Difficulty: Intermediate–Advanced | Prerequisites: All earlier CHEM 2301 material, especially reaction types, stereochemistry, and alkyne/alkene reactivity.
This is the largest section of the exam (27 + 12 = 39 points). The synthesis roadmap tests your ability to connect starting materials to products through a web of reactions, matching reagents to transformations. The multistep synthesis question asks you to design a route from scratch. Both require you to have the full reaction toolkit internalised, not just memorised in isolation, but understood as a connected network. If you cannot quickly recall what each reagent does to an alkyne or alkene, start there.
TL;DR
You need to know the reagents and products for every alkyne and alkene transformation covered this term: halogenation, hydrohalogenation, hydration, hydroboration-oxidation, reduction (catalytic and Lindlar's), ozonolysis, epoxidation, and deuterium-labelled variants. The multistep synthesis question asks you to chain these together, working backwards from the target to find a viable route.
Lindlar's catalyst (H₂, Pd/CaCO₃/Pb)
A "poisoned" palladium catalyst that reduces alkynes to cis-alkenes only. It stops at the double bond and does not continue to the alkane.
In simple terms, Lindlar's is the tool for making a cis double bond from a triple bond.
Catalytic hydrogenation (H₂, Pd or Pt)
Full reduction. An alkyne goes to an alkane; an alkene goes to an alkane. Both π bonds are reduced. Syn addition of hydrogen.
Hydroboration-oxidation (1. BH₃–THF, 2. NaOH/H₂O₂)
Anti-Markovnikov addition of water across a double bond. The OH ends up on the less substituted carbon. Syn addition (both H and OH add from the same face).
Ozonolysis (1. O₃, 2. Zn/H₂O or DMS)
Cleaves a C=C double bond entirely, producing two carbonyl compounds (aldehydes or ketones depending on substitution). With Zn/H₂O (reductive workup), you get aldehydes; with H₂O₂ (oxidative workup), you get carboxylic acids from terminal alkenes.
NaNH₂ (sodium amide)
A very strong base. Deprotonates terminal alkynes to form acetylide anions (RC≡C⁻), which are excellent nucleophiles for alkylation reactions.
HBr (Markovnikov addition)
Adds across a double bond with the H going to the less substituted carbon and Br to the more substituted carbon (Markovnikov selectivity).
NaOD/D₂O
Base-catalysed deuterium exchange. Replaces acidic hydrogens (e.g. terminal alkyne H, α-hydrogens adjacent to carbonyls) with deuterium.
D₂, Pd (deuterium with palladium catalyst)
Full catalytic reduction using deuterium gas instead of hydrogen. Adds D atoms across the π bond(s) with syn stereochemistry.
D₂, Lindlar's catalyst
Partial reduction of an alkyne to a cis-alkene using deuterium. Delivers two D atoms syn across the triple bond, stopping at the alkene stage.
mCPBA (meta-chloroperoxybenzoic acid)
An epoxidation reagent. Converts an alkene to an epoxide with retention of the alkene’s geometry (cis-alkene gives cis-epoxide; trans gives trans).
The roadmap question presents a multistep transformation with blanks (numbered boxes). Each blank is either a reagent/condition or a product. You match letters from a bank to each box. Each letter is used once.
The reaction sequence from the exam (reconstructed):
The roadmap starts with a terminal alkyne and runs through a series of transformations. Here are the key reaction patterns you must recognise:
Alkyne halogenation
Terminal alkyne + Br₂ → 1,2-dibromoalkene (one equivalent) or 1,1,2,2-tetrabromoalkane (two equivalents)
The exam shows Br₂ addition as the first step
Alkyne deprotonation and alkylation
Terminal alkyne + 2 eq. NaNH₂ → dianion (removes both the terminal H and, if possible, a propargylic H)
With only 1 eq. NaNH₂, you get the mono-acetylide, which can be alkylated with an alkyl halide to extend the carbon chain
Partial reduction to cis-alkene
Internal alkyne + H₂/Lindlar's catalyst → cis-alkene
Internal alkyne + D₂/Lindlar's catalyst → cis-alkene with deuterium labels (syn addition of D₂)
Full reduction
Alkyne + D₂/Pd → fully deuterated alkane (all π bonds reduced, D atoms added)
Hydrohalogenation of alkynes
Alkyne + 2 eq. HBr → geminal dibromide (Markovnikov addition, both Br atoms on the more substituted carbon)
Hydroboration-oxidation
Alkene + 1. BH₃–THF, 2. NaOH/H₂O₂ → anti-Markovnikov alcohol (syn addition)
Ozonolysis
Alkene + 1. O₃, 2. Zn/H₂O → two carbonyl fragments
Base-catalysed deuterium exchange
NaOD/D₂O replaces terminal alkyne H with D, and replaces α-H adjacent to C=O with D
Epoxidation
Alkene + mCPBA (or similar peracid) → epoxide. Geometry is retained: a cis-alkene gives a cis-epoxide.
Acid-catalysed hydration (Markovnikov)
Alkyne + H₂SO₄/H₂O/Hg²⁺ → enol → ketone (Markovnikov, via keto–enol tautomerism)
Terminal alkyne → methyl ketone
Oxymercuration-demercuration
H₂O₂, 2. NaOH: this is the oxidation step of hydroboration. Combined with the BH₃ step, it delivers anti-Markovnikov alcohol.
The exam problem:
Synthesise a cis-epoxide from propyne (CH₃C≡CH) and any carbon-based starting materials of four carbons or fewer.
Target: cis-2,3-dimethyloxirane (a cis-epoxide with methyl groups on both carbons).
Retrosynthetic analysis (working backwards):
A cis-epoxide comes from epoxidation of a cis-alkene (mCPBA preserves geometry).
A cis-alkene comes from partial reduction of an internal alkyne (H₂/Lindlar's catalyst).
An internal alkyne (but-2-yne, CH₃C≡CCH₃) comes from alkylation of propyne's acetylide anion with a methyl halide.
The acetylide anion comes from deprotonation of propyne with NaNH₂.
Forward synthesis (the answer, graded at 2 pts per condition, 2 pts per product):
Step 1: CH₃C≡CH + NaNH₂ → CH₃C≡C⁻ Na⁺ (deprotonation to form the acetylide)
Step 2: CH₃C≡C⁻ + CH₃–LG (methyl halide) → CH₃C≡CCH₃ (alkylation to extend the chain, forming but-2-yne)
Step 3: CH₃C≡CCH₃ + H₂/Lindlar's catalyst → cis-CH₃CH=CHCH₃ (partial reduction to the cis-alkene)
Step 4: cis-CH₃CH=CHCH₃ + mCPBA → cis-2,3-dimethyloxirane (epoxidation with retention of geometry)
Key strategic points:
Always think backwards from the target. Ask: "What functional group is in the product, and what reaction makes it?"
The cis geometry of the epoxide is the clue that you need a cis-alkene, which points to Lindlar's reduction.
The internal alkyne must come from alkylation of an acetylide, because you start with a terminal alkyne and need to add one more carbon.
Only primary alkyl halides work for acetylide alkylation (methyl, ethyl, propyl). Secondary and tertiary halides undergo elimination instead.
Students confuse Lindlar's catalyst (partial reduction to cis-alkene) with Na/NH₃ (dissolving metal reduction to trans-alkene). Both reduce alkynes, but to opposite geometric isomers.
Students forget that acetylide alkylation only works with primary (ideally methyl or primary) alkyl halides. Attempting it with a secondary halide gives elimination, not substitution.
Students mix up hydroboration-oxidation (anti-Markovnikov, syn) with acid-catalysed hydration (Markovnikov). Both add water, but to different carbons and with different stereochemistry.
Students assume ozonolysis and epoxidation are the same because both use oxygen-based reagents. Ozonolysis cleaves the double bond; epoxidation preserves the carbon skeleton and adds an oxygen bridge.
⚠️ The roadmap section is worth 27 points (3 per box, 9 boxes). Each box is independent, so a wrong answer in one box does not cost you points elsewhere. Attempt every box.
⚠️ Multistep synthesis is 12 points, graded at 2 points per correct condition and 2 points per correct intermediate/product. Show your work even if you are unsure of one step, because partial credit is available for each correct piece.
⚠️ Pay close attention to stereochemistry. A cis-epoxide requires a cis-alkene, which requires Lindlar's catalyst. A trans-epoxide would need Na/NH₃ reduction instead. The geometry of your target determines the entire synthetic route.
⚠️ When deuterium (D) appears in the product, work out where each D came from. D₂/Pd gives full reduction with D; D₂/Lindlar's gives cis-addition of D; NaOD/D₂O exchanges acidic H for D.
Fill in the blank: H₂ + Lindlar's catalyst reduces an alkyne to a ____-alkene.
cis
True or false: Acetylide anions can be alkylated with 2-bromopropane to form a longer internal alkyne.
False. 2-Bromopropane is a secondary halide and will undergo elimination, not S₂ substitution.
Fill in the blank: The reagent for anti-Markovnikov addition of water to an alkene is ________, followed by ________.
BH₃–THF, followed by NaOH/H₂O₂
True or false: Ozonolysis of a tetrasubstituted alkene produces two ketones.
True.
Fill in the blank: To make a cis-epoxide from an internal alkyne, the two key steps are ________ (to get the cis-alkene) followed by ________ (to form the epoxide).
H₂/Lindlar's catalyst, followed by mCPBA (or another peracid)
Q: Starting from 1-butyne, show a two-step synthesis of cis-2-butene.
A: Step 1: 1-butyne + NaNH₂ → butynide anion. But actually, 1-butyne is already a 4-carbon terminal alkyne. To get cis-2-butene (an internal alkene), you would need the internal alkyne but-2-yne first. So: 1-butyne is not the right starting material for cis-2-butene. Starting from but-2-yne: but-2-yne + H₂/Lindlar's catalyst → cis-2-butene (one step).
Q: What reagents convert propyne into propanal (CH₃CH₂CHO)?
A: This requires anti-Markovnikov hydration of the triple bond. Use hydroboration of the alkyne: 1. disiamylborane (or 9-BBN for alkynes), 2. H₂O₂/NaOH. This places the oxygen on the terminal carbon, giving the aldehyde rather than the methyl ketone.
Q: Why does acid-catalysed hydration of a terminal alkyne give a methyl ketone rather than an aldehyde?
A: Acid-catalysed hydration follows Markovnikov's rule. Water adds to the more substituted carbon of the alkyne, producing an enol with OH on the internal carbon. Tautomerisation of this enol gives the methyl ketone.
Q: Design a synthesis of trans-2,3-dimethyloxirane from propyne. How does this differ from the cis-epoxide synthesis?
A: You need a trans-alkene instead of a cis-alkene. Replace Lindlar's reduction with Na/NH₃ (dissolving metal reduction), which gives the trans-alkene. Then epoxidise with mCPBA. The route: propyne → (NaNH₂, then CH₃I) → but-2-yne → (Na/NH₃) → trans-2-butene → (mCPBA) → trans-2,3-dimethyloxirane.
Q: In the roadmap, what does NaOD/D₂O do to a terminal alkyne?
A: It replaces the terminal alkyne hydrogen with deuterium via base-catalysed exchange: RC≡CH → RC≡CD.
The synthesis roadmap is essentially a test of every reaction in the course combined. Each box references a reaction you learned in an earlier unit. If you are weak on a particular transformation (say, hydroboration-oxidation), go back to that unit's notes and practise it in isolation before trying to place it in a roadmap.
Multistep synthesis connects forward to Organic Chemistry II, where retrosynthetic analysis becomes even more central. The habit of thinking backwards from the target is one of the most important skills you will carry forward.
Deuterium labelling problems connect to NMR spectroscopy: D is NMR-silent for ¹H NMR, so knowing where D ends up helps you predict which signals disappear.
Related Terms / Search Tags
synthesis roadmap, multistep synthesis, retrosynthetic analysis, Lindlar's catalyst, catalytic hydrogenation, hydroboration-oxidation, ozonolysis, epoxidation, mCPBA, acetylide alkylation, NaNH2, HBr addition, Markovnikov, anti-Markovnikov, deuterium labelling, D2O exchange, NaOD, cis-epoxide, trans-epoxide, dissolving metal reduction, Na NH3, partial reduction alkyne, CHEM 2301, Pomerantz, Exam 4