Corrected exercises: amines (202-SNU-RE)

The four isomers are propan-1-amine $CH_{3}CH_{2}CH_{2}NH_{2}$ and propan-2-amine $\left(CH_{3}\right)_{2}CHNH_{2}$, both PRIMARY; $N$-methylethanamine $CH_{3}CH_{2}NHCH_{3}$, SECONDARY; and $N,N$-dimethylmethanamine $\left(CH_{3}\right)_{3}N$, TERTIARY. The counting is radically different from that of alcohols, and this is the most common pitfall of the chapter. For an alcohol, you look at the CARBON bearing the $-OH$ group and count how many other carbons are bonded to it. For an amine, you look at the NITROGEN and count how many carbons are bonded to it. The consequence is immediate: propan-2-amine is a PRIMARY amine, whereas propan-2-ol is a SECONDARY alcohol, even though the skeleton is the same. You must never carry the reasoning over from one chapter to the other.

Propan-1-amine boils at $48$ degrees, $N$-methylethanamine at $37$ and $N,N$-dimethylmethanamine at $3$. The ranking follows the number of hydrogens on nitrogen, hence the number of hydrogen-bond DONOR sites: two for the primary amine, one for the secondary, ZERO for the tertiary. The case at $3$ degrees is the most instructive: $N,N$-dimethylmethanamine does have an electron-rich nitrogen, hence an excellent ACCEPTOR, but it has no $N-H$ to offer, so two of its molecules cannot associate with each other. It is left with dipole-dipole interactions and London forces, hence a drop of $45$ degrees compared with its primary isomer. This is exactly the reasoning used for ethers versus alcohols, and for ketones versus alcohols: what matters is not the presence of the heteroatom but the presence of a hydrogen BONDED to it.

Nitrogen is clearly less electronegative than oxygen, $3.04$ against $3.44$. The $N-H$ bond is therefore less polarized than the $O-H$ bond, the hydrogen carries a smaller partial positive charge, and the resulting hydrogen bond is noticeably weaker, about $15$ kJ/mol against $25$ for an alcohol. For a comparable molar mass, the amine therefore boils lower, here $48$ against $97$ degrees. Solubility follows a different logic and must not be deduced from the boiling point. Nitrogen remains an excellent hydrogen-bond ACCEPTOR, with its lone pair readily available, and water supplies donors in abundance. All short-chain amines, including tertiary ones, which have no $N-H$ at all, are therefore very soluble in water. This is the principle already met for ethers and ketones: a molecule can be very soluble in water and still boil low, because solubility only requires a complementary partner, whereas boiling requires association with itself.

Apply $k=A\exp\left(-\frac{E_{a}}{RT}\right)$ with $RT=8.314\times 298=2478$ J/mol. This gives $\frac{25000}{2478}=10.09$, so $k=1.0\times 10^{13}\times \mathrm{e}^{-10.09}=4.2\times 10^{8}$ s$^{-1}$. Nitrogen therefore inverts about FOUR HUNDRED MILLION times per second, and the lifetime of a given configuration is $\frac{1}{k}=2.4$ nanoseconds. The comparison with the peptide bond is revealing: with its $80$ kJ/mol, the peptide bond rotated only $0.095$ times per second, a conformation that stays stable for about ten seconds. A factor of only three on the activation barrier produces here a factor of more than $4\times 10^{9}$ on the rate, a reminder of how steep the exponential dependence of the Arrhenius law is. A barrier of $25$ kJ/mol is, in practice, no barrier at all at room temperature.

$N$-ethyl-$N$-methylpropan-1-amine has a nitrogen bonded to three DIFFERENT groups, ethyl, methyl and propyl, plus its lone pair: it formally meets the condition for a stereogenic center, the lone pair acting as the fourth substituent. The two forms are indeed mirror images of each other and non-superimposable. But they interconvert four hundred million times per second, as part d) has just established: no sooner had one been isolated than it would have turned back into a racemic mixture within a few nanoseconds. No separation technique, crystallization or chromatography, works on that time scale. Resolution is therefore impossible, not because the molecule is not chiral, but because its chirality has no LIFETIME. To make resolution possible, inversion must be prevented, and there are two ways to do it. The first is to replace the lone pair by a fourth substituent, that is, to move to a QUATERNARY AMMONIUM salt $R_{4}N^{+}$ with four different groups: there is then no lone pair left to flip, and such salts can indeed be resolved. The second is to lock the nitrogen into a rigid, strained ring system, where passing through the PLANAR transition state would be geometrically impossible: some substituted aziridines are configurationally stable for this reason.

The equilibrium is $RNH_{2}+H_{2}O\rightleftharpoons RNH_{3}^{+}+HO^{-}$, with constant $K_{b}$. We nevertheless prefer to describe the amine by the equilibrium of its conjugate acid, $RNH_{3}^{+}+H_{2}O\rightleftharpoons RNH_{2}+H_{3}O^{+}$, with constant $K_{a}$, for two practical reasons. First, the two constants are linked by $K_{a}K_{b}=K_{w}$, so $pK_{a}+pK_{b}=14.0$ at $25$ degrees Celsius: one follows immediately from the other and no information is lost. Second, the rest of the course, carboxylic acids, phenols, alcohols, enols, leaving groups, is already tabulated in $pK_{a}$; bringing amines onto the same scale lets us compare an amine and a carboxylic acid directly, which we constantly need to predict the direction of an acid-base reaction. Remember the reading: the LARGER the $pK_{a}$ of the conjugate acid, the STRONGER the base.

An alkyl group is electron-donating by the inductive effect. Each additional methyl therefore enriches nitrogen in electron density, which makes its lone pair more available to capture a proton, and also stabilizes the ammonium ion formed by spreading out its positive charge. Both readings lead to the same ranking, regular and monotonic: $NH_{3}$ less basic than $MeNH_{2}$, less than $Me_{2}NH$, less than $Me_{3}N$. This is exactly the top row of the figure, and it is indeed what is measured in the GAS PHASE, where the molecule is alone and the inductive effect is the only one at work. This agreement matters: it shows that the inductive argument is not wrong, merely incomplete as soon as a solvent enters the picture.

The figure shows that $Me_{3}N$, last on the top row, falls to second position on the bottom row: trimethylamine is the one that changes place, and it drops two ranks, from the most basic to second from the least basic. The other three keep their relative order. The second effect at work is the SOLVATION of the ammonium ion by water. An ammonium ion stabilizes itself by offering its $N-H$ hydrogens to hydrogen bonds with the surrounding water molecules: $NH_{4}^{+}$ has four, $MeNH_{3}^{+}$ three, $Me_{2}NH_{2}^{+}$ two, and $Me_{3}NH^{+}$ only one. The more methyl groups, the less solvated the ion formed, hence the less stabilized, hence the less basic the amine. This second effect acts in exactly the OPPOSITE direction to the first. Their competition produces a maximum at the secondary stage, hence the observed ranking $9.25$, $9.80$, $10.64$, $10.73$, in which trimethylamine drops out sharply. The fact that the ranking becomes monotonic again in the gas phase, where there is no solvent, is the experimental PROOF of this explanation: it is an excellent example of how two effects are told apart by removing one of them.

The lone pair on the nitrogen of aniline is not free: it is CONJUGATED with the aromatic ring, into which it delocalizes, producing resonance forms that place the negative charge at the ortho and para positions. Protonating this nitrogen would amount to locking the lone pair into an $N-H$ bond, hence to DESTROYING this resonance stabilization, which is costly. Aniline is therefore a much weaker base than an aliphatic amine, whose lone pair has nothing else to do, hence the gap of $10.64-4.63=6.01$ units, a factor of one million. Two earlier results rest on exactly the same mechanism. The first is the acidity of PHENOL, six orders of magnitude above that of an alcohol, because conjugation this time stabilizes the conjugate base. The second is the non-basicity of the AMIDE, where the nitrogen lone pair is committed toward a carbonyl. In all three cases the rule is the same: a lone pair already occupied by resonance is no longer available for acid-base chemistry, and it is never the starting species that you should look at but the cost of the change.

The nitro group in para is strongly ELECTRON-WITHDRAWING, by the inductive effect and above all by the resonance effect, since it opens an additional resonance form in which the lone pair of the amine nitrogen is delocalized all the way onto the oxygens of the nitro group. The lone pair is thus even less available than without a substituent, the base is even weaker, and the $pK_{a}$ falls to $1.00$, nearly four orders of magnitude below aniline. The methoxy group in para, by contrast, is a resonance DONOR: it sends electron density back into the ring, which competes with the delocalization of the amine lone pair and makes it slightly more available; the $pK_{a}$ therefore rises to $5.34$. Calculation of the non-protonated fraction at $pH=7.0$. Apply the Henderson-Hasselbalch relation: the fraction in the basic form is $\frac{1}{1+10^{pK_{a}-pH}}=\frac{1}{1+10^{4.63-7.00}}=\frac{1}{1+10^{-2.37}}=\frac{1}{1.00427}=0.99574$, that is $99.6$ percent. Aniline is therefore ALMOST ENTIRELY in the neutral form at neutral pH, which is consistent with its being a weak base and behaving, in water, as a weakly polar and sparingly soluble organic compound. An aliphatic amine under the same conditions would on the contrary be more than $99.7$ percent protonated, hence entirely ionic: two compounds of the same family, two opposite behaviours at the same pH.

The reaction is $CH_{3}CH_{2}CH_{2}NH_{2}+HCl\rightarrow CH_{3}CH_{2}CH_{2}NH_{3}^{+}Cl^{-}$, and the product is propan-1-amine hydrochloride, or propylammonium chloride. The starting amine is a volatile liquid boiling at $48$ degrees Celsius, moderately soluble in water and with an unpleasant fishy smell. The salt is an IONIC SOLID, very soluble in water, odourless, and melting well above $150$ degrees. The justification is the same as for any salt: the ionic compound is held together by electrostatic interactions between ions, much stronger than the hydrogen bonds between neutral molecules, hence the high melting point; and it dissolves in water because this very polar solvent solvates both ions efficiently. The disappearance of the smell follows from the same fact: an ionic solid has practically no vapour pressure, so nothing reaches the nose.

Step $1$: shake the ether solution with dilute aqueous $HCl$. Only the AMINE, which is basic, is protonated and passes into the aqueous phase as the hydrochloride; benzoic acid and naphthalene, which are not basic, stay in the ether. Separate the phases, then make the aqueous phase basic with $NaOH$: the amine, neutral again, separates out and is recovered by extraction with ether. Step $2$: shake the remaining ether phase with aqueous $NaHCO_{3}$. Benzoic acid, with a $pK_{a}$ of $4.20$ and therefore stronger than carbonic acid at $6.35$, is deprotonated and passes into the aqueous phase with effervescence; naphthalene stays in the ether. The aqueous phase is then acidified with $HCl$ to precipitate benzoic acid. Step $3$: evaporate the ether to recover the naphthalene. The procedure is that of the previous chapter, completed by the ACID step that specifically captures the base: we now have both pincers, one for acids and one for bases, and the neutral compound is the one that always stays in the ether.

First reason, SOLUBILITY: most active amines are weakly polar molecules that dissolve poorly in water, whereas their hydrochloride dissolves well, which is essential for preparing a syrup, an injectable solution, or simply for a tablet to disintegrate in the digestive tract. Second reason, FORMULATION: a crystalline solid can be weighed, purified by recrystallization, dosed precisely and compressed, whereas an oil or a volatile liquid lends itself poorly to making a tablet. Third reason, STABILITY: as a salt, nitrogen no longer has a free lone pair, so it is no longer nucleophilic or oxidizable; many free amines turn brown in air by oxidation, which their hydrochloride does not, hence a much longer shelf life. A fourth reason can be added, the disappearance of the often repulsive smell of free amines.

Apply the Henderson-Hasselbalch relation, the neutral fraction being $\frac{1}{1+10^{pK_{a}-pH}}$. In the stomach, $\frac{1}{1+10^{9.00-1.50}}=\frac{1}{1+10^{7.5}}=3.2\times 10^{-8}$, that is $3.2\times 10^{-6}$ percent: the drug is fully ionized there. In the intestine, $\frac{1}{1+10^{2.5}}=3.2\times 10^{-3}$, that is $0.32$ percent. In the blood, $\frac{1}{1+10^{1.6}}=2.45\times 10^{-2}$, that is $2.5$ percent. The ratio between intestine and stomach is $10^{5}$, a hundred thousand times more neutral form in the intestine. Absorption therefore takes place mainly in the INTESTINE, not in the stomach. The justification is the nature of the cell membrane: it is a LIPID bilayer, hence a nonpolar medium that only a neutral, lipophilic molecule can cross by passive diffusion. A charged species is repelled and must wait for a transporter. The result is counterintuitive and worth remembering: BASES are absorbed in the intestine, whereas weak acids, neutral in an acidic medium, are absorbed as early as the stomach.

At $pH=4.5$, the neutral fraction is $\frac{1}{1+10^{9.00-4.50}}=\frac{1}{1+10^{4.5}}=3.2\times 10^{-5}$, that is $3.2\times 10^{-3}$ percent. This is a thousand times more than at $pH=1.5$, which is not nothing, but it is still ONE HUNDRED TIMES LESS than in the intestine. The conclusion of part d) is therefore unchanged: the intestine remains by far the main site of absorption. The calculation also illustrates a handy property of the logarithmic scale: as long as the pH stays several units below the $pK_{a}$, the neutral fraction is simply multiplied by ten each time the pH rises by one unit, which allows you to reason in your head. It would be wrong to conclude that an antacid has no effect on drug absorption: for other molecules, notably weak acids and compounds whose very dissolution depends on pH, the effect can be considerable, and this is a very real drug interaction.

The expected product is butan-1-amine, by a simple $S_{N}2$ of ammonia on the primary halide. In reality, as soon as a molecule of butan-1-amine is formed, it is a BETTER nucleophile than the starting ammonia, since its butyl group is electron-donating by the inductive effect, as exercise $2$ established. It therefore competes for the remaining 1-bromobutane and gives dibutylamine, which is even more nucleophilic and gives tributylamine, which finally gives tetrabutylammonium bromide. We thus obtain all FOUR compounds, primary, secondary and tertiary amine and quaternary salt, in proportions that are hard to control and practically inseparable since their properties are similar. This is the fatal flaw of the method: each product is more reactive than its precursor, exactly the situation already met with Friedel-Crafts alkylation. The only remedy, without changing method, is to use a HUGE excess of ammonia, often twenty equivalents or more, so that a molecule of 1-bromobutane statistically has far more chances of meeting a molecule of ammonia than a molecule of amine already formed. This works but is inelegant, and impractical if the halide is the precious reagent.

Step $1$: treat phthalimide with $KOH$, which gives potassium phthalimide. Step $2$: add 1-bromobutane; the anion, a nitrogen nucleophile, undergoes an $S_{N}2$ reaction and gives $N$-butylphthalimide. Step $3$: release the amine, either by hydrolysis in acidic or basic medium under heat, or more cleanly with hydrazine, the Ing-Manske reaction, which gives butan-1-amine and an insoluble cyclic hydrazide that is easy to filter off. The $pK_{a}$ of $8.30$ is explained exactly like that of the 1,3-dicarbonyl compounds in the chapter on aldehydes and ketones: the hydrogen sits on a nitrogen flanked by TWO carbonyls, so the anion formed delocalizes its charge onto both oxygens in addition to the nitrogen. An ordinary base is therefore enough to deprotonate it quantitatively, whereas an exotic reagent would be needed to pull a proton off ammonia. Finally, the method can never give a secondary amine because after the first alkylation, the nitrogen of $N$-butylphthalimide bears NO hydrogen at all: there is nothing left to deprotonate, no second anion can form, and polyalkylation is structurally impossible. This is the whole point of the method: phthalimide is an ammonia fitted with an alkylation counter locked at one.

First method, reduction of a NITRILE with $LiAlH_{4}$ in anhydrous ether, then hydrolysis: $R-C\equiv N$ gives $R-CH_{2}-NH_{2}$, by double addition of hydride across the triple bond. Since the nitrile itself comes from a halide by the action of $NaCN$, the complete sequence converts $R-X$ into $R-CH_{2}-NH_{2}$ and GAINS one carbon. Catalytic hydrogenation under pressure can also be used. Second method, reduction of an aromatic NITRO compound, with $H_{2}$ over Raney nickel or palladium, or with a metal in acidic medium, tin or iron in $HCl$: nitrobenzene gives aniline. This route is the only really practical one for aromatic amines, since nucleophilic substitution is impossible on an $sp^{2}$ ring carbon, as the previous chapters showed; nitrogen is therefore installed by an ELECTROPHILIC substitution, nitration, and then reduced.

Pentanamide $CH_{3}\left(CH_{2}\right)_{3}CONH_{2}$, which has FIVE carbons, gives butan-1-amine $CH_{3}\left(CH_{2}\right)_{3}NH_{2}$, which has only FOUR. The overall equation is $CH_{3}\left(CH_{2}\right)_{3}CONH_{2}+Br_{2}+4\,NaOH\rightarrow CH_{3}\left(CH_{2}\right)_{3}NH_{2}+Na_{2}CO_{3}+2\,NaBr+2\,H_{2}O$. The missing carbon is the carbonyl carbon: it leaves as carbonate. The mechanism explains this surprising result: nitrogen is first brominated, then deprotonated, and the anion expels the bromide ion at the same time as the alkyl group MIGRATES from carbon to nitrogen. An isocyanate $R-N=C=O$ forms, which sodium hydroxide hydrolyzes to a carbamate, which then decarboxylates. The migration takes place with complete RETENTION of configuration at the migrating carbon, a point often tested: if the $R$ group carried a stereocenter, it is preserved.

Carbon balance. KEEP the number of carbons: direct alkylation of ammonia, the Gabriel synthesis, reduction of an aromatic nitro compound, and reductive amination of an aldehyde or ketone. GAIN one carbon: the nitrile route, since the cyanide ion brings its own carbon. LOSE one carbon: the Hofmann rearrangement, which eliminates the carbonyl carbon. Three syntheses of butan-1-amine, which has four carbons. From a substrate with FOUR carbons, 1-bromobutane: Gabriel synthesis, in three steps as in part b); this is the cleanest route. From a substrate with THREE carbons, 1-bromopropane: action of $NaCN$ to give butanenitrile, then reduction with $LiAlH_{4}$; the missing carbon is gained. From a substrate with FIVE carbons, pentanamide: Hofmann rearrangement with $Br_{2}$ and $NaOH$, as in part d); the surplus carbon is lost. Knowing how to choose among these three routes according to the number of carbons of the available starting material is exactly what a retrosynthesis problem asks, which is why this balance is worth memorizing in this form.

Benzenesulfonyl chloride $C_{6}H_{5}SO_{2}Cl$ reacts like an acyl chloride, by nucleophilic substitution at sulfur. With a PRIMARY amine, we obtain a sulfonamide $C_{6}H_{5}SO_{2}NHR$, which keeps a hydrogen on nitrogen; this compound, a precipitate at first, DISSOLVES in sodium hydroxide by forming its sodium salt. With a SECONDARY amine, we obtain $C_{6}H_{5}SO_{2}NR_{2}$, with no hydrogen at all on nitrogen: the precipitate stays INSOLUBLE in sodium hydroxide. With a TERTIARY amine, there is no lasting reaction, nitrogen having no hydrogen to lose to stabilize the product; only the amine is recovered, soluble in acidic medium. The three classes are thus distinguished by two successive observations: whether or not a precipitate forms, then whether or not this precipitate dissolves in sodium hydroxide.

The sulfonyl group $-SO_{2}-$ is a much more powerful electron-withdrawing group than a carbonyl, for two cumulative reasons. First the inductive effect: sulfur carries TWO doubly bonded oxygens, hence a considerable partial positive charge. Then the resonance effect: the anion formed after deprotonation delocalizes its charge onto these two oxygens, whereas that of an ordinary amide has only one. The conjugate base is therefore much better stabilized, and the $pK_{a}$ of a primary sulfonamide falls to around $10$, against about $17$ for an ordinary amide, a gap of seven orders of magnitude. At $pK_{a}=10$, the compound is comparable to a phenol: sodium hydroxide, whose conjugate acid is water at $15.7$, therefore deprotonates it quantitatively and the ionic salt formed dissolves. The reasoning is exactly that of phthalimide in exercise $4$ or of the 1,3-dicarbonyl compound in the chapter on ketones: count the electron-withdrawing groups able to share the charge.

With an ALIPHATIC primary amine, a diazonium ion $R-N_{2}^{+}$ does form, but it is extremely unstable and decomposes immediately, even at $0$ degrees, releasing GASEOUS NITROGEN and leaving a carbocation. A vigorous evolution of gas is therefore observed, and a mixture of alcohols, alkenes and halides derived from the carbocation is recovered, with its usual rearrangements. With a SECONDARY amine, there is no hydrogen available to eliminate water and form the $N=N$ double bond: the reaction stops at the $N$-NITROSAMINE $R_{2}N-N=O$, a yellow oil that can be separated. With an AROMATIC primary amine, we obtain a diazonium salt $Ar-N_{2}^{+}$ that is STABLE between $0$ and $5$ degrees Celsius in aqueous solution.

The difference lies entirely in the stability of the diazonium ion formed, and hence in what the residual carbocation would be. In the aliphatic case, the loss of $N_{2}$ would generate an ordinary alkyl carbocation, an accessible species: the driving force is the gain in entropy and stability represented by the release of a molecule of dinitrogen, one of the most stable in chemistry, and nothing opposes it. Decomposition is therefore immediate and uncontrollable, and the result is a mixture of no preparative value; the reaction serves only to reveal a primary amine qualitatively by its gas evolution. In the aromatic case, the loss of $N_{2}$ would require forming a PHENYL cation, whose vacancy would sit in an $sp^{2}$ orbital lying in the plane of the ring, hence perpendicular to the $\pi$ system and unable to draw any stabilization from it: this is the same argument that makes aryl halides inert. This very high barrier stabilizes the aromatic diazonium salt just enough for it to be reacted in the cold with the reagent of your choice, and these are the Sandmeyer reactions and their variants, treated in detail in the series on aromatic compounds. The diazonium group is the only group in aromatic chemistry that can be replaced by almost anything, including a simple hydrogen.

Nitrites, used as preservatives in cured meats where they inhibit botulism and fix the pink colour, are converted into nitrous acid by gastric acidity, whose pH is close to $1.5$. This nitrous acid then meets SECONDARY amines in the food bolus, coming notably from the breakdown of proteins and amino acids, and forms the $N$-nitrosamines described in part c). These compounds are metabolically activated by the liver into alkylating agents that modify DNA, hence their established carcinogenic character, particularly associated with digestive cancers. Vitamin C, or ascorbic acid, plays a protective role because it is a powerful REDUCING agent: it reduces nitrous acid to nitric oxide before it has time to meet an amine. This is a KINETIC competition, and it is the reason why manufacturers are required to add ascorbic acid or sodium erythorbate to nitrite-cured products. Note that the reasoning is the same as for ethanol as the antidote to methanol, seen in the chapter on alcohols: the dangerous reagent is not removed, it is offered a faster reaction.

Step $1$, EXHAUSTIVE METHYLATION: treat butan-2-amine with an excess of iodomethane. Since the amine is primary, its nitrogen carries two hydrogens; THREE equivalents of $CH_{3}I$ are therefore needed to reach the quaternary salt, passing successively through the secondary amine, the tertiary amine, then the quaternary ammonium iodide. In practice a mild base such as $K_{2}CO_{3}$ is added to neutralize the $HI$ formed and regenerate the free amine at each step. Note that polyalkylation, the fatal flaw in preparing amines in exercise $4$, is here the very goal. Step $2$, ANION EXCHANGE: treat with moist silver oxide $Ag_{2}O$. The iodide precipitates as silver iodide, which is very insoluble, so the exchange is driven to completion and the quaternary ammonium hydroxide remains in solution. The role of $Ag_{2}O$ is thus simply to replace iodide, a poor base, by the HYDROXIDE ion, which will be the base of the elimination. Step $3$, HEATING, between $100$ and $200$ degrees Celsius, which triggers the elimination and releases the alkene together with trimethylamine.

The hydrogen removed is the one at the BOTTOM of the projection, at $270$ degrees, carried by the rear carbon. The justification is geometric and comes down to one word: the elimination is $E2$, hence concerted, and it requires the $C-H$ bond being broken and the leaving group to be ANTI-PERIPLANAR, that is, at $180$ degrees from each other in the Newman projection. The $NMe_{3}^{+}$ group occupies the position at $90$ degrees on the front carbon; the only hydrogen on the rear carbon located exactly opposite is the one at the bottom. This alignment is necessary so that the orbital of the $C-H$ bond being broken and the antibonding orbital of the $C-N$ bond, which is also breaking, can overlap and form the $\pi$ bond of the future alkene. The two possible alkenes are but-1-ene, obtained by removing a hydrogen from carbon $1$ as in the figure, and but-2-ene, obtained by removing a hydrogen from carbon $3$. It is BUT-1-ENE, the less substituted, that is by far the major product, at $95$ percent.

The leaving group $NMe_{3}^{+}$ is ENORMOUS: three methyls radiate around nitrogen, making it an obstacle out of all proportion to a bromide ion. Two consequences follow. First, the base must approach a beta hydrogen, and this hydrogen is all the more accessible as the carbon bearing it is less substituted: carbon $1$ is a methyl, unhindered and carrying three equivalent hydrogens, whereas carbon $3$ is a $CH_{2}$ flanked by an additional methyl. Second, and this is the stronger point, the ANTI-PERIPLANAR conformation required costs much more on the substituted side: to align a hydrogen of carbon $3$ with the bulky ammonium group, the other substituents must be brought into very crowded gauche positions, which raises the transition state. On the side of the terminal methyl, rotation is free and one of the three hydrogens is always in the anti position at no cost. Statistics finally work in the same direction, with three hydrogens available against two.

The $NMe_{3}^{+}$ group carries a full positive charge and is therefore an extremely powerful electron-WITHDRAWING group by the inductive effect. It strongly acidifies the hydrogens on the neighbouring carbons, so that the transition state of the elimination acquires a clear CARBANION character on the beta carbon, in other words a marked $E1cb$ character: the breaking of the $C-H$ bond is much further advanced than that of the $C-N$ bond. We must then ask where this partial carbanion is most stable, and this is where the INVERTED scale of carbanions comes in. An alkyl group is electron-donating by the inductive effect: it stabilizes a positive charge and DESTABILIZES a negative charge. The order of stability of carbanions is therefore the exact opposite of that of carbocations, primary more stable than secondary, itself more stable than tertiary. The partial carbanion thus develops preferentially on the LEAST substituted carbon, that is, the terminal methyl, and the less substituted alkene is obtained. The two reasons, steric and electronic, therefore push in the same direction, exactly like the two factors in the hierarchy of acid derivatives from the previous chapter. With a bromide, whose leaving group is small and neutral, neither effect operates and we fall back on Zaitsev, governed by product stability.

For the ammonium salt, the ratio of the more substituted to the less substituted alkene is $\frac{5}{95}=0.0526$. For the bromide, it is $\frac{80}{20}=4.00$. The reversal factor is $\frac{4.00}{0.0526}=76$: by changing only the leaving group, everything else being identical, the product ratio is altered by a factor of seventy-six and we go from an overwhelming majority on one side to a clear majority on the other. The general conclusion is this: the orientation of an elimination is NOT a property of the carbon substrate, contrary to what the usual wording of Zaitsev's rule suggests. It results from the competition between the THERMODYNAMIC stability of the product, which favours the more substituted alkene, and the KINETIC accessibility of the transition state, which favours the most unhindered and most acidic hydrogen. A small neutral leaving group and a lightly hindered base let the first factor win, hence Zaitsev; a bulky, charged leaving group, or a very hindered base such as tert-butoxide, tip the balance to the second, hence Hofmann. There is therefore not a rule and its exception, but a single trade-off whose outcome can be predicted.

First reason, OXIDATION. Aniline is a very electron-rich compound, because of the nitrogen lone pair delocalized into the ring, and concentrated nitric acid is a powerful oxidant. A significant part of the substrate is therefore simply oxidized to coloured polymeric products, the tars observed, which destroys the yield whatever the rest. Second reason, PROTONATION. The nitration medium contains concentrated sulfuric acid, and aniline is a base whose conjugate acid has a $pK_{a}$ of $4.63$: it is therefore massively protonated to the anilinium ion. The $-NH_{3}^{+}$ group has no free lone pair, carries a full positive charge, and thereby becomes a powerful electron-withdrawing group, hence a strongly DEACTIVATING, META-directing group, exactly like the nitro group. A mixture is thus obtained, because two species coexist in the medium and react in parallel: the protonated fraction, by far the larger but poorly reactive, directs meta; the fraction that remained neutral, tiny but extremely reactive, directs ortho and para. The result, about $47$ percent meta, is the product of this competition and is useless in synthesis.

The reaction is $C_{6}H_{5}NH_{2}+\left(CH_{3}CO\right)_{2}O\rightarrow C_{6}H_{5}NHCOCH_{3}+CH_{3}COOH$. The product is $N$-phenylacetamide, universally called acetanilide. This is a classic nucleophilic acyl substitution: the nitrogen of aniline attacks one carbonyl of the anhydride, and the tetrahedral intermediate expels an ethanoate ion, which is indeed a step down the reactivity ladder of the previous chapter.

The nitrogen lone pair is no longer free: it is now engaged in conjugation with the amide CARBONYL. Two consequences each settle one of the problems. On the one hand, nitrogen is no longer BASIC: a protonated amide has a $pK_{a}$ close to $-0.5$, as the previous chapter established, which means that very concentrated sulfuric acid would be needed to protonate it and it therefore passes through the nitration medium intact. There is no longer any deactivating meta-directing species in the flask, and the orientation problem disappears. On the other hand, since the lone pair is shared between the ring and the carbonyl, it is much less available to enrich the ring: the acetamido group remains an ACTIVATING group and remains an ortho-para director, but is markedly more moderate than amino. The ring is therefore no longer rich enough to be oxidized by nitric acid, and the tars disappear. The same effect settles both difficulties, which is the whole elegance of this protection: the group was not removed, its lone pair was only TEMPERED.

The acetamido group is much BULKIER than the amino group: it carries a carbonyl and a methyl, whereas amino carries only two small hydrogens. The ortho positions, immediately adjacent, are therefore crowded and the electrophile approaches them poorly; the para position, on the opposite side of the ring, remains entirely clear. The ortho-para orientation predicted by electronic effects thus remains valid, but the split between the two shifts massively toward PARA for purely steric reasons, hence the more than $90$ percent observed. This steric control is an additional benefit of the protection, distinct from those in part c), and it is what makes the sequence truly preparative: not only is the right positional isomer obtained, it is obtained pure.

To complete the sequence, the amide is hydrolyzed, either with aqueous $NaOH$ at reflux or with $H_{3}O^{+}$ at reflux: the acetyl group is removed and para-nitroaniline is recovered. Note that this prolonged reflux is necessary because the amide is the least reactive derivative on the ladder, which is precisely the property that made it useful as a protecting group. The complete sequence is therefore: acetylate, nitrate, hydrolyze. Synthesis of sulfanilamide. Start from acetanilide and treat it with chlorosulfonic acid $HSO_{3}Cl$, which installs a sulfonyl chloride group in para; the product is 4-acetamidobenzenesulfonyl chloride. Ammonia is then reacted with it, converting this sulfonyl chloride into a sulfonamide, as in the Hinsberg test of exercise $5$. Finally the acetamide is hydrolyzed in dilute acidic medium, which releases the aromatic amine: the product is sulfanilamide, $H_{2}N-C_{6}H_{4}-SO_{2}NH_{2}$. Protection is indispensable from the FIRST step, the sulfonation: without it, aniline would be protonated in the very acidic medium and the sulfonyl group would be poorly placed, when aniline is not simply destroyed. It is also what allows the two nitrogens of the target molecule to be told apart at the end. This compound, derived from Domagk's Prontosil, earned the Nobel Prize in Medicine in $1939$; it acts as a COMPETITIVE INHIBITOR of bacterial folic acid synthesis, by posing as para-aminobenzoic acid, whose shape and size it imitates. This is the notion of competitive inhibition already met with ethanol, the antidote to methanol.

Three resonance forms are written for the ion $C_{6}H_{5}-N\equiv N^{+}$: the positive charge can sit on the terminal nitrogen, on the internal nitrogen with an $N=N$ double bond, and delocalize into the aromatic ring at the ortho and para positions. This extended delocalization means that the positive charge is concentrated nowhere, hence that the electrophile is not very electron-hungry: it is a WEAK electrophile, out of all proportion to the nitronium ion or an acylium carbocation. The practical implication is direct and must be stated: a weak electrophile can only attack a STRONGLY ACTIVATED ring. Benzene, toluene, halobenzenes and, all the more, nitrobenzene do not react at all. Only phenols, preferably as the phenoxide, and aromatic amines are electron-rich enough for the reaction to occur in the cold.

The reaction is $C_{6}H_{5}N_{2}^{+}+C_{6}H_{5}OH\rightarrow C_{6}H_{5}-N=N-C_{6}H_{4}-OH+H^{+}$, and coupling takes place at the PARA position of the hydroxyl group. The mechanism is an ordinary electrophilic aromatic substitution, with formation of the arenium ion and then loss of a proton, which restores aromaticity. The para position is chosen for the two usual reasons: the $-OH$ group is a donor, hence an ortho-para director, and of these two positions the para is by far the less hindered since the incoming group is very bulky. If the para position is already occupied, coupling takes place at ortho, with a lower yield.

A pH that is too LOW is ruled out on the side of the nucleophilic partner. Phenol itself is only a moderate activating group, whereas the PHENOXIDE ion, which carries a full negative charge, is an extraordinarily powerful activating group: it is the one that reacts in practice. Phenol has a $pK_{a}$ of $10.0$, so a pH close to $9$ or $10$ is needed to have a significant fraction of phenoxide. Below $pH=8$, the phenoxide concentration becomes too low and the reaction is too slow. A pH that is too HIGH is ruled out on the side of the electrophilic partner. In a clearly basic medium, the hydroxide ion adds to the terminal nitrogen of the diazonium and first converts it into a diazohydroxide $Ar-N=N-OH$, then into the DIAZOTATE ion $Ar-N=N-O^{-}$. This species is neutral and then anionic: it is no longer electrophilic at all, and the conversion is practically irreversible. Beyond $pH=10$, the reagent is therefore destroyed faster than it couples. For an aromatic amine, the window shifts to $pH=4$ to $7$. The upper limit is the same, that of the diazotate, but it leaves more margin; the lower limit is different: it is the FREE AMINE that is the activating group, and its anilinium ion, with a $pK_{a}$ of $4.63$, is on the contrary a deactivating meta director, as exercise $7$ showed. Below $pH=4$ the amine is too protonated to react. In both cases the reasoning is identical: enough base is needed to activate the nucleophile and not so much that the electrophile is destroyed, which is exactly the structure of the bell-shaped imine curve in the chapter on ketones.

The colour of a molecule depends on the energy gap between its highest occupied orbital and its lowest vacant orbital: the smaller this gap, the longer the wavelength absorbed. This gap shrinks as the conjugated system EXTENDS. Phenol and aniline have only a single aromatic ring, a short conjugated system whose absorption stays in the ultraviolet: they are therefore colourless to the eye. The azo compound links two rings through the $-N=N-$ bridge, whose double bond is itself conjugated with both rings: the delocalized system extends from one end of the molecule to the other, the energy gap drops, and absorption moves into the VISIBLE range. The colour perceived is the complement of the one absorbed, giving the characteristic yellows, oranges and reds. The shade can be tuned by adding a donor on one ring and an acceptor on the other, which extends the delocalization further and shifts the absorption toward the red: this is the design principle of the whole range of azo dyes.

The proportion of the acid form is $\frac{1}{1+10^{pH-pK_{a}}}$. At $pH=2.46$, one unit BELOW the $pK_{a}$, we get $\frac{1}{1+10^{-1}}=\frac{1}{1.1}=0.909$, that is $90.9$ percent of the acid form, hence red. At $pH=4.46$, one unit ABOVE, we get $\frac{1}{1+10^{1}}=\frac{1}{11}=0.0909$, that is $9.1$ percent of the acid form and therefore $90.9$ percent of the basic form, yellow. The transition range thus extends approximately from $2.5$ to $4.5$, that is $pK_{a}\pm 1$, which is the general rule for colour indicators: the eye stops distinguishing the minority colour when it falls below about ten percent. Methyl orange is thus the indicator for titrations of a WEAK base by a strong acid, whose equivalence point is acidic, and it would on the contrary be unsuitable for titrating a weak acid with a strong base, where phenolphthalein is used. The colour change itself is explained by the chemistry of this chapter: protonation changes the delocalization of the azo system by making it adopt a quinoid structure, which shifts the absorption and hence the shade.

The Hinsberg test sorts the three classes in two observations. A precipitate that dissolves in sodium hydroxide signals a PRIMARY amine: the sulfonamide formed keeps a hydrogen on nitrogen, a hydrogen made acidic by the sulfonyl group as exercise $5$ established, and sodium hydroxide deprotonates it to a soluble salt. W and X are therefore primary. A precipitate that stays insoluble signals a SECONDARY amine: the sulfonamide has no hydrogen left on nitrogen, hence nothing to give up to sodium hydroxide. Y is secondary. The absence of reaction signals a TERTIARY amine: nitrogen has no hydrogen, and the addition product it might form cannot stabilize itself by losing a proton. Z is tertiary. Since $C_{3}H_{9}N$ has only one secondary amine and one tertiary amine, Y and Z are already identified at this stage, whereas W and X remain to be told apart.

The two primary isomers are propan-1-amine and propan-2-amine. Reduction of a NITRILE with $LiAlH_{4}$ converts $R-C\equiv N$ into $R-CH_{2}-NH_{2}$: nitrogen necessarily ends up on a TERMINAL carbon, since the nitrile carbon carried only one carbon substituent. W can therefore only be PROPAN-1-AMINE, obtained by reducing propanenitrile $CH_{3}CH_{2}CN$. REDUCTIVE amination, for its part, places nitrogen exactly where the starting carbonyl was, as the chapter on ketones showed: propanone has its carbonyl on the central carbon, so the amine obtained has its nitrogen on that same carbon. X is PROPAN-2-AMINE. The reasoning to remember is general: each method of preparation imposes its own constraint on the POSITION of the functional group created, and it is often this constraint, rather than a chemical test, that completes a structure elucidation.

Y, the only secondary amine of the formula, is $N$-METHYLETHANAMINE $CH_{3}CH_{2}NHCH_{3}$. The yellow oil obtained with nitrous acid is the corresponding nitrosamine, $N$-methyl-$N$-ethylnitrosamine $CH_{3}CH_{2}N\left(CH_{3}\right)NO$: a secondary amine lacks the second hydrogen that would allow water to be eliminated and the $N=N$ double bond to form, so the reaction stops at this stage. Z, the only tertiary amine, is $N,N$-DIMETHYLMETHANAMINE $\left(CH_{3}\right)_{3}N$, better known as trimethylamine. The ionic solid obtained with an excess of iodomethane is tetramethylammonium iodide $\left(CH_{3}\right)_{4}N^{+}I^{-}$: only one equivalent is needed here, since nitrogen is already tertiary, and the quaternary salt obtained has no free lone pair left, so the reaction stops for good.

First method, NITROUS ACID in the cold. Y, a secondary amine, gives the yellow nitrosamine oil described in part c), clearly visible because it separates from the aqueous medium. Z, a tertiary amine, gives nothing of the kind: it is simply protonated by the acidic medium and stays in solution, with no gas evolution or precipitate. Second method, the action of an ACYL CHLORIDE such as ethanoyl chloride. Y, which has a hydrogen on nitrogen, gives a neutral, isolable amide, insoluble in dilute acidic medium. Z, which has none, cannot form an amide: at best it gives an unstable acylammonium salt that decomposes, and the amine is recovered unchanged, extractable by simple basification. A third, purely physical method can be added: the boiling points, $37$ degrees Celsius for Y against $3$ for Z, a gap of $34$ degrees explained by the presence or absence of an $N-H$ hydrogen-bond donor, as exercise $1$ established.

For X, propan-2-amine: exhaustive methylation requires three equivalents of $CH_{3}I$, $Ag_{2}O$ supplies the hydroxide, and heating triggers the elimination. The two beta carbons are the two terminal methyls, strictly equivalent, and the only possible alkene is PROPENE. The Hofmann and Zaitsev rules therefore predict the same product, and the experiment cannot tell them apart: this is a degenerate case. For pentan-2-amine, the situation is quite different. The carbon bearing nitrogen has two different neighbours: on one side carbon $1$, a methyl; on the other carbon $3$, a $CH_{2}$ continued by an ethyl chain. Removing a hydrogen from carbon $1$ gives PENT-1-ENE, a monosubstituted alkene; removing a hydrogen from carbon $3$ gives pent-2-ene, a disubstituted alkene. Zaitsev's rule would predict pent-2-ene; the Hofmann rule, which applies here because the leaving group is the bulky charged ammonium, predicts PENT-1-ENE, and this is indeed the overwhelming major product, for the two reasons, steric and electronic, of exercise $6$. It is therefore this second substrate, not the first, that constitutes a discriminating experiment. Remember the method: before invoking an orientation rule, check that several products are actually possible.

$L$-DOPA is an alpha-amino acid: its carboxyl group and its amino group are carried by the same carbon. The decarboxylation studied in the chapter on carboxylic acids required an electron-withdrawing group in the BETA position, typically a carbonyl, so that a six-centre cyclic transition state could form and an enol be released. Here the situation is different: there is no carbonyl in beta, and spontaneous decarboxylation of an amino acid indeed does not occur in a test tube at $37$ degrees. What makes it possible in the body is a COFACTOR, pyridoxal phosphate, derived from vitamin $B_{6}$, which first forms an IMINE with the amino group, exactly as in the chapter on aldehydes and ketones. This imine, conjugated with the pyridine ring, plays precisely the role of the missing electron-withdrawing group: it accepts the pair of electrons released by the loss of $CO_{2}$ and stabilizes the intermediate carbanion. The principle is therefore the same, a neighbouring electron sink, but it is supplied by the enzyme instead of being present in the substrate. Further along the same pathway, hydroxylation and then methylation convert dopamine into noradrenaline and then adrenaline.

The neutral fraction is $\frac{1}{1+10^{pK_{a}-pH}}=\frac{1}{1+10^{10.6-7.4}}=\frac{1}{1+10^{3.2}}=\frac{1}{1585.9}=6.31\times 10^{-4}$, that is $0.063$ percent. In other words, at physiological pH, $99.94$ percent of dopamine circulates as the ammonium ion, hence CHARGED. The blood-brain barrier is a wall of tightly joined cells whose membrane is a lipid bilayer, a nonpolar medium that a charged species cannot cross by passive diffusion. Giving dopamine to a patient is therefore useless: it stays in the blood, where it moreover produces unwanted cardiovascular effects, and never reaches the brain. $L$-DOPA, on the other hand, is an AMINO ACID, and the brain has active transporters specialized in the passage of neutral amino acids, the $LAT1$ system. It therefore crosses the barrier not by diffusion but by a dedicated transport, and is then decarboxylated on the spot to dopamine by the enzyme of part a). In practice an inhibitor of this same enzyme that cannot cross the barrier, carbidopa, is administered so that decarboxylation does not occur prematurely in the circulation.

The figure shows about $0.3$ percent free base at $pH=5.5$, a value practically on top of the axis, and about $76$ percent at $pH=8.5$. Check by calculation. At $pH=5.5$: $\frac{1}{1+10^{8.0-5.5}}=\frac{1}{1+10^{2.5}}=\frac{1}{317.2}=3.15\times 10^{-3}$, that is $0.315$ percent. At $pH=8.5$: $\frac{1}{1+10^{-0.5}}=\frac{1}{1.316}=0.760$, that is $76.0$ percent. The ratio is $\frac{0.760}{0.00315}=241$. Three pH units are thus enough to multiply the proportion of absorbable form by nearly two hundred and fifty, which illustrates the steepness of the logarithmic scale: this is the same remark as for the intestinal absorption of exercise $3$.

The lining of the mouth, like any biological membrane, lets only the NEUTRAL, lipophilic form through by passive diffusion. With cigar smoke at $pH=8.5$, three quarters of the nicotine is in the free base form: it crosses the lining of the mouth abundantly, and there is no need to inhale, which explains the traditional practice of cigar and pipe smoking. Cigarette smoke, produced from tobaccos and additives that make it acidic, is at $pH=5.5$: only $0.3$ percent free base remains, far too little for absorption through the mouth to be enough. The smoker must therefore INHALE, that is, carry the smoke down to the pulmonary alveoli, where two factors combine: the exchange surface there is immense, of the order of $100$ square metres, and the pH there is that of blood, $7.4$, which at once releases a much larger fraction of free base, about $20$ percent according to the figure. This detour through the lung is the chemical reason why the cigarette is both more quickly addictive, alveolar absorption reaching the brain within seconds, and far more harmful, since it forces tars deep into the respiratory tract.

The two nitrogens have neither the same hybridization nor the same environment. The one in the five-membered ring, the pyrrolidine, is an ordinary $sp^{3}$ nitrogen of an aliphatic tertiary amine: its lone pair occupies a non-conjugated $sp^{3}$ orbital, fully available, hence a conjugate $pK_{a}$ of $8.0$, slightly lowered compared with the $9.8$ of trimethylamine because of the electron-withdrawing effect of the neighbouring aromatic ring. The one in the six-membered ring is a PYRIDINE nitrogen, $sp^{2}$ and part of an aromatic ring. Its lone pair is not involved in the aromaticity, unlike that of a pyrrole, but it occupies an $sp^{2}$ orbital, which has one third $s$ character against only one quarter for an $sp^{3}$. The electrons of an orbital with high $s$ character are held closer to the nucleus, hence less available to capture a proton. To this is added the electron-withdrawing effect of the aromatic ring. The $pK_{a}$ thus falls to $3.1$, nearly five units lower. The consequence for the partition studied is simple: at all the pH values considered, between $5.5$ and $8.5$, this pyridine nitrogen is overwhelmingly NON-protonated, since the pH is several units above its $pK_{a}$. It therefore does not contribute to the charge of the molecule, and the reasoning of parts c) and d), carried out on the pyrrolidine nitrogen alone, is legitimate.

The volatile fraction is the fraction in the NEUTRAL amine form, the protonated form being an ionic salt with practically zero vapour pressure. It is $\frac{1}{1+10^{9.80-2.00}}=\frac{1}{1+10^{7.8}}=1.6\times 10^{-8}$, that is about $1.6\times 10^{-6}$ percent. In other words, in a medium at $pH=2$, trimethylamine is entirely converted into the trimethylammonium ion, an ionic, nonvolatile and therefore odourless species, since a molecule must reach the olfactory receptors through the air to be smelled. Lemon juice, whose citric acid brings the pH to around $2$, thus acts by a simple acid-base reaction and not by any masking of the smell: it does not cover the odorous molecule, it CONVERTS it into a salt that no longer evaporates. Vinegar does the same job. Note that the same reasoning explains the repulsive smell of putrescine and cadaverine, diamines coming from the decarboxylation of ornithine and lysine during the decomposition of proteins, and why these smells fade in an acidic medium.

$n=0.200\times 25.0\times 10^{-3}=5.00\times 10^{-3}$ mol, one amine capturing one proton. $M=\frac{0.365}{5.00\times 10^{-3}}=73.0$ g/mol. A saturated amine with one nitrogen is written $C_{n}H_{2n+3}N$: $14.03\,n+17.03=73.0$ gives $n=4$. Formula: $C_{4}H_{11}N$, molar mass 73.14 g/mol.

At the half-equivalence point, half of the amine is protonated: $[B]=[BH^{+}]$ and $pH=pK_{a}=10.7$. An aromatic amine would have a $pK_{a}$ close to 5: this one is aliphatic.

Primary: butan-1-amine, butan-2-amine, 2-methylpropan-1-amine, 2-methylpropan-2-amine. Secondary: N-methylpropan-1-amine, N-methylpropan-2-amine, N-ethylethanamine. Tertiary: N,N-dimethylethanamine. That makes 8 amines, of which 3 are secondary.

The Hinsberg test points to a secondary amine. The only one whose nitrogen carries two identical groups is N-ethylethanamine, or diethylamine, $(CH_{3}CH_{2})_{2}NH$.

At the equivalence point, the solution contains the diethylammonium ion at $\frac{5.00\times 10^{-3}}{0.0750}=0.0667$ mol/L, a weak acid: $pH=\frac{1}{2}(pK_{a}-\log C)=\frac{1}{2}(10.7+1.18)=5.94$. Methyl red, which changes between 4.4 and 6.2, brackets this pH; phenolphthalein would change well before the equivalence point.