Corrected exercises: alcohols, phenols and ethers (202-SNU-RE)

The class is read on the carbon that CARRIES the hydroxyl group, by counting how many other CARBONS are bonded to it: only one for a primary alcohol, two for a secondary one, three for a tertiary one. Butan-$1$-ol, $CH_{3}CH_{2}CH_{2}CH_{2}OH$, carries the $-OH$ on a terminal carbon bonded to a single carbon: PRIMARY. Butan-$2$-ol carries it on a carbon bonded to two carbons: SECONDARY. $2$-Methylpropan-$2$-ol carries it on a carbon bonded to three methyl groups: TERTIARY. Watch out for the classic trap: you do not count hydrogens, and methanol is conventionally set apart, as a special case of primary alcohol.

The alcohol has a hydrogen carried by the oxygen: it is therefore both a DONOR and an ACCEPTOR of hydrogen bonds, and the butan-$1$-ol molecules associate with one another in a network that has to be broken to vaporize the liquid. The ether has no hydrogen on its oxygen: it is an ACCEPTOR only, so two ethoxyethane molecules cannot bond to each other by hydrogen bonding. All that is left are London dispersion forces and a weak dipole-dipole interaction, exactly as in pentane, hence the almost identical $35$ and $36$ °C. The right conclusion to keep: what matters is not the PRESENCE of an oxygen, it is the presence of an $O-H$.

The two compounds have the same molecular formula and the same functional group, hence the same ability to form hydrogen bonds: the difference is not there. It lies in the SHAPE. Butan-$1$-ol is a stretched-out, almost cylindrical chain, whose large side surface allows extensive contact with its neighbours and therefore strong London forces. $2$-Methylpropan-$2$-ol is rolled into a ball around its central carbon: for a given volume, a sphere offers the smallest possible surface, hence the least contact. The London forces collapse and the boiling point drops by $36$ degrees. This is the same rule as for branched alkanes in the hydrocarbons chapter.

An alcohol molecule has two opposite personalities: the polar $-OH$ HEAD, which hydrogen-bonds with water and therefore loves water, and the nonpolar hydrocarbon TAIL, which can only disrupt the hydrogen-bond network of water without offering anything in return. As long as the tail is short, the head wins and miscibility is total. From four carbons on, the balance tips and solubility drops; at eight carbons, the tail dominates so much that the compound is practically insoluble and behaves like a hydrocarbon. A useful rule of thumb: about one $-OH$ for every four carbons at most if the compound is to remain water-soluble.

Each additional hydroxyl group adds a donor site AND an acceptor site, and therefore multiplies the anchoring points between molecules. Ethane-$1,2$-diol has two and propane-$1,2,3$-triol has three: the network becomes so dense that it takes heating well beyond the boiling point of water to break it, hence $197$ and $290$ °C, and also the syrupy viscosity of the second. Ethane-$1,2$-diol, known as ethylene glycol, is the ANTIFREEZE of the cooling system: added to water, it strongly lowers the freezing point by a colligative effect while raising the boiling point, which protects the engine at both extremes, a harsh winter and a hot summer. It is, however, very toxic if swallowed, for a reason that exercise $10$ will clarify: the body oxidizes it to oxalic acid.

The equilibrium is $C_{6}H_{5}OH+H_{2}O\rightleftharpoons C_{6}H_{5}O^{-}+H_{3}O^{+}$, with constant $K_{a}=\frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}$. Since $pK_{a}=-\log K_{a}$, we have $K_{a}(\text{phenol})=1.0\times 10^{-10}$ and $K_{a}(\text{ethanol})=1.0\times 10^{-16}$. The ratio is $10^{16.0-10.0}=10^{6}$: phenol is ONE MILLION times more acidic than ethanol. This is not a nuance, it is a change of category, and this is why phenol must never be called an aromatic alcohol.

The phenoxide ion carries its negative charge on an oxygen bonded directly to the ring. The lone pair can therefore conjugate with the $\pi$ system: we draw a first form with the charge on the oxygen, then three forms in which the charge has moved onto the ring, at the ORTHO, ORTHO and PARA positions, which makes FOUR resonance forms in all (counting convention: one form per position of the charge, without distinguishing the two Kekule drawings of the ring). The charge is no longer localized on a single atom, it is spread over four atoms: the ion is markedly stabilized, hence easier to form, hence the corresponding acid is stronger. The ethoxide ion $CH_{3}CH_{2}O^{-}$ has no neighbouring $\pi$ system: its charge stays entirely on the oxygen, with no resonance stabilization whatsoever. The gap of six $pK_{a}$ units therefore comes entirely from the RESONANCE of the conjugate base, never from the starting acid. This is the general principle: we always compare the stability of the CONJUGATE BASES.

An alkyl group is a DONOR by the inductive effect. Each additional methyl therefore pushes electron density toward the charged oxygen of the alkoxide, which worsens a negative charge that is already there and DESTABILIZES the anion. The tertiary alcohol, surrounded by three donor groups, has the least stable conjugate base, so it is the weakest acid, with $pK_{a}=18.0$. In the aromatic series the reasoning is the same but with more powerful effects. The $-NO_{2}$ group in the para position is strongly WITHDRAWING, by the inductive effect and above all by the resonance effect: it opens a further resonance form that puts the charge on the oxygens of the nitro group, so it greatly stabilizes the anion and makes the $pK_{a}$ drop to $7.15$. The para methyl, on the contrary, is a donor: it destabilizes slightly, hence $10.3$. Pushed to the extreme with three nitro groups, we get picric acid at $pK_{a}=0.38$, that is a PHENOL more acidic than ethanoic acid, which finishes showing that the functional group alone decides nothing.

The rule is that an acid-base reaction is driven in the direction that produces the WEAKER acid, so an acid of a given $pK_{a}$ reacts with the conjugate base of an acid of larger $pK_{a}$. Against $NaOH$, the acid to consider is water, $pK_{a}=15.7$: phenol ($10.0$) and benzoic acid ($4.20$) are both stronger, so BOTH dissolve by forming their sodium salt. Against $NaHCO_{3}$, the acid to consider is carbonic acid, $pK_{a1}=6.35$: benzoic acid at $4.20$ is stronger, the reaction takes place and gas is seen escaping, $CO_{2}$; phenol at $10.0$ is WEAKER, the reaction does not take place and nothing happens. Hence the discriminating test, a classic of the laboratory: a compound that dissolves in sodium hydroxide BUT does not effervesce with hydrogen carbonate is a phenol; if it effervesces, it is a carboxylic acid. Note that $4$-nitrophenol, at $7.15$, still stays on the phenol side for this test, though only just.

Let $x=[H_{3}O^{+}]$ at equilibrium. With $K_{a}=\frac{x^{2}}{0.10-x}$ and the approximation $x\ll 0.10$, we get $x=\sqrt{K_{a}\times 0.10}=\sqrt{1.0\times 10^{-10}\times 0.10}=\sqrt{1.0\times 10^{-11}}=3.16\times 10^{-6}$ mol/L, so $pH=-\log(3.16\times 10^{-6})\approx 5.50$. Check of the approximation: $\frac{3.16\times 10^{-6}}{0.10}=3.2\times 10^{-5}$, that is $0.0032$ percent, far below the usual threshold of $5$ percent: the approximation is amply justified. The result deserves a comment, because it often surprises: a $0.10$ mol/L phenol solution has a pH of $5.5$, barely acidic, whereas an ethanoic acid solution of the same concentration is at $2.88$. Phenol means acidic COMPARED TO AN ALCOHOL, not acidic in absolute terms.

Acid hydration follows Markovnikov's rule: the proton attaches to the carbon with more hydrogens, which generates the most stable carbocation, here the SECONDARY carbocation rather than the primary one. Water then attacks this carbocation and a proton is lost: the major product is butan-$2$-ol. The counterexample is $3$-methylbut-$1$-ene: protonation gives a secondary carbocation, but a hydride from the neighbouring carbon, which is tertiary, immediately migrates to give a more stable TERTIARY carbocation. The major product is therefore not the expected $3$-methylbutan-$2$-ol but $2$-methylbutan-$2$-ol. This is the flaw of the method: as soon as a carbocation is free, it rearranges if that benefits it, and the alcohol obtained has a skeleton that is no longer that of the starting material.

Boron, less electronegative than hydrogen, carries the partial POSITIVE charge: it is therefore the electrophile, and it attaches to the LESS crowded carbon, that is the terminal carbon. Oxidation by hydrogen peroxide in basic medium then replaces the boron by $-OH$, keeping the position. The product is butan-$1$-ol. The two characteristics to quote are: ANTI-Markovnikov addition, and SYN addition, because the hydrogen and the boron arrive together on the same face of the plane of the double bond, through a four-centre cyclic transition state. There is never a carbocation, hence never a rearrangement.

The mercuric ion attacks the double bond but does NOT release an open carbocation: it forms a bridged MERCURINIUM ion, in which mercury bridges the two carbons. The positive charge is indeed more developed on the more substituted carbon, which directs the attack of water there and does give the Markovnikov orientation; but since no carbon carries a free, planar electron vacancy, no hydride or alkyl migration is possible. The final reduction by $NaBH_{4}$ replaces the mercury by a hydrogen. The lesson is general and also holds for the bromonium bridge in the addition of dibromine: a BRIDGED intermediate directs like a carbocation but does not rearrange like one.

By acid hydration, the proton attaches to carbon $2$ to generate the tertiary carbocation at carbon $1$, and water arrives there: we get $1$-methylcyclohexanol, a tertiary alcohol. By hydroboration-oxidation, boron attaches to the less substituted carbon, that is carbon $2$, and the $-OH$ takes its place: we get $2$-methylcyclohexanol. The stereochemistry is visible here because the ring freezes the faces: the hydrogen and the boron added SYN, so after oxidation the $-OH$ and the methyl end up on opposite sides of the ring. The product is TRANS-$2$-methylcyclohexanol, as a racemic mixture since the starting alkene was achiral and both faces were equally accessible. The two methods therefore give not only different positions but different classes of alcohol, tertiary on one side and secondary on the other.

The organomagnesium compound brings a carbanion $CH_{3}CH_{2}^{-}$ that adds to the carbonyl carbon; we must therefore count the carbons at the end and subtract the two that are supplied. For propan-$1$-ol, with three carbons and a primary alcohol, we need a one-carbon carbonyl that becomes primary: METHANAL $HCHO$. For butan-$2$-ol, with four carbons and secondary, we need a two-carbon ALDEHYDE: ethanal. For $2$-methylbutan-$2$-ol, with five carbons and tertiary, we need a three-carbon KETONE: propanone. The rule to memorize is therefore: methanal gives primary, an aldehyde gives secondary, a ketone gives tertiary. As for ethanol as solvent, it is a deal-breaker: the carbanion is an extremely strong base, the conjugate base of an alkane with $pK_{a}$ above $50$, and ethanol is an acid with $pK_{a}$ $16.0$. The acid-base reaction is therefore favoured by more than thirty orders of magnitude and instantaneous: the organomagnesium compound is destroyed into ethane and magnesium ethoxide before it has had time to meet the carbonyl. This is why we work in anhydrous ether or tetrahydrofuran, solvents without an acidic hydrogen, and why a mere trace of moisture makes the procedure fail.

A is butan-$2$-ol: four carbons in a chain, the $-OH$ on the second. B is $3,3$-dimethylbutan-$2$-ol: the main chain has four carbons, the $-OH$ is on carbon $2$, and carbon $3$ carries the two methyls that the figure shows pointing down. Both are indeed secondary alcohols. The order of reactivity is TERTIARY > SECONDARY > PRIMARY, and it can be read directly on the $E1$ mechanism: the slow step is the departure of water, which generates a carbocation, so the more stable this carbocation, the lower the barrier. A primary alcohol cannot form a reasonable carbocation and actually goes through a concerted $E2$ mechanism, which requires harsher conditions.

Butan-$2$-ol can lose a hydrogen from carbon $1$ or from carbon $3$. Toward carbon $1$ we obtain but-$1$-ene; toward carbon $3$ we obtain but-$2$-ene, in its two configurations $Z$ and $E$. ZAITSEV's rule predicts the MORE SUBSTITUTED alkene, therefore but-$2$-ene, and among its two configurations the $(E)$, which is less crowded. The enthalpies of hydrogenation demonstrate it instead of asserting it: hydrogenating these three alkenes leads to the same butane, so their difference in reaction enthalpy measures exactly their difference in starting stability. But-$1$-ene releases $127$ kJ/mol, so it started from higher; $(E)$-but-$2$-ene releases only $116$, it started from lower, it is more stable by $11$ kJ/mol. The $(Z)$ is intermediate at $120$, the steric strain between the two methyls on the same side costing it $4$ kJ/mol relative to the $(E)$. The stability order $(E)$-but-$2$-ene > $(Z)$-but-$2$-ene > but-$1$-ene is exactly the order of the observed proportions.

The major product is $2,3$-dimethylbut-$2$-ene. The mechanism unfolds in four stages. The oxygen is first protonated, which turns the poor leaving group $HO^{-}$ into the excellent leaving group $H_{2}O$. This water leaves and a SECONDARY carbocation remains on carbon $2$. This carbocation has as its immediate neighbour carbon $3$, which carries two methyls and no hydrogen: a METHYL then migrates with its electron pair toward the vacancy, a so-called $1,2$ shift. The charge ends up on carbon $3$, now surrounded by three carbons: the carbocation has become TERTIARY, hence markedly more stable. Loss of a proton in accordance with Zaitsev finally delivers the tetrasubstituted alkene $2,3$-dimethylbut-$2$-ene. The point to understand is that this rearrangement is PROOF: it would not be possible without a free carbocation, so observing it confirms the $E1$ mechanism on its own. A concerted mechanism could never have moved a methyl.

The equation is $2\,CH_{3}CH_{2}OH\rightarrow CH_{3}CH_{2}OCH_{2}CH_{3}+H_{2}O$, with $H_{2}SO_{4}$ at $140$ °C. The mechanism is a nucleophilic substitution $S_{N}2$: one ethanol molecule is protonated, and a SECOND ethanol molecule, acting as the nucleophile through its oxygen, attacks the carbon and expels water. Two limits follow. First, if two different alcohols are mixed in the hope of a mixed ether, each can protonate and each can attack: we obtain both symmetrical ethers AND the mixed ether, that is three products in similar proportions, which makes purification absurd. Second, with a secondary or tertiary alcohol, the carbon attacked is too crowded for an $S_{N}2$ and the substrate prefers to ionize and then eliminate: we recover the alkene, even at $140$ °C. The method is therefore usable only for a SYMMETRICAL ether from a PRIMARY alcohol, and the Williamson synthesis must be used in all other cases.

First way, distil the alkene as it forms: light alkenes boil far lower than the starting alcohol, since they no longer have hydrogen bonding, so they leave the flask as soon as they appear. Removing a product shifts the equilibrium in the forward direction, according to Le Chatelier's principle. Second way, trap the water formed, either by working in the presence of concentrated sulfuric acid, which is itself a dehydrating agent, or by using a Dean-Stark water separator. Note that these two ways act on the EQUILIBRIUM, not on the rate, whereas temperature acts on both at once, which explains the role of the $140$ versus $180$ °C threshold in the previous question.

With PCC, we obtain butanal, that is the ALDEHYDE. With dichromate in aqueous medium, we obtain butanoic acid, that is the CARBOXYLIC ACID. The difference comes down to water. In aqueous medium, the aldehyde formed partially hydrates to a geminal diol $R-CH(OH)_{2}$, and this diol still has a hydrogen on the carbon bearing the hydroxyls: it can therefore be oxidized a second time, all the way to the acid. PCC works in strictly anhydrous dichloromethane: without water, no hydrate, hence no second oxidation, and the reaction stops cleanly at the aldehyde. The formula to remember: it is not the oxidant that is milder, it is the MEDIUM that is dry.

Butan-$2$-ol, a secondary alcohol, gives butanone, a ketone, and stops there: the carbonyl carbon no longer carries any hydrogen, so nothing more can be removed without breaking a carbon-carbon bond. $2$-Methylpropan-$2$-ol, a tertiary alcohol, does NOT react. The reason is structural and can be said in one sentence: oxidizing an alcohol consists in removing the hydrogen from the oxygen AND a hydrogen carried by the carbon that bears the $-OH$, in order to create the $C=O$ double bond. In a tertiary alcohol, however, this carbon carries three carbons and therefore ZERO hydrogens. Oxidation is impossible, except under very violent conditions that break the skeleton and have no preparative interest.

The orange is that of the dichromate ion $Cr_{2}O_{7}^{2-}$, in which chromium is in the $+VI$ oxidation state; the green is that of the $Cr^{3+}$ ion, in the $+III$ state. The colour change therefore simply signals that the oxidant has been reduced, so that it has found something to oxidize. Among the four alcohols of formula $C_{4}H_{10}O$, butan-$1$-ol and $2$-methylpropan-$1$-ol are primary and butan-$2$-ol is secondary: these THREE give a positive test. $2$-Methylpropan-$2$-ol, tertiary, leaves the solution orange: negative test. The Jones test therefore distinguishes the tertiary from the other three, but does not separate primary from secondary, which is precisely what the Lucas test of the next question does: the two tests are complementary and are performed together.

The mechanism is an $S_{N}1$ nucleophilic substitution. The alcohol is first protonated by $HCl$, water leaves and a carbocation remains, then the chloride ion attaches to it to give the alkyl chloride. The rate is therefore entirely governed by the stability of the carbocation: immediate for a tertiary alcohol, slow for a secondary, zero in the cold for a primary, which would have to go through an inaccessible primary carbocation. $ZnCl_{2}$ is a LEWIS ACID: it coordinates to the lone pair of the oxygen, which further weakens the $C-O$ bond and speeds up the departure of the leaving group, exactly as $AlCl_{3}$ does in Friedel-Crafts reactions. The cloudiness, finally, is a matter of solubility and not of colour: the starting alcohol, with four carbons or fewer, is soluble in the aqueous mixture, whereas the alkyl chloride formed is not and separates as fine droplets that scatter light. What we observe is indeed the appearance of a SECOND PHASE, and this is why the test stops being readable beyond six carbons, where the alcohol itself is already no longer soluble.

Electron balance. In ethanol, the carbon bearing the $-OH$ is bonded to two hydrogens, one oxygen and one carbon, which is an oxidation state of $-1$; in ethanoic acid, this same carbon is bonded to two oxygens and one carbon, which is $+3$. Each ethanol molecule therefore gives up $4$ electrons, and three molecules give up $12$. On the other side, each chromium goes from $+VI$ to $+III$, that is $3$ electrons, and the dichromate ion contains two, so it gains $6$ electrons; two dichromate ions gain $12$. The equation is balanced. Along the way we check the other balances: $17$ oxygens on each side, $34$ hydrogens on each side, and a charge of $+12$ on each side. Calculation. The concentration in the exhaled air is $\frac{0.80}{2100}=3.81\times 10^{-4}$ g/L, which represents $\frac{3.81\times 10^{-4}}{46.07}=8.27\times 10^{-6}$ mol of ethanol per litre of air. The stoichiometry requires two dichromates for three ethanols, so $\frac{2}{3}\times 8.27\times 10^{-6}=5.51\times 10^{-6}$ mol of dichromate consumed per litre of exhaled air. This is a tiny amount, about half a milligram of chromium, and that is precisely what makes the device disposable and inexpensive; it is also why the device has to be calibrated on a well-defined volume of blown air, otherwise the reading means nothing.

A good leaving group is a species that copes well with the negative charge it carries away, in other words a WEAK BASE. The measure of this weakness is the $pK_{a}$ of the conjugate acid: the lower it is, the weaker the base, hence the better the leaving group. The conjugate acid of $HO^{-}$ is water, at $15.7$, which makes it a strong base and therefore a dreadful leaving group. The conjugate acid of $TsO^{-}$ is para-toluenesulfonic acid, at $-2.8$: the charge is delocalized there over the three oxygens of the sulfonate group, the anion is very stable and leaves without difficulty. The gap is $15.7-(-2.8)=18.5$ $pK_{a}$ units, that is about $18.5$ ORDERS OF MAGNITUDE in basicity. The iodide ion does even better with $-10$, its charge being diluted over a very large polarizable atom. This is why the prior transformation of the $-OH$ is not a laboratory refinement but an obligation.

In both cases, the first step is the same: the proton attaches to the oxygen and turns $HO^{-}$, a poor leaving group, into $H_{2}O$, an excellent leaving group. What follows is different. Butan-$1$-ol is primary: no primary carbocation is conceivable, so the bromide ion must attack from the back at the same time as water leaves, that is by $S_{N}2$. The transition state is crowded and the reaction requires CONCENTRATED $HBr$ and prolonged HEATING; the product is $1$-bromobutane. $2$-Methylpropan-$2$-ol is tertiary: water leaves spontaneously and leaves behind a stable tertiary carbocation, which the chloride ion captures at once. This is an $S_{N}1$, and it is so fast that shaking the alcohol with concentrated $HCl$ at room temperature is enough to see a second phase of $2$-chloro-$2$-methylpropane appear within seconds. This is exactly the basis of the Lucas test.

First step. Tosyl chloride reacts with the oxygen of the alcohol, and it is the $O-H$ bond that is broken, never the $C-O$ bond. The asymmetric carbon is therefore not involved at any point: its spatial configuration is entirely RETAINED. The product is $(S)$-butan-$2$-yl tosylate. Pyridine serves to neutralize the $HCl$ released, which would otherwise protonate the medium and cause side reactions. Be careful, the letter of the descriptor can change without anything moving in space, since the Cahn-Ingold-Prelog priorities are recalculated on the new substituents; we must therefore speak of retained configuration, and check the descriptor separately. Second step. The ethanoate ion now attacks the carbon from the back, by $S_{N}2$, and the tosylate leaves: the carbon undergoes WALDEN INVERSION, like an umbrella turning inside out. The product is $(R)$-butan-$2$-yl ethanoate. The net result of the sequence is therefore an inversion, and that is the whole point of the method: it gives a PREDICTABLE stereochemistry, which protonation by $HX$ on a secondary alcohol never guarantees.

$SOCl_{2}$ in pyridine and $PBr_{3}$ convert the alcohol into an alkyl chloride or bromide with INVERSION of configuration, because in both cases the halide nucleophile attacks from the back an intermediate in which the oxygen has been turned into a good leaving group, a chlorosulfite and a dibromophosphite respectively. The decisive advantage is that no free carbocation appears. On $2,2$-dimethylpropan-$1$-ol, known as neopentyl alcohol, this changes everything. This substrate is primary, so $HCl$ should do an $S_{N}2$; but its neighbouring carbon carries three methyls, a wall that almost totally blocks the approach from the back, and the $S_{N}2$ becomes thousands of times too slow. The substrate then prefers to ionize anyway, and the primary carbocation thus formed immediately rearranges by migration of a methyl into a tertiary carbocation: we recover a mixture of products with a modified skeleton, with a lot of alkene. With $PBr_{3}$, the reaction remains a clean substitution and the neopentyl skeleton is entirely preserved, even though it stays slow.

He reads an optical rotation that is practically ZERO. The substrate is a TERTIARY alcohol, so protonation is followed by the departure of water and the formation of a tertiary carbocation. A carbocation is PLANAR: the central carbon is $sp^{2}$, its three substituents lie in a plane and its vacancy is a $p$ orbital perpendicular to it, accessible from BOTH sides with the same probability. The bromide ion therefore attacks as often from one face as from the other and delivers the two enantiomers in equal amounts: the product is a RACEMIC mixture, whose opposite rotations cancel exactly. All the stereochemical information of the starting material was lost the instant the carbocation formed. In practice a very slight excess of the inversion product is observed, because the leaving group that has just departed still slightly shields the face it occupied, but the order of magnitude remains racemization. This is the basic difference with c): an $S_{N}2$ preserves the stereochemical information, an $S_{N}1$ destroys it.

Two reasons for choosing it. First, ether is APROTIC: it has no acidic hydrogen, so it does not destroy the Grignard reagent by an acid-base reaction, unlike water or an alcohol. Second, the two lone pairs of its oxygen coordinate to magnesium, which is a Lewis acid: two ether molecules come to surround the metal, which SOLVATES and stabilizes the reagent and keeps it in solution instead of letting it precipitate. Two hazards. Its boiling point is only $35$ °C and its vapours are denser than air: they creep along the bench and can reach an ignition source several metres away, hence a major risk of fire. In addition, on contact with air and light, ether slowly autoxidizes on the carbon next to the oxygen and forms PEROXIDES, which accumulate at the bottom of the bottle and are explosive on impact or heating. This is the reason for the absolute rule: never distil an old ether to DRYNESS, because it is precisely the peroxides, which are scarcely volatile, that concentrate in the residue.

First disconnection: phenoxide ion $C_{6}H_{5}O^{-}$ and bromoethane. The carbon attacked is primary, and the $S_{N}2$ proceeds without difficulty. Second disconnection: ethoxide ion and bromobenzene. It is ruled out, because the carbon to be attacked belongs to the aromatic ring, where neither the $S_{N}2$ nor the $S_{N}1$ is possible, for the reasons detailed in d). It remains to form the phenoxide. The equilibrium $C_{6}H_{5}OH+HO^{-}\rightleftharpoons C_{6}H_{5}O^{-}+H_{2}O$ has the constant $K=10^{pK_{a}(H_{2}O)-pK_{a}(\text{phenol})}=10^{15.7-10.0}=10^{5.7}\approx 5\times 10^{5}$: the deprotonation is complete, and sodium hydroxide is enough. For ethanol, the same constant is $10^{15.7-16.0}=10^{-0.3}\approx 0.50$: the equilibrium stays split, a good part of the alcohol is not deprotonated, and the remaining $HO^{-}$ ions also attack the haloalkane, giving an alcohol instead of the ether. An alkoxide must therefore be prepared by an IRREVERSIBLE reaction: sodium metal, $2\,CH_{3}CH_{2}OH+2\,Na\rightarrow 2\,CH_{3}CH_{2}O^{-}Na^{+}+H_{2}$, or sodium hydride $NaH$, whose dihydrogen escapes and cannot take the proton back. The rule to remember: a phenol is deprotonated by sodium hydroxide, an alcohol requires sodium or $NaH$; and the oxygen of the ether always comes from the partner whose carbon cannot be attacked.

The products are $2$-iodo-$2$-methylpropane and methanol. The mechanism begins with protonation of the ether oxygen, which becomes a good leaving group in the form of an alcohol. The bond that breaks is then the one on the side that gives the more stable carbocation: here the TERTIARY side, hence a tert-butyl carbocation, captured by the iodide ion. Methanol leaves as a neutral molecule. This is therefore an $S_{N}1$, and the cleavage takes place on the tertiary side, never on the methyl side. With ethoxyethane, neither side is favoured since the two sides are identical and their carbons bonded to the oxygen are primary: cleavage takes place by $S_{N}2$, the iodide attacking a primary carbon. We first obtain one molecule of iodoethane and one molecule of ethanol; with excess $HI$ and heat, the ethanol formed is in turn converted, so that the final balance is $CH_{3}CH_{2}OCH_{2}CH_{3}+2\,HI\rightarrow 2\,CH_{3}CH_{2}I+H_{2}O$. The general lesson: the cleavage mechanism is decided as always by the class of the carbon, $S_{N}1$ if one side can form a stable carbocation, $S_{N}2$ otherwise.

After protonation of the oxygen, two bonds could in principle break: the one to the methyl, and the one to the aromatic ring. The second is impossible for both available mechanisms. An $S_{N}2$ is ruled out there because the backside attack on an $sp^{2}$ carbon of the ring would require passing THROUGH the aromatic plane, which is geometrically impracticable, and because the negatively charged $\pi$ cloud repels the nucleophile. An $S_{N}1$ is ruled out because it would require a phenyl cation, in which the vacancy would be in an $sp^{2}$ orbital lying IN the plane of the ring, therefore perpendicular to the $\pi$ system and unable to draw the slightest stabilization from it: this is a cation of extreme instability. This is exactly the argument that makes aryl halides inert, seen in the alkyl halides chapter. Only the cleavage on the methyl side remains, by $S_{N}2$ on the most accessible carbon there is: the iodide carries off the methyl as iodomethane, and the aromatic fragment keeps the oxygen, giving phenol. We can therefore obtain neither iodobenzene nor methanol.

The amount of methoxybenzene is $n=\frac{10.8}{108.14}=9.99\times 10^{-2}$ mol, and the stoichiometry is one to one for each product. The maximum mass of phenol is $9.99\times 10^{-2}\times 94.11\approx 9.40$ g, and that of iodomethane $9.99\times 10^{-2}\times 141.94\approx 14.18$ g. Check of conservation. The reactants consumed weigh $10.8$ g of methoxybenzene plus $9.99\times 10^{-2}\times 127.91\approx 12.77$ g of $HI$, that is $23.57$ g in total; the products weigh $9.40+14.18=23.58$ g. The two totals agree to within rounding, which was expected since the molar masses themselves are conserved: $108.14+127.91=236.05$ and $94.11+141.94=236.05$. This check is worth doing systematically in an exam, because it detects in a few seconds an unbalanced equation or a molar mass copied wrongly. Finally, note that most of the mass produced is iodine: this is why this cleavage, very effective in the laboratory, has no industrial future.

First method, the action of a PERACID, typically meta-chloroperbenzoic acid denoted $mCPBA$, which transfers its terminal oxygen to the double bond in a single concerted step. Since the $\pi$ bond is never broken in two stages, the two carbons do not rotate relative to each other: the addition is SYN and STEREOSPECIFIC, a $cis$ alkene giving the $cis$ epoxide and a $trans$ alkene the $trans$ epoxide. Second method, in two steps: we first add $Br_{2}$ in the presence of water to obtain a halohydrin, then treat with a base. The base removes the proton of the alcohol, and the alkoxide thus formed attacks by INTRAMOLECULAR $S_{N}2$ the carbon bearing the bromine, which leaves. Like any $S_{N}2$, this ring closure requires the oxygen and the bromine to be anti, which fixes the geometry of the product.

An $sp^{3}$ carbon would like angles of $109.5$ degrees and an $sp^{3}$ oxygen similar angles. But a three-membered ring is necessarily a triangle, so its internal angles are $60$ degrees. The deviation is enormous, of the order of $50$ degrees, and it forces the orbitals to overlap at an angle instead of pointing at each other: the bonds are weakened and the ring stores about $114$ kJ/mol of strain, called Baeyer strain, to which is added the torsional strain due to the eclipsed hydrogens. This energy is already stored in the molecule and will be RELEASED as soon as the ring opens: opening is therefore strongly exothermic, which considerably lowers the transition state. This is why an epoxide reacts with nucleophiles where ethoxyethane, with no strain at all, stays inert. It is the same reasoning as for cyclopropane in the hydrocarbons chapter.

It is carbon a, the LESS substituted, that is attacked. In basic medium, there is no acid to protonate the oxygen: the nucleophile must force the departure of an alkoxide, a poor leaving group, and the only way to achieve this is a classic concerted $S_{N}2$, with backside attack. An $S_{N}2$ obeys steric crowding above all: it chooses the most open carbon, here carbon a, which carries only hydrogens, and not carbon b flanked by its two methyls. Methoxide therefore attaches to a, the ring opens and the resulting alkoxide is protonated by the methanol of the medium during workup. The product is $1$-methoxy-$2$-methylpropan-$2$-ol: the methoxy group on the primary carbon, the hydroxyl group on the tertiary carbon.

It is now carbon b, the MORE substituted, that is attacked, and the product is $2$-methoxy-$2$-methylpropan-$1$-ol: the positions of the methoxy and the hydroxyl are exactly swapped compared to c). The explanation lies in the prior protonation. The acid protonates the ring oxygen, which becomes an excellent leaving group, and this protonation weakens the two $C-O$ bonds so much that the ring begins to open BEFORE the nucleophile has even finished arriving. The transition state therefore acquires a strong carbocation character, with a notable partial positive charge on the carbon that supports it best, that is the TERTIARY carbon b. It is this partial charge that attracts methanol, a weak and neutral nucleophile. The ELECTRONIC factor then outweighs the steric factor and reverses the regioselectivity. A point that is often misunderstood deserves emphasis: despite this carbocation character, there is never a free carbocation, the attack is always from the back and the inversion at the attacked carbon is preserved. This pair of questions is the absolute classic of the chapter: same substrate, same nucleophile, two different products depending on the medium alone.

With $mCPBA$, we obtain the epoxide of cyclohexene; the acid hydrolysis that follows opens the ring by attack of water from the back, hence ANTI, and the two hydroxyls end up on opposite sides of the ring: this is TRANS-cyclohexane-$1,2$-diol. With cold dilute $KMnO_{4}$, or with $OsO_{4}$, the oxidant attaches to the double bond through a five-membered cyclic intermediate that delivers its two oxygens on the same side: the addition is SYN and gives CIS-cyclohexane-$1,2$-diol. On chirality, the $trans$ is CHIRAL: it has no internal plane of symmetry and exists as two enantiomers $(R,R)$ and $(S,S)$; since the starting cyclohexene was achiral and its two faces equivalent, we obtain a RACEMIC mixture, optically inactive by EXTERNAL compensation, but separable in principle by resolution. The $cis$, for its part, has an internal plane of symmetry that exchanges its two asymmetric carbons: it is a MESO compound, achiral despite having two stereogenic centres, and it is optically inactive by INTERNAL compensation, hence inseparable since there is nothing to separate. Two methods, two stereochemistries, and two situations of optical activity of completely different natures.

The degree of unsaturation is calculated by $\frac{2\times 4+2-10}{2}=\frac{10-10}{2}=0$, the oxygen being ignored since it is divalent. No unsaturation: neither double bond nor ring. All the molecules are therefore saturated chains, and the oxygen can occupy only two positions, at the end of a bond as $-OH$, or between two carbons as $-O-$. There are four alcohols: butan-$1$-ol, butan-$2$-ol, $2$-methylpropan-$1$-ol and $2$-methylpropan-$2$-ol. There are three ethers: ethoxyethane, $1$-methoxypropane and $2$-methoxypropane. That makes SEVEN constitutional isomers in all. Note that there is no stereoisomer to add, because butan-$2$-ol would be chiral but that is not an additional CONSTITUTIONAL isomer.

Sodium reacts only with an acidic hydrogen, therefore only with the $O-H$ of an alcohol, according to $2\,ROH+2\,Na\rightarrow 2\,RONa+H_{2}$. The effervescence of dihydrogen is therefore the alcohol versus ether test. A and C are alcohols, B is an ether. Lucas reagent then sorts the classes by the rate of the $S_{N}1$: immediate cloudiness for A, so it is a TERTIARY alcohol, hence $2$-methylpropan-$2$-ol, the only tertiary one on the list. Cloudiness in five minutes for C, so it is a SECONDARY alcohol, hence butan-$2$-ol, the only secondary one on the list; PCC converts it into butanone, and a ketone indeed does not reduce Tollens reagent, which confirms. For B, the only information is that it is not an alcohol: this leaves the THREE possible ethers, and the experiment described does not allow a choice among them. This is the point to stress: the absence of effervescence is a NEGATIVE result, which eliminates a family without designating a member, and a physical measurement, boiling point or spectrum, would be needed to decide. The honest answer is therefore “B is an ether of formula $C_{4}H_{10}O$, not determined further”.

D is an alcohol, since it gives hydrogen with sodium. It is neither tertiary nor secondary, since the Lucas test stays negative in the cold: it is therefore PRIMARY, which is confirmed by oxidation with PCC to an aldehyde, which does reduce Tollens reagent, giving the characteristic silver mirror. The list contains two primary alcohols: butan-$1$-ol and $2$-methylpropan-$1$-ol. The classic chemical tests do not distinguish them, because they have the same functional group and the same class. The simplest is a boiling point measurement, $118$ °C for butan-$1$-ol against $108$ for $2$-methylpropan-$1$-ol, branching lowering the boiling point as in exercise $1$. An equally good answer would be to invoke a spectrum, branching giving a very different signature, but the idea to retain is that at some stage wet-chemistry tests run out and a physical method must be used.

The task is to add ONE carbon to the chain, which is called a homologation, and going through a Grignard reagent is the reference route. Step $1$: $PBr_{3}$ transforms propan-$1$-ol into $1$-bromopropane. $PBr_{3}$ is preferred to $HBr$ to avoid any rearrangement, in accordance with exercise $6$. Step $2$: magnesium in anhydrous ether, which gives propylmagnesium bromide. Step $3$: addition to METHANAL, then acid hydrolysis. The propyl carbanion adds to the carbon of methanal, and after hydrolysis we obtain butan-$1$-ol, a primary alcohol with four carbons. A variant through the nitrile can also be proposed: $1$-bromopropane, then $NaCN$ giving butanenitrile, then acid hydrolysis to butanoic acid, then reduction by $LiAlH_{4}$ to butan-$1$-ol. It is longer but avoids the Grignard reagent, which may be what is asked.