation is slow and irreversible once the Amadori rearrangement has been completed, a hemoglobin molecule can only become more glycated with time: its glycation level integrates the whole glycemic history since it was made. Now hemoglobin is enclosed in the red blood cell, whose lifetime is about $120$ days, and the population of cells is constantly renewed. At any given moment, the blood therefore contains cells of all ages, from zero to $120$ days, and the mean level measured reflects the mean blood glucose weighted over about two to three months, the most recent weeks weighing a little more. A fasting blood glucose, for its part, informs only on the instant of the sample and can be normalized by fasting the evening before, which explains why $HbA1c$ has become the reference marker. Numerical application: $g=1.59\times 7.0-2.59=11.13-2.59=8.54$ mmol/L. This value is clearly above the usual $5.5$ mmol/L, which is consistent with a patient exceeding the diagnostic threshold of $6.5$ percent. One can appreciate the elegance of the whole: a clinical biology test whose basis is entirely the carbonyl chemistry of this chapter, hemiacetal, open form and imine.

Step 1: triphenylphosphine, a nucleophile, attacks bromomethane by SN2 and gives methyltriphenylphosphonium bromide, $Ph_{3}P^{+}-CH_{3}\ Br^{-}$. Step 2: a very strong base, butyllithium, removes a proton from the methyl, made acidic by the neighbouring positive phosphorus, and gives the ylide $Ph_{3}P^{+}-CH_{2}^{-}$.

The $CH_{2}$ of the ylide takes exactly the place of the oxygen: methylenecyclohexane is obtained, a cyclohexane bearing an exocyclic double bond $C=CH_{2}$. Formula: $C_{7}H_{12}$.

The Grignard reagent gives 1-methylcyclohexan-1-ol. Acid dehydration goes through the tertiary carbocation and follows Zaitsev's rule: the double bond forms INSIDE the ring, where it is trisubstituted. The main product is 1-methylcyclohexene, and not methylenecyclohexane, whose exocyclic double bond is only disubstituted. The Wittig reaction, for its part, places the double bond exactly where the carbonyl was, without going through a carbocation.

The formation of the $P=O$ bond of triphenylphosphine oxide, one of the strongest bonds in chemistry, close to 575 kJ/mol. It is this that drives the cleavage of the oxaphosphetane.

$\frac{96.17}{98.14+276.30}=\frac{96.17}{374.44}=25.7$%. Three quarters of the mass engaged ends up as triphenylphosphine oxide, a heavy waste product that must be separated from the product. The reaction is very reliable about the position of the double bond, but poor from the green chemistry standpoint.

$CH_{2}=CH-C(=O)-CH_{3}\leftrightarrow{}^{+}CH_{2}-CH=C(-O^{-})-CH_{3}$. The positive charge appears on the carbonyl carbon in the usual form, and on the terminal $CH_{2}$ in the conjugated form. Two electrophilic sites: the carbonyl carbon and the carbon in the beta position, two atoms further along.

1,2-Addition, directly on the carbonyl: the alkoxide formed is hydrolyzed to 2-methylbut-3-en-2-ol, $CH_{2}=CH-C(OH)(CH_{3})_{2}$, a tertiary alcohol that keeps its $C=C$ double bond.

The methyl adds to the carbon in the beta position; the electrons shift as far as the oxygen and an enolate is obtained. Hydrolysis gives the enol, which at once tautomerizes to a ketone: $CH_{3}-CH_{2}-CH_{2}-CO-CH_{3}$, pentan-2-one. The carbonyl is intact, and it is the $C=C$ double bond that has disappeared.

The 1,4-addition consumes the $C=C$ $\pi$ bond and keeps the $C=O$ bond. Now a $C=O$ bond is much stronger than a $C=C$ bond: the product that retains the carbonyl is lower in energy. Reversible or soft nucleophiles, cuprates, enolates, amines, therefore end up in 1,4-addition; hard, fast nucleophiles, Grignard reagents and hydrides, attack the carbonyl, which carries the strongest positive charge.

$NaBH_{4}$ attacks the carbonyl: cyclohex-2-en-1-ol, an allylic alcohol that keeps its double bond. $H_{2}$ over palladium hydrogenates the $C=C$ bond much faster than the $C=O$: cyclohexanone.