Corrected exercises: resonance and reaction mechanisms (202-SNU-RE)

In a resonance structure, ONLY electrons move, and only the mobile ones: $\pi$ electrons and lone pairs. The nuclei stay nailed in place, the total charge is conserved, and no atom of the second period exceeds an octet. The arrow ↔ does NOT describe two species that interconvert: there is only ONE molecule, the hybrid, a weighted average of the resonance structures and more stable than any one of them. Resonance structures are partial drawings of a reality that cannot be drawn in a single stroke of the pencil.

The two structures swap the roles of the oxygens: $CH_{3}-C(=O)-O^{-}$ ↔ $CH_{3}-C(-O^{-})=O$. If either structure were the real molecule, we would measure one short bond (121 pm) and one long bond (136 pm), as in the neutral acid. Instead both bonds measure 127 pm, halfway between: each bond is half double, and each oxygen carries half a charge. Since the structures are EQUIVALENT, the hybrid is perfectly symmetrical, and this delocalization is precisely what stabilizes the ion.

Three independent double bonds would release $3\times 120=360$ kJ/mol. Real benzene releases only 208: $360-208=152$ kJ/mol are missing, because benzene starts from a LOWER energy than the hypothetical triene. These 152 kJ/mol are the aromatic stabilization energy, the experimental signature of the delocalization of the six $\pi$ electrons over the whole ring, consistent with six equal bonds of 140 pm, intermediate between single (154) and double (134).

Moving an H means moving an ATOM: this is no longer a resonance structure but another molecule linked to the first by a real reaction. The phenomenon that swaps an H and a double bond in this way is called TAUTOMERISM (an equilibrium, arrows ⇌, not ↔). Giving five bonds to a carbon violates the octet: ten electrons around C, forbidden without discussion. Two archetypal errors: resonance moves neither atoms nor the eight-electron limit.

The inductive effect arises from the difference in electronegativity between two bonded atoms: it polarizes the $\sigma$ bond and propagates from bond to bond, weakening by a factor of about 7 with each bond crossed. The resonance effect arises from the delocalization of a lone pair or a $\pi$ pair: it requires conjugation, but it is then transmitted over the whole extent of the system without fading. Classification: $-CH_{3}$ is a donor by inductive effect alone; $-NO_{2}$ is an acceptor by both effects, which makes it the most powerful of the usual acceptors; $-Cl$ is an acceptor by inductive effect but a DONOR by resonance effect since it has delocalizable lone pairs, and the two oppose each other, the inductive effect winning on acidity and the resonance effect on orientation; $-OH$ and $-NH_{2}$ are in the same situation, inductive acceptors and resonance donors, but for them the resonance effect clearly wins, nitrogen being less electronegative than oxygen, which makes $-NH_{2}$ an even stronger donor than $-OH$.

The stronger the acid, the smaller its $pK_{a}$. All four halogenated acids are more acidic than ethanoic acid, in the order $F>Cl>Br>I$: $pK_{a}$ of 2.59, 2.86, 2.90, 3.18. The acidity factor between ethanoic acid and chloroethanoic acid is $10^{4.76-2.86}=10^{1.90}=79$: a single chlorine makes the acid 79 times stronger. The explanation is the electron-withdrawing inductive effect of the halogen, which pulls electrons along the chain and stabilizes the negative charge of the conjugate base, the carboxylate. A stabilized anion means a strong acid. The order follows exactly the decreasing electronegativity of the halogens, which confirms that we are dealing with a pure inductive effect: here the electron-donating resonance effect of chlorine cannot express itself, for lack of conjugation with the carboxyl group.

$10^{4.76-0.65}=10^{4.11}=1.3\times 10^{4}$: three chlorines make the acid thirteen thousand times stronger. The effect is therefore practically additive on the $pK_{a}$ scale, each chlorine removing roughly 1.3 to 1.9 units: $4.76\rightarrow 2.86\rightarrow 1.29\rightarrow 0.65$. This additivity is the signature of independent inductive effects that accumulate. With a $pK_{a}$ of 0.65, trichloroethanoic acid is stronger than phosphoric acid and approaches the strong mineral acids: three chlorine atoms are enough to turn vinegar into a laboratory acid. It is the most striking demonstration that a substituent with no direct link to the acidic site can entirely govern its behaviour.

The series speaks for itself: 2.86 for chlorine in position 2, 4.05 in position 3, 4.52 in position 4, against 4.82 for unsubstituted butanoic acid. The effect collapses as the chlorine moves away from the carboxyl group, and it has practically disappeared in position 4. The acidity ratio between isomer 2 and isomer 4 is $10^{4.52-2.86}=10^{1.66}=46$: two bonds farther away, the acidity is divided by 46. The rule to remember fits in one sentence: the inductive effect is powerful but short-sighted, it loses roughly a factor of 7 per bond crossed and becomes negligible beyond three bonds, whereas the resonance effect carries as far as the conjugation allows.

$\Delta pK_{a}=16-10=6$, and the scale is logarithmic: $\frac{K_{a}(phenol)}{K_{a}(ethanol)}=10^{6}$, exactly one million as stated. The reason lies in the conjugate bases: the ethoxide ion $CH_{3}CH_{2}O^{-}$ concentrates all its charge on its single oxygen; the phenoxide ion, on the other hand, DELOCALIZES the charge into the ring by resonance. A spread-out charge is a stabilized charge: since the conjugate base is more stable, the acid gives up its proton more readily.

Four resonance structures: the charge on oxygen (the major structure, with the charge on the most electronegative atom), then three structures where it enters the ring, on the two ORTHO carbons and on the PARA carbon. Never at meta: the conjugation gives no access to it. This ortho-ortho-para trio is a signature that will come back throughout aromatic chemistry.

The ethanoate ion has only TWO resonance structures against four, but they are EQUIVALENT and each places the charge on an OXYGEN: two ideal positions sharing the charge half and half. The phenoxide ion has more structures, but three of them put the charge on CARBONS, which are barely electronegative: minor structures. Quality beats quantity: ethanoate is better stabilized, ethanoic acid more acidic. Factor against ethanol: $10^{16-4.8}=10^{11.2}\approx 1.6\times 10^{11}$, a hundred billion.

This time resonance acts BEFORE the reaction: in aniline, the lone pair of nitrogen is already delocalized into the ring (resonance structures placing a negative charge at ortho and para, the nitrogen becoming positive). This pair, busy taking part in resonance, is less AVAILABLE to capture a proton: the base is weakened by a factor $10^{10.7-4.6}\approx 10^{6}$. And protonating the nitrogen would DESTROY this delocalization, a cost that saturated cyclohexylamine never has to pay. Resonance stabilizes whichever species it can: the conjugate base for phenol, the starting base for aniline, and the acidity shifts accordingly.

A carbocation is stabilized by the electron-donating inductive effect of neighbouring alkyl groups, which partly make up for the electron deficiency, and by hyperconjugation, that is, the overlap of a neighbouring $C-H$ bond with the vacant orbital. Both effects grow with the number of alkyl groups: tertiary more stable than secondary, more stable than primary, more stable than methyl. Radicals follow EXACTLY the same order, tertiary on top, for the same reasons: a radical is also deficient, with a single electron where two would be needed, and so it benefits just as much from electron donation. The gaps are simply smaller, a radical being less demanding than a cation.

The more acidic an alkane, the more stable the carbanion it releases. Now the $pK_{a}$ values increase: methane 48, ethane 50, propane 51, 2-methylpropane 53. The stability ranking of carbanions is therefore methyl more stable than primary, more stable than secondary, more stable than tertiary: exactly the INVERSE of the order for cations and radicals. The explanation is immediate once the question is posed correctly. An alkyl group is an electron donor; donating electrons to a center that lacks them relieves it, but donating electrons to a center that already has too many makes it worse. The same substituent stabilizes the cation and destabilizes the anion, without changing its nature: it is the charge of the center that changes sign, not the substituent.

The benzylic and allylic carbanions are not localized: their pair is delocalized, over the aromatic ring in one case, over the double bond in the other. A charge spread over several atoms is much more stable than a concentrated charge, hence a gain of about seven $pK_{a}$ units, that is a factor $10^{7}$ on the acidity constant. The difference with the mechanism in a) is one of kind: hyperconjugation and the inductive effect merely SHIFT a little electron density along bonds that stay in place, whereas resonance really SPREADS the charge over several atoms, as bond length measurements confirm. This is why stabilization by delocalization is always an order of magnitude greater than inductive stabilization, and why it decides as soon as it is possible.

Acetylene carries its hydrogen on an $sp$ carbon, whose hybrid orbital has 50% $s$ character against 25% for an $sp^{3}$ carbon. An $s$ orbital is spherical and much closer to the nucleus: a pair occupying it is strongly held, hence stabilized, with no delocalization involved. The acetylide ion is thus stabilized by twenty-five $pK_{a}$ units relative to an $sp^{3}$ carbanion, which makes it by far the most stable of the simple carbanions. The general rule all this suggests is the following: an electron-DONATING group stabilizes an electron-poor center, a carbocation or a radical, and destabilizes an electron-rich center, a carbanion; an electron-WITHDRAWING group does exactly the opposite. It is enough to know the sign of the charge of the intermediate to flip a whole stability scale at once, without learning anything new. The same reasoning explains why a nitro group strengthens an acid and weakens the basicity of an amine.

Increasing stability: methyl < primary < secondary < tertiary. Two effects add up: the electron-donating INDUCTIVE effect of the alkyl groups, which push electron density toward the depleted carbon, and HYPERCONJUGATION: each neighbouring $C-H$ bond partly overlaps the empty $p$ orbital and delocalizes a little of its electrons into it. Count of neighbouring $C-H$ bonds: ethyl cation 3, isopropyl cation 6, tert-butyl cation 9: the hierarchy is literally arithmetic.

$CH_{2}=CH-CH_{2}^{+}$ ↔ $^{+}CH_{2}-CH=CH_{2}$: two EQUIVALENT structures, hence a real charge of $+\frac{1}{2}$ on each terminal carbon and a $\pi$ bond spread over the three carbons. The benzyl cation does even better: four resonance structures (charge on the $CH_{2}$, then at ortho, ortho and para of the ring), the aromatic trio already met with the phenoxide ion.

Benzyl $\approx$ tert-butyl > isopropyl > ethyl >> phenyl. The trap: phenyl and benzyl look alike by name, but the BENZYL cation carries its vacancy on an $sp^{3}$ carbon that has become $sp^{2}$, conjugated with the ring: stabilized by resonance; the PHENYL cation carries its vacancy on a carbon OF the ring, in an $sp^{2}$ orbital pointing outward, PERPENDICULAR to the $\pi$ system: no resonance is possible, and it is one of the worst carbocations in chemistry. A neighbouring aromatic ring helps; being the hole in the ring itself does not.

Markovnikov protonation of 3-methylbut-1-ene $CH_{2}=CH-CH(CH_{3})_{2}$ gives the secondary cation $CH_{3}-\overset{+}{C}H-CH(CH_{3})_{2}$. The H of the neighbouring carbon (the one bearing the two methyls) slides with its pair onto the positive center: a 1,2-hydride shift, and the charge lands on THE carbon of the two methyls: $CH_{3}-CH_{2}-\overset{+}{C}(CH_{3})_{2}$, a TERTIARY one. $Br^{-}$ attaches there: the major product is 2-bromo-2-methylbutane, and not the “expected” 2-bromo-3-methylbutane. Lesson: a mechanism passing through a carbocation is never frozen: the intermediate ALWAYS takes the rearrangement that stabilizes it, and the final product betrays the most stable carbocation, not the first one formed.

SN2: ONE step, concerted: the nucleophile attacks the carbon FROM BEHIND, opposite the leaving group, through a single transition state in which one bond forms while the other breaks. Consequence: INVERSION of configuration (the umbrella flips inside out, Walden inversion): an (R) substrate gives an (S) product. Rate: $v=k[RX][Nu]$, second order, both partners being in the single step. The backside attack requires an accessible carbon: methyl and primary go fast, secondary crawls, tertiary never: three alkyl groups physically block the back.

SN1: TWO steps. First the ionization, slow: the leaving group leaves alone and leaves a CARBOCATION; then the nucleophile captures this cation, fast. Rate: $v=k[RX]$, first order: the nucleophile does not appear, it arrives after the slow step. Since the carbocation is PLANAR, the nucleophile attacks both faces: an optically pure substrate gives a racemized mixture. Everything rests on the formation of the cation: tertiary (or allylic, benzylic) is mandatory; a primary never ionizes.

The one-hump curve is the SN2: one elementary act, a single transition state. The two-hump curve is the SN1: the TROUGH between the humps is the carbocation, an intermediate: a REAL species, short-lived but existing (sometimes observable). The PEAKS are transition states: energy saddle points, not species, unobservable in principle. The first hump of the SN1 is the higher one: ionization is the rate-determining step, and anything that stabilizes the carbocation speeds up the whole reaction.

1-Bromobutane: primary, with $HO^{-}$ a good nucleophile: SN2, product butan-1-ol (with inversion, invisible here for lack of a stereocenter); doubling $[HO^{-}]$ DOUBLES the rate, and doubling $[RX]$ does too ($v=k[RX][Nu]$). 2-Bromo-2-methylpropane: tertiary, with water, a weak nucleophile: SN1, product 2-methylpropan-2-ol; doubling the water concentration changes NOTHING in the rate (water waits for the carbocation), doubling $[RX]$ doubles it. This insensitivity to the nucleophile is, incidentally, the classic experimental test for identifying an SN1.

First step: the electron-rich double bond attacks a bromine molecule and polarizes it; one bromine atom attaches by bonding simultaneously to both carbons, forming a positively charged three-membered ring, the bromonium ion, while the other atom leaves as a bromide ion. Second step: the bromide ion attacks one of the two carbons of the bridge. It can only do so from behind, on the side opposite the bridge, exactly as in a second-order nucleophilic substitution, because the bridging bromine completely occupies one face of the former double bond and blocks any approach from that side. The two bromine atoms therefore end up on opposite sides of the plane: it is an ANTI addition, or anti-periplanar addition, and it is stereospecific.

Start from (E)-but-2-ene, whose two methyls lie on opposite sides of the double bond. Anti attack places the two bromines on opposite sides; writing out the Newman projection or the wedge-and-dash representation, we obtain a compound whose two asymmetric carbons have opposite configurations, $2R,3S$. This molecule has an internal plane of symmetry that exchanges the two halves: it is achiral, it is MESO 2,3-dibromobutane, optically inactive despite its two asymmetric carbons. (Z)-But-2-ene, whose two methyls are on the same side, leads by the same anti attack to the pairs $2R,3R$ and $2S,3S$, formed in strictly equal amounts since the bromide ion attacks either carbon of the bridge indifferently: it is a RACEMIC mixture, also optically inactive, but for an entirely different reason, the cancellation between two enantiomers and not an internal symmetry.

This result proves that the two carbons cannot rotate freely relative to each other between the two steps: the geometric information carried by the starting double bond, $E$ or $Z$, is found intact in the product. This is the very definition of a stereospecific reaction, and only a bridged, rigid intermediate can explain it. If the addition went through an open carbocation, the carbon-carbon bond would become single and free to rotate, the two conformations would equilibrate before the bromide attacks, and both alkenes would give the SAME mixture of meso and racemic. Nothing of the kind is observed: each alkene gives a single, different product. Experiment therefore refutes the open carbocation and requires the bridge.

In the presence of water, which is in very large excess, water becomes the nucleophile that opens the bridge in place of the bromide ion: a bromohydrin is obtained. The regioselectivity is the remarkable point. Water attaches to the MOST substituted carbon, whereas a nucleophile attacking an ordinary substrate always chooses the least hindered carbon. The reason is that the bromonium bridge is not symmetrical: the most substituted carbon better supports a partial positive charge, so its bond to bromine is longer and weaker, and it carries more carbocation character. The nucleophile is drawn to it despite the crowding. The product is therefore 1-bromo-2-methylpropan-2-ol, with the hydroxyl group on the tertiary carbon and the bromine on the primary carbon, in agreement with Markovnikov-type orientation. This unexpected regioselectivity, which contradicts steric hindrance, is a strong second argument for the bridge: a free carbocation would also explain it, but it would not account for the stereochemistry of b). Only the unsymmetrical bridge explains both observations at once.

Substrate $A$. Doubling $[A]$ at constant hydroxide doubles the rate: the order with respect to $A$ is 1. Doubling $[HO^{-}]$ at constant $[A]$ leaves the rate unchanged: the order with respect to hydroxide is 0. The law is $v=k[A]$, with $k=\dfrac{1.0\times 10^{-4}}{0.10}=1.0\times 10^{-3}\;\text{s}^{-1}$, a first-order constant whose unit is indeed an inverse time. Substrate $B$. Doubling $[B]$ doubles the rate, doubling $[HO^{-}]$ also doubles it: both partial orders are 1, the overall order is 2. The law is $v=k[B][HO^{-}]$, with $k=\dfrac{4.0\times 10^{-5}}{0.10\times 0.10}=4.0\times 10^{-3}\;\text{L}\cdot\text{mol}^{-1}\cdot\text{s}^{-1}$. The units are here the quickest check that the two orders have not been mixed up.

Substrate $A$ follows a first-order nucleophilic substitution: the slow step is the ionization, which involves only the substrate, and the nucleophile appears only in the fast step that follows. Such behaviour presupposes a stable carbocation, hence a tertiary substrate, or possibly a secondary one stabilized by delocalization. Substrate $B$ follows a second-order substitution: the slow step is bimolecular, the attack and the departure occurring simultaneously, which presupposes a lightly hindered carbon, hence a primary substrate. The zero order with respect to hydroxide obviously does not mean that it takes no part: without it, there would be no product. It means that its concentration does not influence the RATE, because it comes in after the rate-determining step. A rate law only informs about what happens up to the highest transition state, and that is precisely what makes it a tool for mechanistic diagnosis: it ignores everything that comes afterward.

The optical purity, or enantiomeric excess, is the ratio of the measured rotation to that of the pure enantiomer: $\dfrac{6.0}{13.5}\times 100=44.4\,\%$. This excess of 44.4% in favour of the inverted enantiomer breaks down as follows: if $x$ is the percentage of inverted product and $y$ that of product with retained configuration, then $x+y=100$ and $x-y=44.4$. Solving: $x=\dfrac{100+44.4}{2}=72.2\,\%$ and $y=\dfrac{100-44.4}{2}=27.8\,\%$. The mixture therefore contains 72.2% inverted product and 27.8% retained product, which amounts to saying that it consists of 55.6% racemic mixture and 44.4% purely inverted product.

A pure first-order substitution should give exactly 50% of each enantiomer, since the carbocation is planar and can be attacked equally on both faces. We are far from that: inversion clearly wins. The picture to correct is that of a completely free, symmetrical carbocation. In reality, the leaving group does not move away instantly: it stays for a while in the vicinity of the carbon, as an intimate ion pair, and it partly shields the face it has just left. The nucleophile is therefore statistically favoured on the opposite side, hence an excess of inversion. It is not a total inversion either, because a fraction of the ion pairs separates completely before the attack and gives a racemic mixture. The percentage of inversion thus measures the lifetime of the carbocation: the more stable it is and the better solvated, the longer it lives, the more the ion pair dissociates, and the closer one gets to complete racemization. In a more ionizing solvent, such as water in place of an ethanol-water mixture, one should therefore expect more extensive racemization, with an enantiomeric excess clearly below 44%.

Step 1 (slow): the electron-rich $\pi$ bond attacks the H of HBr: the arrow starts from the MIDDLE of the double bond toward H, while a second arrow slides the pair of the H-Br bond onto bromine. Result: a carbocation and a $Br^{-}$ ion. Step 2 (fast): an arrow starts from a lone pair of $Br^{-}$ toward the positive carbon. Curved arrows ALWAYS follow the electrons, never the atoms: this is the grammar of the whole course.

If H attaches to the terminal $CH_{2}$ carbon, the charge falls on the central carbon: ISOPROPYL cation, secondary. If H attaches to the center, the charge falls at the end: propyl cation, PRIMARY. The secondary one, stabilized by six $C-H$ bonds in hyperconjugation, forms almost exclusively (lower curve of the figure: lower barrier AND more stable intermediate). $Br^{-}$ captures it: major product, 2-bromopropane. Modern Markovnikov rule: the addition goes through THE MOST STABLE CARBOCATION. The rhyme “H goes to the carbon with the most hydrogens” is only the consequence, and it fails as soon as stability decides otherwise (rearrangements, conjugated groups).

Protonation of 2-methylbut-2-ene $CH_{3}-C(CH_{3})=CH-CH_{3}$ can create a TERTIARY cation (charge on the carbon with the two methyls, H received by the CH) or a secondary one: the tertiary wins without contest. Water attacks, a deprotonation concludes: 2-methylbutan-2-ol, the tertiary alcohol. Every acid-catalyzed hydration obeys Markovnikov: the OH ends up on the MOST SUBSTITUTED carbon.

Orange bromine water DECOLOURIZES instantly: bromine adds across the double bond (one $Br_{2}$ molecule per $\pi$ bond), giving colourless 1,2-dibromopentane. Quantitatively: $n=\frac{5.0}{70}=0.071$ mol of pent-1-ene ($M=70$ g/mol), which consume $0.071\times 160=11.4$ g of $Br_{2}$. An alkane has neither a $\pi$ bond nor an electron-rich site to offer: without UV light, no reaction, and the colour stays: it is THE one-minute test for telling an alkane from an alkene.

E2: concerted, second order ($v=k[RX][base]$): the base pulls off an H on the NEIGHBOURING carbon while the leaving group leaves and the double bond forms; strict geometric requirement: H and leaving group ANTI-PERIPLANAR (at 180°, so that the orbitals line up into a $\pi$ bond). It competes with the SN2: same substrate, same kinetics, the base choosing between attacking the carbon (SN2) or the H (E2). E1: ionization first, the same carbocation as the SN1, then loss of an $H^{+}$; first order; it inevitably accompanies the SN1 on tertiary substrates, the two products coexisting.

Removing an H from carbon 1 gives but-1-ene; from carbon 3, but-2-ene, in (Z) and (E) versions. Zaitsev's rule: the MAJOR alkene is the most SUBSTITUTED: here but-2-ene dominates, and above all its (E) form. The figure explains the rule instead of imposing it: but-2-ene (116-120) is lower in energy than but-1-ene (127), and the E2 transition state, where the double bond is already half formed, inherits this advantage: the path to the stable alkene has the lowest barrier.

In the (Z), the two methyls on the same side crowd each other sterically: 4 kJ/mol of strain that the (E), with opposite methyls, does not pay. At equilibrium: $\frac{[E]}{[Z]}=e^{4000/(8.314\times 298)}\approx 5$: five to one, 83% (E). The gap 127 versus 116 comes from SUBSTITUTION: each alkyl group on the double bond stabilizes it by hyperconjugation (neighbouring $C-H$ bonds that delocalize into the $\pi$ system): disubstituted beats monosubstituted, exactly the carbocation argument recycled for alkenes.

Tertiary + concentrated strong base: E2 (the SN2 is blocked by crowding, and the strong base does not wait for the carbocation). Tertiary + warm water: mostly SN1 with a little E1, through the carbocation. Primary + $CH_{3}S^{-}$: clean SN2 (excellent nucleophile, mediocre base, clear backside). Primary + tert-butoxide: E2: too bulky to attack the carbon, the base falls back on the H atoms. The logic in one sentence: the substrate decides whether a carbocation is possible (1 or 2 steps), then the reagent chooses its target: the CARBON if it is a good nucleophile, a HYDROGEN if it is a good base or too big to approach.

Nitrogen: 5 valence electrons, no lone pair, 4 bonds (one with C, two with the doubly bonded oxygen, one with the other): $5-0-4=+1$. Singly bonded oxygen: $6-6-1=-1$. The molecule is neutral, but it carries two opposite formal charges: it is a permanent zwitterion, hence its high dipole moment.

The two structures swap the roles of the two oxygens: each N-O bond is single in one structure and double in the other, an average order of $\frac{1+2}{2}=1.5$. The $-1$ charge is shared between the two oxygens: $-\frac{1}{2}$ on each. The nitrogen keeps its $+1$ charge in both structures.

Both structures respect the octet and carry a single charge, but the first places the negative charge on oxygen, which is more electronegative than carbon: it is the major structure. The second remains useful: it explains why the enolate can react through its CARBON, in particular with carbon electrophiles.

$H_{2}C=O$ is major: octet everywhere, no charge. $H_{2}C^{+}-O^{-}$ is minor: the carbon has only six electrons, but the negative charge is on oxygen. $H_{2}C^{-}-O^{+}$ is to be rejected: the oxygen has only six electrons AND the negative charge is on the least electronegative atom. The minor structure $H_{2}C^{+}-O^{-}$ shows that the carbonyl carbon is electrophilic: this is where nucleophiles attack.

In $^{+}O\equiv C-N^{2-}$: oxygen has one lone pair and three bonds, $6-2-3=+1$; nitrogen has three lone pairs and one bond, $5-6-1=-2$. Three units of charge in all, on two atoms, with a $-2$ charge on a single atom and a positive charge on the most electronegative atom: a negligible structure. The two others carry only a $-1$ charge; the most representative is $^{-}O-C\equiv N$, where the charge is on oxygen, closely followed by $O=C=N^{-}$.

From strongest to weakest: 4-nitro (3.44), 3-nitro (3.45), 3-chloro (3.83), 4-chloro (3.98), 3-methoxy (4.09), benzoic (4.20), 4-methoxy (4.47). Only one acid is weaker than benzoic acid: 4-methoxybenzoic acid.

$10^{4.20-3.44}=10^{0.76}\approx 5.8$: the nitro group in para makes the acid nearly six times stronger.

The oxygen of the methoxy group is electronegative: by inductive effect, it is an acceptor. But it carries lone pairs that delocalize into the ring: by resonance effect, it is a donor, and this effect reaches the ortho and para positions. In para, the resonance donation dominates: it enriches the carbon bearing the carboxylate and destabilizes the negative charge, and the acid is weakened (4.47). In meta, the resonance donation does not reach this carbon and only the inductive withdrawal remains: the acid is strengthened (4.09). Same substituent, two opposite signs.

Chlorine is an inductive acceptor and a weak resonance donor. In meta, it is closer to the carboxyl group, and its inductive effect acts alone. In para, it is farther away, and its small resonance donation offsets part of the inductive effect: isomer 4 is less acidic (3.98 against 3.83).

A nitro group in meta lowers the $pK_{a}$ by $4.20-3.45=0.75$. Two nitro groups in meta: $4.20-2\times 0.75=2.70$. The tabulated value, 2.82, which the statement does not give, is close: additivity is a good approximation, and this is the idea behind the Hammett equation.

The tertiary carbocation captures water, and deprotonation gives 2-methylpropan-2-ol. Two intermediates, the carbocation and the oxonium ion, which are troughs; three transition states, which are peaks.

$E_{a1}=80-0=80$ kJ/mol; $E_{a2}=52-45=7$ kJ/mol; $E_{a3}=-5-(-20)=15$ kJ/mol. The rate-determining step is the protonation: it is the one that leads to the highest point of the profile. Anything that stabilizes the carbocation lowers this peak.

$\Delta_{r}H=-50$ kJ/mol: the reaction is exothermic. From the alcohol, one must climb to the highest point: $80-(-50)=130$ kJ/mol. Dehydration, the reverse reaction, is therefore much slower under the same conditions and requires heating.

Step 1 is endothermic, $+45$ kJ/mol: its transition state is late and resembles the carbocation. Step 2 is strongly exothermic, from $+45$ to $-20$: its transition state, early, also resembles the carbocation, its reactant. The carbocation is thus the right model for the two peaks that frame it.

$\frac{v_{tert}}{v_{sec}}=e^{25000/(8.314\times 298)}=e^{10.1}\approx 2.4\times 10^{4}$: the hydration of 2-methylpropene is about twenty-four thousand times faster. $H_{3}O^{+}$ is consumed in step 1 and regenerated in step 3: it does not appear in the overall equation, it only lowers the barrier.

The central carbon is bonded to three $NH_{2}$ groups; the positive charge can be carried by each of the three nitrogens, the carbon forming a double bond with it: three resonance structures, all equivalent. The $+1$ charge is shared: $+\frac{1}{3}$ on average per nitrogen. Each C-N bond is double in one structure out of three: average order $\frac{2+1+1}{3}=\frac{4}{3}\approx 1.33$. The ion is planar and perfectly symmetrical.

Protonating the doubly bonded nitrogen gives the symmetrical guanidinium ion, stabilized by three equivalent resonance structures. Protonating an $NH_{2}$ would give an $-NH_{3}^{+}$ ion whose charge stays localized, with no resonance. One always protonates where the conjugate acid is best stabilized.

$10^{13.6-10.6}=10^{3}=1,000$: guanidine is a thousand times more basic than an ordinary amine. The gap comes from the conjugate acid: methylammonium keeps its charge on a single nitrogen, guanidinium spreads it over three.

In urea, the lone pairs of the nitrogens are delocalized toward the carbonyl oxygen: they are barely available, and the nitrogen is not basic. Protonation takes place on the OXYGEN, and the ion formed is stabilized by the two nitrogens. But the neutral molecule was already stabilized by this resonance, which protonation does not gain: the basicity remains very weak, 13.5 $pK_{a}$ units below guanidine.

$\frac{[BH^{+}]}{[B]}=10^{12.5-7.4}=10^{5.1}\approx 1.26\times 10^{5}$: one hundred and twenty-six thousand protonated forms for one neutral form. The arginine side chain is therefore positively charged throughout the body, which allows it to form salt bridges with the carboxylate groups of proteins and with the phosphates of DNA.

It is the hydrogen of the $OH$ carried by carbon 3. This group is an enol, but a special enol: its $C_{2}=C_{3}$ double bond is conjugated with the carbonyl $C_{1}=O$. An isolated enol has a $pK_{a}$ close to 10; it is the conjugation with the carbonyl that gains six units.

Structure 1: the charge on the oxygen of C3. Structure 2: $O=C_{3}$ forms and the charge moves to C2. Structure 3: $C_{2}=C_{1}$ forms and the charge arrives on the oxygen of the carbonyl. Three structures, 2 of which place the charge on an oxygen. This is the same sharing between two oxygens as in a carboxylate, but through an intervening C=C double bond: hence “vinylogous”.

$10^{10.0-4.2}=10^{5.8}\approx 6.3\times 10^{5}$: about six hundred thousand times more acidic than phenol, whose charge is spread mostly over carbons.

Blood: $\frac{[A^{-}]}{[AH]}=10^{7.4-4.2}=10^{3.2}\approx 1,585$, that is $\frac{1585}{1586}\approx 99.9$% ascorbate. Stomach: $\frac{[A^{-}]}{[AH]}=10^{1.5-4.2}=10^{-2.7}\approx 0.0020$, that is only $0.20$%: vitamin C is there almost entirely in its acid form.

The anion carried by the oxygen of C2 can only delocalize its charge toward C3, never as far as the oxygen of the carbonyl: no resonance structure equivalent to that of a carboxylate, hence an ordinary enol acidity. The second acidity is weaker still: a proton must be removed from an ion that is already negative, which costs electrostatic energy, and the charge can no longer reach the carbonyl already occupied by the first.