Alkene and Alkyne Interconversions

Introduction

An alkene and an alkyne can be interconverted while retaining the carbon skeleton, but the reagents determine whether the product is a cis alkene, trans alkene or fully saturated alkane. Alkene-to-alkyne conversion commonly passes through a vicinal dihalide and two eliminations. Alkyne-to-alkene conversion requires partial reduction, not ordinary exhaustive hydrogenation.

Core explanation

To make an alkyne from an alkene, first add Br₂ or Cl₂ across the C=C bond. Each former alkene carbon receives one halogen, making a vicinal dihalide. Treatment with a sufficiently strong base removes two equivalents of HX through successive eliminations. The first produces a vinylic halide with C=C and a halogen directly on an alkene carbon; the second produces C≡C. The second elimination can require stronger base because the vinylic C–X bond is harder to eliminate from than an ordinary alkyl halide. Conditions and equivalents must therefore support two eliminations.

For ethene, Br₂ addition gives 1,2-dibromoethane. Double dehydrohalogenation gives ethyne, HC≡CH. The two carbon atoms remain throughout. For an unsymmetrical substrate, choosing which beta H is removed and whether the alkyne can form at the desired position requires closer analysis. If the product is a terminal alkyne and excess very strong base is used, its acidic terminal H may also be removed to form an acetylide; aqueous work-up is then needed to restore HC≡C–.

The reverse direction offers three common outcomes. H₂ over an ordinary active metal catalyst such as Pd/C can reduce an alkyne all the way to an alkane, passing through an alkene. A deactivated Lindlar catalyst permits hydrogenation to stop at an alkene and adds H from the same face of the triple bond, giving a cis alkene from a suitable internal alkyne. Sodium or lithium in liquid ammonia gives a trans alkene through a stepwise electron-and-proton sequence. Thus an internal alkyne can be a common precursor to opposite alkene geometries.

But-2-yne, CH₃C≡CCH₃, is a useful stereochemical example. H₂/Lindlar gives cis-but-2-ene, corresponding to the Z arrangement for this simple case. Na/liquid NH₃ gives trans-but-2-ene, the E arrangement. H₂/Pd-C without a stopping strategy gives butane. The carbon count and double-bond position are the same in the two partial reductions; only geometry differs. A target specifying E or Z cannot be satisfied by writing the generic word "hydrogenation."

Terminal alkynes do not have cis/trans alkene pairs when reduced to terminal alkenes because one resulting double-bond carbon has two hydrogens. For example, propyne can reduce to propene, but propene has no E/Z geometry. Apply the stereochemical choice only when each alkene carbon will have two different substituents.

An alkyne can also add halogens or hydrogen halides, giving halogenated alkenes after one equivalent or more saturated products after excess reagent, but these are addition routes rather than partial reduction to hydrocarbon alkene. In a conversion problem, read both product's functional group and reagent equivalents. A single reagent name does not specify whether the reaction stops at the first addition.

The interconversion route is valuable in synthesis because a triple bond can be assembled at a chosen position and later reduced stereoselectively. However, alkene-to-alkyne through halogenation and double elimination adds steps and may be sensitive to other groups. Choose it when triple-bond chemistry or stereochemical control justifies that complexity.

Step-by-step reasoning

For alkene → alkyne, add X₂ across the double bond, label the vicinal dihalide, and remove two HX equivalents with an adequately strong base. For alkyne → alkene, decide whether target is cis or trans: choose H₂/Lindlar for cis or dissolving metal/liquid ammonia for trans. If target is alkane, ordinary catalytic hydrogenation may suffice. Check terminal-alkyne acidity and E/Z eligibility.

Visual explanation

Draw CH₃C≡CCH₃ in the centre with three arrows: H₂/Lindlar → cis-CH₃CH=CHCH₃, Na/NH₃(l) → trans-CH₃CH=CHCH₃, and excess H₂/Pd → CH₃CH₂CH₂CH₃. On the other side, draw alkene → vicinal dibromide → alkyne as a two-stage upward path in bond order.

Real-world analogy

Think of bond order as a track with three stations: single, double and triple. Ordinary hydrogenation may travel from triple through double to single without stopping. A controlled catalyst exits at the double-bond station and sets one geometry; dissolving-metal conditions exit there by another route and set the opposite geometry. Reaching the same station is not enough if the platform orientation matters.

Real-world example

To prepare cis-but-2-ene from but-2-yne, a chemist uses a partial hydrogenation catalyst such as Lindlar's. To prepare trans-but-2-ene from the same alkyne, dissolving-metal reduction provides the complementary geometry. The example shows how one carbon skeleton can yield two stereoisomeric alkene targets by choosing different reduction conditions.

Why?

Why can standard H₂/Pd fail when an alkene is the desired product? The alkene formed in the first reduction step can remain on an active catalyst surface and undergo a second hydrogenation to the alkane. A less active poisoned catalyst can limit that further reduction, allowing the alkene to be isolated.

Common misconception

"Any alkyne hydrogenation gives a cis alkene." H₂/Pd may continue to an alkane, and metal/liquid-ammonia reduction gives trans alkene from suitable internal alkynes. The precise reagent and stopping conditions determine both bond order and geometry.

Worked example

Question: Choose a route from but-2-yne to trans-but-2-ene without fully reducing it.

Reasoning: The target is an internal trans alkene. Ordinary H₂/Pd risks butane, and Lindlar hydrogenation gives cis alkene. A dissolving-metal reduction provides trans geometry while stopping at the double bond.

Answer: Treat but-2-yne with sodium or lithium in liquid ammonia to obtain trans-but-2-ene.

Quick check

1. What intermediate appears between alkene halogenation and alkyne formation by elimination? Answer: A vicinal dihalide forms first; successive HX eliminations then produce the alkyne.

Exam focus

Show both eliminations for dihalide-to-alkyne and note a possible terminal acetylide under excess strong base. For partial reduction, specify Lindlar cis versus metal/ammonia trans; distinguish both from complete H₂/Pd reduction to alkane. Apply E/Z only when the alkene can possess geometric isomers.

Advanced insight

Partial reduction is a stereochemical design tool. An alkyne's approximately linear geometry can be transformed into one of two alkene configurations by selecting a syn surface-hydrogenation route or a stepwise dissolving-metal route. The alkyne must be placed at the correct skeletal position first, because reduction controls geometry but does not move the double bond along the chain.

Summary

Alkenes can become alkynes through halogen addition to a vicinal dihalide followed by two dehydrohalogenations. Alkynes become alkenes by controlled partial reduction: Lindlar gives cis and dissolving metal/liquid ammonia gives trans for suitable internal substrates. Ordinary catalytic hydrogenation can continue to alkane. Reagent choice controls both bond order and stereochemistry.

Practice questions

1. What product follows H₂/Lindlar treatment of but-2-yne? Answer: cis-But-2-ene, the Z isomer in this symmetric example. 2. What product follows Na/liquid NH₃ treatment of but-2-yne? Answer: trans-But-2-ene, the E isomer. 3. What may form if but-2-yne is hydrogenated fully over Pd/C? Answer: Butane, after the alkene intermediate undergoes further reduction. 4. Why does propene not have cis/trans stereoisomers? Answer: Its terminal double-bond carbon has two identical H substituents.