Difficulty: Intermediate | Prerequisites: Electrophilic addition to alkenes (Doc 1), alkyne structure and bonding.
Alkynes undergo many of the same electrophilic additions as alkenes, but with an extra twist: the triple bond can react with one or two equivalents of reagent, and hydration of an alkyne produces a carbonyl compound rather than an alcohol. Hydroboration-oxidation, covered alongside alkynes here, is the main anti-Markovnikov method for adding water across a double or triple bond. Together, these reactions give you control over whether the oxygen ends up on the more or less substituted carbon, and whether you get a ketone or an aldehyde from an alkyne.
Alkynes react with HX (2 eq.) to give geminal dihalides and with X2 (2 eq.) to give tetrahalides. Mercury-catalysed hydration of an internal alkyne gives a ketone via keto-enol tautomerism. Hydroboration-oxidation of alkenes gives anti-Markovnikov alcohols; on alkynes it gives aldehydes. These are the complementary regiochemistry tools to the Markovnikov reactions from the alkene unit.
Alkyne
A hydrocarbon containing a carbon-carbon triple bond. Alkynes have two pi bonds available for electrophilic addition, so they can react with up to two equivalents of reagent.
Enol
An unstable intermediate with an OH group attached directly to a C=C double bond. Think of it as a vinyl alcohol.
Keto-enol tautomerism
The equilibrium interconversion between a ketone (or aldehyde) and its enol form. The keto form is almost always favoured at equilibrium. In simple terms, the enol rearranges to the carbonyl because the C=O bond is more stable than the combination of C=C plus O-H.
Hydroboration-oxidation
A two-step reaction sequence (1. BH3, 2. H2O2/NaOH) that adds water across a double or triple bond with anti-Markovnikov regiochemistry and syn stereochemistry. No carbocation intermediate forms.
Anti-Markovnikov addition
Addition where the hydrogen ends up on the more substituted carbon and the OH (or other group) on the less substituted carbon. The opposite of Markovnikov's rule.
Syn addition
Both new groups add to the same face of the double bond. Hydroboration is a syn addition because H and BH2 are delivered simultaneously from one face.
Aldehyde
A carbonyl compound with at least one H on the carbonyl carbon (RCHO). Hydroboration-oxidation of a terminal alkyne gives an aldehyde.
Ketone
A carbonyl compound with two carbon groups flanking the C=O (RCOR'). Mercury-catalysed hydration of an internal alkyne gives a ketone.
Alkynes react with HBr or HCl (2 equivalents) to give geminal dihalides (both halogens on the same carbon)
The first equivalent adds following Markovnikov's rule, forming a vinyl halide
The second equivalent adds to the same carbon (again Markovnikov), giving the geminal product
Alkynes react with Br2 or Cl2 (2 equivalents) to give tetrahalides (four halogen atoms, two per carbon of the original triple bond)
Reagents: HgSO4, H2SO4, H2O
Follows Markovnikov's rule: water adds to the more substituted carbon of the triple bond
The initial product is an enol (vinyl alcohol), which is unstable
The enol tautomerises to the ketone (keto form), which is the isolated product
Hydration happens only once (one equivalent of water adds), because after tautomerism the product is a ketone, not another alkene
Tautomers are constitutional isomers that interconvert rapidly by proton transfer
The enol form has C=C and O-H; the keto form has C=O and C-H
Equilibrium strongly favours the keto form for simple ketones and aldehydes
This is the reason alkyne hydration gives a ketone rather than an enol as the final product
Reagents: Step 1, BH3; Step 2, H2O2 and NaOH
Product: alcohol with anti-Markovnikov regiochemistry (OH on the less substituted carbon)
Stereochemistry: syn addition (H and OH end up on the same face)
Mechanism: BH3 adds across the double bond in a single concerted step (no carbocation intermediate), so no rearrangements occur
The boron goes to the less substituted carbon (less steric hindrance), and hydrogen goes to the more substituted carbon
Oxidation step (H2O2/NaOH) replaces the B with OH, retaining the stereochemistry
Same reagents as for alkenes: Step 1, BH3; Step 2, H2O2 and NaOH
The initial product is an enol (OH on the less substituted carbon, anti-Markovnikov)
The enol tautomerises to an aldehyde (not a ketone)
This is the key distinction from mercury-catalysed hydration: mercury gives a ketone (Markovnikov), hydroboration-oxidation gives an aldehyde (anti-Markovnikov)
Reaction | Reagents | Product | Regiochemistry | Notes |
|---|---|---|---|---|
HX addition to alkyne (2 eq.) | HBr or HCl, 2 eq. | Geminal dihalide | Markovnikov | Both halogens on same C |
X2 addition to alkyne (2 eq.) | Br2 or Cl2, 2 eq. | Tetrahalide | N/A | Four halogens total |
Hydration of alkyne | HgSO4, H2SO4, H2O | Ketone | Markovnikov | Via enol intermediate |
Hydroboration-oxidation of alkene | 1) BH3, 2) H2O2/NaOH | Alcohol | Anti-Markovnikov | Syn addition, no carbocation |
Hydroboration-oxidation of alkyne | 1) BH3, 2) H2O2/NaOH | Aldehyde | Anti-Markovnikov | Via enol, then tautomerism |
Keto-enol tautomerism is not just an exam topic. It underpins the chemistry of many biological processes, including DNA base-pair mutations (a tautomeric shift in a base can cause a mispairing during replication). Hydroboration-oxidation is widely used in synthetic chemistry whenever a chemist needs to place an alcohol group with anti-Markovnikov selectivity.
Students often assume that hydration of an alkyne gives an alcohol. It does not. The initial enol tautomerises to a ketone (or aldehyde), so the isolated product is always a carbonyl compound.
Confusing the products of mercury-catalysed hydration (ketone, Markovnikov) with hydroboration-oxidation (aldehyde, anti-Markovnikov). The reagent set determines the regiochemistry, which determines ketone vs aldehyde.
Forgetting that hydroboration-oxidation has syn stereochemistry. Because BH3 adds in a single concerted step, there is no carbocation and therefore no possibility of rearrangement.
Treating keto-enol tautomers as resonance structures. They are not. Tautomers are different compounds (constitutional isomers) in equilibrium; resonance structures are different depictions of the same compound.
⚠️ The contrast between Markovnikov hydration (ketone) and anti-Markovnikov hydroboration-oxidation (aldehyde) for alkynes is a very common exam question.
⚠️ Be able to show the enol intermediate and draw the tautomerism arrow-pushing to the keto form.
⚠️ Know that hydroboration-oxidation gives syn addition and anti-Markovnikov regiochemistry, and be ready to draw the stereochemical outcome on a ring.
True or False: Hydration of a terminal alkyne with HgSO4/H2SO4/H2O gives an aldehyde. (False, it gives a ketone. Hydroboration-oxidation of a terminal alkyne gives the aldehyde.)
Fill in the blank: The enol form of a ketone has a ________ group directly bonded to a C=C. (Hydroxyl / OH.)
True or False: Hydroboration-oxidation proceeds through a carbocation intermediate. (False, it is a concerted addition with no carbocation.)
Fill in the blank: Adding 2 equivalents of HBr to an alkyne produces a ________ dihalide. (Geminal.)
True or False: Keto-enol tautomers are resonance structures. (False, they are constitutional isomers in equilibrium.)
Q: What is the product of treating 1-butyne with HgSO4, H2SO4 and H2O? Explain the intermediate involved.
A: 2-Butanone (methyl ethyl ketone). Water adds across the triple bond following Markovnikov's rule, placing OH on C-2. The initial product is an enol, which tautomerises to 2-butanone.
Q: How would you convert 1-pentyne into pentanal? Name the reagents and the regiochemistry.
A: Treat 1-pentyne with 1) BH3, then 2) H2O2/NaOH. Hydroboration-oxidation gives anti-Markovnikov addition, placing OH on C-1. The enol at C-1 tautomerises to the aldehyde (pentanal).
Q: What product forms when propene undergoes hydroboration-oxidation? Compare to acid-catalysed hydration of propene.
A: Hydroboration-oxidation gives 1-propanol (anti-Markovnikov, OH on the less substituted carbon). Acid-catalysed hydration gives 2-propanol (Markovnikov, OH on the more substituted carbon).
Q: Why does hydroboration-oxidation give syn addition while halogenation gives anti addition?
A: In hydroboration, BH3 delivers both B and H from the same face of the alkene in a single concerted step (syn). In halogenation, the bromonium ion blocks one face, so the nucleophile must attack from the opposite face (anti).
Hydroboration-oxidation is the direct counterpart to acid-catalysed hydration covered in the alkene electrophilic addition notes. Whenever the exam asks you to make an alcohol on the less substituted carbon, hydroboration-oxidation is the answer.
Keto-enol tautomerism will come back in later chapters when you study enolate chemistry, aldol reactions and alpha-halogenation of carbonyl compounds. The concept that an enol can act as a nucleophile is central to those topics.
The pKa and protonation-state question from the review (tyrosine sidechain, pKa 10.1, physiological pH 7.4) ties into acid-base reasoning: when pH < pKa, the protonated form dominates, so the tyrosine phenol is protonated at physiological pH.
Alkyne reactions, electrophilic addition alkynes, HBr alkyne, geminal dihalide, tetrahalide, mercury-catalysed hydration, HgSO4, keto-enol tautomerism, tautomers, enol, vinyl alcohol, hydroboration-oxidation, BH3, H2O2 NaOH, anti-Markovnikov addition, syn addition, aldehyde from alkyne, ketone from alkyne, anti-Markovnikov alcohol, CHM 255, organic chemistry, Purdue