Lesson 13 · Synthesis and Retrosynthesis

Synthesis and Retrosynthesis

This lesson adds no new reactions. It teaches you how to chain the ones you already know into a route. The key skill is planning backward from the target: the target tells you what the last reaction likely was, while the starting material tells you almost nothing.

Rule to remember

Plan backward, verify forward. Never start at the starting material and try reactions at random.

Learning goals

Tell working forward (synthesis) apart from working backward (retrosynthesis), and know when to use each.
Use the carbon count to decide right away whether you need a C−C bond-forming step.
Recognize common functional group conversions on sight and name a reagent for each.
Break a bond you know how to make (a disconnection), and judge whether the forward reaction would actually work.
Find the step that sets the stereochemistry before choosing a precursor.
Check a planned route forward for regiochemistry, stereochemistry, rearrangements, and competing reactions at every step.

Key terms (click to reveal)

Part 01

Two directions, one problem

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

Backward wins because it is a smaller search. From any starting material there are dozens of reactions you could run, and almost none lead anywhere useful. From the target, the functional group narrows the last step to two or three options right away. An alcohol usually comes from an alkene or a carbonyl. A cis alkene comes from an alkyne with Lindlar. The target tells you what the final reaction was.

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 trying reactions at random. 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 sorts every synthesis problem into one of two types.

Same carbon count

Functional group problem

The skeleton is already right. Your whole job is converting functional groups, sometimes with a stereochemistry or regiochemistry requirement. Most problems are this type.

Different carbon count

Skeleton problem

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

In this course, the C−C bond-forming reaction to recognize on sight is acetylide alkylation. Grignard and organolithium additions to carbonyls, and enolate alkylation, join the list if your course covers them. Because these reactions are few, they are the least flexible part of any route, so plan them first and fit everything else around them.

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

When planning, find your target in the left column and read across to see what could make it. When writing the answer, use the forward reagents. The Watch for column is where most marks are lost.

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 a rearrangement would ruin the skeleton	FGI
Alcohol (anti-Markovnikov)	Alkene	1. BH₃·THF 2. H₂O₂, OH⁻	Syn addition, which matters if new stereocenters form	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 helps with 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 with pyridine	Inverts the carbon. To keep the configuration, make a tosylate (TsCl) instead	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 the alkene instead	FGI

Part 04

Disconnections

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

Put simply, you are asking which two pieces could come together to make this bond. The imagined pieces are called synthons. The real reagents that play those roles are the synthetic equivalents: a carbanion synthon becomes an acetylide or a Grignard reagent 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 (2-pentyne), disconnect a bond next to the triple bond. One choice gives CH₃CH₂C≡C⁻ plus CH₃Br. That acetylide comes from 1-butyne and NaNH₂, so the forward route is 1. NaNH₂, 2. CH₃Br. You could also disconnect the other side, giving propynyl anion plus CH₃CH₂Br. Both work. Synthesis problems often have more than one correct answer.

Part 05

Reagent limitations make disconnections fail

This is where synthesis problems get hard. A disconnection can look fine on paper and still be useless, because the forward reaction does not work.

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

Realistic

R−C≡C⁻ + CH₃CH₂Br

Primary halide with an open backside, so the SN2 runs cleanly.

Not realistic

R−C≡C⁻ + (CH₃)₃CBr

Tertiary halide, so E2 wins and the C−C bond never forms.

Some disconnections can be rescued by swapping which piece 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 at the same bond, because an alkyne carbon can be the nucleophile but never the electrophile. What you can do is disconnect a different bond: if the other side of the alkyne is methyl or primary, break that one instead. Knowing which disconnections can be flipped 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 sets it. Find that step before planning the rest, because it usually decides which precursor you need.

Need cis

H₂, Lindlar

From an alkyne. Both H atoms add to the same face on the poisoned catalyst. This is the only 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 mechanism

Concerted or surface additions give syn (BH₃, OsO₄, H₂ on a metal). Three-membered ring intermediates opened from the back give anti (Br₂, epoxide opening).

These steps stack. Lindlar on 2-butyne gives cis-2-butene, and OsO₄ then adds two OH groups syn, giving the meso diol. Use Na/NH₃ instead and the same syn addition gives the chiral pair. Same 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 route forward: starting material → B → A → target. This catches errors that backward planning hides. Going backward, you tend to assume each arrow works. Going forward, you have to say what each 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 win with this substrate?

Will another functional group in the molecule react too?

Does this step overshoot, like a 1° alcohol going 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 targets, same starting material, and the target's functional group alone picks the reagent.

Target: CH₃CH₂CH₂CHO

Aldehyde → hydroboration

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

Target: CH₃CH₂COCH₃

Methyl ketone → mercury

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

Two steps: 1-butyne to 1-butanol

The target, CH₃CH₂CH₂CH₂OH, is a primary alcohol. Four carbons 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 the end carbon? Anti-Markovnikov hydration of an alkene. So 1-butanol ⇒ 1-butene. What makes that terminal alkene? Partial reduction of an alkyne. So 1-butene ⇒ 1-butyne, which is your starting material.

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, not butane, because the poisoned catalyst stops at the alkene. 2. BH₃·THF, then H₂O₂ and OH⁻, puts the OH on the end carbon. The carbon count stays at four, neither step can rearrange, and no other group is present to interfere. The route holds.

Three steps: when the skeleton has to grow

Target: 3-hexanone (six carbons). Starting material: 1-butyne (four carbons). The count changed, so a C−C bond-forming step is required, and that anchors the plan.

Backward: a ketone can come from hydrating an alkyne. 3-hexanone has its carbonyl at C3, so the alkyne must be 3-hexyne, CH₃CH₂C≡CCH₂CH₃. It is symmetric, so hydration can only give one ketone. Now disconnect a bond next to the triple bond: 3-hexyne ⇒ CH₃CH₂C≡C⁻ plus CH₃CH₂Br. That acetylide is just deprotonated 1-butyne, which is where you started.

Forward and verified: 1. NaNH₂ deprotonates 1-butyne to its acetylide. 2. CH₃CH₂Br is a primary halide, so the SN2 runs cleanly and gives 3-hexyne. 3. H₂O, H₂SO₄, HgSO₄ hydrates the symmetric alkyne to 3-hexanone. The carbon count goes from four to six at step 2 and stays there. 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 molecule 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 single reaction you have memorized.

Step 1

Count carbons

Compare the starting material and the 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 a primary alcohol with the OH on the end carbon. Precision narrows the last step.

Step 3

Ask what reaction directly makes it

There are usually two or three options. If the target specifies stereochemistry, that alone rules out most of them.

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 precursor becomes the new target. Stop when you reach what you were given, or something easily made from it.

Step 6

Run the whole pathway forward

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

Watch out

Common mistakes

Starting from the starting material

Trying reactions forward at random is a wide search full of 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 converting functional groups on a skeleton that was the wrong size from the start.

Making an unrealistic disconnection

A disconnection that needs SN2 on a tertiary carbon is not a plan. Every disconnection must match a forward reaction that actually works.

Never verifying forward

Backward planning hides regiochemistry errors because you assume each arrow works. Going forward forces you to say what each reagent really does.

Ignoring stereochemistry until the end

If the target is a cis alkene, the route has to end in a Lindlar reduction, and that shapes everything before it. 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, you need PCC.

Forgetting other functional groups

A reagent that makes your target group can destroy another group elsewhere in the molecule. Check the whole structure at every forward step.

Installing the functional group before building the skeleton

C−C bond-forming reactions are sensitive to what else is present. Build the carbon skeleton first, 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 end up on the end carbon. That means anti-Markovnikov hydration, which needs an alkene. Lindlar takes the alkyne to the alkene without over-reducing to butane, and hydroboration-oxidation then puts the OH where you need it. Plain H₂/Pd would give butane, a dead end.

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

Counting carbons first tells you this is not just a functional group problem. The alkyne is already there; the only thing missing is two carbons, and acetylide alkylation adds them. Forward: 1. NaNH₂, 2. CH₃CH₂Br. 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

It fails: the alkylation is SN2 and tert-butyl bromide is tertiary, so E2 wins. Fix it by breaking the other bond: 3,3-dimethyl-1-butyne, then 1. NaNH₂, 2. CH₃I.

A disconnection has to match a reaction that actually works. With Williamson you can swap which piece is the nucleophile. With acetylide alkylation you cannot, because an alkyne carbon is never the electrophile. But this target has two bonds next to the triple bond. The other one splits into the tert-butyl acetylide plus CH₃I, and methyl iodide is a perfect SN2 partner. The bulky tert-butyl group now sits on the nucleophile, where it does no harm.

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 add two OH groups syn with OsO₄. Syn addition to a cis alkene gives the meso diol.

Two stereochemistry-setting steps are stacked here. The reduction sets cis or trans, and the dihydroxylation adds syn. Na/NH₃ would give trans-2-butene, and syn addition to that gives the chiral (2R,3R) and (2S,3S) pair instead of the meso compound. Decide which step sets the stereochemistry before you pick 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. Each backward arrow looked reasonable, but the forward check catches the regiochemistry: HBr protonates the alkene to give the more stable 3° carbocation, so Br ends up on the 3° carbon. The target needs Br on the less substituted alkene carbon, which takes anti-Markovnikov addition (HBr with peroxides) on that same alkene.

Practice workbook

Synthesis and Retrosynthesis Practice

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

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's functional group names the last reaction: alcohol from an 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. In this course that mostly means acetylide alkylation; Grignard and organolithium additions join the list if your course covers them.
Disconnections must be realistic. An SN2 step onto a tertiary halide is not a plan; it is an elimination.
Find the stereochemical control point before choosing precursors, since it usually decides them.
Always check the finished route forward for regiochemistry, stereochemistry, rearrangements, 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