Organic Chemistry 202-SNU-RE • CEGEP, Montreal

Revision sheet: organic synthesis and retrosynthetic analysis (202-SNU-RE)

Q: How do you carry out a retrosynthesis step by step?

You start from the target and work backward. At each step, you look in its formula for a bond that a known reaction can form, write the two pieces with a double-shafted arrow, and check that the sum of their carbons gives back that of the target. You start again on each piece until you reach catalogue chemicals, then turn the tree over to write the synthesis with single arrows and conditions.

Q: Why can you not nitrate before acylating an aromatic ring?

Because the nitro group depletes the ring so much that no Friedel-Crafts reaction takes place on it afterwards: the second step simply does not happen. By acylating first, you keep a ring that is rich enough, and the acetyl group installed directs the nitration to the meta position. The general rule is to place every Friedel-Crafts step as early as possible in the sequence.

Q: How do you calculate the overall yield of a synthesis?

You multiply the yields of all the steps, you never take their average. Three steps at eighty percent give zero point eight cubed, that is fifty-one point two percent, and not eighty. The check to remember is that the overall yield is always lower than the smallest of the step yields: any answer that violates this inequality is wrong.

Q: What is the difference between a synthon and a synthetic equivalent?

The synthon is the idealized fragment produced by the disconnection, with the charge assigned to it: it exists in no bottle. The synthetic equivalent is the real reagent that plays this role in the laboratory. The methyl carbanion has methylmagnesium bromide as its equivalent, the acetyl cation has ethanoyl chloride. Writing a synthon in a synthesis sequence is a mistake.

Q: When should you protect a functional group?

Only when no selective reagent does the job. You look first for the mildest reagent that suffices, for example sodium borohydride rather than lithium aluminum hydride when an ester has to survive. Protection costs two steps, the one that puts the group on and the one that takes it off, that is about ten points of overall yield, so it is justified only as a last resort.

Q: Why does the Friedel-Crafts reaction fail on aniline?

Because aluminum chloride is a Lewis acid and the lone pair on nitrogen is a Lewis base: the two combine at once. Nitrogen then carries a positive charge and becomes a strongly withdrawing group, so the most activated ring in the repertoire deactivates itself. The remedy is to acetylate the nitrogen first to an amide, which is far less basic.

On this chapter, the exam does not ask you to recite reactions: it asks you to ORDER them. The difference shows at first glance on a paper, because the mistakes are not mechanism mistakes but organization mistakes. A reversed order on an aromatic ring, an overall yield computed as an average, a disconnection that no reaction can close, two carbons that appeared along the way: each time the mechanism written is right, and the answer is wrong.

The habit has three moves. You work BACKWARD from the target and, at each step, look for a bond that a known reaction can form. You COUNT the carbons, the fastest check there is. Then, for every pair of steps, you ask “can I swap them?”, and answer on two levels, orientation first, feasibility second.

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The thread of the chapter

A synthesis is not a list of reactions, it is an ORDER: the same reagents arranged differently give another product or nothing at all, and their yields do not add up, they multiply.

This chapter is part of 202-SNU-RE, Organic Chemistry

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Retrosynthetic analysis
Double-shafted arrow
Retrosynthetic disconnection
Synthon
Synthetic equivalent
Functional group interconversion
Order of synthesis steps
Overall yield of a synthesis
Chemoselectivity
Protecting group
Carbon chain extension
Aldol addition

The essentials

The two arrows, and the direction of reading

• The RETROSYNTHETIC arrow is written DOUBLE-SHAFTED, read “could come from” and runs from the target to its precursors. It asserts nothing, it proposes.
• The pieces it writes are SYNTHONS: idealized fragments carrying a charge, which exist in no bottle. The SYNTHETIC EQUIVALENT is the real reagent that plays this role: $CH_{3}^{-}$ is written $CH_{3}MgBr$ , $C_{6}H_{5}CH_{2}^{+}$ is written benzyl bromide, $CH_{3}CO^{+}$ is written ethanoyl chloride.
• The ANALYSIS goes down the tree with double-shafted arrows, on scratch paper. The SYNTHESIS climbs back up it with single arrows, with real reagents, conditions and step numbers. These are two ways of writing, and the paper hands in only the second.
• At every carbon-carbon disconnection you must CHOOSE which of the two pieces carries the negative charge: that choice, and only that choice, decides which one will be the Grignard reagent.

At the top, two double-shafted arrows go down to two competing sets of precursors, and both counts come to 5 carbons. At the bottom, the same route with a SINGLE arrow: that is the one that goes on the paper.

Writing “could come from” above the first double-shafted arrow takes three seconds and is often worth the half-point for rigour, because the marker then knows you are not confusing analysis with sequence.

The three disconnections that pay off, and the one that cuts nothing

• Carbon-halogen disconnection: it gives the alcohol, and is closed by $PBr_{3}$ , by concentrated $HBr$ or through the tosylate.
• Carbon-oxygen disconnection: that of an ester gives the acid and the alcohol, closed by esterification; that of an ether gives the alkoxide and the haloalkane, closed by the Williamson synthesis.
• Carbon-carbon disconnection NEXT TO a carbonyl: Grignard reagent on an aldehyde or ketone, Grignard reagent on $CO_{2}$ for an acid, nitrile to lengthen by one carbon, aldol addition when the two pieces are identical.
• The FGI, functional group interconversion, is the step that cuts NOTHING: it changes the functional group or the oxidation level and leaves the skeleton intact. Oxidation ladder: alkane, alcohol, aldehyde or ketone, acid, $CO_{2}$ , the ketone being on the same rung as the aldehyde.
• A disconnection is acceptable only if EACH piece has a real synthetic equivalent, and profitable only if it leads back to catalogue chemicals, if possible in two pieces of comparable size.

An FGI goes up or down one rung without touching the skeleton: PCC stops at the aldehyde, dichromate runs on to the acid, and only

LiAlH_{4}

brings the acid back down to the alcohol.

Detaching a single carbon from a ten-carbon target hardly advances the problem; cutting in the middle divides it in two. Among acceptable disconnections, take the most CENTRAL one.

What the order decides, and what it forbids

• On an aromatic ring, the group that must DIRECT is installed FIRST. Donors ( $-OH$ , $-NH_{2}$ , $-OR$ , alkyl groups): ortho and para. Acceptors ( $-NO_{2}$ , $-CN$ , $-COR$ , $-COOH$ , $-SO_{3}H$ ): meta. Halogens: weakly deactivating, but ortho and para directing.
• A Friedel-Crafts step goes as EARLY as possible: it fails on any ring carrying a group more deactivating than a halogen, and on aniline, whose nitrogen binds $AlCl_{3}$ and becomes a deactivating ammonium.
• Harsh conditions, hot acid, strong oxidizing agent, hydrogenation under pressure, go in while the molecule is still robust; fragile or basic groups arrive LAST.
• Test to apply to every pair of consecutive steps: “can I swap them?”. Two levels of answer, orientation first, feasibility second, and it is enough for one of them to refuse for the order to be constrained.

The same ring, two first groups: the donor sends the next one ortho and para, the acceptor meta. This is why the group that must direct is installed first.

The directing effects themselves are revised on the aromatic compounds sheet; here they serve only one purpose, ORDERING the steps.

Overall yield, and the price of a protection

• The overall yield is the PRODUCT of the step yields, $R=\prod R_{i}$ , because each step works only on what the previous one has left it.
• A consequence to keep in mind as a check: the overall yield is always LOWER than the smallest of the step yields. Three steps at $80$ percent give $0.80^{3}=0.512$ , that is $51.2$ percent.
• The yield applies to MOLES, never to masses: you convert, apply, convert back.
• Chemoselectivity: look first for a SELECTIVE reagent, $NaBH_{4}$ rather than $LiAlH_{4}$ , PCC rather than aqueous dichromate. Protect only if none is selective, and protection costs two steps, a factor close to $0.90$ on the overall yield.

Three steps at

80

percent:

80

remain, then

64

, then

51.2

percent, and the overall yield falls below the red line of the smallest step yield.

The question to ask before protecting is always the same: have I exhausted the list of selective reagents?

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Where to disconnect, and which reaction closes it

Read the row of the bond you want to undo. The last column gives the SYNTHESIS reaction that will close it: a red cell means that none does, and that the disconnection must be abandoned.

bond disconnected	real pieces	closing reaction
$C-X$	alcohol	$PBr_{3}$ (FGI)
Example: 1-bromobutane from butan-1-ol: $4+0=4$ carbons, the skeleton does not move.
$C-O$ of an ester	acid $+$ alcohol	esterification
Example: butyl ethanoate: ethanoic acid and butan-1-ol, $2+4=6$ .
$C-O$ of an ether	alkoxide $+$ $R-X$	Williamson
Example: methoxyethane: methanolate and bromoethane, $1+2=3$ .
$C-C$ next to an $OH$	carbonyl compound $+$ $RMgX$	addition then $H_{3}O^{+}$
Example: pentan-2-ol: ethanal and $C_{3}H_{7}MgBr$ , $2+3=5$ .
$C-C$ of the $COOH$	$RMgX$ $+$ $CO_{2}$	carbonation
Example: 3-methylbutanoic acid: four-carbon Grignard reagent and $CO_{2}$ , $4+1=5$ .
symmetrical $C-C$	the same aldehyde twice	aldol addition
Example: 2-ethylhex-2-enal: two butanal, $4+4=8$ .
$C-C$ far from any functional group	none	none unacceptable disconnection
Example: putting a methyl group on carbon 3 of butanoic acid: $4+1=5$ by the count, and yet no reagent does it. Same form, other result: the $C_{2}-C_{3}$ bond of the same acid is, for its part, alpha to the carboxyl: the enolate exists and the disconnection works, $2+3=5$ . What to do: Climb back to a bond NEXT TO a functional group: here $C_{1}-C_{2}$ , Grignard reagent then $CO_{2}$ .
acyl $R-CO^{-}$	none	none synthon with no equivalent
Example: pentan-3-one: the anion $CH_{3}CH_{2}CO^{-}$ does not exist, even though $2+3=5$ seemed to add up. Same form, other result: the carbon of a carbonyl is naturally $\delta^{+}$ : $CH_{3}CO^{+}$ , for its part, does have an equivalent, ethanoyl chloride. What to do: Reverse the charges on the SAME bond: $CH_{3}CH_{2}^{-}$ , hence $C_{2}H_{5}MgBr$ , on propanenitrile.

The carbon count appears in every example on purpose: it is done in your head, before any mechanism, and it eliminates impossible proposals at a stroke.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Believing the order of the steps is free

3 points, it is the synthesis question of the exam

What not to write

“We nitrate benzene, then acylate with $CH_{3}COCl$ and $AlCl_{3}$ : the nitro group directs meta, so we do get the intended ketone.”

What to write

“Nitrobenzene is too deactivated to undergo any Friedel-Crafts reaction: we ACYLATE first, then nitrate, and it is the acetyl group, a meta director, that places the nitro group at position 3.”

Two columns, the same two reagents, two outcomes: on the left, acylation first gives the meta-nitrated ketone; on the right, nitration first leaves a ring too deactivated for Friedel-Crafts.

Why: The first group installed commands both the RATE and the POSITION of the second. Swapping two steps therefore does not give the same product, and sometimes gives none.

2. Computing the overall yield as an average

2 points, and a starting mass underestimated by more than a third

What not to write

“Three steps at $80$ percent: on average, the overall yield is $80$ percent.”

What to write

“The yields MULTIPLY: $0.80^{3}=0.512$ , that is $51.2$ percent, and the overall yield is always lower than the smallest of the step yields.”

Left: the real product after

2

4

6

8

steps at

90

percent. Right: the announced average, fixed at

90

. The gap keeps growing, and at eight steps it exceeds

46

points.

Why: Each step works only on what the previous one has left it, and lost material never comes back. The gap from the average grows with the number of steps.

3. Reading the double-shafted arrow as a reaction arrow

1 point of rigour per arrow, and a sequence judged unworkable

What not to write

“The tree gives $C_{6}H_{5}^{-}$ and butanal, so we put $C_{6}H_{5}^{-}$ and butanal in the flask.”

What to write

“The tree is written in SYNTHONS, which exist in no bottle. We then climb back up it with single arrows: phenylmagnesium bromide on butanal, then acidic hydrolysis.”

Why: The double shaft says “could come from”, not “gives”. Handing in the analysis as it stands means giving a string of impossible reactions to read, and the marker marks it as such.

4. Doing a Friedel-Crafts reaction on aniline

2 points, and two protection steps forgotten in the plan

What not to write

“Aniline carries $-NH_{2}$ , the most powerful activating group in the repertoire: acylation will work even better than on benzene.”

What to write

“ $AlCl_{3}$ is a Lewis acid and binds to the lone pair on nitrogen: the ring then carries a strongly deactivating ammonium. We first acetylate the nitrogen to an amide, acylate the ring, then hydrolyze.”

On the left,

AlCl_{3}

binds to the lone pair on nitrogen and the ring carries a deactivating ammonium. On the right, the acetylated nitrogen no longer binds it: we acylate, then hydrolyze the amide.

Why: The reagent that is supposed to attack the ring begins by deactivating it. It is the case where feasibility, not orientation, decides the order, and it comes up every session on an amine target.

5. Believing the shortest synthesis is the best

1 point, and the wrong plan kept for the rest of the problem

What not to write

“Route B has only two steps against four for route A: it surely gives more.”

What to write

“ $0.92\times 0.88\times 0.95\times 0.90=0.692$ against $0.85\times 0.78=0.663$ : the route that is twice as long gives more. We compare PRODUCTS, not numbers of steps.”

The route that is twice as long gives more,

69.2

against

66.3

percent: the number of steps decides nothing, the product of the yields decides.

Why: The number of steps decides nothing by itself. It weighs on cost, time and the number of chances of going wrong, arguments that must then be NAMED instead of invoking a yield.

6. Forgetting that a reagent sees ALL the functional groups

the whole question: the product announced is not the one obtained

What not to write

“Ethyl 4-oxopentanoate carries a ketone: we reduce it with $LiAlH_{4}$ and obtain ethyl 4-hydroxypentanoate.”

What to write

“ $LiAlH_{4}$ does not sort: it would also take the ester down and we would isolate pentane-1,4-diol. With $NaBH_{4}$ in cold methanol, the ketone alone is reduced.”

The same substrate, two reducing agents:

NaBH_{4}

touches only the ketone and keeps the ester,

LiAlH_{4}

also takes the ester down and gives pentane-1,4-diol.

Why: A reagent does not know what you are aiming at. Before writing a step, you confront it with every functional group present, and choose the mildest reagent that suffices.

7. Proposing a disconnection that no reaction closes

3 points, the proposed synthesis is unworkable from end to end

What not to write

“For 3-methylbutanoic acid, we start from butanoic acid and attach a methyl group to its carbon 3.”

What to write

“No reagent functionalizes a bare carbon. We disconnect $C_{1}-C_{2}$ : Grignard reagent of 1-bromo-2-methylpropane, added to solid $CO_{2}$ , then acidic hydrolysis, $4+1=5$ .”

Why: You disconnect only bonds that a reaction can FORM. The criterion is checked before any calculation: each piece must have a real synthetic equivalent.

8. Losing or gaining carbons along the way

2 points, and the product named is not the one in the scheme

What not to write

“Pentan-2-ol comes from propanone and propylmagnesium bromide, then hydrolysis.”

What to write

“ $3+3=6$ and not $5$ : we would obtain 2-methylpentan-2-ol, tertiary. A SECONDARY alcohol with five carbons requires an aldehyde: ethanal and $C_{3}H_{7}MgBr$ , $2+3=5$ .”

In red,

3+3=6

carbons and a Grignard reagent on a ketone: the alcohol is tertiary. In green, the aldehyde gives the intended secondary alcohol with five carbons.

Why: Two independent checks catch this mistake in a second, the carbon COUNT and the CLASS of the alcohol: Grignard reagent on a ketone gives a tertiary alcohol, on an aldehyde a secondary one, on methanal a primary one.

Which method to choose

Where to disconnect, faced with a target

Look at the SHAPE of the target, not its name: which functional group does it carry, and which bond borders that group?

Three numbered disconnections on 3-methylbutanoic acid: 1 is closed by Grignard reagent plus

CO_{2}

, 2 by an enolate, 3 is closed by nothing, because it falls on a carbon that no functional group borders.

If an $OH$ or a $C=O$ , with a chain attached to the functional carbon → disconnect the neighbouring $C-C$ bond: carbonyl compound plus Grignard reagent

Example: pentan-2-ol: ethanal and $C_{3}H_{7}MgBr$ , $2+3=5$

the carbonyl compound is chosen on the CLASS aimed at, aldehyde for a secondary alcohol, ketone for a tertiary one
If a carboxylic acid at the end of a chain → disconnect $C_{1}-C_{2}$ : Grignard reagent then $CO_{2}$ , or $NaCN$ then hydrolysis

Example: 3-methylbutanoic acid: $4+1=5$

both routes lengthen the chain by EXACTLY one carbon
If an ester or an ether → disconnect the $C-O$ bond: esterification, or the Williamson synthesis

Example: butyl ethanoate: $2+4=6$
If a halogen, or a functional group at the wrong oxidation level → this is not a disconnection but an FGI: the skeleton does not move

Example: 1-bromobutane from butan-1-ol by $PBr_{3}$ , $4+0=4$
If the disconnection would give two IDENTICAL pieces → think aldol addition: the target comes from a single aldehyde engaged twice

Example: 2-ethylhex-2-enal: two butanal, $4+4=8$

a single aldehyde, hence a single possible combination: the difficulty of crossed aldol reactions disappears
If the bond aimed at borders NO functional group → the disconnection is unacceptable: climb back towards the nearest functional group

Example: a methyl group on carbon 3 of an acid: no reagent puts it there

If no branch applies, it is almost always because an FGI is missing upstream: bring the molecule to a functional group that opens a disconnection, then start again.

In what order to place the steps

Take the steps TWO by TWO and ask each time: can I swap them?

If one of the two is a Friedel-Crafts reaction → place it as EARLY as possible, before any group more deactivating than a halogen

Example: acylate then nitrate: the reverse order does not react at all
If the target is the PARA or ORTHO isomer → install the ortho-para directing group first

Example: brominate then acylate gives the ketone at position 4
If the target is the META isomer → install the meta directing group first

Example: acylate then nitrate gives the nitro group at position 3
If a step uses a reagent that would attack another functional group of the target → place it BEFORE that group appears, and protect only as a last resort

Example: esterify while the group is still a nitro group, reduce to the amine afterwards
If a step is harsh, hot acid, strong oxidizing agent, hydrogenation under pressure → place it while the molecule is still robust

Example: oxidize the methyl group of toluene to carboxyl LAST
If the two steps are independent, with no orientation or incompatibility involved → the order is free: you then decide on yield and price

Example: benzocaine: esterify then reduce, or reduce then esterify, both work

It is enough for one of the two levels, orientation or feasibility, to refuse the swap for the order to be constrained, and you must then JUSTIFY it on the paper, not merely respect it.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing a retrosynthetic analysis, then the synthesis

When to use it: the question says “propose a synthesis of X from Y”, or gives a target and a catalogue of starting materials

1 Write the structural formula of the target, COUNT its carbons and compute its degree of unsaturation, then do the same for the imposed starting material. The difference in carbons tells you at once whether a carbon-carbon disconnection will be needed or just FGIs.
2 Spot the functional groups, and circle the bonds that a reaction in the course can form: carbon-halogen, carbon-oxygen of an ester or ether, carbon-carbon next to a carbonyl.
3 Write the retrosynthetic step with a DOUBLE-SHAFTED arrow, give the two synthons with their charge, then their real synthetic equivalent, and check the carbon count under the arrow.
4 Start again on each precursor until you reach the imposed starting material or a catalogue chemical. When two branches are acceptable, keep both and decide at the end.
5 TURN THE TREE OVER: rewrite the sequence from top to bottom, single arrows, real reagents, conditions, numbered steps. It is this writing, and this one alone, that the marker grades as the synthesis.
6 Check the order pair by pair, “can I swap these two steps?”, and justify in one line each constrained order, on orientation or on feasibility.
7 Conclude with the overall yield, the product of the step yields, and with the choice of route if there were several.

Concluding sentence

“The target has ten carbons and the starting material six: the carbon-carbon disconnection alpha to the alcohol divides them into $6+4$ , and the addition of phenylmagnesium bromide to butanal, followed by acidic hydrolysis, closes this bond.”

The trap: handing in the analysis tree as it stands, double shafts and synthons included, instead of the synthesis sequence: it is the costliest mistake of the chapter, because it turns sound reasoning into a string of impossible reactions.

Marking: in general 1 point per justified retrosynthetic step, 1 point for the conditions of each step, 1 point for the argued order and 1 point for the overall yield

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The carbon count, under every arrow

Add the carbons of the two pieces and compare with the target. The check is done in your head, before any mechanism, and it needs no additional data.

a $3+3$ written under a target that has five carbons rules out the disconnection without your having to look at the reagent: you would obtain a six-carbon molecule.
The class of the alcohol obtained

A Grignard reagent gives a TERTIARY alcohol on a ketone, a SECONDARY one on an aldehyde, a PRIMARY one on methanal. You read the class aimed at on the target and work back to the carbonyl compound.

aiming at a secondary alcohol starting from a ketone is impossible, whatever the Grignard reagent: the mistake is visible without any calculation.
The overall yield against the smallest yield

Compare the announced overall yield with the smallest of the step yields: it must be strictly lower. This check catches every disguised average.

announcing $80$ percent after three steps at $90$ , $80$ and $70$ is wrong without even redoing the calculation, since $80>70$ . The true result is $50.4$ percent.
The swap, pair by pair

Take the sequence you wrote and mentally swap each pair of consecutive steps. If the product changes or becomes impossible, the order is constrained and must be justified in one line.

if the swap changes nothing, you have a freedom: pointing it out often earns the discussion mark, whereas keeping quiet loses it.
Each reagent against all the functional groups present

Reread the molecule at each step and ask, group by group, whether the reagent written could touch it. An alcohol, an amine and an ester are the three that trap most often.

$LiAlH_{4}$ written on a molecule that still carries an ester: the answer key expects the diol, not the intended alcohol, and the whole question falls.

The typical problem, taken apart

Synthesis of 4-bromoaniline from benzene

4-Bromoaniline, $C_{6}H_{6}BrN$ , is an intermediate for azo dyes. The only chemicals available from the catalogue are benzene, aniline, dibromine, fuming nitric acid and the usual reducing agents.

Propose a synthesis, justify the order of the steps, then calculate the mass of target obtained from $40.0$ g of benzene, knowing that the three steps have yields of $85$ , $72$ and $90$ percent.

Molar masses in g/mol: C $12.01$ ; H $1.008$ ; N $14.01$ ; Br $79.90$ .

The situation of the example: we start from benzene, we aim at 4-bromoaniline, bromine and amine in para. The whole job is to find the steps and their ORDER.

Step 1

Target $C_{6}H_{6}BrN$ : one benzene ring, one bromine and one amine in PARA. Starting material: benzene, six carbons. Count: $6+0=6$ , no carbon to add or remove.

Why

The carbon count rules out any carbon-carbon disconnection at the outset. The whole problem is therefore a matter of FGIs and ORDER, which says at once where the marks will be won or lost.

Step 2

First retrosynthetic step, an FGI: $-NH_{2}$ could come from $-NO_{2}$ by reduction. Precursor: 1-bromo-4-nitrobenzene.

Why

This is the counter-intuitive step of the problem. Aniline is in the catalogue and seems closer to the target, but its ring is so activated that dibromine puts THREE bromines on it; taking the amine back to the nitro group moves the work onto a deactivated ring, hence a manageable one.

Step 3

Second step: the nitro group comes from nitration of bromobenzene. Third step: the bromine comes from bromination of benzene. The tree stops, we have reached the imposed starting material.

Why

It is the bromine that must DIRECT, so it is installed first: weakly deactivating but ortho and para directing, it sends the nitro group to position 4.

Step 4

Swap test on the bromination-nitration pair: by nitrating first, we get nitrobenzene, a META director, and the bromine would go to position 3. The order is therefore constrained by orientation.

Why

This is the line the marker looks for. A correct synthesis whose order is not justified loses the argumentation mark, because nothing proves it was not found by chance.

Step 5

Swap test on the nitration-reduction pair: by reducing first, we would hold 4-bromoaniline ahead of time, whose amine is a strong activator and would direct nitration to position 2. The order is constrained here too.

Why

Both pairs are constrained for the same underlying reason, orientation, but they must be tested separately: an order is established only when every consecutive pair has been examined.

Step 6

Synthesis sequence, tree turned over. (1) $Br_{2}$ with $FeBr_{3}$ . (2) $HNO_{3}$ and $H_{2}SO_{4}$ , then separation of the para isomer. (3) $Sn$ or $Fe$ in $HCl$ , then neutralization with carbonate.

Why

You climb back up the tree with SINGLE arrows, with the real reagents and the conditions. The ortho-para separation is an operation in its own right: it costs yield and keeping quiet about it distorts the rest of the calculation.

Step 7

Choice of reducing agent: $Sn/HCl$ or $Fe/HCl$ , and not $H_{2}$ over palladium, which would also cleave the carbon-bromine bond by hydrogenolysis.

Why

This is the chemoselectivity question of the problem. A selective reagent exists, so nothing is protected: protection would have cost two steps for an identical result.

Step 8

Overall yield: $0.85\times 0.72\times 0.90=0.5508$ , that is $55.1$ percent. Then $M(\text{benzene})=78.11$ g/mol and $M(\text{target})=6\times 12.01+6\times 1.008+79.90+14.01=172.02$ g/mol.

Why

The yield applies to MOLES: you therefore need both molar masses before applying anything, and this is where the paper that multiplies the masses directly goes wrong.

Step 9

$n(\text{benzene})=\frac{40.0}{78.11}=0.5121$ mol, hence $n(\text{target})=0.5121\times 0.5508=0.2821$ mol and $m=0.2821\times 172.02=48.5$ g.

Why

Three lines, one per conversion: moles, yield, mass. Writing them separately earns the method mark even if the final value is missed.

Step 10

Checks. $55.1<72$ : the overall yield is indeed lower than the smallest yield. And the target is $\frac{172.02}{78.11}=2.20$ times heavier than benzene, so $40.0\times 2.20\times 0.551\approx 48.5$ g.

Why

Two checks independent of the main calculation, both doable in ten seconds. The second is the more useful: it catches a molar mass error as well as a yield error.

The conclusion, written out

“We brominate benzene with $Br_{2}/FeBr_{3}$ , nitrate the bromobenzene with $HNO_{3}/H_{2}SO_{4}$ and separate the para isomer, then reduce the nitro group with $Sn/HCl$ . The order is imposed twice by orientation: the bromine, an ortho-para director, must precede the nitration, and the amine, an activating group, must appear last. The overall yield is $0.85\times 0.72\times 0.90=55.1$ percent, and $40.0$ g of benzene give $48.5$ g of 4-bromoaniline.”

The classic mistake on this problem: starting from aniline because it is in the catalogue and looks more like the target: dibromine puts three bromines on it in a single step, and the route is dead before it has started.

Learn by heart

• The DOUBLE-SHAFTED arrow is read “could come from”. It lives on scratch paper; the paper carries only single arrows.
• A synthon is not a reagent: $CH_{3}^{-}$ is written $CH_{3}MgBr$ , $C_{6}H_{5}CH_{2}^{+}$ is written benzyl bromide.
• Three profitable disconnections at the CEGEP level: $C-X$ , $C-O$ of an ester or ether, $C-C$ next to a carbonyl.
• The overall yield is the PRODUCT of the yields, and it is always LOWER than the smallest of them.
• The group that must DIRECT is installed first; a Friedel-Crafts step goes as EARLY as possible.
• Friedel-Crafts fails as soon as a group deactivates more than a halogen, and on ANILINE, whose nitrogen binds $AlCl_{3}$ .
• Grignard reagent on a ketone: TERTIARY alcohol. On an aldehyde: SECONDARY. On methanal: PRIMARY.
• Protecting costs TWO steps, about ten points of overall yield: look first for a selective reagent.

Frequently asked questions

How do you carry out a retrosynthesis step by step?

You start from the target and work backward. At each step, you look in its formula for a bond that a known reaction can form, write the two pieces with a double-shafted arrow, and check that the sum of their carbons gives back that of the target. You start again on each piece until you reach catalogue chemicals, then turn the tree over to write the synthesis with single arrows and conditions.

Why can you not nitrate before acylating an aromatic ring?

Because the nitro group depletes the ring so much that no Friedel-Crafts reaction takes place on it afterwards: the second step simply does not happen. By acylating first, you keep a ring that is rich enough, and the acetyl group installed directs the nitration to the meta position. The general rule is to place every Friedel-Crafts step as early as possible in the sequence.

How do you calculate the overall yield of a synthesis?

You multiply the yields of all the steps, you never take their average. Three steps at eighty percent give zero point eight cubed, that is fifty-one point two percent, and not eighty. The check to remember is that the overall yield is always lower than the smallest of the step yields: any answer that violates this inequality is wrong.

What is the difference between a synthon and a synthetic equivalent?

The synthon is the idealized fragment produced by the disconnection, with the charge assigned to it: it exists in no bottle. The synthetic equivalent is the real reagent that plays this role in the laboratory. The methyl carbanion has methylmagnesium bromide as its equivalent, the acetyl cation has ethanoyl chloride. Writing a synthon in a synthesis sequence is a mistake.

When should you protect a functional group?

Only when no selective reagent does the job. You look first for the mildest reagent that suffices, for example sodium borohydride rather than lithium aluminum hydride when an ester has to survive. Protection costs two steps, the one that puts the group on and the one that takes it off, that is about ten points of overall yield, so it is justified only as a last resort.

Why does the Friedel-Crafts reaction fail on aniline?

Because aluminum chloride is a Lewis acid and the lone pair on nitrogen is a Lewis base: the two combine at once. Nitrogen then carries a positive charge and becomes a strongly withdrawing group, so the most activated ring in the repertoire deactivates itself. The remedy is to acetylate the nitrogen first to an amide, which is far less basic.

Practise it

Corrected exercises: Synthesis and retrosynthetic analysis, 202-SNU-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-snu-synthesis-retrosynthesis. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: organic synthesis and retrosynthetic analysis (202-SNU-RE)

The essentials

The two arrows, and the direction of reading

The three disconnections that pay off, and the one that cuts nothing

What the order decides, and what it forbids

Overall yield, and the price of a protection

The rules in table form

Where to disconnect, and which reaction closes it

The mistakes that cost marks

1. Believing the order of the steps is free

2. Computing the overall yield as an average

3. Reading the double-shafted arrow as a reaction arrow

4. Doing a Friedel-Crafts reaction on aniline

5. Believing the shortest synthesis is the best

6. Forgetting that a reagent sees ALL the functional groups

7. Proposing a disconnection that no reaction closes

8. Losing or gaining carbons along the way

Which method to choose

Where to disconnect, faced with a target

In what order to place the steps

How the answer is expected to be written

Writing a retrosynthetic analysis, then the synthesis

Check before you hand in

The typical problem, taken apart

Synthesis of 4-bromoaniline from benzene

Learn by heart

Frequently asked questions

How do you carry out a retrosynthesis step by step?

Why can you not nitrate before acylating an aromatic ring?

How do you calculate the overall yield of a synthesis?

What is the difference between a synthon and a synthetic equivalent?

When should you protect a functional group?

Why does the Friedel-Crafts reaction fail on aniline?

Corrected exercises: Synthesis and retrosynthetic analysis, 202-SNU-RE

See also

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