Hydrocarbons · Lesson 10 of 14
Alkyne Acidity, Addition, and Polymerisation
“Explain terminal-alkyne acidity and track the products formed when two pi bonds react.”
• Distinguish terminal acidic hydrogens from other alkyne hydrogens. • Relate carbon s character to conjugate-base stabilisation. • Predict successive addition and hydration products. • Connect ethyne polymerisation to conducting chains and aromatic rings. • Plan a multistage conversion while preserving carbon accounting.
Physical trends are familiar, but one hydrogen is different
Alkynes have broadly similar size-dependent physical trends to alkenes and alkanes. The lower members are gases, intermediate members liquids, and higher members solids. They are weakly polar, generally less dense than water, poorly soluble in it, and more soluble in organic solvents. Boiling point and density usually increase along the homologous series. Commercial acetylene may have a noticeable odour from impurities; pure ethyne should not be identified by smelling a sample.
The distinctive chemical feature of a terminal alkyne is its hydrogen directly bonded to an carbon. The hybrid orbital has 50% s character, holding electron density relatively close to the nucleus. Carbon therefore stabilises the negative charge left after proton removal better than comparable or carbons. Ethyne is consequently appreciably more acidic than ethene or ethane, though it is still a weak acid in everyday aqueous terms.
Sodamide supplies the strongly basic amide ion; the product contains a propynide anion. Sodium metal also reacts with terminal acetylene hydrogens, forming acetylide salts with hydrogen evolution. This is a metal reaction, rather than the same proton-transfer step as amide treatment. Formal replacement of both ethyne hydrogens accounts for mono- and disodium ethynide formulas in the source.
Problem
Compare but-1-yne and but-2-yne in their reactions with sodamide.
- 1.But-1-yne contains a C≡C–H group. Removal of that hydrogen leaves negative charge on an sp carbon and produces a but-1-ynide salt.
- 2.But-2-yne has no hydrogen directly attached to a triple-bond carbon. Its methyl hydrogens are attached to sp3 carbon and do not show the same terminal-alkyne reaction under this comparison.
- 3.The diagnostic distinction is therefore terminal versus internal alkyne, not simply whether the molecule contains hydrogen somewhere.
Within this elementary comparison, ethyne is more acidic than propyne, and both are far more readily deprotonated at a terminal carbon than an internal alkyne lacking that site. Alkyl electron donation helps explain the ethyne/propyne difference. Do not extend this short trend into a claim that all hydrogens on all alkynes have identical acidity.
Two pi bonds allow successive additions
An alkyne can add two molecules of hydrogen, halogen, or hydrogen halide. The first equivalent makes an alkene derivative; the second can consume the remaining pi bond. Product identity therefore depends on the reagent amount as well as the starting structure. For simple unsymmetrical alkynes, the usual hydrogen-halide addition follows Markovnikov orientation, represented in the source through a vinylic-cation intermediate.
Bromine solution is decolourised here too. The test therefore detects reaction associated with unsaturation, but cannot by itself distinguish an alkene from an alkyne. Hydrogen-halide addition is different from bromination: successive HBr additions to a terminal alkyne normally place both bromines on the same carbon, giving a geminal dihalide.
Problem
Find the products of adding one and two HBr molecules to propyne.
- 1.The first addition gives , named 2-bromoprop-1-ene. The hydrogen joins the terminal carbon and bromine the substituted carbon.
- 2.The second gives , 2,2-dibromopropane. The two bromines occupy the same carbon, so the product is geminal.
- 3.For ethyne, the corresponding sequence is bromoethene followed by 1,1-dibromoethane. Do not confuse the geminal hydrogen-halide product with a vicinal dihalide from alkene bromination.
Hydration first makes an enol
Alkynes do not simply react with ordinary water by standing in it. On warming with dilute sulphuric acid and mercuric sulphate catalyst, water adds across the triple bond. The initially drawn product has hydroxyl attached to a double-bonded carbon, an enol. It rearranges by a proton transfer and double-bond shift into a more stable carbonyl compound. This is tautomerisation, not insertion of an extra carbon atom.
Problem
Compare the carbonyl groups formed in their mercury-catalysed hydration.
- 1.Ethyne gives ethenol initially. Its hydroxyl-bearing carbon retains a hydrogen, and tautomerisation makes ethanal.
- 2.Propyne gives an enol whose hydroxyl-bearing carbon is connected to two carbon groups after rearrangement. The carbonyl product is propanone.
- 3.The location of the hydroxyl group in the initial addition determines the carbonyl position. Retain the complete skeleton throughout the rearrangement.
Ethyne can make a chain or an aromatic ring
Linear polymerisation of ethyne gives polyethyne or polyacetylene, with an alternating unsaturated backbone. Its simplest repeat is CH=CH. Under special conditions, especially appropriate doping, the conjugated polymer can conduct electricity; this is not a claim that every untreated sample behaves like a metal. Thin conducting films can be light electrodes and other electrical materials.
Passing ethyne through a red-hot iron tube at about 873 K instead allows three molecules to form benzene. This cyclic trimerisation builds a six-carbon ring. It is an important bridge from an aliphatic starting material into aromatic chemistry.
Problem
Plan a conversion using the reactions developed so far.
- 1.Neutralise ethanoic acid to sodium ethanoate. Soda-lime decarboxylation gives methane, and controlled chlorination gives chloromethane.
- 2.Use Wurtz coupling to join methyl groups into ethane. Chlorinate to chloroethane, then use alcoholic potassium hydroxide to eliminate HCl and make ethene.
- 3.Add bromine to form 1,2-dibromoethane. Carry out the two elimination stages, alcoholic potassium hydroxide followed by sodamide, to reach ethyne.
- 4.Finally trimerise ethyne in a red-hot iron tube to benzene. This long route illustrates reaction connections; it is not a claim that every step gives a pure product quantitatively or that it is the most efficient industrial route.
Terminal alkyne acidity, unsaturation tests, and hydration answer different questions. An internal alkyne still undergoes addition even though it lacks the terminal acidic hydrogen. A polymerisation equation also needs its product pattern specified: a chain and a six-membered ring are different outcomes.
Quiz
Which hydrogen is removed most readily by sodamide in but-1-yne?
Why does sp carbon stabilise a negative charge better in this comparison?
What is the final product of mercury-catalysed propyne hydration?
What forms after two normal HBr additions to propyne?
How many ethyne molecules combine to give one benzene molecule?
Practice Problems
- Write a sodamide reaction for but-1-yne and explain the contrasting behaviour of but-2-yne.
- Compare ethyne, ethene, and ethane in acidity using carbon hybridisation.
- Show products after one and two bromine equivalents react with propyne.
- Write ethyne and propyne hydration sequences, including enols and carbonyl products.
- Compare the repeat units of polyethyne and polythene.
- Give a reagent-labelled route from ethanoic acid to benzene and explain where its carbon count decreases or increases.
Key Takeaways
• Only a hydrogen directly attached to terminal sp carbon shows the characteristic terminal-alkyne acidity. • Two pi bonds allow sequential additions; reagent amount controls how far addition proceeds. • Hydration produces an enol that rearranges into a carbonyl compound. • Ethyne can form either a conjugated polymer chain or benzene under different conditions.