Back to the on-screen lesson ·
An alkyne's two pi bonds can each add a reagent, so equivalents and conditions control the product: Lindlar's catalyst gives the cis alkene, sodium in ammonia the trans, excess hydrogen the alkane, and hydration gives an enol that becomes a ketone.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will count the hydrogen an alkyne needs, match conditions to products, predict and order alkyne hydration, choose conditions for a cis alkene, and tabulate equivalents and products.
You know how alkenes add reagents, with regiochemistry and stereochemistry. An alkyne's triple bond has two pi bonds, so it can add twice.
An equivalent is one mole of reagent per mole of substrate. Lindlar's catalyst is palladium deliberately poisoned so that hydrogenation stops at the alkene. An enol is a compound with an OH on a C=C; tautomerization moves a hydrogen to turn it into its keto form.
Each pi bond of a triple bond can take one molecule of reagent:
Another way: table
Controlling an alkyne's reduction.
| Conditions | Product |
|---|---|
| H2, Lindlar's catalyst | cis alkene |
| Na in liquid NH3 | trans alkene |
| excess H2, Pd | alkane |
Water adds to an alkyne the same way it adds to an alkene, but the product has its OH on a carbon that is still doubly bonded: an enol. Enols are much less stable than their keto forms, because a C=O is a stronger bond than a C=C. A proton moves from the oxygen to the neighbouring carbon, the C=C becomes C–C and the C–O becomes C=O. So hydration of an alkyne is a route to ketones, not alcohols, and the ketone's carbonyl sits on the more substituted carbon.
An alkyne can undergo only one reaction step. It can add twice.
All hydrogenations give the alkane. Lindlar's catalyst stops at the cis alkene.
Sodium in ammonia gives the cis alkene. It gives trans.
Alkyne hydration gives an alcohol. The enol becomes a ketone.
The target is the cis alkene: stop after one H2, added from one face.
Stop, and syn.
Lindlar's catalyst adds one H2 syn and does not reduce the alkene further.
Choose the catalyst.
So hydrogenation over the poisoned catalyst gives (Z)-pent-2-ene.
Cis from syn addition.
How many H2 per propyne?
Two.
How much H2 in total?
6 mol of H2, on ordinary palladium.
How many moles of H2 are needed to turn $6$ mol of but-2-yne, CH3C≡CCH3, completely into butane on a palladium catalyst?
Answer:
Match each set of conditions to the product it gives from but-2-yne.
| (Z)-but-2-ene | (E)-but-2-ene | butane | 2,2,3,3-tetrabromobutane | |
|---|---|---|---|---|
| H2 with Lindlar's catalyst | ||||
| sodium in liquid ammonia | ||||
| excess H2 with palladium | ||||
| two equivalents of Br2 |
Water adds to propyne, HC≡CCH3, with an acid and mercury(II) catalyst. What is the organic product that is isolated?
Put the steps of the acid-catalysed hydration of propyne in order.
Number the steps in order (write the number in the box):
Many insect pheromones contain a cis C=C. A chemist has the matching alkyne. Which conditions give the cis alkene?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Propyne reacts with each reagent below. Give the equivalents of reagent added per propyne and the kind of product.
| equivalents added | product | |
|---|---|---|
| excess HBr | ||
| H2 with Lindlar's catalyst | ||
| excess H2 with palladium |
You can control what an alkyne becomes. Tell someone why hydration of propyne gives propanone. Next unit: how spectra reveal structure.
8. Your turn: 3 mol of propyne to propane, step 3