Corrected practice final exam in organic chemistry (202-SNU-RE)

The product is $(R)$-$2$-azidobutane. The azide ion is an excellent nucleophile but a weak base, the substrate is secondary, and the solvent is POLAR APROTIC: every condition for an $S_{N}2$ reaction is met. The attack comes from the back side, hence the WALDEN INVERSION at the stereocenter. The role of dimethyl sulfoxide is decisive: it solvates the sodium cation very well through its negatively charged oxygen, but having no acidic hydrogen it cannot form a hydrogen bond with the ANION. The azide therefore stays bare, not wrapped in solvent, and its nucleophilicity is at its maximum. Note that the descriptor changes from $S$ to $R$, but this change of letter must be checked independently of the spatial inversion, since the Cahn-Ingold-Prelog priorities are recalculated on the new substituents.

The product is butan-$2$-ol, and the polarimeter would read an optical rotation of practically ZERO. There is no strong nucleophile and no base here: water and ethanol are weak nucleophiles, and the medium is protic and warm, which favours ionization. This is therefore a SOLVOLYSIS by an $S_{N}1$ mechanism. The secondary carbocation formed is PLANAR, so it can be attacked from both faces with the same probability, and the two enantiomers are obtained in equal amounts: a racemic mixture, optically inactive. In practice a slight excess of the inversion product is observed, because the bromide that has just left still shields a little the face it occupied, but the order of magnitude remains racemization. A little alkene would also be obtained through a competing $E1$ reaction.

Both treatments give an $E2$ elimination, since both reagents are strong bases. Treatment $4$, with ethanolate, which is small and unhindered, follows ZAITSEV'S RULE: the major product is but-$2$-ene, the more substituted and therefore more stable alkene, mainly in its $(E)$ form; but-$1$-ene accounts for only $20$ percent. Treatment $3$, with $2$-methylpropan-$2$-olate, which is very bulky with its three methyl groups, gives on the contrary $73$ percent BUT-$1$-ENE, the less substituted alkene. The explanation is steric and concerns the BASE, not the substrate: a bulky base can no longer reach the hydrogen on carbon $3$, which sits on a $CH_{2}$ that is crowded, whereas it easily reaches the hydrogens of the terminal methyl group, which are more exposed and three times as numerous. This is exactly the trade-off of the Hofmann elimination, except that the bulk is now supplied by the base and not by the leaving group: in both cases, once access to the hydrogen becomes the limiting factor, the stability of the product no longer decides.

Methanol is a polar PROTIC solvent. Its hydroxyl hydrogens form hydrogen bonds with the azide anion and surround it with a solvation shell that must be stripped off, at least partially, before the nucleophile can attack. This desolvation costs energy and adds to the activation barrier, so the effective nucleophilicity of azide collapses, typically by a factor of a thousand to ten thousand compared with dimethyl sulfoxide. Methanol would also favour ionization of the substrate, and therefore the competing $S_{N}1$ pathway, which would degrade the stereochemistry as well. The general rule to remember is that for an $S_{N}2$ reaction with an anionic nucleophile, one always chooses a polar APROTIC solvent.

The proton adds to a terminal carbon of the diene, which generates an ALLYLIC carbocation, stabilized by resonance. This cation has two resonance forms that place the positive charge on two different carbons, carbon $2$ and carbon $4$. The bromide can therefore attack either one: at carbon $2$ it gives $3$-bromobut-$1$-ene, the $1,2$ addition product; at carbon $4$ it gives $1$-bromobut-$2$-ene, the $1,4$ addition product. There are thus not two intermediates but a SINGLE one, whose charge is delocalized, and two possible points of attack. This is what makes the exercise instructive: the competition is not over the formation of the intermediate but over the second step.

KINETIC control describes the situation in which the major product is the one that forms the FASTEST, that is, the one with the lowest transition state; it prevails when the reaction is irreversible under the conditions used. THERMODYNAMIC control describes the situation in which the major product is the most STABLE, regardless of its rate of formation; it requires the reaction to be reversible and the equilibrium to have time to establish itself. At $-80$ degrees, the thermal energy is too low to cross the barrier back: the system is trapped on the first product formed, the $1,2$ addition, whose transition state is the lowest because the bromide attacks the nearest and most charged carbon. That is kinetic control. At $40$ degrees, the reaction becomes reversible: each product can give back the carbocation, and the system converges on the more stable one, the $1,4$ addition, whose internal double bond is disubstituted and therefore more stable than a terminal double bond. That is thermodynamic control. Experiment $3$ is the decisive proof of this reading: it shows that at high temperature the kinetic product CONVERTS into the thermodynamic product, which is possible only if the reaction is truly reversible. Without it, one could not tell a change of regime from a change of mechanism.

Position $1$ of naphthalene is the KINETIC position: the arenium ion that forms there is better stabilized by resonance, so its transition state is lower and the reaction is faster. But this position is crowded by the hydrogen at position $8$, which sits on the other ring and points straight at the incoming group, which destabilizes the PRODUCT. Position $2$ is the THERMODYNAMIC position: it forms more slowly, since its intermediate is less well stabilized, but the product obtained is uncrowded and therefore more stable. At $80$ degrees the sulfonation is practically irreversible and the $1$ isomer is recovered; at $160$ degrees it becomes reversible, equilibrium is established, and the $2$ isomer wins. Sulfonation lends itself particularly well to this demonstration because it is the only clearly reversible electrophilic aromatic substitution, a property that is in fact exploited to use it as a temporary blocking group.

The difference comes down to one word: REVERSIBILITY. In the three experiments above, the two products can interconvert, so the temperature can tip the system from one regime to the other and one can truly speak of kinetic or thermodynamic control. Nothing of the kind applies in the other two cases. The Zaitsev or Hofmann elimination is IRREVERSIBLE: an alkene that has formed does not give back the starting substrate under the reaction conditions, and the observed ratio reflects only the relative rates of formation, that is, the heights of the two transition states. The opening of an epoxide is irreversible in the same way, the ring strain released ruling out any return. In these two cases, the trade-off is therefore played ENTIRELY at the level of the transition state, between a steric factor and an electronic factor, and not between rate and product stability. Confusing the two situations is the most common conceptual error of the course: one speaks of kinetic or thermodynamic control ONLY when the products can interconvert, and this must be checked before using these words.

The order of decreasing acidity is: carboxyl $2.20$, then ammonium $9.11$, then phenol $10.07$. The carboxyl group is markedly more acidic than an ordinary carboxylic acid, $2.20$ against $4.8$, a difference of $2.6$ units and therefore a factor of $400$. The reason is the INDUCTIVE effect of the ammonium group carried by the immediately neighbouring carbon: it bears a full positive charge, so it strongly withdraws electrons and stabilizes the negative charge of the carboxylate formed. A better stabilized conjugate base means a stronger acid. This is the same reasoning as that applied to the chloroethanoic acids, with a far more powerful electron-withdrawing group than a halogen, since it is a full charge and not a mere polarization.

What must be compared is what happens to the CHARGE on deprotonation, not the groups taken in isolation. The ammonium ion goes from a charge of $+1$ to a charge of $0$: it GETS RID of a charge, which the solvent and the molecule welcome. The phenol goes from a charge of $0$ to a charge of $-1$: it CREATES a charge, which costs more. With comparable intrinsic stabilization, a deprotonation that neutralizes is therefore always easier than one that charges. A second factor adds to this, the electrostatic environment: in the zwitterionic form, the phenolate that would form would sit in a molecule that already carries a negative carboxylate, which destabilizes it by repulsion, whereas deprotonating the ammonium removes a positive charge in an environment that is already negative. The two effects go in the same direction and are enough to reverse the ranking that an isolated phenol and an isolated amine would give.

At $pH=1$, we are below all three $pK_{a}$ values: everything is protonated. The carboxyl group is in the $COOH$ form, the nitrogen is $NH_{3}^{+}$, the phenol is $OH$. The net charge is $+1$. At $pH=7$, we are above the first $pK_{a}$ but below the other two: the carboxyl group is deprotonated to $COO^{-}$, the nitrogen remains $NH_{3}^{+}$, the phenol remains $OH$. This is the ZWITTERION and the net charge is $0$. At $pH=12$, we are above all three $pK_{a}$ values: the carboxyl group is $COO^{-}$, the nitrogen is neutral $NH_{2}$, and the phenol is deprotonated to the phenolate $O^{-}$. The net charge is $-2$.

The isoelectric point is the pH at which the average net charge is zero. From part c), the form of zero charge is the zwitterion, and it is bracketed by the deprotonation of the carboxyl group below and that of the ammonium above. These are therefore the two $pK_{a}$ values to average: $pI=\frac{2.20+9.11}{2}=5.66$. The $pK_{a}$ of the phenol must not be involved, since it lies beyond and does not bound the neutral form. The rule is general: first identify the form of zero charge, then average the two $pK_{a}$ values that bracket it, never the first and the last mechanically.

The group reprotonated first is the PHENOLATE. The reasoning is that of any acid-base reaction: dissolved carbon dioxide gives carbonic acid, with $pK_{a1}=6.35$, and it can only protonate bases whose conjugate acid is WEAKER, that is, with a $pK_{a}$ above $6.35$. The phenolate, whose conjugate acid is at $10.07$, and the amine, whose conjugate acid is at $9.11$, both meet the condition; the carboxylate, at $2.20$, does not and stays deprotonated. Between the two candidates, the STRONGER base captures the proton first, hence the one whose conjugate acid has the higher $pK_{a}$: the phenolate at $10.07$, ahead of the amine at $9.11$. Bubbling $CO_{2}$ thus takes the molecule from charge $-2$ to charge $-1$, then to charge $0$ if it is continued, but never beyond.

It is the AMINE group that reacts, giving the corresponding amide, $N$-acetyltyrosine. Two conditions must be met for a group to attack an anhydride: it must be nucleophilic, and it must be available in unprotonated form. At $pH=9$, the amine is partly deprotonated since the pH is close to its $pK_{a}$ of $9.11$, so a notable fraction has its lone pair available. And an amine is a much better nucleophile than an alcohol or a phenol, nitrogen being less electronegative than oxygen and therefore holding its electrons less tightly. The other groups are ruled out for distinct reasons. The carboxylate is already deprotonated and negatively charged; it could attack, but it would give an unstable mixed anhydride that would revert, and it is in any case a mediocre nucleophile because its charge is delocalized over two oxygens. The phenol, at $pH=9$, is still mostly in its neutral form since its $pK_{a}$ is $10.07$, and a neutral phenol is a very weak nucleophile. This natural selectivity of nitrogen is what makes the specific acetylation of amines in the presence of alcohols possible, a technique used both to protect aniline before nitration and to label proteins.

The product is the secondary alcohol obtained by reducing the KETONE alone, the alkene and the ester remaining intact. First argument, that of the REAGENT: $NaBH_{4}$ is a mild reducing agent that reduces only aldehydes and ketones; it is too weakly nucleophilic to attack an ester, and it does not act at all on a carbon-carbon double bond, which is not electrophilic. Second argument, that of the SUBSTRATE: in the ester, the oxygen of the alkoxy group donates its lone pair by resonance to the carbonyl carbon, which markedly lowers its positive charge and makes it much less electrophilic than the carbon of a ketone, which has no donor neighbour. The two arguments agree, and a good exam answer gives both: a selectivity is always explained on the side of the reagent AND on the side of the substrate.

An excess of $LiAlH_{4}$ reduces the ketone AND the ester. The ketone becomes a secondary alcohol; the ester becomes a PRIMARY alcohol, with release of ethanol from the alkoxy part. The alkene is still not touched, because a hydride does not add to an unactivated $C=C$ bond. The product is therefore a diol that still carries the terminal double bond, $CH_{2}=CH-CH_{2}-CH(OH)-CH_{2}-CH_{2}OH$. Note that the molecule uses up three hydrides, one for the ketone and two for the ester, and that an excess is used anyway for the reasons seen in the chapter on ketones.

Catalytic hydrogenation over palladium reduces the $C=C$ double bond and that bond alone: the product is $CH_{3}CH_{2}CH_{2}-CO-CH_{2}-COOCH_{2}CH_{3}$. The two carbonyl groups are spared because hydrogenating a $C=O$ bond demands much harsher conditions: a more active catalyst such as Raney nickel, and high pressure and temperature. The reason is twofold: the $C=O$ bond is stronger than the $C=C$ bond, and above all a carbonyl adsorbs less well on the metal surface than an alkene, whose $\pi$ electrons coordinate easily to palladium. With this pair of treatments we therefore have a complementary pair of tools: $NaBH_{4}$ reduces the carbonyl without touching the alkene, and $H_{2}$ over palladium reduces the alkene without touching the carbonyl. Knowing how to choose between the two is exactly what an exam tests.

An excess of Grignard reagent attacks the ketone, which gives a tertiary alcohol, and also attacks the ester, which likewise gives a tertiary alcohol after DOUBLE addition. It is impossible to stop at the intermediate ketone derived from the ester because that ketone is MUCH more electrophilic than the ester it comes from: it no longer has a donor heteroatom to stabilize its carbonyl by resonance, and it is less hindered. It therefore reacts with a second equivalent faster than the ester itself is consumed, and this second addition is final since a ketone has no leaving group. This is the general situation of a sequence in which the intermediate is more reactive than the substrate: it can never be isolated. To obtain the ketone, one would have to use a deliberately milder reagent, a dialkylcuprate or a Weinreb amide. Finally, note that the Grignard reagent would in any case be destroyed by the acidic hydrogens discussed in part e).

Between the ketone and the ester, the molecule has a $CH_{2}$ group located ALPHA to BOTH carbonyl groups: this is a $1,3$-dicarbonyl motif. Its hydrogens therefore have a $pK_{a}$ close to $11$, since the anion formed delocalizes its charge onto both oxygens at once. Ethanolate, whose conjugate acid is at $16$, removes them QUANTITATIVELY, and a stabilized enolate forms. In a catalytic amount and with no other partner, one will mostly observe a rapid exchange of these hydrogens with the solvent, along with a possible intramolecular or intermolecular condensation of the Claisen or aldol type if conditions allow. The point to see is that none of the previous questions could be asked in a basic medium without taking this acidic site into account: the molecule is not only electrophilic at three places, it is also NUCLEOPHILIC at one.

By decreasing electrophilicity: the KETONE first, the ESTER next, and the ALKENE last, the latter not being electrophilic at all but nucleophilic. The general rule that the whole exercise illustrates is this: in a polyfunctional molecule, the product of a reaction is decided by the MEETING of the power of the reagent with the hierarchy of reactivity of the functional groups present. A mild reagent touches only the most reactive group and one obtains a selective transformation; a powerful reagent touches everything it can reach. Predicting a product therefore always means answering two questions in order: which functional groups is this reagent able to attack, then which of the groups present is the most reactive. And when the natural hierarchy does not give the desired result, there remains the solution of the next exercise: protect.

First reason, the most immediate: as soon as a molecule of Grignard reagent formed, its carbanion would attack the carbonyl of ANOTHER molecule of $4$-bromoacetophenone still present in the medium. One would obtain a disordered polymerization instead of the desired reagent, which is why a Grignard reagent is never prepared on a substrate that itself carries an electrophilic group. Second reason, independent of the first: even supposing the Grignard reagent formed, propanone would still have to be added, but the medium would already contain a ketone, that of the substrate, more reactive or at least competitive. No selectivity is possible. These two reasons come down to the same principle: a Grignard reagent is incompatible with any electrophilic functional group present in the same molecule or in the same flask.

The ketone is protected as a cyclic ACETAL, a dioxolane. The reaction is treatment with an excess of ethane-$1,2$-diol, with a trace of para-toluenesulfonic acid as catalyst, removing the water formed by means of a Dean-Stark apparatus. The diol is preferred to two molecules of a simple alcohol for two reasons: closing a five-membered ring is entropically favourable, which makes the protection more complete, and the later deprotection nevertheless remains easy since the equilibrium reverses in the presence of an excess of water. Recall that the acetal is a full ACETAL and not a hemiacetal, which is essential here: a hemiacetal keeps an $-OH$ and would reopen in a basic medium.

Step $1$: ethane-$1,2$-diol, catalytic para-toluenesulfonic acid, removal of water; the ketone is converted into the dioxolane and the aryl bromide is intact. Step $2$: magnesium in anhydrous ether; the aryl bromide gives the Grignard reagent, the protection having removed the only electrophilic group in the molecule. Step $3$: addition of propanone, then mild acid hydrolysis at the end of the reaction; the aryl carbanion adds to the carbonyl of propanone and, after protonation, the desired tertiary alcohol is obtained. Step $4$: acid hydrolysis, for example dilute $H_{3}O^{+}$ at room temperature, with an excess of water; the acetal reopens and gives back the ketone. The product is $2$-(4-acetylphenyl)propan-$2$-ol. Note that the final hydrolysis of step $3$ and the deprotection of step $4$ can be carried out in a single operation if sufficiently acidic conditions are chosen, which is in fact the usual practice.

An acetal no longer contains any carbonyl: its two oxygens are engaged in $C-O$ single bonds, exactly as in an ether, and the central carbon no longer carries any acidic hydrogen or any notable partial positive charge. There is therefore nothing for a carbanion to attack, and nothing for a base to remove. The acetal is thus perfectly inert toward Grignard reagents, organolithiums, hydrides and all strong bases. It is precisely this inertness that defines a good protecting group: it must go on easily, survive the intended transformation, and come off easily under conditions that do not damage the product. The acetal meets all three conditions, and the fact that it is sensitive only to an aqueous ACID medium, the opposite regime from that of the Grignard reagent, is what makes the strategy robust.

The strategy would work in principle, but it is clearly worse, and one must be able to say why. The positive point is that $NaBH_{4}$ would indeed reduce the ketone without touching the aryl bromide, and that a final oxidation with PCC or with the Jones reagent would restore the ketone from the secondary alcohol. The flaw is decisive: the alcohol obtained carries an ACIDIC hydrogen, with a $pK_{a}$ close to $16$. Now a Grignard reagent is the conjugate base of a hydrocarbon with a $pK_{a}$ above $50$: it would therefore be destroyed immediately by an acid-base reaction with this alcohol, releasing a hydrocarbon, before it had met anything. This alcohol would therefore have to be protected too, for example as a silyl ether, which adds two steps and ruins the point of the method. Second flaw, the final oxidation would have to be carried out in the presence of the TERTIARY alcohol created in the previous step; it is admittedly not oxidizable, having no hydrogen on its carbon, which saves the manoeuvre, but the margin is thin. The lesson is that a good protecting group does not merely mask the targeted reactivity: it must introduce NO new functional group liable to interfere, and that is exactly what the acetal does, leaving only an ether-type motif.

For $4$-BROMOANILINE: brominate first, then nitrate, then reduce. The bromine installed on benzene is an ORTHO-PARA director, despite its deactivating character, and the para position is by far the favoured one for steric reasons: nitration therefore gives mainly $1$-bromo-$4$-nitrobenzene, whose reduction provides $4$-bromoaniline. For $3$-BROMOANILINE: nitrate first, then brominate, then reduce. The nitro group is strongly deactivating and a META director, so bromination of nitrobenzene gives $1$-bromo-$3$-nitrobenzene, whose reduction provides $3$-bromoaniline. The two syntheses use exactly the same three reactions; only their order differs, and it entirely decides which isomer is obtained. The underlying rule is that each substituent already present DIRECTS the next: the order is therefore chosen so that the desired directing group is in place at the right moment.

Because the amino group would survive neither of the two remaining steps. In the strongly acidic nitration medium, aniline is protonated to the anilinium ion; now $-NH_{3}^{+}$ is a powerful deactivator and a META director, so an uncontrollable mixture would result, not to mention oxidation of the very electron-rich ring by nitric acid, which produces tars. In bromination, the amino group is such a powerful activator that the reaction does not stop at the monosubstituted product: bromine water gives $2,4,6$-tribromoaniline directly, even in the cold and without a catalyst. The nitrogen must therefore be introduced in a DEACTIVATED form, that is, the nitro group, all the ring chemistry carried out, and the conversion to an amine left to the very end. If an amine were absolutely needed at an intermediate stage, it would have to be protected by acetylation, which tempers its lone pair without removing it.

We go through the alkene. Step $1$: concentrated sulfuric acid at $180$ degrees Celsius, which dehydrates butan-$1$-ol to but-$1$-ene. Step $2$: hydration in aqueous $H_{2}SO_{4}$, which follows Markovnikov's rule and puts the hydroxyl group back on carbon $2$: butan-$2$-ol is obtained. No reaction moves a hydroxyl group directly from one carbon to the next, because such a transformation would require breaking a $C-O$ bond and forming another on a neighbouring carbon without passing through any intermediate, which no mechanism allows. The universal strategy is therefore to ERASE the positional information by forming an alkene, whose double bond covers both carbons at once, then to reinstall it where one wants by choosing the method of addition. Acid hydration for Markovnikov, hydroboration for anti-Markovnikov: this pair of tools makes it possible to target either carbon at will.

The pairing that works is $2$-methylpropan-$2$-olate with IODOMETHANE. The Williamson synthesis is an $S_{N}2$ reaction, so the carbon attacked must be accessible: here it is that of iodomethane, the most open there is, while the bulk is carried by the alkoxide, a mere nucleophile that attacks through a small oxygen atom. The reverse pairing, methoxide on $2$-iodo-$2$-methylpropane, would fail: the carbon to be attacked would be tertiary, surrounded by three methyl groups, and no $S_{N}2$ is possible there. But methoxide remains a strong base and the substrate has hydrogens in the beta position: an $E2$ ELIMINATION would therefore win, giving $2$-methylpropene and methanol instead of the ether. Hence the rule: in a Williamson synthesis, the bulk always goes on the ALKOXIDE side, and the halide must be primary or methyl.

Silver oxide has only one role: to exchange the iodide anion for the HYDROXIDE anion, the precipitation of silver iodide driving the exchange to completion. This exchange makes sense only on a QUATERNARY ammonium salt, the only species whose anion is a free counter-ion. If the starting amine were treated with $Ag_{2}O$ before any methylation, nothing useful would happen: the amine is not a salt, and it would at most be slightly basified. And if the methylation were stopped at an intermediate stage, one would obtain a neutral secondary or tertiary amine, which carries no charge and therefore has no leaving group: subsequent heating would cause no elimination, for lack of a nucleofuge. The whole sequence rests on the fact that the neutral group $-NR_{2}$ is a very poor leaving group whereas $-NR_{3}^{+}$ is excellent, the positive charge turning a strong base into a neutral amine at the moment of departure. Exhaustive methylation is therefore not a preparation, it is the ACTIVATION STEP of the leaving group, and it must necessarily come before everything else.

The general rule is this: in organic chemistry, each transformation changes the reactivity of the molecule for the next one. An installed substituent can DIRECT the next attack, as in a); it can MAKE IT IMPOSSIBLE by deactivating or protonating, as in b); it can ERASE a piece of positional information that will be reinstalled afterwards, as in c); it can decide which partner may play which role, as in d); or it can be the ACTIVATION without which nothing happens, as in e). A synthesis is therefore never a list of reactions to be carried out in any order: it is a SEQUENCE, in which each step prepares the state in which the next must find the molecule. This is precisely what a retrosynthesis problem tests, and it is why one solves it by working backward from the target rather than forward from the reagent.

The $M+1$ peak at $9.9$ percent gives $\frac{9.9}{1.1}=9$ carbons. The mass $150$ is EVEN, so the number of nitrogens is zero or even; the absence of an $N-H$ band and the mass balance will lead us to rule it out. Nine carbons weigh $108$, leaving $150-108=42$ units. With two oxygens, that is $32$, $10$ hydrogens remain: the formula $C_{9}H_{10}O_{2}$ weighs exactly $150$ and fits. The degree of unsaturation is $\frac{2\times 9+2-10}{2}=5$, which suggests an aromatic ring, which uses up four, plus one additional unsaturation. The absence of an $M+2$ cluster rules out chlorine and bromine.

The band at $1740$ cm$^{-1}$, accompanied by a strong band at $1235$ corresponding to the $C-O$ bond, and the total absence of any band between $3200$ and $3600$, which rules out alcohol and acid, point to an ESTER. The position $1740$ is very close to the reference value $1735$: the carbonyl is therefore NEITHER conjugated, which would have lowered it toward $1720$, NOR carried by a phenol, which would have raised it toward $1755$. We deduce valuable structural information even before looking at the NMR: the aromatic ring, whose presence is signalled by the bands at $1600$ and $1500$ and by the $C-H$ at $3035$, is attached directly NEITHER to the carbonyl NOR to the ester oxygen. It must therefore be separated from the functional group by at least one $sp^{3}$ carbon.

The total of the integrations is $3+2+5=10$ hydrogens, in agreement with the formula. The multiplet at $7.35$ integrating for $5$ hydrogens indicates a MONOsubstituted aromatic ring. The singlet at $2.10$ integrating $3$ is a methyl with no neighbour, whose shift corresponds exactly to a $CH_{3}$ carried by a carbonyl. The singlet at $5.10$ integrating $2$ is a $CH_{2}$ with no neighbour. Its shift, very far downfield for a methylene, is explained by the ACCUMULATION of two independent deshielding effects: it is carried by a carbon directly bonded to the ester OXYGEN, which already accounts for about $4$ ppm, and it is moreover BENZYLIC, and therefore subject to the ring current of the aromatic nucleus, which adds about $1$ ppm. A $CH_{2}$ that is only benzylic would come out near $2.6$ ppm, a $CH_{2}$ that is only bonded to an ester oxygen near $4.1$: it is indeed their sum that gives $5.10$. This additive reasoning on effects is very profitable on an exam.

Let us count the expected signals. A MONOsubstituted benzene has four inequivalent carbons: the carbon bearing the substituent, the two ortho carbons equivalent to each other, the two meta carbons equivalent to each other, and the para carbon. To these are added the carbonyl carbon, that of the $CH_{2}$ and that of the $CH_{3}$, for SEVEN signals in all. The spectrum shows exactly seven, which confirms it. The assignment is immediate: $170.9$ ppm for the ester carbonyl, a typical value in the $160$-$185$ range; $136.0$ for the substituted aromatic carbon; $128.5$, $128.2$ and $128.1$ for the aromatic carbons bearing hydrogen; $66.3$ for the $CH_{2}$ bonded to oxygen, a value well within the $50$-$90$ range; and $21.0$ for the methyl. The structure is BENZYL ETHANOATE, $C_{6}H_{5}CH_{2}-O-CO-CH_{3}$, also called benzyl acetate, an ester with a jasmine scent widely used in perfumery. All the data agree, including the remark in part b) that the ring had to be separated from the functional group by an $sp^{3}$ carbon.

The fragment at $43$ is the ACYLIUM cation $CH_{3}CO^{+}$, obtained by cleavage of the bond between the carbonyl and the oxygen; it is stabilized by delocalization of the charge between carbon and oxygen and is the most commonplace fragment of any acetylated compound. The fragment at $91$ corresponds to $C_{7}H_{7}^{+}$, obtained by BENZYLIC cleavage of the bond between the $CH_{2}$ and the oxygen; it immediately rearranges to the TROPYLIUM ion, a seven-carbon ring carrying six delocalized $\pi$ electrons, hence AROMATIC in the Hückel sense with $n=1$. This aromaticity gives it exceptional stability, and that is why this peak generally dominates the spectrum. The fragment at $108$ corresponds to the loss of $42$ units, that is, one molecule of KETENE $CH_{2}=C=O$, by a rearrangement with hydrogen transfer; what remains is benzyl alcohol, of mass $108$, exactly the alcohol from which the ester derives. Note the consistency of the whole: the three fragments separately point to the three pieces of the molecule, the acetyl group, the benzyl group and the parent alcohol.

Dibromine is polarized as it approaches the $\pi$ cloud of the alkene, and the bromine bearing the partial positive charge attacks, forming a bridged BROMONIUM ion, in which the bromine atom bridges the two carbons. This intermediate is cyclic and completely blocks one face of the former double bond. The bromide ion released must therefore attack from the OTHER face, in an $S_{N}2$ fashion on one of the two carbons of the bridge. The addition is consequently ANTI, that is, the two bromines add on opposite sides of the plane of the former double bond. This is a STEREOSPECIFIC addition: the stereochemistry of the product is entirely determined by that of the reactant.

Osmium tetroxide adds to the double bond forming a five-membered cyclic ester, in which the two oxygens are bonded to the same osmium atom and therefore necessarily arrive on the SAME side. The subsequent reductive hydrolysis releases the two hydroxyl groups without touching the carbon-oxygen bonds. The addition is therefore SYN, the two $-OH$ groups ending up on the same side of the plane of the former double bond. Cold dilute permanganate proceeds through the same type of cyclic intermediate and gives the same stereochemistry. This is also a stereospecific addition, but of the opposite sense to the previous one.

Reagent I on $(E)$-but-$2$-ene: the two methyl groups are on opposite sides, and anti addition places the two bromines also on opposite sides; the product is MESO $2,3$-dibromobutane, which has an internal plane of symmetry exchanging its two stereocenters. Reagent I on $(Z)$-but-$2$-ene: the two methyl groups are on the same side, and anti addition this time leads to the pair of enantiomers $(R,R)$ and $(S,S)$, obtained in equal amounts since the bromonium ion can form on either face of the achiral alkene: it is a RACEMIC mixture. Reagent II on $(E)$-but-$2$-ene: syn addition reverses the result, and one obtains the racemic mixture of the butane-$2,3$-diols $(R,R)$ and $(S,S)$. Reagent II on $(Z)$-but-$2$-ene: one obtains MESO butane-$2,3$-diol. The table of the four cases is therefore organized very simply: for a given stereochemistry of addition, the two alkenes give opposite results, and for a given alkene, the two reagents give opposite results.

The polarimeter would read ZERO in all four cases, but for two reasons of an entirely different nature, and that is the heart of the question. For the two MESO products, the cancellation is INTERNAL: each molecule contains two stereocenters of opposite configuration, and the internal plane of symmetry makes the molecule superimposable on its mirror image. It is therefore ACHIRAL, individually inactive, and no manipulation can ever draw optical activity from it. For the two RACEMIC mixtures, the cancellation is EXTERNAL: each molecule taken separately is indeed chiral and optically active, but the sample contains exactly as much $(R,R)$ as $(S,S)$, whose opposite rotations cancel. The distinction is essential and comes up regularly on exams: a zero reading never proves by itself that a compound is achiral.

The products that can be separated are the two RACEMIC MIXTURES, that is, the $2,3$-dibromobutane from $(Z)$-but-$2$-ene and the butane-$2,3$-diol from $(E)$-but-$2$-ene. Each contains two real enantiomers that can in principle be separated, which is called a RESOLUTION. The two meso products cannot be separated because there is nothing to separate: they are single pure compounds, not mixtures. The method of resolution rests on the fact that two enantiomers have identical physical properties in an achiral medium, but different ones in a CHIRAL medium. The mixture is therefore reacted with an enantiomerically pure chiral reagent, which turns the pair of enantiomers into a pair of DIASTEREOMERS, which have different solubilities and melting points and are separated by fractional crystallization; the auxiliary reagent is then removed. On a diol, one would typically use a chiral acid to form two diastereomeric esters. Modern methods more often use chromatography on a chiral stationary phase, which exploits exactly the same principle but without a chemical reaction. It must be remembered that it is IMPOSSIBLE to resolve a racemate by achiral means alone.

First reaction, SAPONIFICATION: the triglyceride, which is a triester, is hydrolyzed in a basic medium to glycerol and three potassium carboxylates. Each triglyceride molecule therefore consumes exactly three hydroxide ions. This reaction counts the ESTER groups, and so indirectly measures the SIZE of the molecule: for a given mass, the shorter the chains, the more molecules per gram and the more potassium hydroxide is needed. Second reaction, the ADDITION of diiodine across the carbon-carbon double bonds, each $C=C$ bond taking up exactly one molecule of $I_{2}$. This reaction counts the UNSATURATIONS. The two assays are therefore complementary and independent: one gives the size, the other the degree of unsaturation.

The saponification value is $190$ mg of potassium hydroxide per gram of fat, that is, $0.190$ g per gram. One mole of triglyceride needs three moles of potash, so the relation is $SV=\frac{3\times M(KOH)\times 1000}{M}$, where the factor $1000$ converts grams to milligrams. The factor $3$ comes directly from the stoichiometry: three ester groups per molecule, hence three potash. We get $M=\frac{3\times 56.11\times 1000}{190}=\frac{168330}{190}=886$ g/mol. This is an AVERAGE molar mass, since a natural fat is never a pure compound but a mixture of triglycerides with varied chains; the value therefore measures a weighted average, which is more than enough to characterize an oil.

For $100$ g of fat, the amount of substance is $\frac{100}{886}=0.1129$ mol. The mass of diiodine taken up being $86$ g, this represents $\frac{86}{253.81}=0.3389$ mol of diiodine. The average number of double bonds per molecule is therefore $\frac{0.3389}{0.1129}=3.00$. The value is remarkably close to a whole number, which is an excellent sign of consistency: the fat contains on average exactly THREE double bonds per molecule. Since it has three chains, this strongly suggests that each chain carries a single one, rather than an uneven distribution, even though the assay alone cannot decide between the two hypotheses. Note that a non-integer result would have been nothing unusual, being an average over a mixture.

An average molar mass of $886$ g/mol and three double bonds distributed one per chain point to TRIOLEIN, the triester of glycerol with three oleic acids, whose exact molar mass is $885.5$ g/mol. The main fatty acid is therefore OLEIC ACID, with eighteen carbons and one double bond of cis configuration, written $C18{:}1$. The corresponding fat is very probably OLIVE OIL, whose actual iodine value lies between $80$ and $88$ and whose saponification value is around $190$, in excellent agreement with the measurements.

An unchanged saponification value means that the average molar mass is the same, so the LENGTH of the chains has not changed: we are still dealing with eighteen-carbon chains. The iodine value falling from $86$ to $5$ means that the double bonds have practically disappeared: the average number goes from $3.00$ to about $0.17$. The size has therefore been kept and the unsaturation removed, which corresponds to a HYDROGENATION of the chains: oleic acid has become stearic acid, with eighteen carbons and no double bond. The physical consequence is immediate. The straight saturated chains stack in parallel with maximum contact, so the London forces are strong and the melting point high: the sample is SOLID at $20$ degrees Celsius, it is a fat. The first sample, whose cis chains carry a kink that prevents tight stacking, is on the contrary LIQUID: it is an oil. The pair of indices thus makes it possible to follow an industrial hydrogenation without ever isolating anything.

First advantage, the measurement is NON-DESTRUCTIVE and continuous: the solution is placed in the polarimeter cell and the rotation is read at regular intervals on the same sample, which directly gives a complete kinetic curve. A chemical assay would on the contrary require withdrawing a sample, quenching the reaction and titrating at each point, which consumes material and introduces an error at each withdrawal. Second advantage, the measurement is SPECIFIC to the transformation being followed: sucrose rotates by $+66.5$ degrees and the equimolar mixture of glucose and fructose by $-19.7$, so the signal changes not only in magnitude but in SIGN, which makes the progress of the reaction readable without complicated calibration and without ambiguity. A third advantage can be added: polarimetry does not disturb the equilibrium, whereas assaying the reducing power with Fehling's solution would consume the glucose formed and shift the reaction.

Classic two-step route: benzene is first sulfonated with fuming sulfuric acid, which gives benzenesulfonic acid; an alkali fusion follows, that is, heating with solid sodium hydroxide at about $300$ degrees Celsius, then acidification, which replaces the sulfonate group by a hydroxyl and releases phenol. An alternative four-step route goes through nitration, reduction to aniline, diazotization and then hydrolysis of the diazonium salt in hot water. The first is shorter and is the one expected here. Note that no direct nucleophilic substitution is conceivable on benzene, an aromatic carbon undergoing neither $S_{N}1$ nor $S_{N}2$: a particular leaving group, sulfonate or diazonium, must therefore be installed, and that is the whole trick.

The nitration is written $C_{6}H_{5}OH+HNO_{3}\rightarrow HO-C_{6}H_{4}-NO_{2}+H_{2}O$. The hydroxyl group is a resonance donor, hence strongly ACTIVATING and an ORTHO-PARA director: $2$-nitrophenol and $4$-nitrophenol are obtained, and no meta isomer. Dilute, cold nitric acid is used for two reasons. First because phenol is already very activated: it does not need sulfuric acid to generate enough nitronium ion, and the nitrating mixture would lead to polynitration that is hard to stop, as far as picric acid. Second because the very electron-rich phenol ring is easily OXIDIZED by concentrated nitric acid, which produces tars and destroys the yield. The mildness of the conditions is therefore a necessity here and not a convenience, exactly as for aniline.

The ortho isomer carries its nitro group immediately next to the hydroxyl group, at a distance that allows an INTRAMOLECULAR hydrogen bond between the phenol hydrogen and an oxygen of the nitro group, closing a six-membered pseudo-ring. This hydrogen is then used up inside the molecule and is no longer available to link two molecules together. The para isomer, whose two groups are diametrically opposed, cannot form any internal bond; its molecules therefore associate with one another through a network of INTERmolecular hydrogen bonds that must be broken to vaporize the compound. Hence a much greater volatility for the ortho isomer, which melts at $45$ degrees Celsius and distils with steam, against $114$ degrees for the para isomer, which stays in the flask. This is exactly the reasoning used on the dimer of carboxylic acids, applied in the opposite direction: what counts is not the number of possible hydrogen bonds, but whether they are internal or external.

Next step, the REDUCTION of the nitro group to an amine: dihydrogen over palladium, or tin in hydrochloric acid followed by basification, which gives $4$-aminophenol. Last step, ACYLATION with one equivalent of ethanoic anhydride, in a slightly buffered aqueous medium: this gives $N$-($4$-hydroxyphenyl)ethanamide, that is, paracetamol. The complete synthesis from phenol therefore takes three steps: nitration, reduction, acylation.

Both functional groups are nucleophilic, but NITROGEN is much more so than the phenol oxygen, for two cumulative reasons. First, nitrogen is less electronegative than oxygen: it holds its electrons less tightly, so its lone pair is more available to attack an electrophilic carbon. Second and above all, the lone pair of the phenol is CONJUGATED with the aromatic ring, where it is permanently delocalized, which makes it distinctly less accessible; the lone pair of the aromatic amine is too, but the amine remains the better nucleophile of the two all the same. The difference in rate is large enough for the acylation to be selective with a single equivalent of anhydride and under mild conditions. Note that the selectivity would disappear if an excess of anhydride and a basic catalyst were used, which would acetylate the phenol as well: the observed selectivity is KINETIC and therefore depends on the conditions, which must be specified.

The amount of phenol is $\frac{50.0}{94.11}=0.5313$ mol. The stoichiometry is one to one at each step, so the theoretical amount of paracetamol would be the same. The overall yield is the PRODUCT of the three yields: $0.60\times 0.95\times 0.88=0.5016$, that is, $50.2$ percent. The amount actually obtained is therefore $0.5313\times 0.5016=0.2665$ mol, a mass of $0.2665\times 151.17=40.3$ g. The expected comment concerns the multiplication of yields: three steps none of which is poor, the best even reaching $95$ percent, leave barely half of the product in the end. This is the central difficulty of any multistep synthesis, and the reason why one always tries to reduce the NUMBER of steps as much as to improve each one: a ten-step synthesis at $90$ percent per step yields only $35$ percent.

The degree of unsaturation is $\frac{2\times 10+2-12}{2}=5$: four for the aromatic ring and one for the carbonyl. The two obvious fragments are, on one side, the BENZYL or phenylethyl group coming from the aromatic ring, and on the other, the METHYL KETONE group $CH_{3}CO-$. The most natural bond to cut is the one that separates them, that is, the bond between carbon $2$ and carbon $3$ of the chain, or the one between carbon $3$ and carbon $4$, depending on the method considered. Recognizing the methyl ketone motif is decisive here, because it immediately points to the acetoacetic ester synthesis of the table.

The bond cut is the one between the carbon alpha to the ketone and the benzyl group. Step $1$: treat ethyl $3$-oxobutanoate with sodium ethanolate; with a $pK_{a}$ of only $10.7$, the deprotonation is quantitative and gives an enolate stabilized over two carbonyl groups. Step $2$: add benzyl bromide; the anion undergoes an $S_{N}2$ reaction on this primary, benzylic and particularly reactive substrate. Step $3$: hydrolyze the ester in hot aqueous acid, then heat; the resulting beta-keto acid decarboxylates spontaneously through a six-membered cyclic transition state, and $4$-phenylbutan-$2$-one is obtained. Count: benzyl bromide contributes SEVEN carbons, the acetoacetic ester synthesis adds THREE according to the table, which makes ten, exactly the target. Checking by counting carbons is the quickest test of a retrosynthesis proposal and should be done systematically.

This time the bond cut is the one between the two central carbons of the chain. Step $1$: condense BENZALDEHYDE with PROPANONE in dilute basic medium, then heat; after aldol addition and then dehydration (aldol condensation), $4$-phenylbut-$3$-en-$2$-one, known as benzylideneacetone, is obtained. This crossed aldol reaction is workable because benzaldehyde has NO alpha hydrogen: it cannot give an enolate and is condemned to the role of electrophile, while propanone alone can supply the nucleophile. Only one combination is therefore possible instead of four. Step $2$, indispensable: hydrogenate the $C=C$ double bond of the product with $H_{2}$ over palladium. This choice of reagent is crucial and is the whole point of the question: dihydrogen over palladium reduces the alkene without touching the carbonyl, whereas a hydride such as $NaBH_{4}$ would do exactly the opposite and ruin the synthesis.

The bond cut is the one between the carbonyl and the chain. Step $1$: prepare $2$-phenylethylmagnesium bromide from $(2$-bromoethyl$)$benzene and magnesium in anhydrous ether. Step $2$: add ETHANAL, then hydrolyze in acid medium; $4$-phenylbutan-$2$-ol, a secondary alcohol, is obtained. Step $3$, indispensable: oxidize this secondary alcohol to a ketone, with PCC or with dichromate in acid medium. The choice of oxidant is immaterial here, a ketone being impossible to overoxidize, unlike a primary alcohol, where PCC would be required. A Grignard reagent can never deliver a ketone directly from an aldehyde: it always gives an alcohol, which is why an oxidation step is added almost systematically to this route.

A Friedel-Crafts route would consist in acylating benzene with $3$-oxobutanoyl chloride or an equivalent, in order to install the chain directly on the ring. It is less convenient for two reasons. First, acylation would place the carbonyl DIRECTLY on the ring, which would give a conjugated aromatic ketone, whereas the target has its carbonyl three carbons away from the ring: one would therefore have to acylate with a four-carbon acyl chloride and then selectively reduce the aromatic carbonyl to $CH_{2}$ by Clemmensen or Wolff-Kishner, in the presence of the other carbonyl, which would have to be protected. Second, a direct Friedel-Crafts alkylation with a three-carbon halide would be ruled out, since it proceeds through a carbocation and would rearrange, as the chapter on aromatic compounds established. The route therefore exists, but it requires protection, acylation, selective reduction and deprotection, where the other three take two or three steps.

The ACETOACETIC route is the best. It has three steps, two of which are often done without isolating the intermediate, it starts from cheap and readily available reagents, and above all none of its steps poses a selectivity problem: the deprotonation is quantitative, the alkylation takes place on a very reactive benzylic halide, and the decarboxylation is spontaneous and irreversible since it releases a gas. The ALDOL route is also short, only two steps, and starts from even cheaper reagents; its drawback is the hydrogenation step, which requires pressure equipment and a choice of reagent where an error is fatal, and the fact that the dehydration must be carried through completely, or undehydrated aldol will be left. The GRIGNARD route is the weakest of the three: it has three steps, one of them an oxidation, it requires strictly anhydrous conditions, and its starting reagent, $(2$-bromoethyl$)$benzene, is considerably more expensive than benzyl bromide. The deciding criterion in synthesis is never the number of steps alone: it is the combination of the number of steps, the cost of the starting reagents, and the number of chances to go wrong on selectivity.

PARTLY CORRECT. The first sentence is strictly true: a catalyst lowers the activation barrier of both directions of the reaction in the same proportion, so it speeds up the approach to equilibrium without ever changing its position, which depends only on thermodynamics. The conclusion, however, is FALSE, and for a reason that has nothing to do with catalysis. CONCENTRATED sulfuric acid is not only a catalyst: it is also a powerful DEHYDRATING AGENT, which takes up the water formed by the esterification. By removing a product from the medium, it shifts the equilibrium in the forward direction according to Le Chatelier's principle, and the yield really does increase. Correction: the catalyst as such shifts nothing, but concentrated sulfuric acid plays TWO distinct roles here, and it is the second that improves the yield. A trace of para-toluenesulfonic acid, a pure non-dehydrating catalyst, would speed up the reaction without changing the yield at all.

FALSE, and the error concerns the second species. The three conditions cited, cyclic, planar and fully conjugated, are necessary but NOT sufficient: the fourth and last of Hückel's conditions is missing, the numerical one, namely having $4n+2$ delocalized $\pi$ electrons. The tropylium ion $C_{7}H_{7}^{+}$ has SIX $\pi$ electrons spread over seven carbons, which gives $4n+2=6$ so $n=1$: it is indeed AROMATIC, which explains its exceptional stability and the intensity of the peak at $m/z=91$ in mass spectrometry. The cyclopentadienyl cation $C_{5}H_{5}^{+}$ has only FOUR $\pi$ electrons, that is, $4n$ with $n=1$: it is therefore ANTIAROMATIC, that is, less stable than its open-chain analogue, and is in fact extremely unstable. It is the cyclopentadienyl ANION $C_{5}H_{5}^{-}$ that is aromatic, with its six $\pi$ electrons, and its aromaticity explains the remarkably low $pK_{a}$ of cyclopentadiene, close to $16$ whereas an alkane exceeds $50$. Confusing the cation with the anion is the error here, and it completely reverses the conclusion.

PARTLY CORRECT. The first sentence is correct: an $S_{N}2$ reaction proceeds by back-side attack and gives a clear, predictable inversion, whereas an $S_{N}1$ goes through a planar carbocation that can be attacked from both faces, hence racemization. The conclusion, however, is FALSE because it reverses an implication. A carbocation does produce a racemate, but a racemate does not prove a carbocation. There is a second, entirely different source of racemic mixtures: the CREATION of a stereocenter from achiral reactants in an achiral medium. Three examples from the course are enough to establish it. The addition of $HCN$ to butan-$2$-one gives a racemic cyanohydrin although the mechanism is a nucleophilic addition without the slightest carbocation, simply because the carbonyl is planar and can be attacked from both faces. The hydroboration-oxidation of an achiral alkene gives a racemic alcohol although it is syn and concerted. The opening of an achiral epoxide by a nucleophile also gives a racemate by $S_{N}2$. Correction: a racemic mixture signals a PLANAR INTERMEDIATE OR SUBSTRATE that can be attacked from both sides, which includes the carbocation but is not limited to it.