Lesson 13 · Synthesis and Retrosynthesis

Synthesis and Retrosynthesis

This lesson does not add reactions. It teaches you how to connect the ones you already know into a sequence. The skill is planning backward from the target, because the target tells you what the last reaction had to be, while the starting material tells you almost nothing.

Rule to remember

Plan backward, verify forward. Never start by staring at the starting material and trying reactions at random.

Learning goals

Tell the difference between working forward (synthesis) and working backward (retrosynthesis), and know when to use each.
Use carbon count to decide immediately whether you need a C−C bond-forming step.
Recognize the common functional group interconversions on sight and name a reagent for each direction.
Make a disconnection at a bond you know how to form, and judge whether it is chemically realistic.
Identify the stereochemical control point in a route before choosing a precursor.
Verify a planned route forward, checking regiochemistry, stereochemistry, rearrangement, and competing reactions at every step.

Key terms (click to reveal)

Part 01

Two directions, one problem

Synthesis runs forward: starting material to target. Retrosynthesis runs backward: target to a simpler precursor, one step at a time, using the open double arrow (⇒) meaning "could be made from." You solve backward and you write the answer forward.

The reason backward wins is a search problem. From any starting material there are dozens of reactions you could run, and almost none of them lead anywhere useful. But from the target, the functional group narrows the last step to two or three candidates immediately. An alcohol had to come from an alkene or a carbonyl. A cis alkene had to come from an alkyne over Lindlar. The target hands you the answer to the question "what was the final reaction?"

Two diagrams side by side: forward search from a starting material branching into many dead ends, and backward search from a target narrowing to a small number of viable precursors

Forward from the starting material is a wide search. Backward from the target is a narrow one.

The biggest mistake

Looking at the starting material and randomly trying reactions. That is guessing, not planning. Start at the target every time.

Part 02

First move: count the carbons

Before anything else, count the carbons in the starting material and in the target. That one number splits every synthesis problem into two very different kinds.

Same carbon count

Functional group problem

The skeleton is already correct. Your entire job is moving between functional groups, possibly with a stereochemistry or regiochemistry requirement attached. Most problems are this kind.

Different carbon count

Skeleton problem

You need a C−C bond-forming step, and there are only a few of those. Find where it goes first, then arrange the functional group changes around it.

The C−C bond-forming reactions worth recognizing on sight are acetylide alkylation, Grignard and organolithium additions to carbonyls, and enolate alkylation if your course covers it. Because these are scarce, they are the least flexible part of any route, which makes them the natural anchor: place the C−C step first and let everything else fall around it.

A decision diagram: count carbons in starting material and target, one branch labeled same count leading to functional group interconversion, the other labeled different count leading to carbon-carbon bond forming reactions

Carbon count is the first branch point. It decides what kind of problem you are solving.

Part 03

Let the target name the last reaction

Read this table right to left when you are planning and left to right when you are writing the answer. The Watch for column is where most lost marks live.

If the target is	It could come from	Forward reagents	Watch for	Move
Alcohol (Markovnikov)	Alkene	1. Hg(OAc)₂, H₂O 2. NaBH₄	Use this over H₃O⁺ when rearrangement would ruin the skeleton	FGI
Alcohol (anti-Markovnikov)	Alkene	1. BH₃·THF 2. H₂O₂, OH⁻	Syn addition, so this is also your stereochemical control point	FGI
1° alcohol	Aldehyde, acid, or ester	NaBH₄ (aldehyde) or LiAlH₄ (acid, ester)	NaBH₄ is too weak for acids and esters	Reduction
2° alcohol	Ketone	NaBH₄	Leaves esters untouched, which is useful for selectivity	Reduction
Alkene (Zaitsev)	Alcohol	H₂SO₄, heat	E1, so rearrangements are possible	FGI
cis alkene	Alkyne	H₂, Lindlar catalyst	Only route to cis; do not use plain Pd/C	Reduction
trans alkene	Alkyne	Na, NH₃ (l)	Only route to trans	Reduction
Alkyl halide	Alcohol	PBr₃ or SOCl₂	Inverts the carbon; use TsCl instead if configuration must be kept	FGI
Aldehyde	1° alcohol, or terminal alkyne	PCC, or 1. R₂BH 2. H₂O₂/OH⁻	Jones would overshoot the 1° alcohol to the acid	Oxidation
Carboxylic acid	1° alcohol	CrO₃, H₂SO₄, H₂O (Jones)	PCC stops one step short	Oxidation
Ketone	2° alcohol, or terminal alkyne	PCC or Jones, or H₂O/H₂SO₄/HgSO₄	From a terminal alkyne you get a methyl ketone specifically	Oxidation
Internal alkyne (longer chain)	Terminal alkyne + alkyl halide	1. NaNH₂ 2. R−X	SN2, so the halide must be methyl or 1°	C−C bond
Ether	Alcohol + alkyl halide	1. NaH 2. R′−X	SN2, so put the bulk on the alkoxide side	FGI
1,2-diol (syn)	Alkene	OsO₄, or cold dilute KMnO₄	Hot KMnO₄ cleaves instead of hydroxylating	FGI

Part 04

Disconnections

A disconnection is breaking a bond you know how to make, in your head, to see what would build it. For an internal alkyne R−C≡C−R′, a useful disconnection is into the acetylide R−C≡C⁻ and the halide R′−X, because acetylides form C−C bonds by SN2.

Simpley, you are asking what molecules could make this bond. The fragments you imagine are synthons, and the real reagents that play those roles are the synthetic equivalents: a carbanion synthon becomes an acetylide or a Grignard in practice.

2-pentyne shown with a wavy line through the bond next to the alkyne, a retrosynthetic double arrow leading to the propynyl anion plus ethyl bromide, and below it the forward reaction with sodium amide then ethyl bromide

Break the bond next to the alkyne. The pieces are an acetylide and a primary halide.

Worked example: to make CH₃CH₂C≡CCH₃, disconnect one bond next to the triple bond. That gives CH₃CH₂C≡C⁻ plus CH₃Br. The acetylide comes from 1-butyne with NaNH₂, so the forward route reads: 1. NaNH₂, 2. CH₃Br. Notice you could have disconnected the other side instead, giving propynyl anion plus CH₃CH₂Br. Both are legitimate, which is normal: a good problem usually has more than one correct answer.

Part 05

Reagent limitations make disconnections fail

This is where synthesis problems get hard. A disconnection can look perfectly symmetric on paper and still be worthless, because the forward reaction does not run.

Acetylide alkylation is the standard example. The reaction is SN2, so the halide must be methyl or primary. Disconnecting to an acetylide plus (CH₃)₃CBr looks fine as a bond analysis, but tert-butyl bromide is tertiary: backside attack is blocked, the acetylide acts as a base instead, and you get elimination rather than your product. That disconnection is not chemically realistic.

Realistic

R−C≡C⁻ + CH₃CH₂Br

Primary halide, unhindered backside attack, SN2 proceeds cleanly.

Not realistic

R−C≡C⁻ + (CH₃)₃CBr

Tertiary halide, so E2 elimination wins and you never form the C−C bond.

Some disconnections can be rescued by swapping which fragment is the nucleophile. Williamson ether synthesis is the classic case: put the bulk on the alkoxide and keep the halide methyl or primary. Acetylide alkylation cannot be rescued that way, because an alkyne carbon can be the nucleophile but never the electrophile. Knowing which reactions are reversible in that sense saves real time.

Part 06

Find the stereochemical control point

When a target specifies cis or trans, or a particular stereoisomer, one step in the route is doing that work. Find it before you plan the rest, because it usually dictates which precursor you need.

Need cis

H₂, Lindlar

From an alkyne. Syn delivery on the poisoned surface. There is no other route to cis in this course.

Need trans

Na, NH₃ (l)

From an alkyne. Anti addition through radical intermediates. Again, the only route.

Need syn or anti

Pick by intermediate

Concerted additions give syn (BH₃, OsO₄, H₂). Bridged intermediates give anti (Br₂, epoxide opening).

These stack. Reducing 2-butyne with Lindlar and then dihydroxylating with OsO₄ adds two OH groups syn to a cis alkene, which gives the meso diol. Choose Na/NH₃ for the reduction instead and the same syn addition now gives the chiral pair. Same reagent in the second step, different answer, because the first step set the geometry.

Part 07

Plan backward, verify forward

Once you have Target ⇐ A ⇐ B ⇐ starting material, reverse it and walk the whole thing forward: starting material → B → A → target. This step catches errors that backward planning hides, because working backward you tend to assume each arrow behaves, while working forward you have to say what the reagent actually does.

Does this reagent actually give that product?

Markovnikov or anti-Markovnikov?

Syn or anti? Does a stereocenter invert?

Could a carbocation rearrange here?

Would SN2 or E2 dominate with this substrate?

Will another functional group in the molecule react too?

Does this step overshoot, like 1° alcohol past the aldehyde?

Do the carbon counts still match at every stage?

A four-box route drawn with retrosynthetic arrows pointing right to left from target to starting material, and below it the same boxes with forward arrows left to right, each forward arrow annotated with a check mark and a verification question

Plan right to left. Then check left to right, one reagent at a time.

Part 08

Worked examples

One step: the target picks the reagent

Starting material: CH₃CH₂C≡CH. Two different targets, same substrate, and the functional group alone decides the answer.

Target: CH₃CH₂CH₂CHO

Aldehyde → hydroboration

1. R₂BH 2. H₂O₂, OH⁻. Anti-Markovnikov hydration puts the oxygen on the terminal carbon, and the enol tautomerizes to the aldehyde.

Target: CH₃CH₂COCH₃

Methyl ketone → mercury

H₂O, H₂SO₄, HgSO₄. Markovnikov hydration puts the oxygen on the internal carbon, giving the methyl ketone after tautomerization.

Two steps: 1-butyne to 1-butanol

Target is CH₃CH₂CH₂CH₂OH, a primary alcohol. Carbon count: four in, four out, so this is a pure functional group problem, no C−C bond needed.

Working backward: what makes a primary alcohol with the OH on a terminal carbon? Anti-Markovnikov hydration of an alkene. So 1-butanol ⇐ 1-butene. And what makes a terminal alkene? Partial reduction of an alkyne. So 1-butene ⇐ 1-butyne, which is where you started.

Retrosynthetic sequence showing 1-butanol from 1-butene from 1-butyne with open double arrows, and below it the forward route with hydrogen over Lindlar catalyst then borane and hydrogen peroxide with hydroxide

1-butanol ⇐ 1-butene ⇐ 1-butyne. Then check it forward.

Forward and verified: 1. H₂ with Lindlar catalyst gives 1-butene rather than butane, because the poisoned catalyst stops at the alkene. 2. BH₃·THF, then H₂O₂ and OH⁻, delivers OH to the terminal carbon. Carbon count still four, no rearrangement possible in either step, and no other functional group present to interfere. The route holds.

Three steps: when the skeleton has to grow

Target: 2-hexanone (six carbons). Starting material: 1-butyne (four carbons). The count changed, so a C−C bond-forming step is mandatory, and that anchors the plan.

Backward: a methyl ketone comes from Markovnikov hydration of a terminal alkyne, so 2-hexanone ⇐ 1-hexyne. A six-carbon terminal alkyne from a four-carbon one means adding two carbons at the far end, which acetylide alkylation cannot do to a terminal position without destroying it. So instead alkylate 1-butyne at its terminal carbon with a two-carbon halide and accept that the result is internal, or rethink. Here the clean route is to alkylate acetylene itself: 2-hexanone ⇐ 1-hexyne ⇐ acetylide of 1-hexyne, built from acetylene plus 1-bromobutane.

Forward from acetylene: 1. NaNH₂, then CH₃CH₂CH₂CH₂Br, giving 1-hexyne. 2. H₂O, H₂SO₄, HgSO₄, giving 2-hexanone. The halide is primary so the SN2 works, and the terminal alkyne is preserved through the alkylation because only one end is deprotonated. This is the pattern worth internalizing: build the skeleton first, install the functional group last.

Pattern to keep

Skeleton first, functional group last. C−C bond-forming reactions are picky about what else is in the molecule, so run them early while the substrate is still simple.

Part 09

The framework, in order

Run every synthesis problem through these six steps in this order. The order matters more than any individual reaction you have memorized.

Step 1

Count carbons

Compare starting material and target. Did the skeleton change? This decides whether you need a C−C bond-forming reaction at all.

Step 2

Identify the target functional group

Name it precisely: not just alcohol, but primary alcohol with the OH on the terminal carbon. Precision here narrows the last step.

Step 3

Ask what reaction directly makes it

Usually two or three candidates. If the target specifies stereochemistry, that requirement eliminates most of them immediately.

Step 4

Work backward one step

Write the precursor with a retrosynthetic arrow. One step only. Do not try to see the whole route at once.

Step 5

Repeat until you reach the starting material

Each new precursor becomes the new target. Stop when you land on what you were given, or on something trivially made from it.

Step 6

Run the whole pathway forward

Check every step for regiochemistry, stereochemistry, rearrangement, competing elimination, and other reactive groups. This is where you catch the errors.

Watch out

Common mistakes

Starting from the starting material

Trying reactions forward at random is a wide search with mostly dead ends. The target tells you what the last reaction was; the starting material tells you almost nothing.

Skipping the carbon count

If you never check, you can spend the whole problem doing functional group changes on a skeleton that was the wrong size from the start.

Making an unrealistic disconnection

A bond analysis that requires SN2 on a tertiary carbon is not a plan. Every disconnection has to correspond to a forward reaction that actually runs.

Never verifying forward

Backward planning hides regiochemistry errors, because you assume each arrow works. Walking forward forces you to state what each reagent really does.

Ignoring stereochemistry until the end

If the target is cis, the alkyne reduction step was decided before you planned anything else. Find the stereochemical control point early.

Overshooting an oxidation

Jones takes a 1° alcohol past the aldehyde to the acid. If the target is the aldehyde, PCC was the required reagent and the route fails without it.

Forgetting other functional groups

A reagent that solves your target group may destroy another one elsewhere in the molecule. Check the whole structure at every forward step, not just the site you care about.

Installing the functional group before building the skeleton

C−C bond-forming reactions are sensitive to what else is present. Build the carbon framework early, then install the functional group.

Checkpoint

Try it before you scroll on

Convert 1-butyne into 1-butanol. Plan it backward first, then write the forward route.

Show answer

Backward: 1-butanol ⇐ 1-butene ⇐ 1-butyne. Forward: 1. H₂, Lindlar. 2. BH₃·THF, then H₂O₂/OH⁻.

The target is a primary alcohol, so the OH must land on the terminal carbon: that means anti-Markovnikov hydration, which needs an alkene. Lindlar takes the alkyne to the alkene without overreducing to butane, and hydroboration-oxidation then puts the OH where you need it. Plain H₂/Pd would give butane and dead-end the route.

Your target is 2-pentyne and your starting material is propyne. What is the carbon count telling you, and what is the disconnection?

Show answer

Propyne has 3 carbons, 2-pentyne has 5, so you need a C−C bond-forming step. Disconnect to propynyl anion plus CH₃CH₂Br.

Counting carbons first is what tells you this is not a functional group problem. The alkyne is already there, so the only missing piece is two carbons, and acetylide alkylation is the tool. Forward: 1. NaNH₂, 2. CH₃CH₂Br. The ethyl bromide is primary, so the SN2 works.

A classmate proposes making 4,4-dimethyl-2-pentyne by treating propynyl anion with tert-butyl bromide. Why is this a bad disconnection, and can it be fixed?

Show answer

The alkylation is SN2 and tert-butyl bromide is tertiary, so E2 elimination dominates. This disconnection cannot be rescued by swapping roles, because the alkyne carbon cannot be the electrophile.

A disconnection has to correspond to a reaction that actually works. Unlike Williamson synthesis, where you can reverse which fragment is nucleophile and which is electrophile, acetylide alkylation only runs one direction. When a disconnection forces a hindered SN2, you need a different bond to break, or a different route entirely.

Convert 2-butyne into (2R,3S)-2,3-butanediol, or explain what stereochemical decision the route hinges on.

Show answer

Reduce to cis-2-butene with H₂/Lindlar, then syn dihydroxylate with OsO₄. Syn addition to a cis alkene gives the meso diol.

This is a route with two stereochemical control points stacked together. The reduction sets cis or trans, and the dihydroxylation adds syn. Choosing Na/NH₃ instead would give trans-2-butene, and syn addition there gives the chiral (2R,3R) and (2S,3S) pair rather than the meso compound. Identify which step sets the stereochemistry before you commit to a precursor.

You plan: 2-methyl-2-butanol, H₂SO₄ and heat, then HBr, aiming for 2-bromo-3-methylbutane. Run it forward and check it.

Show answer

The route fails. Dehydration gives 2-methyl-2-butene, and HBr adds Markovnikov to give 2-bromo-2-methylbutane, not the target.

This is why you verify forward. The backward plan looked reasonable at each arrow, but the forward check catches the regiochemistry: HBr protonates to give the more stable tertiary carbocation, so bromide ends up on the tertiary carbon. If you actually wanted the less substituted bromide, you would need an anti-Markovnikov route, not this one.

Practice workbook

Synthesis and Retrosynthesis Practice

One-step, two-step, and multistep routes, plus disconnection problems and broken routes you have to find the error in.

Open the practice set

Lesson summary

Synthesis runs forward, retrosynthesis runs backward. Solve backward, write forward.
Count carbons first. Same count means a functional group problem; a different count means you need a C−C bond-forming step.
The target functional group names the last reaction. Alcohol from alkene or carbonyl, cis alkene from Lindlar, trans alkene from Na/NH₃, aldehyde from PCC or hydroboration, methyl ketone from HgSO₄ hydration.
A disconnection breaks a bond you know how to make. Acetylide alkylation, Grignard addition, and organolithium addition are the C−C bonds available to you.
Disconnections must be chemically realistic. An SN2 step onto a tertiary halide is not a plan, it is elimination.
Find the stereochemical control point before choosing precursors, since it usually decides them.
Always verify the finished route forward, checking regiochemistry, stereochemistry, rearrangement, and competing reactions at every step.
Build the skeleton first and install the functional group last.

Course complete

That is all thirteen lessons

You started with bond-line drawings and finished planning multistep routes. If a reaction ever feels like a memorized string of reagents, go back to the lesson that explains its mechanism: that is where the pattern lives.

Back to the course map

Lesson 12: Alcohols and Ethers Practice set