Corrected exercises: organic synthesis and retrosynthetic analysis (202-SNU-RE)

It is read “could come from” and links a TARGET to its PRECURSORS: it goes back in time. Three differences from a reaction arrow. Direction: it starts from what you want, not from what you have. Pieces: they are SYNTHONS, idealized fragments carrying a charge, which exist in no bottle. Status: it asserts nothing, it proposes a hypothesis to be validated. Hence the two shafts. The expected habit: you analyze on scratch paper with double-shafted arrows, then CLIMB BACK UP the tree in the opposite direction to write the synthesis, with single arrows, real reagents and conditions. A paper that hands in the analysis as it stands gives the reader a string of impossible reactions, and the examiner marks it as such.

The SYNTHON is the idealized fragment produced by the disconnection, with the charge assigned to it; the SYNTHETIC EQUIVALENT is the REAL reagent that plays its role. $CH_{3}^{-}$: methylmagnesium bromide. $C_{6}H_{5}CH_{2}^{+}$: benzyl bromide, whose carbon is $\delta^{+}$ and carries an excellent leaving group. $^{-}CN$: sodium cyanide, where the anion really exists. $CH_{3}CO^{+}$: ethanoyl chloride with $AlCl_{3}$, or ethanoic anhydride. The pitfall is mixing up the two columns: writing $CH_{3}^{-}$ in a synthesis sequence is a mistake, $CH_{3}MgBr$ is expected. Second point, decisive for what follows: at every carbon-carbon disconnection you must CHOOSE which piece carries the negative charge, and that choice decides which one will be the Grignard reagent.

The target is $C_{6}H_{5}-CH(OH)-CH_{2}-CH_{2}-CH_{3}$, that is $C_{10}H_{14}O$. Degree of unsaturation: $DU=\frac{2\times 10+2-14}{2}=4$, that is the three double bonds of the ring plus the ring itself. Oxygen does not enter the formula for the degree of unsaturation, so the hydroxyl group contributes nothing. This is the check that lets you, faced with an unknown molecular formula, assert the presence of a benzene ring without having seen the structure: $DU=4$ with at least six carbons is almost always its signature. Above all, remember that the target has TEN carbons, because the whole job of the next questions is to find them again in the two pieces of each disconnection.

Disconnection 1, between the ring and the carbon bearing the $OH$: the synthons are $C_{6}H_{5}^{-}$ and the cation of the functional carbon, whose synthetic equivalent is BUTANAL, a carbonyl being the universal equivalent of a carbocation carrying a future $OH$. The bond is closed by adding phenylmagnesium bromide to butanal, then acidic hydrolysis. Count: $6+4=10$ ✓. Disconnection 2: synthons $CH_{3}CH_{2}CH_{2}^{-}$, equivalent propylmagnesium bromide, and BENZALDEHYDE; same reaction; $7+3=10$ ✓. The two routes are equally good and the choice is made on the price of the reagents. The carbon count is the fastest check of a retrosynthesis: it is done in your head, before any mechanism, and it eliminates impossible proposals at a stroke.

Disconnection 3 separates an eight-carbon fragment, $C_{6}H_{5}CH(OH)CH_{2}-$, from a two-carbon one, so $8+2=10$. Closing it would require a carbanion, or a carbocation, in the BETA position of an alcohol. No reaction in the course makes one: a Grignard reagent needs a halogen on the targeted carbon, and the neighbouring alcohol would destroy it by an acid-base reaction as soon as it formed; a carbocation there would be stabilized neither by the ring, too far away, nor by the oxygen, which is not on the right carbon. General criterion: a disconnection is acceptable only if EACH piece has a real synthetic equivalent AND a known reaction joins the two back together. Failing to check this is costly, because you sometimes notice it three steps later.

Butyl ethanoate is $CH_{3}-COO-CH_{2}CH_{2}CH_{2}CH_{3}$, that is $C_{6}H_{12}O_{2}$. You disconnect the carbon-oxygen bond on the ACYL side, the one that esterification can close: the pieces are ethanoic acid, two carbons, and butan-1-ol, four carbons, and $2+4=6$ ✓. You close it by a Fischer esterification. Watch out for the other carbon-oxygen disconnection, on the ALKYL side: it would give the ethanoate ion and a butyl carbocation, and would be closed by an $S_{N}2$ reaction between sodium ethanoate and 1-bromobutane. It exists, but it starts from an alkyl halide instead of an alcohol, so it costs one more step upstream. Two disconnections, two real routes, a trade-off based on what you have available.

1-Bromobutane comes from butan-1-ol: you disconnect the carbon-bromine bond and close it with $PBr_{3}$, or with hot concentrated $HBr$, or through the tosylate. Count: $4+0=4$, the skeleton has not moved by a single carbon. This is the important point: it is not a true disconnection but a functional group INTERCONVERSION, a step that changes the functional group without touching the chain. It is almost always done to prepare the next one, here to be able to make a Grignard reagent, which requires a halogen and not an alcohol. Counting the carbons at each step tells the two apart immediately: a true disconnection splits them between two pieces, an interconversion leaves them all on the same side.

2-Phenylbutan-2-ol is $C_{6}H_{5}-C(CH_{3})(OH)-CH_{2}CH_{3}$, that is $C_{10}H_{14}O$, $DU=4$. It is a TERTIARY alcohol whose functional carbon carries three DIFFERENT groups: THREE carbon-carbon disconnections are therefore acceptable. Disconnecting the ethyl group leaves acetophenone, to be joined back with ethylmagnesium bromide: $8+2=10$. Disconnecting the methyl group leaves 1-phenylpropan-1-one, with methylmagnesium bromide: $9+1=10$. Disconnecting the phenyl group leaves butan-2-one, with phenylmagnesium bromide: $4+6=10$. All three are closed by the same reaction, addition of the Grignard reagent to the ketone followed by acidic hydrolysis, and all give a tertiary alcohol. The class check doubles the count check: a Grignard reagent on a ketone always gives a tertiary alcohol.

The figure shows the two best branches. You keep ACETOPHENONE and ethylmagnesium bromide. The criterion is not the elegance of the disconnection but the AVAILABILITY of the starting materials: acetophenone is a bulk industrial chemical, a by-product of the cumene process, hence very cheap, and bromoethane is one of the cheapest haloalkanes. The branch butan-2-one plus phenylmagnesium bromide is almost as good. The branch propiophenone plus methylmagnesium bromide is the weakest: propiophenone is not a common catalogue chemical, it would have to be prepared by Friedel-Crafts acylation, which adds a step UPSTREAM. A route is judged on the whole tree, not on its last step.

First condition: a known reaction closes the bond, and each piece has a real synthetic equivalent. Second condition: the disconnection leads back to AVAILABLE compounds, and if possible it splits the molecule into two pieces of comparable size. Detaching a single carbon from a ten-carbon target hardly advances the problem, whereas cutting in the middle divides it in two. So you look for the most CENTRAL disconnection among those that are chemically acceptable, and you favour reactions that create carbon-carbon bonds: they are the only ones that build the skeleton, all the others merely change the functional groups.

You count the bonds from the functional carbon to oxygen. Alkane: zero. Alcohol: one. Aldehyde: two. Acid: three. Carbon dioxide: four. The order is therefore alkane, alcohol, aldehyde, acid, $CO_{2}$. A KETONE sits on the same rung as the aldehyde, and this equality explains two facts. First fact: the same reducing agent, $NaBH_{4}$, takes both of them down one rung to the alcohol. Second fact: no ordinary oxidizing agent takes a ketone up, because it would have to pull off a hydrogen that the ketone does not have on its functional carbon. This is why the oxidation test distinguishes an aldehyde from a ketone so well, and why a tertiary alcohol is not oxidized further.

Steps up from a primary alcohol: to the aldehyde with PCC in dichloromethane; from the aldehyde to the acid with $K_{2}Cr_{2}O_{7}$ in aqueous $H_{2}SO_{4}$ or with $KMnO_{4}$. Steps down: from the acid to the alcohol with $LiAlH_{4}$, $NaBH_{4}$ not being strong enough; from the aldehyde or ketone to the alcohol with $NaBH_{4}$. The difference between PCC and aqueous dichromate lies in the WATER: in aqueous medium the aldehyde hydrates to a geminal diol, whose remaining hydrogen on the carbon is immediately reoxidized, and the reaction runs on to the acid. PCC works in anhydrous medium, the hydrate cannot form, and the oxidation stops. Mixing them up costs you the whole product: you aimed at the aldehyde, you isolate the acid.

2-Phenylethanoic acid is $C_{6}H_{5}-CH_{2}-COOH$, that is $C_{8}H_{8}O_{2}$. NITRILE route: benzyl bromide plus sodium cyanide, an $S_{N}2$ reaction on a primary, benzylic carbon and therefore very fast, giving phenylacetonitrile; then hydrolysis in hot aqueous acid. GRIGNARD route: benzyl bromide plus magnesium in anhydrous ether, then solid carbon dioxide, then acidic hydrolysis. Count in both cases: $7+1=8$ ✓. Both routes lengthen the chain by EXACTLY one carbon, supplied by the cyanide in one case and by the carbon dioxide in the other. Practical difference: the Grignard reagent is destroyed by the slightest trace of water, the nitrile route is more robust but needs a long hydrolysis.

The nitrile opens onto three functional groups. Towards the CARBOXYLIC ACID, by acidic hydrolysis when hot, or basic hydrolysis followed by acidification, the amide being the intermediate. Towards the PRIMARY AMINE, with $LiAlH_{4}$, which gives $R-CH_{2}-NH_{2}$: the nitrile carbon ends up in the chain of the amine. Towards the KETONE, with a Grignard reagent followed by acidic hydrolysis, the Grignard reagent adding to the nitrile carbon to give an imine salt that aqueous acid hydrolyzes. This third route is the only one of the three that lengthens the chain AGAIN, since it brings in the carbon of the Grignard reagent as well. The nitrile is thus both a way to add one carbon and a disguise for three functional groups.

Three steps. One, toluene treated with N-bromosuccinimide and a peroxide, or with dibromine under light: RADICAL bromination of the side chain, which gives benzyl bromide. Two, sodium cyanide: phenylacetonitrile. Three, hot acidic hydrolysis: 2-phenylethanoic acid. Direct oxidation of toluene with hot permanganate is not suitable, and the carbon count is enough to say so: it gives BENZOIC acid, seven carbons, whereas the target has eight. Side-chain oxidation does not stop where you would like, it razes everything that sticks out past the benzylic carbon. To obtain a carboxyl separated from the ring by a $CH_{2}$, you have to build, not oxidize.

$-OCH_{3}$: oxygen donates a lone pair to the ring by resonance, an effect that outweighs its inductive withdrawal, hence strongly ACTIVATING, ORTHO and PARA directing. $-CH_{3}$: inductive donor only, weakly activating, ortho and para. $-Cl$: strong inductive acceptor but resonance donor, hence WEAKLY deactivating and yet ortho and para directing. This is the exception to know, and it is understood by separating the two questions: the inductive effect decides the RATE, resonance decides the POSITION, because only resonance stabilizes the carbocation intermediate of ortho or para attack. $-COCH_{3}$, $-NO_{2}$ and $-SO_{3}H$: acceptors, DEACTIVATING, META directing. The grid in the solution ranks the six groups.

You must ACYLATE FIRST, then nitrate. First reason, orientation: the acetyl group is meta directing, so it sends the nitro group to position 3, which is the target. Second reason, far more compelling: by nitrating first you get nitrobenzene, too deactivated to undergo any Friedel-Crafts reaction at all. The sequence is $CH_{3}COCl$ with $AlCl_{3}$, then $HNO_{3}$ and $H_{2}SO_{4}$. Note the hierarchy: the second reason alone is enough to rule out the reverse order, because even with the right orientation the reaction would not take place. Molar mass of target 1, $C_{8}H_{7}NO_{3}$: $8\times 12.01+7\times 1.008+14.01+3\times 16.00=165.15$ g/mol.

You must BROMINATE FIRST. Bromine directs ortho and para, so it sends the acyl group to the para position, position 4. The reverse order would give 1-(3-bromophenyl)ethanone, the meta isomer, since the acetyl group directs meta: you would get a product, but not this one, and on an exam that is a wrong answer, not an approximate one. If the Friedel-Crafts reaction still works on bromobenzene, it is because a halogen is only WEAKLY deactivating. The limit is sharp: acylation succeeds on a ring carrying any group up to and INCLUDING the halogens and fails as soon as the group deactivates more, nitro, cyano, carbonyl, sulfonic acid, ammonium. The orders of magnitude say it: bromobenzene ten times slower than benzene, nitrobenzene a million times. Target 2, $C_{8}H_{7}BrO$: $199.04$ g/mol.

The two targets differ only in the directing effect of the first group, and it is that effect that imposes everything. RULE: the group that must DIRECT is installed FIRST, and a Friedel-Crafts step goes as early as possible, before any strongly deactivating group. The habit, in front of two steps, is to ask “can I swap them?” and to answer on TWO levels, orientation first, feasibility second. Useful corollary: for a meta product you install a meta director first, for a para product you install an ortho-para director first. A reversed order costs the whole question, even if every reaction written exists, because the product obtained is not the one that was asked for.

Aniline fails for a reason of a different nature. Aluminum chloride is a Lewis acid and the lone pair on nitrogen is a Lewis base: the two combine immediately. Nitrogen then carries a positive charge and becomes a strongly WITHDRAWING group, hence deactivating and meta directing: the most activated ring in the repertoire has deactivated itself on contact with the reagent that was supposed to attack it. The classic remedy is to protect nitrogen as an amide, by acetylating aniline: the amide is far less basic, no longer complexes $AlCl_{3}$ and remains a moderate ortho-para activator; you then deprotect by hydrolysis. Do not confuse the two failures: nitrobenzene is deactivated BEFORE the reaction, aniline deactivates itself BY the reagent.

Route A: $0.92\times 0.88\times 0.95\times 0.90=0.6922$, that is $69.2$ percent. Route B: $0.85\times 0.78=0.663$, that is $66.3$ percent. The overall yield is the PRODUCT of the yields, and the reason is physical: each step works only on what the previous one has left it, and the material lost in the first step never comes back. We write $R=\prod R_{i}$. The mental check to keep: the overall yield is always LOWER than the smallest of the step yields, here $69.2<88$ and $66.3<78$. Any answer that violates this inequality is wrong without having to redo the calculation.

Route A wins, $69.2$ against $66.3$, even though it has TWICE AS MANY steps. This is the surprising result to remember: the number of steps decides nothing by itself, the yields decide. That said, comparing the yields alone does not exhaust the choice of a route: route A costs two more purifications, two more days of work, twice as much solvent and two more chances to go wrong. The gap of $2.9$ points is thin, and a chemist may quite properly keep route B for those reasons, provided he names them. What would be wrong is to keep it in the belief that it gives more.

$n(P)=\frac{50.0}{106.17}=0.4709$ mol. The stoichiometry being one mole for one mole, $n(\text{target})=0.4709\times 0.6922=0.3260$ mol, hence $m=0.3260\times 164.20=53.5$ g. Order-of-magnitude check, to be made systematically: the target is $\frac{164.20}{106.17}=1.55$ times heavier than the starting material and about thirty percent is lost, so we expect $50\times 1.55\times 0.69\approx 53$ g ✓. The classic pitfall is to apply the yield to the MASS of starting material: you would find $50.0\times 0.6922=34.6$ g, which forgets that the two molecules do not have the same molar mass. The yield applies to MOLES.

The first three steps give $0.92\times 0.88\times 0.95=0.7691$. With a last step at $0.72$: $0.7691\times 0.72=0.5538$, that is $55.4$ percent, and route A is then clearly beaten. The threshold is computed directly: we look for $x$ such that $0.7691\,x=0.663$, hence $x=0.862$, that is $86.2$ percent. As soon as this last step falls below $86.2$ percent, route B takes the lead back. This threshold is HIGH, and that is the useful information: the lead of route A hangs by a thread, and a single step that degrades a little is enough to reverse the ranking.

Average reported: $\frac{92+88+95+90}{4}=91.25$ percent. Correct calculation: $n(\text{target})=\frac{100}{164.20}=0.6090$ mol, hence $n(P)=\frac{0.6090}{0.6922}=0.8798$ mol and $m=0.8798\times 106.17=93.4$ g. With his average: $\frac{0.6090}{0.9125}=0.6674$ mol, that is $70.9$ g. He would commit $70.9$ g instead of $93.4$ g and would collect $\frac{70.9}{93.4}=75.9$ percent of what he aimed for, that is about $75.9$ g of target instead of $100$ g. To remember: the average does not always get the RANKING wrong, but it ALWAYS gets the value wrong, and the error grows with the number of steps. Ten steps at $90$ percent: the average says $90$, the product is $0.9^{10}=0.349$, that is $34.9$ percent.

$NaBH_{4}$ in methanol, cold: it reduces aldehydes and ketones, it does not touch esters or acids. You obtain ethyl 4-hydroxypentanoate, which by the way readily closes to a five-membered lactone. $LiAlH_{4}$, for its part, does not sort: it would also take the ester down to the primary alcohol and you would isolate pentane-1,4-diol. The reason for this selectivity is the ELECTROPHILICITY of the targeted carbon: in an ester, the singly bonded oxygen feeds the carbonyl by resonance and depletes it much less than a ketone carbonyl does. Rule: between two groups of the same type, you rank by electrophilicity and choose the mildest reagent that suffices.

PCC in dichloromethane. Aqueous acidic dichromate would not stop at the aldehyde: it would hydrate and the reaction would run on to pent-4-enoic acid. Hot permanganate would do worse, it would CLEAVE the terminal double bond and leave a diacid or an acid plus carbon dioxide. PCC is therefore doubly selective here: it stops at the right rung and it leaves the alkene alone. Vocabulary to hold firmly. CHEMOselectivity chooses between two DIFFERENT functional groups. REGIOselectivity chooses between two POSITIONS of the same transformation, as with Markovnikov. STEREOselectivity chooses between two stereoisomers of the same product, as with the syn addition of hydroboration.

The Grignard reagent is an extremely strong base: its carbon carries a partial negative charge and pulls off any slightly acidic proton. An alcohol is an acid, with a $pK_{a}$ near $16$. As soon as a molecule of Grignard reagent forms, it is destroyed by the hydroxyl of another substrate molecule; and the alkoxide obtained attacks its own bromine by an intramolecular $S_{N}2$ reaction through a five-membered ring, a ring that closes very easily, which gives tetrahydrofuran. Two remedies. PROTECT the alcohol, as a silyl ether or a tetrahydropyranyl ether, then deprotect: cost, two steps. Or REVERSE the roles at the moment of the disconnection and have the ELECTROPHILE carry the oxygen, by offering oxirane to a simple Grignard reagent: this remedy costs NO step, because it is decided on paper.

Three qualities. It goes on in very good yield, under mild conditions, on the targeted group and on that group alone. It is INERT under all the conditions of the steps it must survive. It comes off selectively, under conditions that the rest of the molecule tolerates. The acetal fulfils the contract against a Grignard reagent for a structural reason: it carries NO acidic hydrogen, its two oxygens being bonded to carbons, and it offers NO electrophilic site, its central carbon being saturated and no longer having a double bond to oxygen. No proton to pull off, no carbon to attack. And it comes off by simple aqueous acidic hydrolysis, conditions that a Grignard reagent never meets before the end.

$0.94\times 0.96=0.9024$: protection applies a factor of $90.2$ percent, nearly ten points lost, before even counting the two purifications, the two days and the solvents. You refuse to pay that price in two cases. When a selective reagent exists, and questions a) and b) illustrate it: protecting the ester before reducing the ketone would be a mistake of method, not just a needless expense. And when the synthesis is already long, because each added pair of steps MULTIPLIES the overall yield again: two protections in a six-step route cost it twenty points. The question before protecting is always the same: have I exhausted the list of selective reagents?

Mistake in the CARBON COUNT. Propanone supplies three carbons and propylmagnesium bromide three more: the product would be 2-methylpentan-2-ol, a TERTIARY alcohol with six carbons, whereas the target has five and carries a secondary alcohol. Two independent checks would have been enough. The count: $3+3=6$ and not $5$. The CLASS: a Grignard reagent on a ketone always gives a tertiary alcohol, on an aldehyde a secondary one, on methanal a primary one, so aiming at a secondary alcohol requires an aldehyde. Corrections: ethanal plus propylmagnesium bromide, $2+3=5$; or butanal plus methylmagnesium bromide, $4+1=5$; or, with no disconnection at all, reduction of pentan-2-one by $NaBH_{4}$, by far the shortest if the ketone is available.

Mistake in the DISCONNECTION, that is in the choice of charge. The synthon $CH_{3}CH_{2}CO^{-}$, the acyl anion, has no simple synthetic equivalent at the CEGEP level: no common reagent carries a negative charge on a carbon already doubly bonded to oxygen, because the natural polarity of that carbon is the opposite, it is $\delta^{+}$. The right analysis disconnects the SAME bond but assigns the charges the other way round: $CH_{3}CH_{2}^{-}$, hence ethylmagnesium bromide, and $CH_{3}CH_{2}CO^{+}$, whose equivalent is propanenitrile, which gives an imine salt hydrolyzed to the ketone. Safer variant, in two stages: propanal plus ethylmagnesium bromide gives pentan-3-ol, which is then oxidized with PCC. The wrong choice of charge cannot be recovered.

Mistake of INCOMPATIBLE CONDITIONS. A TERTIARY alcohol heated in concentrated sulfuric acid does not esterify, it DEHYDRATES: the stabilized tertiary carbocation forms immediately, then loses a proton and gives the most substituted alkene, here 2-methylbut-2-ene according to Zaitsev. You would collect an alkene and water, not an ester. Correction: avoid the strongly acidic medium and use a powerful acylating agent under mild conditions, ethanoyl chloride or ethanoic anhydride in the presence of pyridine; no carbocation then forms and the tertiary ester is obtained cleanly. Principle to remember: before writing a step, you check that its conditions are compatible with ALL the functional groups present.

Mistake about the NATURE of the transformation. A catalyst speeds up the approach to equilibrium, it never SHIFTS it, and Fischer esterification is a limited transformation. Starting from an equimolar mixture, the extent of reaction satisfies $\frac{x^{2}}{(1-x)^{2}}=K=4$, hence $\frac{x}{1-x}=2$ and $x=\frac{2}{3}$: the yield levels off at $66.7$ percent, whatever the amount of sulfuric acid and whatever the heating time. To go higher you must shift the equilibrium: a large excess of one reagent, removal of water as it forms with a Dean-Stark apparatus, or a switch to the anhydride or the acyl chloride, whose reactions are COMPLETE. Confusing kinetics with thermodynamics makes you accept an unworkable plan.

Mistake in the DISCONNECTION, the most serious: you disconnect only bonds that a reaction can FORM, and no reaction in the course installs an alkyl group on a non-functionalized carbon in the middle of a chain. There is no reagent that goes and puts a methyl group on carbon 3 of a butanoic acid. The right disconnection is the $C_{1}-C_{2}$ bond: the five carbons divide into a fragment of FOUR carbons and the carboxyl carbon, supplied by carbon dioxide. You start from 1-bromo-2-methylpropane, make its Grignard reagent in anhydrous ether, add it to solid carbon dioxide, and hydrolyze in acidic medium. Count: $4+1=5$ ✓. Variant: $NaCN$ then acidic hydrolysis, same count.

FALSE, and the assignment gives three counterexamples. Target 1 of exercise 4 requires acylating BEFORE nitrating, otherwise the second step does not take place at all. Target 2 requires brominating before acylating, otherwise you get the meta isomer instead of the para: both reactions do take place, but the product is not the right one. And exercise 6 shows the third case, that of a group incompatible with a later reagent, which must be protected or installed later. Correct sentence: the order of the steps is a datum of the problem on the same footing as the reagents, and it is decided on two criteria, orientation and feasibility.

FALSE, it is the PRODUCT of the yields. Numerical counterexample: three steps at $90$, $80$ and $70$ percent have an average of $80$ percent and an overall yield of $0.90\times 0.80\times 0.70=0.504$, that is $50.4$ percent, nearly thirty points lower. The reason is physical: each step works only on what the previous one has left it. The gap grows with the number of steps, ten steps at $90$ percent giving $34.9$ percent. Correct sentence: the overall yield is the product of the step yields, and it is always lower than the smallest of them. Whoever computes an average commits too little material and misses the target.

FALSE, and the difference runs deeper than a direction of reading. The retrosynthetic arrow links a target to its PRECURSORS and is read “could come from”; the pieces it writes are SYNTHONS, fictitious species carrying a charge, which are found in no bottle. A reaction arrow links real species under real conditions. It is to forbid the confusion that it is given TWO shafts. Correct sentence: the retrosynthetic arrow is an analysis tool, not an equation; once the tree is built, you CLIMB BACK UP it in the opposite direction to write the synthesis, with single arrows, real reagents and conditions.

FALSE. It fails on any ring carrying a group more deactivating than a halogen: nitro, cyano, carbonyl, sulfonic acid, ammonium. It also fails on aniline, yet one of the most activated rings, because aluminum chloride binds to the lone pair on nitrogen and turns it into a strongly deactivating ammonium. What is TRUE in the statement is that it still works on halobenzenes, which are weakly deactivated, which makes target 2 of exercise 4 possible. Correct sentence: the Friedel-Crafts reaction requires a ring at least as electron-rich as a halobenzene, and it therefore goes as early as possible in a sequence.

FALSE, and exercise 5 gives the numerical counterexample: four steps at $92$, $88$, $95$ and $90$ percent give $69.2$ percent, against $66.3$ percent for two steps at $85$ and $78$. The route that is twice as long gives more. What is TRUE is that at EQUAL yields the shorter one always wins, and that one step fewer means one purification, one day, one volume of solvent and one chance of error fewer. Correct sentence: the number of steps is one criterion among others, together with the overall yield, the price of the starting materials and the number of chances of going wrong on a selectivity.

Only two steps. First step, a DISCONNECTION: the carbon-oxygen bond of the ester, on the acyl side, which gives a carboxylic acid and ethanol, and which esterification closes. Second step, a FUNCTIONAL GROUP INTERCONVERSION: the amino group comes from the nitro group by reduction, and no bond of the skeleton is touched. As the two steps are independent, they can be chained in BOTH orders, which gives two syntheses from the same catalogue chemical. Route A: 4-nitrobenzoic acid, then ethyl 4-nitrobenzoate, then benzocaine. Route B: 4-nitrobenzoic acid, then 4-aminobenzoic acid, then benzocaine. The tree does not decide between them, it is the CONDITIONS of each step that will.

Step 1: 4-nitrobenzoic acid, large excess of ethanol, a few drops of concentrated sulfuric acid, reflux for several hours, then neutralization and recrystallization; you isolate ethyl 4-nitrobenzoate. Step 2: reduction of the nitro group to the amine with $H_{2}$ over palladium on carbon, or with tin in hydrochloric acid followed by neutralization with carbonate, or with tin(II) chloride. The ester fears none of these reducing agents, and that is exactly the advantage of order A: the harshest step, esterification in hot acidic medium, takes place while the molecule does NOT yet carry an amine. General rule: aggressive conditions as early as possible, fragile or basic groups last.

Two precautions. The aromatic amine is basic, the $pK_{a}$ of the anilinium ion being about $4.6$: in sulfuric acid it is fully protonated, so it consumes a whole equivalent of acid and you must commit an EXCESS so that enough is left to catalyze. And the product is first isolated as an ammonium salt, which must be freed with a weak base, sodium carbonate, without going as far as hydrolyzing the ester; route B is in fact the classic laboratory preparation. On the other hand, replacing the Fischer esterification by an acyl chloride would be a mistake of CHEMOSELECTIVITY: the nitrogen of an amine is more nucleophilic than the oxygen of an alcohol, so you would collect an amide, even a polyamide.

Overall yield: $0.88\times 0.91=0.8008$, that is $80.1$ percent. Amount committed: $n=\frac{25.0}{167.12}=0.1496$ mol. Amount of benzocaine: $0.1496\times 0.8008=0.1198$ mol. Mass: $m=0.1198\times 165.19=19.8$ g. Order-of-magnitude check: the target is very slightly lighter than the starting material, $165.19$ against $167.12$, so the expected mass lies a little below $80$ percent of $25.0$ g, that is a little under $20$ g ✓. The pitfall is the one from exercise 5, applying the yield to masses instead of moles; here the two molar masses are so close that the error would go unnoticed, which makes it all the more dangerous.

If you NITRATE first, the Friedel-Crafts acylation becomes impossible, nitrobenzene being too deactivated, and the sequence stops there. If you ACYLATE first, the acetyl group is a META director, so nitration would place the nitro group at position 3 and the final oxidation would lead to 3-nitrobenzoic acid, not to the para isomer. Neither order works, and the reason is structural: for the nitro group to go para, an ORTHO-PARA director must already be on the ring at the moment of nitration. Toluene supplies it; you nitrate toluene, separate the para isomer, and oxidize the methyl group to carboxyl LAST. This is the rule of exercise 4 again.

2-Ethylhexan-1-ol is $CH_{3}CH_{2}CH_{2}CH_{2}-CH(C_{2}H_{5})-CH_{2}OH$, that is $C_{8}H_{18}O$, and $DU=\frac{2\times 8+2-18}{2}=0$: a saturated, acyclic molecule. Carbon 2 carries an ETHYL group, and this branch in the ALPHA position to the functional group is the signature of an aldol addition, because it is exactly the carbon that the enolate engages. You therefore disconnect the bond between carbons 2 and 3, and obtain two IDENTICAL fragments of FOUR carbons: two molecules of butanal, one playing the enolate and the other the electrophile, $4+4=8$ ✓. The full step takes place in two stages: first an interconversion, the primary alcohol comes from the aldehyde by reduction, then the aldol disconnection on that aldehyde.

A crossed aldol reaction between two aldehydes that each have alpha hydrogens gives FOUR products: each can supply the enolate and each can play the electrophile, which makes two times two combinations, including the two condensations of each aldehyde with itself. Here only one aldehyde is committed: only one combination is possible and only one product forms. The difficulty is not solved, it DISAPPEARS. This is why industry builds symmetrical molecules by aldol addition whenever it can, and also why, in retrosynthesis, a target whose aldol disconnection gives two IDENTICAL fragments is particularly favourable.

The equation is $2\,C_{4}H_{8}O\rightarrow C_{8}H_{14}O+H_{2}O$. Carbons: $2\times 4=8$ on the left, $8$ on the right ✓. Hydrogens: $2\times 8=16$ on the left, $14+2=16$ on the right ✓. Oxygens: $2$ on the left, $1+1=2$ on the right ✓. Atom economy: $\frac{126.19}{2\times 72.10}=\frac{126.19}{144.20}=0.875$, that is $87.5$ percent. The check is immediate: the complement is $144.20-126.19=18.01$ g/mol, exactly the molar mass of the water lost. Do not confuse it with the yield: atom economy is a theoretical limit fixed by the equation, the yield measures what you actually obtain.

Because the target is the SATURATED alcohol: you want to reduce both the carbon-carbon double bond and the carbonyl. Selectivity is not a virtue in itself, it is a means of obtaining ONLY what you aim at, and here you aim at everything. A selective reagent would even be a nuisance: $NaBH_{4}$ would reduce the carbonyl and leave the alkene, giving 2-ethylhex-2-en-1-ol; a catalyst that is too mild would hydrogenate the alkene and leave the aldehyde, giving 2-ethylhexanal. You therefore choose nickel under pressure, which does not sort. General method: you look first at WHAT YOU WANT, and you ask the reagent to be selective only if the target requires it.

Overall yield: $0.96\times 0.90\times 0.98=0.8467$, that is $84.7$ percent, the factor $2$ in step 2 being a stoichiometric ratio and not a loss. The calculation: $n(\text{propene})=\frac{1.000\times 10^{6}}{42.08}=2.376\times 10^{4}$ mol; $n(\text{butanal})=2.376\times 10^{4}\times 0.96=2.281\times 10^{4}$ mol; $n(\text{enal})=\frac{2.281\times 10^{4}}{2}\times 0.90=1.027\times 10^{4}$ mol; $n(\text{alcohol})=1.027\times 10^{4}\times 0.98=1.006\times 10^{4}$ mol; $m=1.006\times 10^{4}\times 130.22=1.310\times 10^{6}$ g, that is $1.31$ tonnes. Independent check: each target consumes two propenes, so the maximum mass ratio is $\frac{130.22}{2\times 42.08}=1.547$, and $1.547\times 0.847=1.310$ tonnes ✓.